CHEM 105 - ELEMENTS OF GENERAL CHEMISTRY Faraday’s laws of electrolysis Question Bank Set 2

CHEM 105 - ELEMENTS OF

GENERAL CHEMISTRY - Faraday’s

laws of electrolysis

Question Bank - Set 2

Liberty University

Question 1

Question

A student performs an electrolysis experiment using a copper(II) sulfate solu-

tion. The student passes a current of 2.5 A through the solution for 30 minutes.

During this time, they observe the mass of copper deposited on the cathode to

be 3.6 g. Determine the charge passed through the solution and the number

of moles of electrons transferred during the experiment. Use Faraday’s laws of

electrolysis.

Solution

Step 1: Calculate the charge passed through the solution. Given: Current,

I= 2.5 A Time, t= 30 minutes = 30 minutes ×60 seconds/minute = 1800

seconds

Using the formula Q=It, where: Q= charge passed (coulombs) I= current

(amperes) t= time (seconds)

Substitute the given values: Q= 2.5 A ×1800 s = 4500 C

Therefore, the charge passed through the solution is 4500 C.

Step 2: Determine the number of moles of electrons transferred during the

experiment. We know that 1 Faraday (F) is equal to the charge of one mole of

electrons, which is 96485 C (1 F = 96485 C).

To find the number of moles of electrons transferred, we use the formula: n

F, where: n= number of moles of electrons transferred Q= charge passed

(coulombs) F= Faraday constant (96485 C)

Substitute the values: n=4500 C

96485 C ≈0.0467 mol

Therefore, the number of moles of electrons transferred during the experi-

ment is approximately 0.0467 mol.

Question 2

Question

An aqueous solution of sodium chloride is subjected to electrolysis using inert

electrodes. If a current of 2.00 A is passed through the solution for 1.00 hour,

what mass of sodium metal is deposited at the cathode? (Assume 100

Solution

Step 1: Determine the quantity of electricity passed through the solution. The

quantity of electricity (Q) can be calculated using the formula:

Q=I×t

where: - Qis the quantity of electricity, - Iis the current in amperes (A), - tis

the time in hours.

Given that I= 2.00 A and t= 1.00 hour, we have:

Q= 2.00 A ×1.00 hr = 2.00 C

Step 2: Calculate the number of moles of electrons passed through the so-

lution. Since 1 Faraday (F) is equal to the charge of 1 mole of electrons (1 F

= 96500 C), the number of moles of electrons (n) can be calculated using the

formula:

n=Q

96500

Substitute Q= 2.00 C into the formula:

n=2.00 C

96500 C/F ≈2.07 ×10−5mol

Step 3: Determine the number of moles of sodium deposited at the cathode.

In the electrolysis of sodium chloride, 2 moles of electrons are required to pro-

duce 2 moles of sodium. Therefore, the number of moles of sodium deposited

at the cathode (nNa) is equal to ncalculated in Step 2:

nNa = 2.07 ×10−5mol

Step 4: Calculate the mass of sodium deposited at the cathode. Using the

molar mass of sodium (MNa = 23.0 g/mol), the mass of sodium deposited can

be determined:

Mass of sodium = nNa ×MNa

Mass of sodium = 2.07 ×10−5mol ×23.0 g/mol

Mass of sodium ≈4.76 ×10−4g

Therefore, approximately 4.76 ×10−4g of sodium metal is deposited at the

cathode.

Question 3

Question

A student is conducting an electrolysis experiment using a copper(II) chloride

solution. If the student passes a current of 2.50 A through the solution for 30

minutes, how much copper will be deposited at the cathode? (Given: Atomic

mass of copper = 63.55 g/mol, Faraday’s constant = 9.65 ×104C/mol)

Solution

Step 1: Calculate the total charge passed through the solution. Given: Current

(I) = 2.50 A, Time (t) = 30 minutes = 1800 seconds.

Using the formula:

Q=I×t

Q= 2.50 A ×1800 s

Q= 4500 C

Step 2: Calculate the number of moles of electrons passed. Since 1 Faraday

(F) is equal to the charge of 1 mole of electrons,

1 F = 9.65 ×104C/mol

So, the number of moles of electrons passed can be calculated as:

Moles of electrons = Q

9.65 ×104

Moles of electrons = 4500

9.65 ×104

Moles of electrons ≈0.0466 mol

Step 3: Calculate the number of moles of copper deposited. From the bal-

anced chemical equation:

2e−+Cu2+ →Cu

1 mole of Cu is deposited for every 2 moles of electrons transferred.

Therefore, the moles of copper deposited is half the moles of electrons passed:

Moles of copper = 1

2×Moles of electrons

Moles of copper = 1

2×0.0466

Moles of copper = 0.0233 mol

Step 4: Calculate the mass of copper deposited. Using the atomic mass of

copper:

Mass of copper = Moles of copper ×Atomic mass of copper

Mass of copper = 0.0233 ×63.55

Mass of copper ≈1.48 g

Therefore, approximately 1.48 grams of copper will be deposited at the cath-

ode.

Question 4

Question

State Faraday’s laws of electrolysis and explain how they are applied in deter-

mining the amount of substance deposited during electrolysis of a solution.

Solution

Faraday’s laws of electrolysis are a set of two laws that describe how the amount

of substance deposited or liberated during electrolysis is related to the amount

of electric charge passed through the electrolytic cell.

Faraday’s First Law: The mass of an element deposited during electrolysis

is directly proportional to the quantity of electricity passed through the cell.

Faraday’s Second Law: The masses of different elements deposited by

the same quantity of electricity are proportional to their chemical equivalent

weights.

Application: To determine the amount of substance deposited during elec-

trolysis of a solution, we can use Faraday’s laws in the following way:

Step 1: Determine the number of moles of electrons transferred during the

electrolysis process using the equation:

moles of electrons = electric charge passed (Coulombs)

where Fis the Faraday constant (96,485 C/mol).

Step 2: Determine the number of moles of the substance deposited by

dividing the number of moles of electrons by the number of electrons involved

in the reaction (from the balanced equation).

Step 3: Calculate the mass of the substance deposited using the formula:

mass of substance = number of moles ×molar mass

By following these steps and applying Faraday’s laws, we can accurately

determine the amount of substance deposited during electrolysis of a solution.

Question 5

Question

During the electrolysis of an aqueous solution of copper sulfate, a current of

5.00 A is passed through the solution for 1.50 hours. If the current efficiency is

(Given: atomic mass of copper = 63.5 g/mol, Faraday constant = 96485 C/mol)

Solution

Step 1: Calculate the total charge passed through the electrolyte. The formula

to calculate the charge passed is:

Q=I·t

where: Q= charge (in coulombs) I= current (in amperes) t= time (in seconds)

Converting time from hours to seconds:

t= 1.50 hours ×3600 seconds/hour = 5400 seconds

Substitute the given values:

Q= 5.00 A ×5400 s = 27000 C

Step 2: Calculate the number of moles of electrons transferred. The Faraday

constant relates the charge to the number of moles of electrons:

1 mol of electrons = 96485 C

Moles of electrons = Q

96485

Substitute the charge value:

Moles of electrons = 27000

96485 = 0.28 mol

Step 3: Calculate the moles of copper deposited. Since the current efficiency

is 80

Moles of copper deposited = 0.28 mol ×0.80 = 0.224 mol

Step 4: Calculate the mass of copper deposited. Using the atomic mass of

copper given:

Mass of copper deposited = Moles of copper deposited×Atomic mass of copper

Mass of copper deposited = 0.224 mol ×63.5 g/mol = 14.24 g

Therefore, the mass of copper deposited at the cathode is 14.24 g.

Question 6

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of molten

lead(II) bromide, PbBr2. If lead is deposited at the cathode, what is the mass

of lead deposited in 30 minutes? (Given: atomic masses of Pb = 207.2 g/mol,

Br = 79.9 g/mol; Faraday constant = 96485 C mol−1)

Solution

Step 1: Find the number of moles of electrons passing through the circuit. Step

2: Use the stoichiometry of the electrolysis reaction to find the moles of lead

deposited. Step 3: Calculate the mass of lead deposited.

Step 1: The current passing through the circuit is I= 2.5Aand the time

is t= 30 min =30

60 hrs = 0.5hrs.

The charge passing through the circuit is given by Q=I·t.

Now, using Faraday’s law of electrolysis:

Q=nF

where nis the number of moles of electrons and Fis the Faraday constant.

Substitute the given values to find n:

n=Q

F=I·t

F=2.5A·0.5hrs

96485 C mol−1

Step 2: The balanced equation for the electrolysis of molten lead(II) bro-

mide is:

Pb2+ + 2Br−→Pb + Br2

This indicates that 2 moles of electrons are required to deposit 1 mole of lead.

So, the moles of lead deposited is half the moles of electrons:

moles of Pb = n

Step 3: Calculate the mass of lead deposited using the moles of lead ob-

tained in Step 2:

mass of Pb = moles of Pb×molar mass of Pb = I·t

2F×molar mass of Pb = 2.5A·0.5hrs

2×96485 C mol−1×207.2g/mol

Question 7

Question

A student is performing an electrolysis experiment using a solution of sodium

chloride (NaCl). The student passes a current of 2.5 A through the solution for

30 minutes. During this time, they observe the formation of chlorine gas (Cl2)

at one electrode and sodium metal (Na) at the other electrode. Calculate the

mass of sodium metal deposited at the cathode during this electrolysis process.

(Given: Faraday constant, F= 96485 C/mol, atomic mass of sodium, MNa =

23 g/mol, 1 hour = 3600 seconds)

Solution

Step 1: Calculate the total charge passed through the electrolyte: The total

charge passed can be calculated using the formula:

Q=It

where Qis the total charge (in coulombs), Iis the current (in amperes), and tis

the time (in seconds). Given I= 2.5 A and t= 30 minutes = 30 ×60 s = 1800 s,

we have:

Q= 2.5 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of electrons passed: One mole of

electrons has a charge of 1 Faraday, which is equal to the charge of 1 mol of

electrons. Hence, the number of moles of electrons can be calculated using the

formula:

Number of moles of electrons = Q

Given F= 96485 C/mol, we have:

Number of moles of electrons = 4500 C

96485 C/mol ≈0.0467 mol

Step 3: Calculate the number of moles of sodium deposited: In the electrol-

ysis of molten sodium chloride, 2 moles of electrons are required to reduce 1

mole of sodium ions (Na+) to form 1 mole of sodium metal (Na). Therefore,

in this case, since 0.0467 moles of electrons are passed, the number of moles of

sodium deposited will be half that amount:

Number of moles of sodium deposited = 0.0467 mol

2= 0.0234 mol

Step 4: Calculate the mass of sodium deposited: The mass of sodium de-

posited can be calculated using the formula:

Mass of sodium deposited = Number of moles ×Molar mass

Given the molar mass of sodium, MNa = 23 g/mol, we have:

Mass of sodium deposited = 0.0234 mol ×23 g/mol = 0.5382 g

Therefore, the mass of sodium metal deposited at the cathode during this

electrolysis process is approximately 0.5382 grams.

Question 8

Question

A student performs an electrolysis experiment using a current of 2.50 A. If

the student plated out 0.500 g of copper, determine the time taken for this

electrolysis to occur. (Given: Faraday constant F= 96,485 C/mol and the

molar mass of copper is 63.55 g/mol.)

Solution

Step 1: Calculate the number of moles of copper plated out.

Moles of copper = Mass of copper plated out

Molar mass of copper =0.500 g

63.55 g/mol

Step 2: Calculate the charge passed through the electrolyte solution.

Q=It

where Qis the charge passed (in coulombs), Iis the current (in amperes), and

tis the time taken (in seconds).

Step 3: Find the number of moles of electrons transferred. Each mole of

copper requires 2 moles of electrons for the reduction process. Therefore,

Moles of electrons = 2 ×Moles of copper

Step 4: Calculate the total charge required to plate out the given amount of

copper. The total charge required is given by:

Q=n×F

where nis the number of moles of electrons transferred, and Fis the Faraday

constant.

Step 5: Equate the two expressions for charge passed and charge required.

It =n×F

Step 6: Solve for t.

t=n×F

Question 9

Question

A solution contains both Cu2+ and Ag+ions. A current of 2.0 A is passed

through the solution for 3.0 hours. If 1.0 g of copper is deposited at the cathode,

what mass of silver is deposited at the cathode? (Given: F = 96500 C/mol,

m(Cu) = 63.5 g/mol, m(Ag) = 107.9 g/mol)

Solution

Step 1: Calculate the charge that passed through the cell. The charge (Q)

passing through the cell can be calculated using the formula:

Q=I·t

where Iis the current in amperes and tis the time in seconds. Given I= 2.0 A,

t= 3.0 hours ×3600 s/hour = 10800 s, we have

Q= 2.0 A ×10800 s = 21600 C

Step 2: Determine the number of moles of copper deposited. The number

of moles of copper deposited (nCu) can be calculated using the formula:

nCu =m

where mis the mass of copper deposited and Mis the molar mass of copper.

Given m= 1.0 g, M= 63.5 g/mol, we have

nCu =1.0 g

63.5 g/mol = 0.0157 mol

Step 3: Using the ratio of charges passed and moles deposited to find the

molar mass of silver. The ratio of charges passed and moles deposited is the

same for both copper and silver because the same current is used. Therefore,

we can equate the two ratios to find the molar mass of silver.

QCu

nCu

=QAg

nAg

where QCu is the charge required to deposit copper and QAg is the charge

required to deposit silver. Given that Q= 21600 C has passed through the cell

and the Faraday constant F = 96500 C/mol, we can substitute:

nCu ·F=Q

nAg ·F

Step 4: Solve for the mass of silver deposited. Substitute the known values

into the formula derived in Step 3 and solve for the mass of silver deposited

(mAg):

21600

0.0157 ·96500 =21600

nAg ·96500

nAg =21600

0.0157 ·96500

nAg = 1.44 ×10−2mol

Finally, to find the mass of silver deposited:

mAg =nAg ·MAg

mAg = 1.44 ×10−2mol ×107.9 g/mol

mAg = 1.55 g

Therefore, 1.55 g of silver is deposited at the cathode.

Question 10

Question

An aqueous solution of silver nitrate (AgNO3) is electrolyzed using inert elec-

trodes. If 96500 C of charge is passed through the solution, calculate the

mass of silver deposited at the cathode. Given: 1 F = 96500 Coulombs/mol,

molar mass of Ag = 107.87 g/mol.

Solution

Step 1: Determine the number of moles of electrons passed through the solution.

- From Faraday’s first law of electrolysis, we know that 1 Faraday (F) of charge

is equivalent to 1 mole of electrons. - Given that 96500 Coulombs of charge is

passed, we can calculate the number of moles of electrons:

Moles of electrons = 96500 C

96500 C/mol = 1 mol

Step 2: Determine the number of moles of silver deposited at the cathode. -

The balanced half-reaction for the deposition of silver at the cathode is: Ag++

1e−→Ag. - From the reaction, we see that 1 mole of silver is deposited for

every 1 mole of electrons passed. - Since 1 mole of electrons is passed, 1 mole

of silver is deposited at the cathode.

Step 3: Calculate the mass of silver deposited. - The molar mass of silver is

given as 107.87 g/mol. - The mass of silver deposited can be calculated as:

Mass of silver = Number of moles×Molar mass = 1 mol×107.87 g/mol = 107.87 g

Therefore, the mass of silver deposited at the cathode is 107.87 g.

Question 11

Question

In an electrolytic cell, a current of 2.50 A is passed through a solution of cop-

per(II) sulfate for 30.0 minutes. If the reaction produces 5.00 g of copper,

calculate the Faraday constant (F) and Faraday’s constant (F).

Solution

Step 1: Find the moles of copper produced. Given: Current, I= 2.50 A Time,

t= 30.0 minutes Mass of copper produced, m= 5.00 g Atomic mass of copper,

MCu = 63.55 g/mol

We first convert time from minutes to seconds: 30.0 minutes = 30.0×

60 seconds = 1800 seconds

We then calculate the moles of copper produced:

moles of Cu = mass of Cu

MCu

=5.00 g

63.55 g/mol = 0.0787 mol

Step 2: Calculate the charge passed through the cell. We use the formula to

find the charge:

Q=It

where Qis the charge, Iis the current, and tis the time. Substitute the given

values:

Q= 2.50 A ×1800 s = 4500 C

Step 3: Find Faraday’s constant (F). The number of moles of electrons

involved in the reaction is equal to the moles of copper produced, which is

0.0787 mol. Since 1 F (Faraday) is the charge carried by one mole of electrons,

we have:

F=Q

where: - nis the number of moles of electrons; - Qis the charge passed. Sub-

stitute the values:

F=4500 C

0.0787 mol = 57161 C/mol

Step 4: Find Faraday constant (F). The Faraday constant (F) is related to

Avogadro’s constant and is given by:

F=F × NA

where NAis Avogadro’s constant (6.022 ×1023 mol−1). Substitute the values:

F= 57161 C/mol ×6.022 ×1023 mol−1= 3.44 ×103C/mol

Therefore, Faraday constant F= 3.44 ×103C/mol.

Question 12

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution with copper electrodes. The student passes a current of 2.50 A for 30

minutes. During this time, copper is deposited on one electrode. Determine the

mass of copper deposited and the volume of gas evolved at the other electrode.

(Take the molar mass of copper as 63.5 g/mol, and assume standard temperature

and pressure for the gas.)

Solution

Step 1: Find the amount of charge passed through the electrolyte.

Charge(Q) = Current ×Time

Given that the current is 2.50 A and the time is 30 minutes, we need to

convert the time to seconds: 30 minutes = 30 ×60 seconds = 1800 seconds.

Therefore, the charge passed through the electrolyte is:

Q= 2.50 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of electrons passed.

Moles of electrons = Charge

Faraday constant

The Faraday constant is 96485 C/mol.

Moles of electrons = 4500 C

96485 C/mol ≈0.0467 mol

Step 3: Determine the mass of copper deposited on the electrode.

Mass of copper = Moles of electrons ×Molar mass of copper

Given that the molar mass of copper is 63.5 g/mol:

Mass of copper = 0.0467 mol ×63.5 g/mol ≈2.97 g

Step 4: Calculate the volume of gas evolved at the other electrode using the

ideal gas law.

Volume of gas = Number of moles of gas ×Universal gas constant ×Temperature

Pressure

At standard temperature and pressure, the temperature is 273 K and the

pressure is 1 atm. Since one mole of electrons produces one mole of gas (by

Faraday’s first law of electrolysis), the number of moles of gas is 0.0467 mol.

Volume of gas = 0.0467 mol ×0.0821 L/mol ·atm ·K−1×273 K

1 atm ≈1.07 L

Question 13

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution. The student plated out 10.0 g of copper in 30 minutes using a current

of 2.00 A. Calculate the number of moles of electrons that flowed through the

cell during this time.

Solution

Step 1: Find the molar mass of copper. Step 2: Calculate the number of moles

of copper deposited. Step 3: Use Faraday’s laws of electrolysis to determine the

number of moles of electrons.

Step 1:

The molar mass of copper is 63.55 g/mol.

Step 2:

Given: Mass of copper plated out, m= 10.0 g

Molar mass of copper, M= 63.55 g/mol

Number of moles of copper deposited:

n=m

M=10.0 g

63.55 g/mol = 0.1577 mol

Step 3:

Given: Time, t= 30 min = 1800 s

Current, I= 2.00 A

We know that 1 mole of electrons carries a charge of 1 F (Faraday). From

Faraday’s laws of electrolysis, the total charge Qpassed through the cell can be

calculated using:

Q=It

The total charge can be related to the number of moles of electrons, ne, through

the equation:

Q=ne×F

where Fis the Faraday’s constant.

Therefore, we have:

Q=ne×F=⇒It =ne×F=⇒ne=It

Substitute the values:

ne=2.00 A ×1800 s

96485 C/mol = 0.0372 mol

So, the number of moles of electrons that flowed through the cell during this

time is 0.0372 moles.

Question 14

Question

A solution contains a mixture of copper(II) sulfate, CuSO4, and silver nitrate,

AgNO3, ions. A current of 2.50 A is passed through the solution for 2.00 hours.

During this time, 2.00 g of copper is deposited at the cathode.

Given:

Ecu =−0.34V

Eag = 0.80V

molar mass of Cu = 63.55g/mol

molar mass of Ag = 107.87g/mol

F= 96500C/mol

Calculate the mass of silver deposited during the electrolysis process.

Solution

Step 1: Calculate the charge passed through the electrolyte. Given that the

current, I, is 2.50 A and the time, t, is 2.00 hours, we can calculate the total

charge passed, Q, using the formula Q=I×t.

Q= 2.50A×2.00hours = 5.00Ah

Step 2: Calculate the number of moles of copper deposited. The number of

moles, n, of copper deposited can be calculated using the formula

n=m

where mis the mass of copper deposited (2.00 g) and Mis the molar mass of

copper (63.55 g/mol).

n=2.00g

63.55g/mol = 0.03151mol

Step 3: Calculate the number of electrons transferred during the deposition

of copper. The number of electrons transferred, ne, during the deposition of

copper can be calculated using Faraday’s laws of electrolysis. For the reduction

of copper(II) ions to copper atoms, the number of electrons transferred can be

determined by balancing the half-reaction:

Cu2+ + 2e−→Cu

From this half-reaction, we can see that 2 electrons are involved in the deposition

of 1 mole of copper. Therefore, the total number of electrons transferred during

the deposition of 0.03151 moles of copper is:

ne= 2 ×0.03151 = 0.06302 moles of electrons

Step 4: Calculate the charge carried by these electrons. The charge car-

ried by the electrons can be calculated by multiplying the number of moles of

electrons by Faraday’s constant, F.

Qe=ne×F= 0.06302mol ×96500C/mol = 6083.63C

Step 5: Calculate the mass of silver deposited. Now, we need to find out

how many moles of silver ions were reduced during the electrolysis process. To

do this, we find the total charge passed during the electrolysis, Q, and use the

standard reduction potential of silver, Eag , to determine the number of moles

of silver ions reduced. Since the total charge passed is the sum of the charge

carried by the copper ions and the silver ions:

Q=QCu2+ +QAg+

QAg+=Q−QCu2+

QAg+= 5.00Ah −6083.63C= 17916.37C

Then, we calculate the number of moles of silver ions, nAg+, reduced using

the formula:

nAg+=QAg+

nAg+=17916.37C

96500C/mol = 0.1858mol

Finally, we calculate the mass of silver deposited using the number of moles

of silver ions reduced and the molar mass of silver:

mAg =nAg+×MAg

mAg = 0.1858mol ×107.87g/mol =20.01 g

Therefore, the mass of silver deposited during the electrolysis process is 20.01

Question 15

Question

Consider an electrolytic cell containing a solution of copper sulfate (CuSO4).

A steady current of 2.5 A is passed through the cell for 20 minutes. If copper

is deposited at the cathode, what mass of copper is deposited?

Given: Faraday’s constant, F= 96,500 C/mol Atomic mass of copper,

Cu = 63.5g/mol

Solution

Step 1: Calculate the total charge passing through the cell. Given that the

current is 2.5 A and time is 20 minutes, first convert the time to seconds:

20 minutes = 20 ×60 = 1200 seconds

The total charge, Q, passing through the cell can be calculated using the

formula:

Q=I×t

Q= 2.5A×1200 s

Q= 3000 C

Step 2: Calculate the number of moles of electrons. Since each mole of

electrons carries a charge of 1 Faraday, we can calculate the number of moles of

electrons using the formula:

n=Q

n=3000 C

96500 C/mol

n≈0.0311 mol

Step 3: Calculate the amount of copper deposited. From the balanced equa-

tion for the deposition of copper, we know that 2 moles of electrons are required

to deposit 1 mole of copper. Therefore, the number of moles of copper deposited

will be half of the number of moles of electrons.

nCu =n

nCu =0.0311 mol

nCu ≈0.0156 mol

Step 4: Calculate the mass of copper deposited. The mass of copper de-

posited can be calculated using the formula:

Mass = nCu ×Molar mass of Cu

Mass = 0.0156 mol ×63.5 g/mol

Mass ≈0.9936 g

Therefore, approximately 0.9936 grams of copper will be deposited.

Question 16

Question

An electrolytic cell contains a solution of silver nitrate (AgNO3). If a current of

2.50 A is passed through the cell for 2.00 hours, what mass of silver is deposited

at the cathode? (Given: 1 F = 96,485 C/mol, mAg = 107.87 g/mol)

Solution

Step 1: Calculate the total charge passing through the cell. Let’s use the formula

Q=I·t, where Qis the charge, Iis the current, and tis the time. Given

I= 2.50 A and t= 2.00 hours, first convert the time to seconds:

t= 2.00 hours ×3600 s/hour = 7200 s

Now, calculate the total charge:

Q= 2.50 A ×7200 s = 18,000 C

Step 2: Determine the number of moles of silver deposited. We know that

1 F = 96,485 C/mol. We can calculate the number of moles of silver deposited

using the equation Q=n·F, where nis the number of moles and Fis Faraday’s

constant.

n=Q

F=18,000 C

96,485 C/mol ≈0.1863 mol

Step 3: Calculate the mass of silver deposited. The molar mass of silver is

107.87 g/mol. We can use this information to find the mass of silver deposited

at the cathode:

Mass of silver = n×Molar mass of silver = 0.1863 mol×107.87 g/mol = 20.09 g

Therefore, approximately 20.09 grams of silver is deposited at the cathode.

Question 17

Question

In an electrolytic cell, a current of 3.0 A is passed for 1.5 hours through a

solution of copper(II) sulfate, CuSO4. Calculate the mass of copper deposited

on the cathode. Given that the molar mass of copper is 63.5 g/mol and the

Faraday constant is 96,500 C/mol.

Solution

Step 1: Calculate the total charge passed through the cell.

Charge (C) = Current (A) ×Time (s)

Step 2: Convert the time from hours to seconds.

1.5 hours = 1.5×60 ×60 seconds

Step 3: Calculate the total charge in Coulombs.

Charge (C) = 3.0 A ×1.5×60 ×60 s

Step 4: Calculate the number of moles of electrons passed through the cell

using Faraday’s laws of electrolysis.

1 coulomb = 1

96500 mol of electrons

Step 5: Calculate the number of moles of electrons.

Moles of electrons = Charge (C)

96500 C/mol

Step 6: Since each copper ion, Cu2+, requires 2 moles of electrons to form

copper metal, the number of moles of copper deposited will be half of the moles

of electrons.

Step 7: Calculate the mass of copper deposited using the number of moles

of copper and the molar mass of copper.

Mass of copper (g) = Moles of copper ×Molar mass of copper (g/mol)

Question 18

Question

A solution of copper(II) sulfate, CuSO4, is electrolyzed using a constant current

of 2.50 A for 2.00 hours. If copper metal forms at the cathode, calculate the

mass of copper deposited.

(The molar mass of copper is 63.5 g/mol, the Faraday constant is 96,485

C/mol, and the charge of an electron is 1.60 ×10−19 C)

Solution

Step 1: Calculate the total charge passing through the circuit. Given: Current,

I= 2.50 A Time, t= 2.00 hours = 7200 s

The total charge Qcan be calculated using the formula Q=I×t. Plugging

in the values, we get Q= 2.50 A ×7200 s = 18000 C.

Step 2: Determine the number of moles of electrons. The number of moles

of electrons can be calculated using the equation n=Q

F, where Fis Faraday’s

constant. Substitute the values to get n=18000 C

96485 C/mol .

n≈0.18625 mol.

Step 3: Find the number of moles of copper deposited. From the balanced

equation in electrolysis of copper(II) sulfate:

Cu2+ + 2e−→Cu

we see that 2 moles of electrons are required to deposit 1 mole of copper. Thus,

the number of moles of copper deposited is half the number of moles of electrons:

nCu =n

2=0.18625 mol

nCu ≈0.09313 mol.

Step 4: Calculate the mass of copper deposited. The mass of copper, m, can

be calculated using the formula m=nCu ×Molar mass of copper. Substitute

the values to get m= 0.09313 mol ×63.5 g/mol.

m≈5.92 g.

Therefore, the mass of copper deposited during the electrolysis process is

approximately 5.92 g.

Question 19

Question

Consider an electrolytic cell where copper metal is to be obtained from a copper

sulfate solution using a current of 5.0 A. If it takes 25 minutes to deposit 10.5

g of copper metal, what is the total charge that passed through the cell during

this time? Assume the Faraday constant is 9.65 ×104C/mol.

Solution

Step 1: Calculate the moles of copper deposited. Given that the molar mass of

copper is 63.5 g/mol, we can calculate the moles of copper deposited:

Moles of copper = Mass of copper deposited

Molar mass of copper

Moles of copper = 10.5 g

63.5 g/mol = 0.165 mol

Step 2: Calculate the total charge passed through the cell. The total charge

passed through the cell can be calculated using Faraday’s laws of electrolysis:

Total charge = Number of moles of electrons ×Faraday constant

Since the reaction to deposit copper involves the transfer of 2 moles of electrons

per mole of copper:

Number of moles of electrons = 2 ×Moles of copper

Number of moles of electrons = 2 ×0.165 = 0.33 mol

Now, we can calculate the total charge:

Total charge = 0.33 mol ×9.65 ×104C/mol = 3.18 ×104C

Therefore, the total charge that passed through the cell during this time is

3.18 ×104C.

Question 20

Question

A solution of silver nitrate AgNO3is electrolyzed using a silver electrode. The

mass of silver electrode decreases by 0.625 g in 20 minutes. Assuming 100

Solution

Step 1: Find the number of moles of silver deposited on the electrode.

Given: Change in mass of silver electrode = 0.625 g Time taken = 20 minutes

= 20/60 hours = 1/3 hours

The molar mass of silver (Ag) = 107.87 g/mol

Number of moles of silver deposited = Change in mass / Molar mass Number

of moles of silver deposited = 0.625 g / 107.87 g/mol

Step 2: Calculate the charge passing through the electrolytic cell.

1 mole of electrons = 1 Faraday of charge = 96485 C

Total charge passed = number of moles x Avogadro’s number x Faraday’s

constant

Step 3: Determine the current flowing through the electrolytic cell.

Current (I) = Total charge passed / time taken

These steps can be used to determine the current flowing through the elec-

trolytic cell when a solution of silver nitrate is electrolyzed using a silver elec-

trode.

Question 21

Question

During the electrolysis of aqueous copper(II) sulfate solution using copper elec-

trodes, if a current of 2.50 A is passed through the solution for 2.00 hours,

calculate the mass of copper deposited at the cathode. (Atomic masses: Cu =

63.5 g/mol, S = 32.1 g/mol, O = 16.0 g/mol)

Solution

Step 1: Find the total charge passed through the electrolyte. Given that the

current is 2.50 A and the time is 2.00 hours, we can use the formula:

Q=I×t

Q= 2.50 A ×2.00 hr ×3600 s/hr

Q= 2.50 ×2.00 ×3600

Q= 18000 C

Step 2: Determine the moles of electrons passed. Each mole of electrons has

a charge of 1 Faraday, which is 96485 C.

Moles of electrons = 18000 C

96485 C/mol

Moles of electrons = 0.1866 mol

Step 3: Calculate the mass of copper deposited. The electrolysis of aqueous

copper(II) sulfate solution using copper electrodes will deposit copper at the

cathode. The balanced equation is:

Cu2+ + 2e−→Cu

From the equation, 2 moles of electrons are required to deposit 1 mole of copper.

Therefore, moles of copper deposited = moles of electrons passed / 2

Moles of copper = 0.1866 mol

Moles of copper = 0.0933 mol

Step 4: Calculate the mass of copper deposited.

Mass of copper = Moles of copper ×Molar mass of copper

Mass of copper = 0.0933 mol ×63.5 g/mol

Mass of copper = 5.92 g

Therefore, the mass of copper deposited at the cathode is 5.92 g.

Question 22

Question

A solution of copper(II) sulfate, CuSO4, is electrolyzed using copper electrodes.

If a current of 0.5 A is passed through the solution for 2 hours, what mass of

copper is deposited at the cathode? (Given: 1 F= 96,485 C/mol)

Solution

Step 1: Write the balanced half-reaction for the reduction that occurs at the

cathode. The reduction half-reaction for the deposition of copper from Cu2+

ions is:

Cu2+ + 2e−→Cu

This reaction requires 2 moles of electrons to deposit 1 mole of copper.

Step 2: Calculate the total charge passed through the circuit. The total

charge passed through the circuit can be calculated using the formula:

Q=I×t

where: - Qis the total charge passed (in coulombs), - I= 0.5Ais the current,

and - t= 2 hours = 2 ×3600 sec = 7200 sec is the time.

Substitute the values:

Q= 0.5A×7200 sec = 3600 C

Step 3: Calculate the number of moles of electrons passed through the cir-

cuit. Since 1 mole of electrons carries a charge of 1 F, the number of moles of

electrons can be calculated as:

Moles of electrons = Q

1F=3600 C

96500 C/mol

Moles of electrons ≈0.0374 mol

Step 4: Calculate the moles of copper deposited at the cathode. Since 2

moles of electrons are needed to deposit 1 mole of copper, the moles of copper

deposited can be calculated as:

Moles of copper = 1

2×Moles of electrons

Moles of copper = 1

2×0.0374 mol ≈0.0187 mol

Step 5: Calculate the mass of copper deposited at the cathode. The molar

mass of copper is 63.55 g/mol. Therefore, the mass of copper deposited can be

calculated as:

Mass of copper = Moles of copper ×Molar mass of copper

Mass of copper = 0.0187 mol ×63.55 g/mol ≈1.186 g

Therefore, approximately 1.186 g of copper is deposited at the cathode dur-

ing the electrolysis of copper(II) sulfate solution.

Question 23

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of silver

nitrate (AgNO3) for 2 hours. If the atomic mass of silver is 107.87 g/mol,

calculate the mass of silver deposited on the cathode. (Given: Faraday constant,

F = 96500C/mol)

Solution

Step 1: Calculate the total charge passed through the cell. The charge passed,

Q=I×t, where Iis the current in amperes and tis the time in seconds. Given

I= 2.5Aand t= 2hours = 2 ×3600s= 7200s,

Q= 2.5A×7200 s= 18000 C

Step 2: Calculate the number of moles of electrons passed through the cell.

One mole of electrons carries 1 Faraday of charge, i.e., 96500 C.

Number of moles of electrons = Q

96500

Step 3: Calculate the number of moles of silver deposited. In the reaction,

Ag++e−→Ag, one mole of silver is deposited per mole of electrons transferred.

Hence, the number of moles of silver deposited is equal to the number of moles

of electrons passed through the cell.

Step 4: Calculate the mass of silver deposited. Given the atomic mass of

silver is 107.87 g/mol,

Mass of silver deposited = Number of moles of silver deposited×Atomic mass of silver

Substitute the values and solve for the mass of silver deposited.

Question 24

Question

A student is performing an electrolysis experiment using a copper (II) sulfate

solution. In one trial, the student passed a current of 0.50 A through the solution

for 30 minutes. The student observed that 0.75 g of copper was deposited

on the cathode. Determine the Faraday constant and Faraday’s First Law of

Electrolysis constant using this information.

Solution

Step 1: Calculate the charge passed through the solution. Given: Current (I)

= 0.50 A Time (t) = 30 minutes = 30 * 60 seconds = 1800 s

The charge passed, Q, can be calculated using the formula:

Q=I×t

Q= 0.50 A ×1800 s

Q= 900 C

Step 2: Convert the mass of copper deposited to moles. Given: Mass of

copper deposited = 0.75 g Molar mass of copper (Cu) = 63.5 g/mol

The number of moles of copper deposited can be calculated using the for-

mula:

Moles = Mass

Molar mass

Moles of Cu = 0.75 g

63.5 g/mol

Moles of Cu = 0.0118 mol

Step 3: Determine the number of electrons involved in the electrolysis pro-

cess. One copper atom requires 2 electrons to form Cu: 1 Cu atom + 2e Cu

Given the moles of copper: Moles of Cu = Moles of electrons

Step 4: Calculate the Faraday constant and Faraday’s First Law of Electrol-

ysis constant. The Faraday constant (F) is equal to the charge of one mole of

electrons:

F=Charge (C)

Moles of electrons

F=900 C

0.0118 mol

F≈76241.52 C/mol

The Faraday’s First Law of Electrolysis constant, k, is calculated as:

k=Faraday constant

96000

k=76241.52 C/mol

96000

k≈0.793 mol/C

Therefore, the Faraday constant is approximately 76241.52 C/mol and Fara-

day’s First Law of Electrolysis constant is approximately 0.793 mol/C.

Question 25

Question

In a certain electrolytic cell, a current of 1.5 A is passed for 3 hours. During

this time, the mass of the substance deposited at the cathode is found to be

2.7 g. If the substance is a divalent metal, determine the atomic weight of the

metal (in g/mol). Assume 100

Solution

Step 1: Calculate the total charge passed through the cell. Given that the

current passed is 1.5 A and the time is 3 hours, we first convert the time to

seconds:

Time (s) = 3 hours ×3600 s/hour = 10800 s

Then, calculate the total charge passed using the formula Q=I·t:

Q= 1.5 A ×10800 s = 16200 C

Step 2: Determine the number of moles of the substance deposited. Since

the metal is divalent, the number of moles deposited (n) can be calculated using

Faraday’s first law of electrolysis:

n=Q

where F= 96500 C/mol is the Faraday constant.

Substitute the values to find the number of moles:

n=16200 C

2×96500 C/mol =16200

193000 mol

Step 3: Calculate the molar mass of the metal. Given that 2.7 g of the metal

is deposited, we can calculate the molar mass (M) using the formula:

M=Mass

Moles =2.7 g

16200

193000 mol

Perform the calculation to find the molar mass in g/mol.

Question 26

Question

During the electrolysis of a certain molten compound, 1.20 grams of the com-

pound is decomposed in 10 minutes using a constant current of 2.50 A. Calculate

the molar mass of the compound.

Solution

Step 1: Calculate the charge passed through the circuit. Given that the current

is 2.50 A and the time is 10 minutes, first convert the time to seconds:

10 minutes ×60 seconds/minute = 600 seconds

The charge passed through the circuit can be calculated using the formula Q=

It:

Q= 2.50 A ×600 s = 1500 C

Step 2: Calculate the number of moles of electrons transferred. For the

compound to be decomposed, 2 moles of electrons must be transferred per mole

of the compound. Therefore, the number of moles of electrons can be calculated

as:

moles of electrons = 1500 C

96500 C/mol =1500

96500 ≈0.0155 mol

Step 3: Determine the moles of the compound decomposed. Since 2 moles

of electrons are required to decompose 1 mole of the compound, the moles of

the compound decomposed is half the moles of electrons:

moles of compound = 0.0155

2= 0.00775 mol

Step 4: Calculate the molar mass of the compound. The molar mass of the

compound can be calculated using the formula:

molar mass = mass of compound

moles of compound =1.20 g

0.00775 mol ≈154.84 g/mol

Therefore, the molar mass of the compound is approximately 154.84 g/mol.

Question 27

Question

An electrolysis experiment was conducted using a solution of copper(II) sulfate,

CuSO4, with copper electrodes. The cell was set up such that the cathode

was the copper strip connected to the negative terminal of the battery. After a

certain amount of time, it was observed that the mass of the copper cathode had

increased. Explain this observation in terms of Faraday’s laws of electrolysis.

Solution

To explain the increase in mass of the copper cathode in the electrolysis exper-

iment using Faraday’s laws of electrolysis, we can break down the process into

the following steps:

Step 1: Determine the reaction occurring at the cathode The re-

action that occurs at the cathode during the electrolysis of copper(II) sulfate

solution is:

Cu2+ + 2e−→Cu

Step 2: Calculate the number of moles of electrons transferred

From the balanced equation, we see that 2 moles of electrons are required for

the reduction of 1 mole of copper ions. So, the number of moles of electrons

transferred is equal to the moles of copper deposited on the cathode.

Step 3: Apply Faraday’s first law of electrolysis Faraday’s first law

states that the mass of a substance produced at an electrode during electrolysis

is directly proportional to the quantity of electricity passed through the cell.

Mathematically, it can be expressed as:

Mass of substance = charge ×molar mass/Faraday’s constant

Step 4: Calculate the mass of copper deposited Given that the cur-

rent passing through the cell is related to the number of moles of electrons

transferred, we can calculate the mass of copper deposited on the cathode using

the formula:

Mass of Cu = Molar mass of Cu ×(Number of moles of electrons transferred)

Step 5: Explain the increase in mass of the copper cathode The in-

crease in mass of the copper cathode observed is due to the deposition of copper

metal from the copper(II) sulfate solution onto the cathode. This deposition

is a result of the reduction of copper ions by the supply of electrons from the

cathode.

Therefore, the increase in mass of the copper cathode is a consequence of the

reduction reaction occurring at the cathode during the electrolysis of copper(II)

sulfate solution.

Question 28

Question

An aqueous solution of copper sulfate is electrolyzed using platinum electrodes.

If 5.0 A of current is passed through the cell for 2.0 hours, what mass of copper

will be deposited at the cathode? (Given: Atomic weight of copper = 63.5

g/mol, Faraday’s constant = 96500 C/mol)

Solution

Step 1: Find the total charge passed through the cell. Given: Current (I) = 5.0

A, Time (t) = 2.0 hours = 7200 s The total charge Q passed through the cell

can be calculated using the formula Q=I×t.Q= 5.0A×7200 s= 36000 C

Step 2: Calculate the number of moles of electrons. Each mole of electrons

corresponds to 1 Faraday of charge, which is equal to 96500 C/mol. Number

of moles of electrons, n, can be determined by dividing the total charge Q by

Faraday’s constant: n=Q

96500 =36000

96500 ≈0.373 mol

Step 3: Determine the number of moles of copper deposited. In the balanced

half-reaction for the electrolysis of copper sulfate, 2 moles of electrons are needed

to deposit 1 mole of copper. So, the number of moles of copper deposited will

be half of the number of moles of electrons: nCu =n

2=0.373

2= 0.1865 mol

Step 4: Calculate the mass of copper deposited. Using the molar mass of

copper (63.5 g/mol), the mass (m) of copper deposited can be found using the

formula m=nCu×Molar mass of copper. m= 0.1865mol×63.5g/mol ≈11.84g

Therefore, approximately 11.84 grams of copper will be deposited at the

cathode.

Question 29

Question

When a current of 2.5 A is passed through an electrolyte of copper(II) sulfate

solution for 20 minutes, 12 g of copper is deposited at the cathode. Determine

the number of moles of electrons required to deposit this much copper.

Solution

To determine the number of moles of electrons required to deposit the given

mass of copper, we first need to calculate the charge passed through the cell

during the electrolysis process.

Step 1: Calculate the charge passed Given: Current, I= 2.5 A

Time, t= 20 minutes = 20 ×60 seconds = 1200 seconds

The charge, Q, passed through the cell is given by:

Q=I×t

Substitute the given values to find the charge passed through the cell.

Step 2: Calculate the number of moles of electrons The equation

relating charge to the number of moles of electrons is:

Q=nF

where nis the number of moles of electrons and Fis the Faraday constant

(9.65 ×104C/mol).

Solving for n, we get:

n=Q

Substitute the calculated charge into the equation to find the number of

moles of electrons.

Step 3: Calculate the mass of copper deposited Given: Mass of cop-

per, m= 12 g

Molar mass of copper, MCu = 63.55 g/mol

The number of moles of copper deposited can be calculated using the for-

mula:

nCu =m

MCu

Substitute the given values to find the number of moles of copper deposited.

Step 4: Determine the number of moles of electrons required Since

the number of moles of electrons and the number of moles of copper deposited

during electrolysis are equal, we have:

n=nCu

Substitute the calculated values for the number of moles of electrons and

copper to find the final answer.

Question 30

Question

A student is conducting an electrolysis experiment using a copper (Cu) electrode

and a silver (Ag) electrode connected to a battery. The student observes that

0.5 grams of copper is deposited on the copper electrode over a period of 10

minutes. Calculate the current passing through the cell, given that the atomic

mass of copper is 63.55 g/mol and the Faraday constant is 96485 C/mol.

Solution

Step 1: Find the number of moles of copper deposited on the copper electrode.

Atomic mass of copper (Cu) = 63.55 g/mol

Moles of Cu deposited = Mass of Cu deposited

Atomic mass of Cu

Moles of Cu deposited = 0.5 g

63.55 g/mol = 0.00788 mol

Step 2: Determine the quantity of charge passing through the cell.

Quantity of charge = Moles of Cu deposited ×Faraday constant

Quantity of charge = 0.00788 mol ×96485 C/mol = 0.761 C

Step 3: Calculate the current passing through the cell.

Current = Quantity of charge

Time

Current = 0.761 C

10 min =0.761 C

600 s = 0.00126 A

Therefore, the current passing through the cell is 0.00126 A.

Question 31

Question

Electrolysis of an aqueous solution of silver nitrate (AgNO3) is carried out using

a silver electrode. If the electrolysis is run at a constant current of 2 A for 30

minutes, calculate the mass of silver deposited on the electrode. Given that the

molar mass of silver is 107.87 g/mol and the Faraday constant is 9.65 ×104

C/mol.

Solution

Step 1: Calculate the total charge passed through the circuit. Given that the

current is 2 A and the time is 30 minutes, we first convert the time to seconds:

30 minutes = 30 ×60 = 1800 seconds

The total charge passed can be determined using the formula:

Q=I×t

where Qis the charge, Iis the current, and tis the time.

Substitute the values to find Q:

Q= 2 A ×1800 s = 3600 C

Step 2: Determine the moles of silver deposited on the electrode. The num-

ber of moles of silver (n) can be calculated using the formula:

n=Q

where Qis the charge passed and Fis Faraday’s constant.

Substitute the values to find n:

n=3600 C

9.65 ×104C/mol ≈0.0373 mol

Step 3: Calculate the mass of silver deposited. The mass of silver deposited

can be determined using the formula:

Mass = n×Molar mass

where the molar mass of silver is 107.87 g/mol.

Substitute the values to find the mass:

Mass = 0.0373 mol ×107.87 g/mol ≈4.02 g

Therefore, the mass of silver deposited on the electrode is approximately

4.02 g.

Question 32

Question

A student is conducting an electrolysis experiment using a solution of copper

sulfate, CuSO4. The student passes a current of 2.50 A through the solution for

30.0 minutes. The student observes that 3.50 g of copper metal is deposited on

the cathode.

What is the standard cell potential for the reaction Cu2+(aq)+2e−→Cu(s)?

Solution

Step 1: Calculate the charge (Q) passing through the electrolyte solution using

the formula Q=I·t.

Given: I= 2.50 A, t = 30.0 minutes = 1800 s

Q= 2.50 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of copper deposited at the cathode

using the formula n=m

Given: m= 3.50 g, M = 63.5 g/mol

n=3.50 g

63.5 g/mol ≈0.0551 mol

Step 3: Determine the number of electrons transferred in the reaction using

the stoichiometry of the reaction.

From the reaction: 1 mol of Cu2+ requires 2 mol of electrons

2 mol ×0.0551 mol = 0.1102 mol of electrons

Step 4: Calculate the Faraday constant (F) using the formula F=Q

F=4500 C

0.1102 mol ≈40787 C/mol

Step 5: Calculate the standard cell potential using the formula E◦=Q

nF .

E◦=4500 C

0.1102 mol ×40787 C/mol ≈1.03 V

Therefore, the standard cell potential for the reaction Cu2+(aq) + 2e−→

Cu(s) is approximately 1.03 V.

Question 33

Question

During the electrolysis of an aqueous solution of copper(II) sulfate, 1.20 grams

of copper is deposited at the cathode. Calculate the total charge passed through

the cell. Given that the Faraday constant is 96485 C/mol.

Solution

Step 1: Calculate the number of moles of copper deposited at the cathode.

Step 2: Use the mole-to-charge relationship to find the total charge passed

through the cell.

Step 1: Calculate the number of moles of copper deposited at the cathode.

The molar mass of copper is 63.55 g/mol. Therefore, the number of moles of

copper deposited can be calculated as:

Number of moles = Mass

Molar mass =1.20 g

63.55 g/mol

Number of moles ≈0.0189 mol

Step 2: Use the mole-to-charge relationship to find the total charge passed

through the cell.

Since 1 mole of electrons has a charge of 96485 C (1 Faraday), we can calculate

the total charge passed through the cell as:

Total charge = Number of moles ×96485 C/mol

Total charge = 0.0189 mol ×96485 C/mol

Total charge ≈1825 C

Therefore, the total charge passed through the cell during the electrolysis is

approximately 1825 Coulombs.

Question 34

Question

Explain Faraday’s laws of electrolysis.

Solution

Faraday’s laws of electrolysis describe the quantitative relationship between the

amount of a substance produced or consumed during electrolysis and the amount

of electricity passed through the electrolyte. There are two laws: 1. The amount

of a substance produced or consumed during electrolysis is directly proportional

to the quantity of electricity passed through the electrolyte. 2. The amounts of

different substances produced or consumed by the same quantity of electricity

are proportional to their equivalent weights.

Step 1: Let Qbe the total charge passing through the electrolyte in coulombs

(C), Ibe the current passing through the electrolyte in amperes (A), and tbe

the time the current flows in seconds (s). The charge passed through the elec-

trolyte can be calculated as:

Q=I×t

Step 2: The amount of a substance produced or consumed during electrol-

ysis is directly proportional to the quantity of electricity passed through the

electrolyte. This relationship is given by Faraday’s law:

m=Q

F×M

where: - mis the mass of the substance produced or consumed in grams (g),

-Qis the total charge passed through the electrolyte in coulombs (C), - Fis

Faraday’s constant (96500 C/mol), - Mis the molar mass of the substance in

grams per mole (g/mol).

Step 3: The second law states that the amounts of different substances

produced or consumed by the same quantity of electricity are proportional to

their equivalent weights. The equivalent weight (E) of a substance is the mass

that would react with or produce 1 mole of electrons. It is related to the molar

mass (M) of the substance by:

E=M

where nis the number of electrons involved in the reaction.

Step 4: From Faraday’s law, we can rewrite it in terms of the equivalent

weight:

m=Q

F×E

This equation shows that the mass of a substance produced or consumed during

electrolysis is directly proportional to the equivalent weight of the substance.

Thus, Faraday’s laws of electrolysis are fundamental principles that govern

the quantitative aspects of electrolysis reactions.

Question 35

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution. The student observes that copper is deposited on the cathode. If the

student uses a current of 2.5 A for 2 hours, calculate the mass of copper de-

posited. The atomic mass of copper is 63.55 g/mol and the Faraday constant is

96,485 C/mol.

(You may assume 100

Question 2

Question

An aqueous solution of sodium chloride is subjected to electrolysis using inert

electrodes. If a current of 2.00 A is passed through the solution for 1.00 hour,

what mass of sodium metal is deposited at the cathode? (Assume 100

Solution

Step 1: Determine the quantity of electricity passed through the solution. The

quantity of electricity (Q) can be calculated using the formula:

Q=I×t

where: - Qis the quantity of electricity, - Iis the current in amperes (A), - tis

the time in hours.

Given that I= 2.00 A and t= 1.00 hour, we have:

Q= 2.00 A ×1.00 hr = 2.00 C

Step 2: Calculate the number of moles of electrons passed through the so-

lution. Since 1 Faraday (F) is equal to the charge of 1 mole of electrons (1 F

= 96500 C), the number of moles of electrons (n) can be calculated using the

formula:

n=Q

96500

Substitute Q= 2.00 C into the formula:

n=2.00 C

96500 C/F ≈2.07 ×10−5mol

Step 3: Determine the number of moles of sodium deposited at the cathode.

In the electrolysis of sodium chloride, 2 moles of electrons are required to pro-

duce 2 moles of sodium. Therefore, the number of moles of sodium deposited

at the cathode (nNa) is equal to ncalculated in Step 2:

nNa = 2.07 ×10−5mol

Step 4: Calculate the mass of sodium deposited at the cathode. Using the

molar mass of sodium (MNa = 23.0 g/mol), the mass of sodium deposited can

be determined:

Mass of sodium = nNa ×MNa

Mass of sodium = 2.07 ×10−5mol ×23.0 g/mol

Mass of sodium ≈4.76 ×10−4g

Therefore, approximately 4.76 ×10−4g of sodium metal is deposited at the

cathode.

Question 3

Question

A student is conducting an electrolysis experiment using a copper(II) chloride

solution. If the student passes a current of 2.50 A through the solution for 30

minutes, how much copper will be deposited at the cathode? (Given: Atomic

mass of copper = 63.55 g/mol, Faraday’s constant = 9.65 ×104C/mol)

Solution

Step 1: Calculate the total charge passed through the solution. Given: Current

(I) = 2.50 A, Time (t) = 30 minutes = 1800 seconds.

Using the formula:

Q=I×t

Q= 2.50 A ×1800 s

Q= 4500 C

Step 2: Calculate the number of moles of electrons passed. Since 1 Faraday

(F) is equal to the charge of 1 mole of electrons,

1 F = 9.65 ×104C/mol

So, the number of moles of electrons passed can be calculated as:

Moles of electrons = Q

9.65 ×104

Moles of electrons = 4500

9.65 ×104

Moles of electrons ≈0.0466 mol

Step 3: Calculate the number of moles of copper deposited. From the bal-

anced chemical equation:

2e−+Cu2+ →Cu

1 mole of Cu is deposited for every 2 moles of electrons transferred.

Therefore, the moles of copper deposited is half the moles of electrons passed:

Moles of copper = 1

2×Moles of electrons

Moles of copper = 1

2×0.0466

Moles of copper = 0.0233 mol

Step 4: Calculate the mass of copper deposited. Using the atomic mass of

copper:

Mass of copper = Moles of copper ×Atomic mass of copper

Mass of copper = 0.0233 ×63.55

Mass of copper ≈1.48 g

Therefore, approximately 1.48 grams of copper will be deposited at the cath-

ode.

Question 4

Question

State Faraday’s laws of electrolysis and explain how they are applied in deter-

mining the amount of substance deposited during electrolysis of a solution.

Solution

Faraday’s laws of electrolysis are a set of two laws that describe how the amount

of substance deposited or liberated during electrolysis is related to the amount

of electric charge passed through the electrolytic cell.

Faraday’s First Law: The mass of an element deposited during electrolysis

is directly proportional to the quantity of electricity passed through the cell.

Faraday’s Second Law: The masses of different elements deposited by

the same quantity of electricity are proportional to their chemical equivalent

weights.

Application: To determine the amount of substance deposited during elec-

trolysis of a solution, we can use Faraday’s laws in the following way:

Step 1: Determine the number of moles of electrons transferred during the

electrolysis process using the equation:

moles of electrons = electric charge passed (Coulombs)

where Fis the Faraday constant (96,485 C/mol).

Step 2: Determine the number of moles of the substance deposited by

dividing the number of moles of electrons by the number of electrons involved

in the reaction (from the balanced equation).

Step 3: Calculate the mass of the substance deposited using the formula:

mass of substance = number of moles ×molar mass

By following these steps and applying Faraday’s laws, we can accurately

determine the amount of substance deposited during electrolysis of a solution.

Question 5

Question

During the electrolysis of an aqueous solution of copper sulfate, a current of

5.00 A is passed through the solution for 1.50 hours. If the current efficiency is

(Given: atomic mass of copper = 63.5 g/mol, Faraday constant = 96485 C/mol)

Solution

Step 1: Calculate the total charge passed through the electrolyte. The formula

to calculate the charge passed is:

Q=I·t

where: Q= charge (in coulombs) I= current (in amperes) t= time (in seconds)

Converting time from hours to seconds:

t= 1.50 hours ×3600 seconds/hour = 5400 seconds

Substitute the given values:

Q= 5.00 A ×5400 s = 27000 C

Step 2: Calculate the number of moles of electrons transferred. The Faraday

constant relates the charge to the number of moles of electrons:

1 mol of electrons = 96485 C

Moles of electrons = Q

96485

Substitute the charge value:

Moles of electrons = 27000

96485 = 0.28 mol

Step 3: Calculate the moles of copper deposited. Since the current efficiency

is 80

Moles of copper deposited = 0.28 mol ×0.80 = 0.224 mol

Step 4: Calculate the mass of copper deposited. Using the atomic mass of

copper given:

Mass of copper deposited = Moles of copper deposited×Atomic mass of copper

Mass of copper deposited = 0.224 mol ×63.5 g/mol = 14.24 g

Therefore, the mass of copper deposited at the cathode is 14.24 g.

Question 6

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of molten

lead(II) bromide, PbBr2. If lead is deposited at the cathode, what is the mass

of lead deposited in 30 minutes? (Given: atomic masses of Pb = 207.2 g/mol,

Br = 79.9 g/mol; Faraday constant = 96485 C mol−1)

Solution

Step 1: Find the number of moles of electrons passing through the circuit. Step

2: Use the stoichiometry of the electrolysis reaction to find the moles of lead

deposited. Step 3: Calculate the mass of lead deposited.

Step 1: The current passing through the circuit is I= 2.5Aand the time

is t= 30 min =30

60 hrs = 0.5hrs.

The charge passing through the circuit is given by Q=I·t.

Now, using Faraday’s law of electrolysis:

Q=nF

where nis the number of moles of electrons and Fis the Faraday constant.

Substitute the given values to find n:

n=Q

F=I·t

F=2.5A·0.5hrs

96485 C mol−1

Step 2: The balanced equation for the electrolysis of molten lead(II) bro-

mide is:

Pb2+ + 2Br−→Pb + Br2

This indicates that 2 moles of electrons are required to deposit 1 mole of lead.

So, the moles of lead deposited is half the moles of electrons:

moles of Pb = n

Step 3: Calculate the mass of lead deposited using the moles of lead ob-

tained in Step 2:

mass of Pb = moles of Pb×molar mass of Pb = I·t

2F×molar mass of Pb = 2.5A·0.5hrs

2×96485 C mol−1×207.2g/mol

Question 7

Question

A student is performing an electrolysis experiment using a solution of sodium

chloride (NaCl). The student passes a current of 2.5 A through the solution for

30 minutes. During this time, they observe the formation of chlorine gas (Cl2)

at one electrode and sodium metal (Na) at the other electrode. Calculate the

mass of sodium metal deposited at the cathode during this electrolysis process.

(Given: Faraday constant, F= 96485 C/mol, atomic mass of sodium, MNa =

23 g/mol, 1 hour = 3600 seconds)

Solution

Step 1: Calculate the total charge passed through the electrolyte: The total

charge passed can be calculated using the formula:

Q=It

where Qis the total charge (in coulombs), Iis the current (in amperes), and tis

the time (in seconds). Given I= 2.5 A and t= 30 minutes = 30 ×60 s = 1800 s,

we have:

Q= 2.5 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of electrons passed: One mole of

electrons has a charge of 1 Faraday, which is equal to the charge of 1 mol of

electrons. Hence, the number of moles of electrons can be calculated using the

formula:

Number of moles of electrons = Q

Given F= 96485 C/mol, we have:

Number of moles of electrons = 4500 C

96485 C/mol ≈0.0467 mol

Step 3: Calculate the number of moles of sodium deposited: In the electrol-

ysis of molten sodium chloride, 2 moles of electrons are required to reduce 1

mole of sodium ions (Na+) to form 1 mole of sodium metal (Na). Therefore,

in this case, since 0.0467 moles of electrons are passed, the number of moles of

sodium deposited will be half that amount:

Number of moles of sodium deposited = 0.0467 mol

2= 0.0234 mol

Step 4: Calculate the mass of sodium deposited: The mass of sodium de-

posited can be calculated using the formula:

Mass of sodium deposited = Number of moles ×Molar mass

Given the molar mass of sodium, MNa = 23 g/mol, we have:

Mass of sodium deposited = 0.0234 mol ×23 g/mol = 0.5382 g

Therefore, the mass of sodium metal deposited at the cathode during this

electrolysis process is approximately 0.5382 grams.

Question 8

Question

A student performs an electrolysis experiment using a current of 2.50 A. If

the student plated out 0.500 g of copper, determine the time taken for this

electrolysis to occur. (Given: Faraday constant F= 96,485 C/mol and the

molar mass of copper is 63.55 g/mol.)

Solution

Step 1: Calculate the number of moles of copper plated out.

Moles of copper = Mass of copper plated out

Molar mass of copper =0.500 g

63.55 g/mol

Step 2: Calculate the charge passed through the electrolyte solution.

Q=It

where Qis the charge passed (in coulombs), Iis the current (in amperes), and

tis the time taken (in seconds).

Step 3: Find the number of moles of electrons transferred. Each mole of

copper requires 2 moles of electrons for the reduction process. Therefore,

Moles of electrons = 2 ×Moles of copper

Step 4: Calculate the total charge required to plate out the given amount of

copper. The total charge required is given by:

Q=n×F

where nis the number of moles of electrons transferred, and Fis the Faraday

constant.

Step 5: Equate the two expressions for charge passed and charge required.

It =n×F

Step 6: Solve for t.

t=n×F

Question 9

Question

A solution contains both Cu2+ and Ag+ions. A current of 2.0 A is passed

through the solution for 3.0 hours. If 1.0 g of copper is deposited at the cathode,

what mass of silver is deposited at the cathode? (Given: F = 96500 C/mol,

m(Cu) = 63.5 g/mol, m(Ag) = 107.9 g/mol)

Solution

Step 1: Calculate the charge that passed through the cell. The charge (Q)

passing through the cell can be calculated using the formula:

Q=I·t

where Iis the current in amperes and tis the time in seconds. Given I= 2.0 A,

t= 3.0 hours ×3600 s/hour = 10800 s, we have

Q= 2.0 A ×10800 s = 21600 C

Step 2: Determine the number of moles of copper deposited. The number

of moles of copper deposited (nCu) can be calculated using the formula:

nCu =m

where mis the mass of copper deposited and Mis the molar mass of copper.

Given m= 1.0 g, M= 63.5 g/mol, we have

nCu =1.0 g

63.5 g/mol = 0.0157 mol

Step 3: Using the ratio of charges passed and moles deposited to find the

molar mass of silver. The ratio of charges passed and moles deposited is the

same for both copper and silver because the same current is used. Therefore,

we can equate the two ratios to find the molar mass of silver.

QCu

nCu

=QAg

nAg

where QCu is the charge required to deposit copper and QAg is the charge

required to deposit silver. Given that Q= 21600 C has passed through the cell

and the Faraday constant F = 96500 C/mol, we can substitute:

nCu ·F=Q

nAg ·F

Step 4: Solve for the mass of silver deposited. Substitute the known values

into the formula derived in Step 3 and solve for the mass of silver deposited

(mAg):

21600

0.0157 ·96500 =21600

nAg ·96500

nAg =21600

0.0157 ·96500

nAg = 1.44 ×10−2mol

Finally, to find the mass of silver deposited:

mAg =nAg ·MAg

mAg = 1.44 ×10−2mol ×107.9 g/mol

mAg = 1.55 g

Therefore, 1.55 g of silver is deposited at the cathode.

Question 10

Question

An aqueous solution of silver nitrate (AgNO3) is electrolyzed using inert elec-

trodes. If 96500 C of charge is passed through the solution, calculate the

mass of silver deposited at the cathode. Given: 1 F = 96500 Coulombs/mol,

molar mass of Ag = 107.87 g/mol.

Solution

Step 1: Determine the number of moles of electrons passed through the solution.

- From Faraday’s first law of electrolysis, we know that 1 Faraday (F) of charge

is equivalent to 1 mole of electrons. - Given that 96500 Coulombs of charge is

passed, we can calculate the number of moles of electrons:

Moles of electrons = 96500 C

96500 C/mol = 1 mol

Step 2: Determine the number of moles of silver deposited at the cathode. -

The balanced half-reaction for the deposition of silver at the cathode is: Ag++

1e−→Ag. - From the reaction, we see that 1 mole of silver is deposited for

every 1 mole of electrons passed. - Since 1 mole of electrons is passed, 1 mole

of silver is deposited at the cathode.

Step 3: Calculate the mass of silver deposited. - The molar mass of silver is

given as 107.87 g/mol. - The mass of silver deposited can be calculated as:

Mass of silver = Number of moles×Molar mass = 1 mol×107.87 g/mol = 107.87 g

Therefore, the mass of silver deposited at the cathode is 107.87 g.

Question 11

Question

In an electrolytic cell, a current of 2.50 A is passed through a solution of cop-

per(II) sulfate for 30.0 minutes. If the reaction produces 5.00 g of copper,

calculate the Faraday constant (F) and Faraday’s constant (F).

Solution

Step 1: Find the moles of copper produced. Given: Current, I= 2.50 A Time,

t= 30.0 minutes Mass of copper produced, m= 5.00 g Atomic mass of copper,

MCu = 63.55 g/mol

We first convert time from minutes to seconds: 30.0 minutes = 30.0×

60 seconds = 1800 seconds

We then calculate the moles of copper produced:

moles of Cu = mass of Cu

MCu

=5.00 g

63.55 g/mol = 0.0787 mol

Step 2: Calculate the charge passed through the cell. We use the formula to

find the charge:

Q=It

where Qis the charge, Iis the current, and tis the time. Substitute the given

values:

Q= 2.50 A ×1800 s = 4500 C

Step 3: Find Faraday’s constant (F). The number of moles of electrons

involved in the reaction is equal to the moles of copper produced, which is

0.0787 mol. Since 1 F (Faraday) is the charge carried by one mole of electrons,

we have:

F=Q

where: - nis the number of moles of electrons; - Qis the charge passed. Sub-

stitute the values:

F=4500 C

0.0787 mol = 57161 C/mol

Step 4: Find Faraday constant (F). The Faraday constant (F) is related to

Avogadro’s constant and is given by:

F=F × NA

where NAis Avogadro’s constant (6.022 ×1023 mol−1). Substitute the values:

F= 57161 C/mol ×6.022 ×1023 mol−1= 3.44 ×103C/mol

Therefore, Faraday constant F= 3.44 ×103C/mol.

Question 12

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution with copper electrodes. The student passes a current of 2.50 A for 30

minutes. During this time, copper is deposited on one electrode. Determine the

mass of copper deposited and the volume of gas evolved at the other electrode.

(Take the molar mass of copper as 63.5 g/mol, and assume standard temperature

and pressure for the gas.)

Solution

Step 1: Find the amount of charge passed through the electrolyte.

Charge(Q) = Current ×Time

Given that the current is 2.50 A and the time is 30 minutes, we need to

convert the time to seconds: 30 minutes = 30 ×60 seconds = 1800 seconds.

Therefore, the charge passed through the electrolyte is:

Q= 2.50 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of electrons passed.

Moles of electrons = Charge

Faraday constant

The Faraday constant is 96485 C/mol.

Moles of electrons = 4500 C

96485 C/mol ≈0.0467 mol

Step 3: Determine the mass of copper deposited on the electrode.

Mass of copper = Moles of electrons ×Molar mass of copper

Given that the molar mass of copper is 63.5 g/mol:

Mass of copper = 0.0467 mol ×63.5 g/mol ≈2.97 g

Step 4: Calculate the volume of gas evolved at the other electrode using the

ideal gas law.

Volume of gas = Number of moles of gas ×Universal gas constant ×Temperature

Pressure

At standard temperature and pressure, the temperature is 273 K and the

pressure is 1 atm. Since one mole of electrons produces one mole of gas (by

Faraday’s first law of electrolysis), the number of moles of gas is 0.0467 mol.

Volume of gas = 0.0467 mol ×0.0821 L/mol ·atm ·K−1×273 K

1 atm ≈1.07 L

Question 13

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution. The student plated out 10.0 g of copper in 30 minutes using a current

of 2.00 A. Calculate the number of moles of electrons that flowed through the

cell during this time.

Solution

Step 1: Find the molar mass of copper. Step 2: Calculate the number of moles

of copper deposited. Step 3: Use Faraday’s laws of electrolysis to determine the

number of moles of electrons.

Step 1:

The molar mass of copper is 63.55 g/mol.

Step 2:

Given: Mass of copper plated out, m= 10.0 g

Molar mass of copper, M= 63.55 g/mol

Number of moles of copper deposited:

n=m

M=10.0 g

63.55 g/mol = 0.1577 mol

Step 3:

Given: Time, t= 30 min = 1800 s

Current, I= 2.00 A

We know that 1 mole of electrons carries a charge of 1 F (Faraday). From

Faraday’s laws of electrolysis, the total charge Qpassed through the cell can be

calculated using:

Q=It

The total charge can be related to the number of moles of electrons, ne, through

the equation:

Q=ne×F

where Fis the Faraday’s constant.

Therefore, we have:

Q=ne×F=⇒It =ne×F=⇒ne=It

Substitute the values:

ne=2.00 A ×1800 s

96485 C/mol = 0.0372 mol

So, the number of moles of electrons that flowed through the cell during this

time is 0.0372 moles.

Question 14

Question

A solution contains a mixture of copper(II) sulfate, CuSO4, and silver nitrate,

AgNO3, ions. A current of 2.50 A is passed through the solution for 2.00 hours.

During this time, 2.00 g of copper is deposited at the cathode.

Given:

Ecu =−0.34V

Eag = 0.80V

molar mass of Cu = 63.55g/mol

molar mass of Ag = 107.87g/mol

F= 96500C/mol

Calculate the mass of silver deposited during the electrolysis process.

Solution

Step 1: Calculate the charge passed through the electrolyte. Given that the

current, I, is 2.50 A and the time, t, is 2.00 hours, we can calculate the total

charge passed, Q, using the formula Q=I×t.

Q= 2.50A×2.00hours = 5.00Ah

Step 2: Calculate the number of moles of copper deposited. The number of

moles, n, of copper deposited can be calculated using the formula

n=m

where mis the mass of copper deposited (2.00 g) and Mis the molar mass of

copper (63.55 g/mol).

n=2.00g

63.55g/mol = 0.03151mol

Step 3: Calculate the number of electrons transferred during the deposition

of copper. The number of electrons transferred, ne, during the deposition of

copper can be calculated using Faraday’s laws of electrolysis. For the reduction

of copper(II) ions to copper atoms, the number of electrons transferred can be

determined by balancing the half-reaction:

Cu2+ + 2e−→Cu

From this half-reaction, we can see that 2 electrons are involved in the deposition

of 1 mole of copper. Therefore, the total number of electrons transferred during

the deposition of 0.03151 moles of copper is:

ne= 2 ×0.03151 = 0.06302 moles of electrons

Step 4: Calculate the charge carried by these electrons. The charge car-

ried by the electrons can be calculated by multiplying the number of moles of

electrons by Faraday’s constant, F.

Qe=ne×F= 0.06302mol ×96500C/mol = 6083.63C

Step 5: Calculate the mass of silver deposited. Now, we need to find out

how many moles of silver ions were reduced during the electrolysis process. To

do this, we find the total charge passed during the electrolysis, Q, and use the

standard reduction potential of silver, Eag , to determine the number of moles

of silver ions reduced. Since the total charge passed is the sum of the charge

carried by the copper ions and the silver ions:

Q=QCu2+ +QAg+

QAg+=Q−QCu2+

QAg+= 5.00Ah −6083.63C= 17916.37C

Then, we calculate the number of moles of silver ions, nAg+, reduced using

the formula:

nAg+=QAg+

nAg+=17916.37C

96500C/mol = 0.1858mol

Finally, we calculate the mass of silver deposited using the number of moles

of silver ions reduced and the molar mass of silver:

mAg =nAg+×MAg

mAg = 0.1858mol ×107.87g/mol =20.01 g

Therefore, the mass of silver deposited during the electrolysis process is 20.01

Question 15

Question

Consider an electrolytic cell containing a solution of copper sulfate (CuSO4).

A steady current of 2.5 A is passed through the cell for 20 minutes. If copper

is deposited at the cathode, what mass of copper is deposited?

Given: Faraday’s constant, F= 96,500 C/mol Atomic mass of copper,

Cu = 63.5g/mol

Solution

Step 1: Calculate the total charge passing through the cell. Given that the

current is 2.5 A and time is 20 minutes, first convert the time to seconds:

20 minutes = 20 ×60 = 1200 seconds

The total charge, Q, passing through the cell can be calculated using the

formula:

Q=I×t

Q= 2.5A×1200 s

Q= 3000 C

Step 2: Calculate the number of moles of electrons. Since each mole of

electrons carries a charge of 1 Faraday, we can calculate the number of moles of

electrons using the formula:

n=Q

n=3000 C

96500 C/mol

n≈0.0311 mol

Step 3: Calculate the amount of copper deposited. From the balanced equa-

tion for the deposition of copper, we know that 2 moles of electrons are required

to deposit 1 mole of copper. Therefore, the number of moles of copper deposited

will be half of the number of moles of electrons.

nCu =n

nCu =0.0311 mol

nCu ≈0.0156 mol

Step 4: Calculate the mass of copper deposited. The mass of copper de-

posited can be calculated using the formula:

Mass = nCu ×Molar mass of Cu

Mass = 0.0156 mol ×63.5 g/mol

Mass ≈0.9936 g

Therefore, approximately 0.9936 grams of copper will be deposited.

Question 16

Question

An electrolytic cell contains a solution of silver nitrate (AgNO3). If a current of

2.50 A is passed through the cell for 2.00 hours, what mass of silver is deposited

at the cathode? (Given: 1 F = 96,485 C/mol, mAg = 107.87 g/mol)

Solution

Step 1: Calculate the total charge passing through the cell. Let’s use the formula

Q=I·t, where Qis the charge, Iis the current, and tis the time. Given

I= 2.50 A and t= 2.00 hours, first convert the time to seconds:

t= 2.00 hours ×3600 s/hour = 7200 s

Now, calculate the total charge:

Q= 2.50 A ×7200 s = 18,000 C

Step 2: Determine the number of moles of silver deposited. We know that

1 F = 96,485 C/mol. We can calculate the number of moles of silver deposited

using the equation Q=n·F, where nis the number of moles and Fis Faraday’s

constant.

n=Q

F=18,000 C

96,485 C/mol ≈0.1863 mol

Step 3: Calculate the mass of silver deposited. The molar mass of silver is

107.87 g/mol. We can use this information to find the mass of silver deposited

at the cathode:

Mass of silver = n×Molar mass of silver = 0.1863 mol×107.87 g/mol = 20.09 g

Therefore, approximately 20.09 grams of silver is deposited at the cathode.

Question 17

Question

In an electrolytic cell, a current of 3.0 A is passed for 1.5 hours through a

solution of copper(II) sulfate, CuSO4. Calculate the mass of copper deposited

on the cathode. Given that the molar mass of copper is 63.5 g/mol and the

Faraday constant is 96,500 C/mol.

Solution

Step 1: Calculate the total charge passed through the cell.

Charge (C) = Current (A) ×Time (s)

Step 2: Convert the time from hours to seconds.

1.5 hours = 1.5×60 ×60 seconds

Step 3: Calculate the total charge in Coulombs.

Charge (C) = 3.0 A ×1.5×60 ×60 s

Step 4: Calculate the number of moles of electrons passed through the cell

using Faraday’s laws of electrolysis.

1 coulomb = 1

96500 mol of electrons

Step 5: Calculate the number of moles of electrons.

Moles of electrons = Charge (C)

96500 C/mol

Step 6: Since each copper ion, Cu2+, requires 2 moles of electrons to form

copper metal, the number of moles of copper deposited will be half of the moles

of electrons.

Step 7: Calculate the mass of copper deposited using the number of moles

of copper and the molar mass of copper.

Mass of copper (g) = Moles of copper ×Molar mass of copper (g/mol)

Question 18

Question

A solution of copper(II) sulfate, CuSO4, is electrolyzed using a constant current

of 2.50 A for 2.00 hours. If copper metal forms at the cathode, calculate the

mass of copper deposited.

(The molar mass of copper is 63.5 g/mol, the Faraday constant is 96,485

C/mol, and the charge of an electron is 1.60 ×10−19 C)

Solution

Step 1: Calculate the total charge passing through the circuit. Given: Current,

I= 2.50 A Time, t= 2.00 hours = 7200 s

The total charge Qcan be calculated using the formula Q=I×t. Plugging

in the values, we get Q= 2.50 A ×7200 s = 18000 C.

Step 2: Determine the number of moles of electrons. The number of moles

of electrons can be calculated using the equation n=Q

F, where Fis Faraday’s

constant. Substitute the values to get n=18000 C

96485 C/mol .

n≈0.18625 mol.

Step 3: Find the number of moles of copper deposited. From the balanced

equation in electrolysis of copper(II) sulfate:

Cu2+ + 2e−→Cu

we see that 2 moles of electrons are required to deposit 1 mole of copper. Thus,

the number of moles of copper deposited is half the number of moles of electrons:

nCu =n

2=0.18625 mol

nCu ≈0.09313 mol.

Step 4: Calculate the mass of copper deposited. The mass of copper, m, can

be calculated using the formula m=nCu ×Molar mass of copper. Substitute

the values to get m= 0.09313 mol ×63.5 g/mol.

m≈5.92 g.

Therefore, the mass of copper deposited during the electrolysis process is

approximately 5.92 g.

Question 19

Question

Consider an electrolytic cell where copper metal is to be obtained from a copper

sulfate solution using a current of 5.0 A. If it takes 25 minutes to deposit 10.5

g of copper metal, what is the total charge that passed through the cell during

this time? Assume the Faraday constant is 9.65 ×104C/mol.

Solution

Step 1: Calculate the moles of copper deposited. Given that the molar mass of

copper is 63.5 g/mol, we can calculate the moles of copper deposited:

Moles of copper = Mass of copper deposited

Molar mass of copper

Moles of copper = 10.5 g

63.5 g/mol = 0.165 mol

Step 2: Calculate the total charge passed through the cell. The total charge

passed through the cell can be calculated using Faraday’s laws of electrolysis:

Total charge = Number of moles of electrons ×Faraday constant

Since the reaction to deposit copper involves the transfer of 2 moles of electrons

per mole of copper:

Number of moles of electrons = 2 ×Moles of copper

Number of moles of electrons = 2 ×0.165 = 0.33 mol

Now, we can calculate the total charge:

Total charge = 0.33 mol ×9.65 ×104C/mol = 3.18 ×104C

Therefore, the total charge that passed through the cell during this time is

3.18 ×104C.

Question 20

Question

A solution of silver nitrate AgNO3is electrolyzed using a silver electrode. The

mass of silver electrode decreases by 0.625 g in 20 minutes. Assuming 100

Solution

Step 1: Find the number of moles of silver deposited on the electrode.

Given: Change in mass of silver electrode = 0.625 g Time taken = 20 minutes

= 20/60 hours = 1/3 hours

The molar mass of silver (Ag) = 107.87 g/mol

Number of moles of silver deposited = Change in mass / Molar mass Number

of moles of silver deposited = 0.625 g / 107.87 g/mol

Step 2: Calculate the charge passing through the electrolytic cell.

1 mole of electrons = 1 Faraday of charge = 96485 C

Total charge passed = number of moles x Avogadro’s number x Faraday’s

constant

Step 3: Determine the current flowing through the electrolytic cell.

Current (I) = Total charge passed / time taken

These steps can be used to determine the current flowing through the elec-

trolytic cell when a solution of silver nitrate is electrolyzed using a silver elec-

trode.

Question 21

Question

During the electrolysis of aqueous copper(II) sulfate solution using copper elec-

trodes, if a current of 2.50 A is passed through the solution for 2.00 hours,

calculate the mass of copper deposited at the cathode. (Atomic masses: Cu =

63.5 g/mol, S = 32.1 g/mol, O = 16.0 g/mol)

Solution

Step 1: Find the total charge passed through the electrolyte. Given that the

current is 2.50 A and the time is 2.00 hours, we can use the formula:

Q=I×t

Q= 2.50 A ×2.00 hr ×3600 s/hr

Q= 2.50 ×2.00 ×3600

Q= 18000 C

Step 2: Determine the moles of electrons passed. Each mole of electrons has

a charge of 1 Faraday, which is 96485 C.

Moles of electrons = 18000 C

96485 C/mol

Moles of electrons = 0.1866 mol

Step 3: Calculate the mass of copper deposited. The electrolysis of aqueous

copper(II) sulfate solution using copper electrodes will deposit copper at the

cathode. The balanced equation is:

Cu2+ + 2e−→Cu

From the equation, 2 moles of electrons are required to deposit 1 mole of copper.

Therefore, moles of copper deposited = moles of electrons passed / 2

Moles of copper = 0.1866 mol

Moles of copper = 0.0933 mol

Step 4: Calculate the mass of copper deposited.

Mass of copper = Moles of copper ×Molar mass of copper

Mass of copper = 0.0933 mol ×63.5 g/mol

Mass of copper = 5.92 g

Therefore, the mass of copper deposited at the cathode is 5.92 g.

Question 22

Question

A solution of copper(II) sulfate, CuSO4, is electrolyzed using copper electrodes.

If a current of 0.5 A is passed through the solution for 2 hours, what mass of

copper is deposited at the cathode? (Given: 1 F= 96,485 C/mol)

Solution

Step 1: Write the balanced half-reaction for the reduction that occurs at the

cathode. The reduction half-reaction for the deposition of copper from Cu2+

ions is:

Cu2+ + 2e−→Cu

This reaction requires 2 moles of electrons to deposit 1 mole of copper.

Step 2: Calculate the total charge passed through the circuit. The total

charge passed through the circuit can be calculated using the formula:

Q=I×t

where: - Qis the total charge passed (in coulombs), - I= 0.5Ais the current,

and - t= 2 hours = 2 ×3600 sec = 7200 sec is the time.

Substitute the values:

Q= 0.5A×7200 sec = 3600 C

Step 3: Calculate the number of moles of electrons passed through the cir-

cuit. Since 1 mole of electrons carries a charge of 1 F, the number of moles of

electrons can be calculated as:

Moles of electrons = Q

1F=3600 C

96500 C/mol

Moles of electrons ≈0.0374 mol

Step 4: Calculate the moles of copper deposited at the cathode. Since 2

moles of electrons are needed to deposit 1 mole of copper, the moles of copper

deposited can be calculated as:

Moles of copper = 1

2×Moles of electrons

Moles of copper = 1

2×0.0374 mol ≈0.0187 mol

Step 5: Calculate the mass of copper deposited at the cathode. The molar

mass of copper is 63.55 g/mol. Therefore, the mass of copper deposited can be

calculated as:

Mass of copper = Moles of copper ×Molar mass of copper

Mass of copper = 0.0187 mol ×63.55 g/mol ≈1.186 g

Therefore, approximately 1.186 g of copper is deposited at the cathode dur-

ing the electrolysis of copper(II) sulfate solution.

Question 23

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of silver

nitrate (AgNO3) for 2 hours. If the atomic mass of silver is 107.87 g/mol,

calculate the mass of silver deposited on the cathode. (Given: Faraday constant,

F = 96500C/mol)

Solution

Step 1: Calculate the total charge passed through the cell. The charge passed,

Q=I×t, where Iis the current in amperes and tis the time in seconds. Given

I= 2.5Aand t= 2hours = 2 ×3600s= 7200s,

Q= 2.5A×7200 s= 18000 C

Step 2: Calculate the number of moles of electrons passed through the cell.

One mole of electrons carries 1 Faraday of charge, i.e., 96500 C.

Number of moles of electrons = Q

96500

Step 3: Calculate the number of moles of silver deposited. In the reaction,

Ag++e−→Ag, one mole of silver is deposited per mole of electrons transferred.

Hence, the number of moles of silver deposited is equal to the number of moles

of electrons passed through the cell.

Step 4: Calculate the mass of silver deposited. Given the atomic mass of

silver is 107.87 g/mol,

Mass of silver deposited = Number of moles of silver deposited×Atomic mass of silver

Substitute the values and solve for the mass of silver deposited.

Question 24

Question

A student is performing an electrolysis experiment using a copper (II) sulfate

solution. In one trial, the student passed a current of 0.50 A through the solution

for 30 minutes. The student observed that 0.75 g of copper was deposited

on the cathode. Determine the Faraday constant and Faraday’s First Law of

Electrolysis constant using this information.

Solution

Step 1: Calculate the charge passed through the solution. Given: Current (I)

= 0.50 A Time (t) = 30 minutes = 30 * 60 seconds = 1800 s

The charge passed, Q, can be calculated using the formula:

Q=I×t

Q= 0.50 A ×1800 s

Q= 900 C

Step 2: Convert the mass of copper deposited to moles. Given: Mass of

copper deposited = 0.75 g Molar mass of copper (Cu) = 63.5 g/mol

The number of moles of copper deposited can be calculated using the for-

mula:

Moles = Mass

Molar mass

Moles of Cu = 0.75 g

63.5 g/mol

Moles of Cu = 0.0118 mol

Step 3: Determine the number of electrons involved in the electrolysis pro-

cess. One copper atom requires 2 electrons to form Cu: 1 Cu atom + 2e Cu

Given the moles of copper: Moles of Cu = Moles of electrons

Step 4: Calculate the Faraday constant and Faraday’s First Law of Electrol-

ysis constant. The Faraday constant (F) is equal to the charge of one mole of

electrons:

F=Charge (C)

Moles of electrons

F=900 C

0.0118 mol

F≈76241.52 C/mol

The Faraday’s First Law of Electrolysis constant, k, is calculated as:

k=Faraday constant

96000

k=76241.52 C/mol

96000

k≈0.793 mol/C

Therefore, the Faraday constant is approximately 76241.52 C/mol and Fara-

day’s First Law of Electrolysis constant is approximately 0.793 mol/C.

Question 25

Question

In a certain electrolytic cell, a current of 1.5 A is passed for 3 hours. During

this time, the mass of the substance deposited at the cathode is found to be

2.7 g. If the substance is a divalent metal, determine the atomic weight of the

metal (in g/mol). Assume 100

Solution

Step 1: Calculate the total charge passed through the cell. Given that the

current passed is 1.5 A and the time is 3 hours, we first convert the time to

seconds:

Time (s) = 3 hours ×3600 s/hour = 10800 s

Then, calculate the total charge passed using the formula Q=I·t:

Q= 1.5 A ×10800 s = 16200 C

Step 2: Determine the number of moles of the substance deposited. Since

the metal is divalent, the number of moles deposited (n) can be calculated using

Faraday’s first law of electrolysis:

n=Q

where F= 96500 C/mol is the Faraday constant.

Substitute the values to find the number of moles:

n=16200 C

2×96500 C/mol =16200

193000 mol

Step 3: Calculate the molar mass of the metal. Given that 2.7 g of the metal

is deposited, we can calculate the molar mass (M) using the formula:

M=Mass

Moles =2.7 g

16200

193000 mol

Perform the calculation to find the molar mass in g/mol.

Question 26

Question

During the electrolysis of a certain molten compound, 1.20 grams of the com-

pound is decomposed in 10 minutes using a constant current of 2.50 A. Calculate

the molar mass of the compound.

Solution

Step 1: Calculate the charge passed through the circuit. Given that the current

is 2.50 A and the time is 10 minutes, first convert the time to seconds:

10 minutes ×60 seconds/minute = 600 seconds

The charge passed through the circuit can be calculated using the formula Q=

It:

Q= 2.50 A ×600 s = 1500 C

Step 2: Calculate the number of moles of electrons transferred. For the

compound to be decomposed, 2 moles of electrons must be transferred per mole

of the compound. Therefore, the number of moles of electrons can be calculated

as:

moles of electrons = 1500 C

96500 C/mol =1500

96500 ≈0.0155 mol

Step 3: Determine the moles of the compound decomposed. Since 2 moles

of electrons are required to decompose 1 mole of the compound, the moles of

the compound decomposed is half the moles of electrons:

moles of compound = 0.0155

2= 0.00775 mol

Step 4: Calculate the molar mass of the compound. The molar mass of the

compound can be calculated using the formula:

molar mass = mass of compound

moles of compound =1.20 g

0.00775 mol ≈154.84 g/mol

Therefore, the molar mass of the compound is approximately 154.84 g/mol.

Question 27

Question

An electrolysis experiment was conducted using a solution of copper(II) sulfate,

CuSO4, with copper electrodes. The cell was set up such that the cathode

was the copper strip connected to the negative terminal of the battery. After a

certain amount of time, it was observed that the mass of the copper cathode had

increased. Explain this observation in terms of Faraday’s laws of electrolysis.

Solution

To explain the increase in mass of the copper cathode in the electrolysis exper-

iment using Faraday’s laws of electrolysis, we can break down the process into

the following steps:

Step 1: Determine the reaction occurring at the cathode The re-

action that occurs at the cathode during the electrolysis of copper(II) sulfate

solution is:

Cu2+ + 2e−→Cu

Step 2: Calculate the number of moles of electrons transferred

From the balanced equation, we see that 2 moles of electrons are required for

the reduction of 1 mole of copper ions. So, the number of moles of electrons

transferred is equal to the moles of copper deposited on the cathode.

Step 3: Apply Faraday’s first law of electrolysis Faraday’s first law

states that the mass of a substance produced at an electrode during electrolysis

is directly proportional to the quantity of electricity passed through the cell.

Mathematically, it can be expressed as:

Mass of substance = charge ×molar mass/Faraday’s constant

Step 4: Calculate the mass of copper deposited Given that the cur-

rent passing through the cell is related to the number of moles of electrons

transferred, we can calculate the mass of copper deposited on the cathode using

the formula:

Mass of Cu = Molar mass of Cu ×(Number of moles of electrons transferred)

Step 5: Explain the increase in mass of the copper cathode The in-

crease in mass of the copper cathode observed is due to the deposition of copper

metal from the copper(II) sulfate solution onto the cathode. This deposition

is a result of the reduction of copper ions by the supply of electrons from the

cathode.

Therefore, the increase in mass of the copper cathode is a consequence of the

reduction reaction occurring at the cathode during the electrolysis of copper(II)

sulfate solution.

Question 28

Question

An aqueous solution of copper sulfate is electrolyzed using platinum electrodes.

If 5.0 A of current is passed through the cell for 2.0 hours, what mass of copper

will be deposited at the cathode? (Given: Atomic weight of copper = 63.5

g/mol, Faraday’s constant = 96500 C/mol)

Solution

Step 1: Find the total charge passed through the cell. Given: Current (I) = 5.0

A, Time (t) = 2.0 hours = 7200 s The total charge Q passed through the cell

can be calculated using the formula Q=I×t.Q= 5.0A×7200 s= 36000 C

Step 2: Calculate the number of moles of electrons. Each mole of electrons

corresponds to 1 Faraday of charge, which is equal to 96500 C/mol. Number

of moles of electrons, n, can be determined by dividing the total charge Q by

Faraday’s constant: n=Q

96500 =36000

96500 ≈0.373 mol

Step 3: Determine the number of moles of copper deposited. In the balanced

half-reaction for the electrolysis of copper sulfate, 2 moles of electrons are needed

to deposit 1 mole of copper. So, the number of moles of copper deposited will

be half of the number of moles of electrons: nCu =n

2=0.373

2= 0.1865 mol

Step 4: Calculate the mass of copper deposited. Using the molar mass of

copper (63.5 g/mol), the mass (m) of copper deposited can be found using the

formula m=nCu×Molar mass of copper. m= 0.1865mol×63.5g/mol ≈11.84g

Therefore, approximately 11.84 grams of copper will be deposited at the

cathode.

Question 29

Question

When a current of 2.5 A is passed through an electrolyte of copper(II) sulfate

solution for 20 minutes, 12 g of copper is deposited at the cathode. Determine

the number of moles of electrons required to deposit this much copper.

Solution

To determine the number of moles of electrons required to deposit the given

mass of copper, we first need to calculate the charge passed through the cell

during the electrolysis process.

Step 1: Calculate the charge passed Given: Current, I= 2.5 A

Time, t= 20 minutes = 20 ×60 seconds = 1200 seconds

The charge, Q, passed through the cell is given by:

Q=I×t

Substitute the given values to find the charge passed through the cell.

Step 2: Calculate the number of moles of electrons The equation

relating charge to the number of moles of electrons is:

Q=nF

where nis the number of moles of electrons and Fis the Faraday constant

(9.65 ×104C/mol).

Solving for n, we get:

n=Q

Substitute the calculated charge into the equation to find the number of

moles of electrons.

Step 3: Calculate the mass of copper deposited Given: Mass of cop-

per, m= 12 g

Molar mass of copper, MCu = 63.55 g/mol

The number of moles of copper deposited can be calculated using the for-

mula:

nCu =m

MCu

Substitute the given values to find the number of moles of copper deposited.

Step 4: Determine the number of moles of electrons required Since

the number of moles of electrons and the number of moles of copper deposited

during electrolysis are equal, we have:

n=nCu

Substitute the calculated values for the number of moles of electrons and

copper to find the final answer.

Question 30

Question

A student is conducting an electrolysis experiment using a copper (Cu) electrode

and a silver (Ag) electrode connected to a battery. The student observes that

0.5 grams of copper is deposited on the copper electrode over a period of 10

minutes. Calculate the current passing through the cell, given that the atomic

mass of copper is 63.55 g/mol and the Faraday constant is 96485 C/mol.

Solution

Step 1: Find the number of moles of copper deposited on the copper electrode.

Atomic mass of copper (Cu) = 63.55 g/mol

Moles of Cu deposited = Mass of Cu deposited

Atomic mass of Cu

Moles of Cu deposited = 0.5 g

63.55 g/mol = 0.00788 mol

Step 2: Determine the quantity of charge passing through the cell.

Quantity of charge = Moles of Cu deposited ×Faraday constant

Quantity of charge = 0.00788 mol ×96485 C/mol = 0.761 C

Step 3: Calculate the current passing through the cell.

Current = Quantity of charge

Time

Current = 0.761 C

10 min =0.761 C

600 s = 0.00126 A

Therefore, the current passing through the cell is 0.00126 A.

Question 31

Question

Electrolysis of an aqueous solution of silver nitrate (AgNO3) is carried out using

a silver electrode. If the electrolysis is run at a constant current of 2 A for 30

minutes, calculate the mass of silver deposited on the electrode. Given that the

molar mass of silver is 107.87 g/mol and the Faraday constant is 9.65 ×104

C/mol.

Solution

Step 1: Calculate the total charge passed through the circuit. Given that the

current is 2 A and the time is 30 minutes, we first convert the time to seconds:

30 minutes = 30 ×60 = 1800 seconds

The total charge passed can be determined using the formula:

Q=I×t

where Qis the charge, Iis the current, and tis the time.

Substitute the values to find Q:

Q= 2 A ×1800 s = 3600 C

Step 2: Determine the moles of silver deposited on the electrode. The num-

ber of moles of silver (n) can be calculated using the formula:

n=Q

where Qis the charge passed and Fis Faraday’s constant.

Substitute the values to find n:

n=3600 C

9.65 ×104C/mol ≈0.0373 mol

Step 3: Calculate the mass of silver deposited. The mass of silver deposited

can be determined using the formula:

Mass = n×Molar mass

where the molar mass of silver is 107.87 g/mol.

Substitute the values to find the mass:

Mass = 0.0373 mol ×107.87 g/mol ≈4.02 g

Therefore, the mass of silver deposited on the electrode is approximately

4.02 g.

Question 32

Question

A student is conducting an electrolysis experiment using a solution of copper

sulfate, CuSO4. The student passes a current of 2.50 A through the solution for

30.0 minutes. The student observes that 3.50 g of copper metal is deposited on

the cathode.

What is the standard cell potential for the reaction Cu2+(aq)+2e−→Cu(s)?

Solution

Step 1: Calculate the charge (Q) passing through the electrolyte solution using

the formula Q=I·t.

Given: I= 2.50 A, t = 30.0 minutes = 1800 s

Q= 2.50 A ×1800 s = 4500 C

Step 2: Calculate the number of moles of copper deposited at the cathode

using the formula n=m

Given: m= 3.50 g, M = 63.5 g/mol

n=3.50 g

63.5 g/mol ≈0.0551 mol

Step 3: Determine the number of electrons transferred in the reaction using

the stoichiometry of the reaction.

From the reaction: 1 mol of Cu2+ requires 2 mol of electrons

2 mol ×0.0551 mol = 0.1102 mol of electrons

Step 4: Calculate the Faraday constant (F) using the formula F=Q

F=4500 C

0.1102 mol ≈40787 C/mol

Step 5: Calculate the standard cell potential using the formula E◦=Q

nF .

E◦=4500 C

0.1102 mol ×40787 C/mol ≈1.03 V

Therefore, the standard cell potential for the reaction Cu2+(aq) + 2e−→

Cu(s) is approximately 1.03 V.

Question 33

Question

During the electrolysis of an aqueous solution of copper(II) sulfate, 1.20 grams

of copper is deposited at the cathode. Calculate the total charge passed through

the cell. Given that the Faraday constant is 96485 C/mol.

Solution

Step 1: Calculate the number of moles of copper deposited at the cathode.

Step 2: Use the mole-to-charge relationship to find the total charge passed

through the cell.

Step 1: Calculate the number of moles of copper deposited at the cathode.

The molar mass of copper is 63.55 g/mol. Therefore, the number of moles of

copper deposited can be calculated as:

Number of moles = Mass

Molar mass =1.20 g

63.55 g/mol

Number of moles ≈0.0189 mol

Step 2: Use the mole-to-charge relationship to find the total charge passed

through the cell.

Since 1 mole of electrons has a charge of 96485 C (1 Faraday), we can calculate

the total charge passed through the cell as:

Total charge = Number of moles ×96485 C/mol

Total charge = 0.0189 mol ×96485 C/mol

Total charge ≈1825 C

Therefore, the total charge passed through the cell during the electrolysis is

approximately 1825 Coulombs.

Question 34

Question

Explain Faraday’s laws of electrolysis.

Solution

Faraday’s laws of electrolysis describe the quantitative relationship between the

amount of a substance produced or consumed during electrolysis and the amount

of electricity passed through the electrolyte. There are two laws: 1. The amount

of a substance produced or consumed during electrolysis is directly proportional

to the quantity of electricity passed through the electrolyte. 2. The amounts of

different substances produced or consumed by the same quantity of electricity

are proportional to their equivalent weights.

Step 1: Let Qbe the total charge passing through the electrolyte in coulombs

(C), Ibe the current passing through the electrolyte in amperes (A), and tbe

the time the current flows in seconds (s). The charge passed through the elec-

trolyte can be calculated as:

Q=I×t

Step 2: The amount of a substance produced or consumed during electrol-

ysis is directly proportional to the quantity of electricity passed through the

electrolyte. This relationship is given by Faraday’s law:

m=Q

F×M

where: - mis the mass of the substance produced or consumed in grams (g),

-Qis the total charge passed through the electrolyte in coulombs (C), - Fis

Faraday’s constant (96500 C/mol), - Mis the molar mass of the substance in

grams per mole (g/mol).

Step 3: The second law states that the amounts of different substances

produced or consumed by the same quantity of electricity are proportional to

their equivalent weights. The equivalent weight (E) of a substance is the mass

that would react with or produce 1 mole of electrons. It is related to the molar

mass (M) of the substance by:

E=M

where nis the number of electrons involved in the reaction.

Step 4: From Faraday’s law, we can rewrite it in terms of the equivalent

weight:

m=Q

F×E

This equation shows that the mass of a substance produced or consumed during

electrolysis is directly proportional to the equivalent weight of the substance.

Thus, Faraday’s laws of electrolysis are fundamental principles that govern

the quantitative aspects of electrolysis reactions.

Question 35

Question

A student is performing an electrolysis experiment using a copper(II) sulfate

solution. The student observes that copper is deposited on the cathode. If the

student uses a current of 2.5 A for 2 hours, calculate the mass of copper de-

posited. The atomic mass of copper is 63.55 g/mol and the Faraday constant is

96,485 C/mol.

(You may assume 100

Solution

Step 1: Calculate the total charge passed through the circuit.

Total charge = current ×time

Total charge = 2.5 A ×2 hours ×3600 s/hour

Total charge = 2.5 C/s ×7200 s

Total charge = 18,000 C

Step 2: Determine the number of moles of electrons that passed through the

circuit.

Number of moles of electrons = Total charge

Faraday constant

Number of moles of electrons = 18,000 C

96,485 C/mol

Number of moles of electrons ≈0.187 mol

Step 3: Calculate the mass of copper deposited.

Mass of copper deposited = Number of moles of electrons×molar mass of copper

Mass of copper deposited = 0.187 mol ×63.55 g/mol

Mass of copper deposited ≈11.9 g

Therefore, the mass of copper deposited on the cathode after applying a

current of 2.5 A for 2 hours is approximately 11.9 g.

CHEM 105 - ELEMENTS OF GENERAL CHEMISTRY Faraday’s laws of electrolysis Question Bank Set 2

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