CHEM 105 - ELEMENTS OF GENERAL CHEMISTRY Faraday’s laws of electrolysis Question Bank Set 1

CHEM 105 - ELEMENTS OF

GENERAL CHEMISTRY - Faraday’s

laws of electrolysis

Question Bank - Set 1

Liberty University

Question 1

Question

A student sets up an electrolytic cell containing a solution of copper(II) sulfate,

CuSO4, and passes a current through the cell for 30 minutes. During this time,

they observe that the mass of the cathode decreases by 0.75 grams. If the

student knows that the equivalent mass of copper is 31.75 grams per coulomb,

calculate the current passing through the cell.

Solution

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

Given that the equivalent mass of copper is 31.75 grams per coulomb, the num-

ber of coulombs required to deposit 0.75 grams of copper is:

0.75 g

31.75 g/C = 0.0236 C

The number of moles of copper deposited is:

0.75 g

63.55 g/mol = 0.0118 mol

Step 2: Use Faraday’s laws of electrolysis to calculate the total charge passed

through the cell. Faraday’s first law states that the mass of substance deposited

at an electrode is directly proportional to the quantity of electricity passed

through the cell. The total charge passed through the cell can be calculated

using the formula:

Charge (C) = Number of moles ×Equivalent mass ×Faraday constant

where the Faraday constant is 96500 C/mol. Therefore, the total charge passed

through the cell is:

0.0118 mol ×31.75 g/C ×96500 C/mol = 35.11 C

Step 3: Calculate the current passing through the cell. Given that the

current passed through the cell in 30 minutes (0.5 hours) is:

35.11 C

0.5 h = 70.22 C/h = 70.22 A

Therefore, the current passing through the cell is 70.22 Amperes.

Question 2

Question

A certain electrochemical cell consists of a copper electrode immersed in a cop-

per(II) sulfate solution and a platinum electrode immersed in a potassium iodide

solution. If a constant current of 2.50 A flows through the cell for 2.00 hours, cal-

culate the mass of copper deposited on the copper electrode. (Given: Faraday’s

constant F= 96,485 C/mol, molar mass of copper MCu = 63.55 g/mol)

Solution

Step 1: Find the total charge passed through the cell. Given that the current is

2.50 A and the time is 2.00 hours, we can calculate the total charge using the

formula:

Q=I×t

Q= 2.50 A ×2.00 hours ×3600 s/hour (to convert hours to seconds)

Q= 18,000 C

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

We know that 1 Faraday (F) is equal to 1 mole of electrons. Therefore, the

number of moles of electrons passed through the cell is:

moles of electrons = Q

moles of electrons = 18,000 C

96,485 C/mol

moles of electrons ≈0.186 mol

Step 3: Calculate the mass of copper deposited on the electrode. Since the

reaction at the copper electrode is Cu(II) ions gaining electrons to form solid

copper, the balanced chemical equation is:

Cu2+ + 2e−

→Cu

From the equation, we see that 2 moles of electrons are required to deposit

1 mole of copper. Therefore, the number of moles of copper deposited is equal

to half the number of moles of electrons passed through the cell. Calculate the

mass of copper deposited using the molar mass of copper (MCu):

mass of copper = 1

2×0.186 mol ×63.55 g/mol

mass of copper ≈5.93 g

Therefore, the mass of copper deposited on the copper electrode is approxi-

mately 5.93 grams.

Question 3

Question

A solution of silver nitrate (AgNO3) is electrolyzed using a current of 2.00 A

for 3.00 hours. If the reduction of Ag+ions occurs at the cathode, determine

the mass of silver deposited. (Hint: The molar mass of silver is 107.87 g/mol)

Solution

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

I= 2.00 A Time, t= 3.00 hours = 10800 seconds

The total charge passed, Q=I×t Q = 2.00 A ×10800 s Q= 21600 C

Step 2: Determine the moles of Ag+ions reduced. From the balanced chem-

ical equation for the reduction of Ag+ions: Ag++e−

→Ag

It is a 1:1 ratio, so the moles of Ag+ions reduced is equal to the total charge

passed, since 1 mole of electrons carries a charge of 1 F (Faraday): Moles of

Ag+ions reduced = Q

96500 mol Moles of Ag+ions reduced = 21600

96500 mol Moles

of Ag+ions reduced = 0.224 mol

Step 3: Calculate the mass of silver deposited. Using the molar mass of

silver (Ag): Mass of Ag deposited = moles ×molar mass Mass of Ag deposited

= 0.224 mol ×107.87 g/mol Mass of Ag deposited = 24.16 g

Therefore, the mass of silver deposited during the electrolysis of the solution

of silver nitrate is 24.16 g.

Question 4

Question

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

trodes. The electrolysis of this solution leads to the deposition of silver metal

on the cathode. If a current of 2.50 A is passed through the solution for 45.0

minutes, calculate the mass of silver deposited on the cathode.

Given: Atomic mass of silver (Ag) = 107.87 g/mol. Faraday’s constant (F)

= 96485 C/mol.

Solution

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

I= 2.50 A Time, t= 45.0 minutes = 2700 seconds

The total charge passing through the solution can be calculated using the

formula:

Q=I×t

Q= 2.50 A ×2700 s

Q= 6750 C

Step 2: Determine the number of moles of silver deposited. The amount

of charge required to deposit one mole of silver is equal to the charge of one

mole of electrons (Faraday’s constant). Therefore, the number of moles of silver

deposited is given by:

moles of Ag = Q

moles of Ag = 6750 C

96485 C/mol

moles of Ag ≈0.07 mol

Step 3: Calculate the mass of silver deposited on the cathode. Given: Atomic

mass of silver (Ag) = 107.87 g/mol

The mass of silver deposited is given by:

Mass of Ag = moles of Ag ×atomic mass of Ag

Mass of Ag = 0.07 mol ×107.87 g/mol

Mass of Ag ≈7.55 g

Therefore, the mass of silver deposited on the cathode is approximately 7.55

grams.

Question 5

Question

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

amount of substance liberated at an electrode during electrolysis and the quan-

tity of electricity passed through the electrolyte. A student performs an electrol-

ysis experiment where a current of 2.5 A is passed through a solution of silver

nitrate for 45 minutes. Calculate the mass of silver deposited on the cathode

during this time. Given: Atomic mass of silver = 107.87 g/mol, and Faraday’s

constant = 96500 C/mol.

Solution

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

Charge = Current ×Time

Charge = 2.5 A ×45 min ×60 s/min

Charge = 2.5 A ×2700 s

Charge = 6750 C

Step 2: Determine the number of moles of silver ions (Ag+) reduced. One

mole of electrons corresponds to the reduction of one mole of silver ions.

Moles of electrons = Charge

Faraday’s constant

Moles of electrons = 6750 C

96500 C/mol

Moles of electrons ≈0.07003 mol

Step 3: Calculate the mass of silver deposited.

Mass of silver = Moles of electrons ×Atomic mass of silver

Mass of silver = 0.07003 mol ×107.87 g/mol

Mass of silver ≈7.56 g

Therefore, the mass of silver deposited on the cathode during this time is

approximately 7.56 g.

Question 6

Question

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

sulfate for 30 minutes. Calculate the mass of copper deposited at the cathode.

Given: Atomic mass of copper = 63.5 u, Faraday’s constant = 96500 C mol−1.

Solution

Let’s start by calculating the total charge passed through the cell, which can be

used to determine the amount of copper deposited.

Step 1: Calculate the total charge passed through the cell The formula

relating charge, current, and time is:

Q=I×t

where: - Qis the total charge passed through the cell, - Iis the current (2.5

A), and - tis the time (30 minutes).

Substitute the given values:

Q= 2.5A×30 min

Convert the time from minutes to seconds (1 min = 60 s):

t= 30 min ×60 s/min = 1800 s

Calculate the total charge:

Q= 2.5A×1800 s

Q= 4500 C

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

Since 1 Faraday is equal to 96500 Cand corresponds to 1 mole of electrons, we

can determine the number of moles of electrons passed through the cell.

1F= 96500 C/mol

Number of moles of electrons = Q

96500

Substitute the value of Q:

Number of moles of electrons = 4500 C

96500

Number of moles of electrons ≈0.0466 mol

Step 3: Calculate the mass of copper deposited From the balanced chemical

equation for the electrolysis of copper sulfate with copper deposited at the cath-

ode, we know that for every mole of electrons, one mole of copper is deposited.

Given the atomic mass of copper as 63.5 u, we can calculate the mass of

copper deposited.

Mass of copper deposited = Number of moles of electrons×Atomic mass of copper

Mass of copper deposited = 0.0466 mol ×63.5g/mol

Mass of copper deposited ≈2.96 g

Therefore, the mass of copper deposited at the cathode is approximately

2.96 grams.

Question 7

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution of

copper(II) sulfate, CuSO4, for 30 minutes. If the atomic mass of copper is 63.5

g/mol, calculate the mass of copper deposited on the cathode during this time.

Solution

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

(I) = 2.5 A, Time (t) = 30 minutes = 1800 seconds. The total charge (Q) passed

is given by Q=I·t. Substituting the given values, we get

Q= 2.5 A ×1800 s = 4500 C.

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

copper(II) sulfate, the mole ratio of electrons to copper ions is 2:1. Therefore,

the number of moles of electrons passed is

Moles of electrons = Q

where Faraday’s constant F= 96485 C/mol. Substituting the value of Q, we

get

Moles of electrons = 4500 C

96485 C/mol ≈0.0466 mol.

Step 3: Calculate the mass of copper deposited. Since the mole ratio of

electrons to copper ions is 2:1, the moles of copper deposited will be half of the

moles of electrons passed.

Moles of copper = 0.5×Moles of electrons = 0.5×0.0466 mol = 0.0233 mol.

The mass of copper deposited can be calculated as follows:

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

Mass of copper = 0.0233 mol ×63.5 g/mol ≈1.48 g.

Therefore, the mass of copper deposited on the cathode during this electrol-

ysis experiment is approximately 1.48 g.

Question 8

Question

A student is conducting an electrolysis experiment where they are passing a

current through a solution of CuSO4. If they observe the deposition of 3.50 g

of copper in 45 minutes, calculate the current passing through the electrolyte.

Assume the Faraday constant is 96,500 C/mol and the molar mass of copper is

63.5 g/mol.

Solution

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

Moles of Cu = Mass of Cu

Molar mass of Cu =3.50 g

63.5 g/mol

Step 2: Calculate the charge required to deposit this amount of copper using

Faraday’s laws of electrolysis.

Q= Moles of Cu ×Faraday constant = 3.50 g

63.5 g/mol×96500 C/mol

Step 3: Determine the current passing through the electrolyte using the

definition of current.

I=Q

Step 4: Substitute the values and solve for the current.

I=3.50 g

63.5 g/mol ×96500 C/mol

45 ×60 s

Question 9

Question

An electrolysis cell contains a solution of silver nitrate, AgNO3, and is connected

to a source of electrical current. If a current of 2.50 A is passed through the cell

for 3.00 hours, calculate the mass of silver (Ag) deposited. (Atomic masses: Ag

= 107.87 g/mol)

Solution

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

I= 2.50 A Time, t= 3.00 hours

The total charge passed through the cell can be calculated using the formula:

Q=I×t

Substitute the given values:

Q= 2.50 A ×3.00 hours

Step 2: Convert the time from hours to seconds for consistency in units.

Since 1 hour is equal to 3600 seconds, we convert 3.00 hours to seconds:

3.00 hours ×3600 s/h = 10800 s

Step 3: Calculate the total charge passed through the cell.

Q= 2.50 A ×10800 s

Step 4: Calculate the moles of electrons transferred using Faraday’s laws.

We know that 1 Faraday (F) is equal to 96485 C/mol.

First, calculate the total charge in coulombs (C):

Q= 2.50 A ×10800 s

Now, convert the charge to moles of electrons:

n=Q

Step 5: Calculate the moles of silver deposited. In the electrolysis of silver ni-

trate (AgNO3), each mole of electrons transferred corresponds to the deposition

of 1 mole of silver (Ag).

Given that the molar mass of silver (Ag) is 107.87 g/mol, we can calculate

the mass of silver deposited.

Step 6: Calculate the mass of silver deposited. Using the relationship:

1 mol Ag = 107.87 g

Therefore, the mass of silver deposited is:

Mass of Ag = n×Molar mass of Ag

Substitute the calculated value of n into the formula.

Step 7: Calculate the final answer. Substitute the calculated values into

the formula to find the mass of silver deposited. Calculate the final answer in

grams, with appropriate units.

Question 10

Question

Consider an electrolytic cell where a constant current of 2.5 A is passed through

a solution of copper sulfate for 30 minutes. If copper metal is deposited at the

cathode, determine the mass of copper deposited. Given: atomic mass of copper

= 63.5 g/mol, Faraday’s constant = 96,485 C/mol.

Solution

Step 1: Determine the charge transferred The charge transferred can be found

using the formula:

Charge = Current ×Time

Charge = 2.5 A ×30 minutes ×60 s/min

Charge = 2.5 A ×1800 s

Charge = 4500 C

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

moles of copper deposited can be determined by dividing the charge by Faraday’s

constant:

Moles of Cu = Charge

Faraday’s constant

Moles of Cu = 4500 C

96485 C/mol

Moles of Cu ≈0.0467 mol

Step 3: Find the mass of copper deposited The mass of copper deposited

can be calculated using the formula:

Mass = Moles ×Molar mass

Mass = 0.0467 mol ×63.5 g/mol

Mass ≈2.97 g

Therefore, approximately 2.97 grams of copper will be deposited at the cath-

ode after 30 minutes with a constant current of 2.5 A flowing through the solu-

tion.

Question 11

Question

An aqueous solution of sodium chloride (NaCl) is electrolyzed using inert elec-

trodes. If a current of 2.50 A is passed through the solution, what mass of

chlorine gas will be produced in 1 hour?

Given: Faraday’s constant = 96485 C/mol Molar mass of chlorine = 35.45 g/mol

Solution

Step 1: Calculate the charge passed through the circuit in 1 hour. Given:

Current, I= 2.50 A Time, t= 1 hour = 3600 s

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

Q=I·t

Substitute the given values:

Q= 2.50 A ×3600 s = 9000 C

Step 2: Determine the number of moles of electrons involved in the reac-

tion. Each mole of electrons corresponds to 1 Faraday of charge (96485 C/mol).

Therefore, the number of moles of electrons (n) can be calculated by dividing

the total charge passed (Q) by Faraday’s constant:

n=Q

Faraday’s constant

Substitute the values:

n=9000 C

96485 C/mol ≈0.0933 mol

Step 3: Determine the moles of chlorine gas produced. Since the stoichiom-

etry of the reaction is 2 moles of electrons per mole of chlorine, the moles of

chlorine gas produced will be half of the moles of electrons:

Moles of chlorine gas = n

2=0.0933 mol

2= 0.0466 mol

Step 4: Calculate the mass of chlorine gas produced. The mass of chlorine

gas produced can be calculated using the molar mass of chlorine:

Mass of chlorine gas = Moles ×Molar mass

Substitute the values:

Mass of chlorine gas = 0.0466 mol ×35.45 g/mol ≈1.65 g

Therefore, approximately 1.65 grams of chlorine gas will be produced in 1

hour.

Question 12

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution

of silver nitrate (AgNO3). If 0.5 g of silver is deposited at the cathode in 10

minutes, determine the number of moles of silver ions reduced.

Solution

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

current is 2.5 A and the time is 10 minutes, we first convert the time to seconds:

10 minutes = 10 ×60 seconds = 600 seconds

The total charge Qpassed is given by:

Q=I×t

Q= 2.5 A ×600 s

Q= 1500 C

Step 2: Calculate the number of moles of silver deposited. 1. Determine the

molar mass of silver, Ag:

Ag = 107.87 g/mol

2. Calculate the number of moles using the given mass of silver:

0.5 g of silver = 0.5 g ×1 mol

107.87 g

0.5 g of silver = 0.00464 mol

Step 3: Determine the number of moles of silver ions reduced. From Fara-

day’s laws of electrolysis, the charge required to deposit one mole of any ion

is equal to its equivalent mass in grams. For silver, the equivalent mass is its

molar mass which is 107.87 g/mol.

Therefore, the number of moles of silver ions reduced is:

Moles of silver ions reduced = Q

Charge per mole

Moles of silver ions reduced = 1500 C

107.87 C/mol

Moles of silver ions reduced ≈13.90 mol

Thus, the number of moles of silver ions reduced is approximately 13.90 mol.

Question 13

Question

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

solution. The student passes a current of 2.50 A through the solution for 30.0

minutes. If the experiment resulted in the deposition of 5.00 grams of copper

onto the cathode, calculate the Faraday constant (F).

Solution

Step 1: Determine the amount of charge passed through the solution. Given

that the current is 2.50 A and the time is 30.0 minutes, we need to convert the

time to seconds:

30.0 minutes = 30.0×60 seconds = 1800 seconds

The amount of charge passed through the solution can be calculated using

the formula:

Q=I×t

where Qis the charge in Coulombs, Iis the current in amperes, and tis the

time in seconds. Substitute the given values into the formula:

Q= 2.50 A ×1800 s = 4500 C

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

tion. One equivalent (or one mole) of electrons has a charge of 1 Faraday, which

is equal to F= 96485 C/mol.

To find the number of moles of electrons, we divide the total charge by the

charge of one mole of electrons:

Moles of electrons = 4500 C

96485 C/mol

Step 3: Calculate the amount of copper deposited using the given mass. The

molar mass of copper is 63.55 g/mol. Use this to find the number of moles of

copper deposited:

Moles of copper = 5.00 g

63.55 g/mol

Step 4: Using Faraday’s laws of electrolysis, relate the number of moles of

electrons to the number of moles of copper. The balanced equation for the

electrolysis of copper(II) sulfate is:

Cu2+ + 2e−

→Cu

For each mole of copper deposited, 2 moles of electrons are required. There-

fore, the moles of electrons should be twice the moles of copper deposited.

Step 5: Calculate the Faraday constant. Since 2 moles of electrons are

required for each mole of copper deposited, we have:

Moles of electrons = 2 ×Moles of copper

Equating this to the ratio calculated in Step 2, we can solve for the Faraday

constant:

2×5.00 g

63.55 g/mol =4500 C

96485 C/mol

Solving for Fgives:

F=2×5.00 g ×96485 C/mol

4500 C ×63.55 g/mol

Question 14

Question

An aqueous solution of copper(II) sulfate (CuSO4) is electrolyzed using graphite

electrodes. If a current of 2.5 A is passed through the solution for 1 hour, what

mass of copper is deposited at the cathode? (Given: atomic masses - Cu = 63.5

g/mol, S= 32.1 g/mol, O= 16.0 g/mol, H= 1.0 g/mol; Faraday constant =

96500 C/mol; 1 hour = 3600 s)

Solution

Step 1: Calculate the total charge passed through the solution. The total charge

can be calculated using the formula:

Q=I×t

where Qis the total charge in Coulombs, Iis the current in Amperes, and tis

the time in seconds. Given I= 2.5Aand t= 1 hour = 3600 s, we have:

Q= 2.5A×3600 s= 9000 C

Step 2: Determine the number of moles of electrons transferred. Since 1 F

(Faraday) of charge is equivalent to 96500 Cof charge, we can calculate the

number of moles of electrons transferred using the formula:

n=Q

where nis the number of moles of electrons transferred, Qis the total charge

in Coulombs, and Fis the Faraday constant. Substitute Q= 9000 Cand

F= 96500 C/mol to get:

n=9000 C

96500 C/mol = 0.0935 mol

Step 3: Determine the number of moles of copper deposited. Since each

mole of Cu2+ ion is reduced to one mole of copper metal, the number of moles

of copper deposited is the same as the number of moles of electrons transferred.

Thus, 0.0935 mol of copper is deposited.

Step 4: Calculate the mass of copper deposited. To find the mass of copper

deposited, we can use the formula:

Mass = Number of moles ×Molar mass

Given the molar mass of copper as 63.5g/mol, we have:

Mass = 0.0935 mol ×63.5g/mol = 5.92 g

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

Question 15

Question

A certain electrolytic cell operates under a constant current of 2.5 A for 3.5

hours. During this time, 1.25 g of aluminum is deposited on the cathode. Cal-

culate the faraday constant F. Given that the molar mass of aluminum is

27 g/mol.

Solution

Step 1: Calculate the charge (in coulombs) passed through the electrolytic cell.

We know that the charge can be calculated using the formula:

Charge (C) = Current (A) ×Time (s)

Given that the current is 2.5 A and the time is 3.5 hours = 3.5×3600 s, we

have:

Charge (C) = 2.5 A ×3.5×3600 s

Charge (C) = 2.5×3.5×3600 C

Step 2: Calculate the number of moles of aluminum deposited. Using the

formula:

Moles = Mass (g)

Molar Mass (g/mol)

Given that the mass of aluminum deposited is 1.25 g and the molar mass is

27 g/mol, we have:

Moles = 1.25 g

27 g/mol

Step 3: Calculate the faraday constant. The faraday constant is the charge

required to deposit one mole of a substance. Since the charge is directly pro-

portional to the moles of substance deposited, we have:

F=Charge (C)

Moles

Substitute the values calculated in Step 1 and Step 2 into the formula above

to find the faraday constant F.

Question 16

Question

An electrolytic cell containing a molten salt with a molar mass of 80.0 g mol−1

is connected to a 6.0 V battery. If the mass of the metal deposited during

electrolysis is 1.6 g, calculate the time taken for the deposition.

Solution

Step 1: Calculate the number of moles of the deposited metal. Given that the

molar mass of the metal is 80.0 g mol−1and the mass of the metal deposited is

1.6 g, we can calculate the number of moles (n) using the formula:

n=Mass

Molar Mass

n=1.6 g

80.0 g mol−1

n= 0.02 mol

Step 2: Determine the charge required to deposit the given amount of metal.

From Faraday’s first law of electrolysis, we know that the amount of substance

deposited at an electrode is directly proportional to the quantity of electricity

passed through it. The formula for charge (Q) in terms of moles of substance

(n) is:

Q=n×F

where Fis the Faraday constant which is ≈96500 C mol−1.

Substitute n= 0.02 mol and F= 96500 C mol−1into the formula:

Q= 0.02 ×96500

Q= 1930 C

Step 3: Calculate the current and time taken for deposition. Given that the

potential difference across the cell is 6.0 V, we can use the formula Q=I×t

to find the current (I) and time (t).

I=Q

t=Q

Substitute Q= 1930 C and I=6.0 V

Rinto the formula:

t=1930 C

6.0 V

t=1930 C ×R

6.0 V

Therefore, the time taken for the deposition will depend on the resistance of

the electrolytic cell.

Question 17

Question

Consider an electrolytic cell containing a solution of silver nitrate, where silver

is deposited on the cathode and the current flowing through the cell is 2.0 A. If

1.0 g of silver is deposited in 30 minutes, determine the Faraday constant.

Solution

Step 1: Find the number of moles of silver deposited. Given: Current, I= 2.0 A

Time, t= 30 min = 1800 s Mass of silver deposited, m= 1.0 g

We know that the charge deposited (Q) is given by:

Q=It

Substitute the given values:

Q= 2.0 A ×1800 s = 3600 C

Step 2: Find the number of moles of silver deposited. The charge on one

mole of electrons is given by the Faraday constant F, which is equal to 96485 C

mol−1. The number of moles of electrons can be calculated using the formula:

n=Q

Where eis the elementary charge (1.6×10−19 C). Therefore,

n=3600 C

1.6×10−19 C

Step 3: Calculate the Faraday constant. The Faraday constant is defined as

the charge on one mole of electrons, so:

F=Q/n

Substitute the calculated values:

F=3600 C

Finally, simplify the expression to find the Faraday constant.

Question 18

Question

When a current of 2.5 A is passed through a solution of a compound of silver

for 2 hours, 10.8 g of silver is deposited at the cathode. Calculate the atomic

weight of silver.

Solution

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

I= 2.5 A Time, t= 2 hours = 2×3600 seconds We know that charge, Q=I×t

Q= 2.5×2×3600

Q= 2.5×7200

Q= 18000 C

Step 2: Calculate the number of moles of silver deposited. Given: Mass of

silver deposited, m= 10.8 g Atomic weight of silver, MNumber of moles of

silver, n=m

MSince 1 F (Faraday) is the charge needed to deposit 1 mole of Ag,

we know that charge for 1 mole of Ag is Q= 1 ×96500 C So, 18000 C of charge

is needed to deposit 1 mole of Ag.

18000

96500 =m/M

M=10.8×96500

18000 = 58.17 g/mol

Therefore, the atomic weight of silver is 107.87 g/mol.

Question 19

Question

In an electrolytic cell, a certain amount of current is passed through a solution

of copper sulfate using copper electrodes. If 2.75 grams of copper is deposited

on the cathode in 25 minutes, determine the current passing through the cell in

amperes.

(Given: Atomic mass of copper = 63.5 g/mol, Faraday constant = 9.65×104

C/mol)

Solution

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

Moles of copper = Mass

Molar mass

Moles of copper = 2.75 g

63.5 g/mol

Step 2: Use Faraday’s laws to determine the charge passed through the cell.

Charge = Moles ×Faraday constant

Step 3: Calculate the current passing through the cell.

Current = Charge

Time

Now, let’s calculate the current passing through the cell in amperes.

Question 20

Question

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

sulfate, CuSO4. If a current of 2.50 A is passed through the solution for 1.50

hours, how many grams of copper will be deposited at the cathode? (Given:

atomic mass of copper = 63.5 g/mol, Faraday constant = 96,500 C/mol)

Solution

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

I= 2.50 A Time, t= 1.50 hours = 1.50 ×3600 s = 5400 s

The total charge passed, Q=I×t

Calculate Q:

Q= 2.50 A ×5400 s = 13,500 C

Step 2: Find the number of moles of copper deposited. Using Faraday’s first

law of electrolysis, 1 mole of electrons produces 1 mole of copper:

1 mol of electrons = 1 Faraday = 96485 C

Number of moles of electrons, n=Q

96500 =13,500 C

96500 C/mol

Calculate n:

n=13,500 C

96500 C/mol ≈0.140 mol

Step 3: Calculate the mass of copper deposited at the cathode. Given:

Atomic mass of copper, MCu = 63.5 g/mol

Mass of copper deposited, m=n×MCu

Calculate m:

m= 0.140 mol ×63.5 g/mol ≈8.89 g

Therefore, approximately 8.89 grams of copper will be deposited at the cath-

ode.

Question 21

Question

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

tion. The student collects the data below:

- Mass of copper deposited on the cathode: 6.25 g - Current passing through

the electrolyte: 2.50 A - Time taken: 45 minutes

Determine the following: a) The charge passed through the electrolyte b)

The number of moles of electrons passed c) The number of moles of copper

deposited d) Calculate the experimental Faraday’s constant value using the

data provided

Given: - Atomic mass of copper = 63.5 g/mol - Faraday’s constant =

96500 C/mol

Solution

a) To determine the charge passed through the electrolyte, we use the formula:

Q=I·t

where: - Qis the charge passed through the electrolyte - Iis the current passing

through the electrolyte - tis the time taken

Step 1: Substitute the given values into the formula.

Q= 2.50 A×(45 min ×60 s/min)

Step 2: Calculate the charge passed through the electrolyte.

Q= 2.50 A×2700 s= 6750 C

Therefore, the charge passed through the electrolyte is 6750 C.

b) To find the number of moles of electrons passed, we use the formula:

Moles of electrons = Q

where: - Qis the charge passed through the electrolyte - Fis Faraday’s constant

Step 1: Substitute the given values into the formula.

Moles of electrons = 6750 C

96500 C/mol

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

Moles of electrons = 6750

96500 mol ≈0.0701 mol

Therefore, the number of moles of electrons passed is approximately 0.0701

mol.

c) The number of moles of copper deposited can be determined by stoichiom-

etry. Since each copper ion (Cu2+) requires 2 moles of electrons to be reduced

to copper, the number of moles of copper deposited is half the number of moles

of electrons passed.

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

Moles of copper deposited = 0.5×Moles of electrons = 0.5×0.0701 mol

Step 2: Simplify the expression.

Moles of copper deposited = 0.0350 mol

Therefore, the number of moles of copper deposited is 0.0350 mol.

d) The experimental Faraday’s constant value can be calculated using the

mass of copper deposited and the number of moles of copper deposited.

Step 1: Calculate the experimental Faraday’s constant value.

Experimental Faraday’s constant = Mass of copper deposited

Moles of copper deposited

Step 2: Substitute the given values into the formula.

Experimental Faraday’s constant = 6.25 g

0.0350 mol

Step 3: Calculate the experimental value of Faraday’s constant.

Experimental Faraday’s constant = 6.25

0.0350 mol/g ≈178.57 mol/g

Therefore, the experimental Faraday’s constant value is approximately 178.57

mol/g.

Question 22

Question

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

tion. The student passes a constant current of 0.5 A through the solution for

30 minutes. During this time, copper(II) ions are reduced at the cathode and

copper metal is deposited. If the student measures that 2 grams of copper is

deposited, calculate the number of moles of electrons that have passed through

the cell.

Solution

Step 1: Find the molar mass of copper. Given that the molar mass of copper is

63.55 g/mol.

Step 2: Calculate the number of moles of copper deposited. Given that 2

grams of copper is deposited.

Number of moles = 2 g

63.55 g/mol = 0.0315 mol

Step 3: Use Faraday’s laws of electrolysis to relate the number of moles of

copper to the number of moles of electrons. The balanced half-reaction for the

reduction of copper(II) ions to copper is:

Cu2+ + 2e−

→Cu

From the half-reaction, 2 moles of electrons are required to deposit 1 mole of

copper.

Step 4: Calculate the number of moles of electrons. Given that 2 moles of

electrons are required to deposit 1 mole of copper.

Number of moles of electrons = 2 ×0.0315 mol = 0.0630 mol

Therefore, the number of moles of electrons that have passed through the

cell is 0.0630 mol.

Question 23

Question

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

chloride (NaCl) for 3.00 hours. If the standard electrode potential for the re-

duction of Na+ions to Na metal is -2.71 V, calculate the mass of sodium metal

deposited at the cathode.

Solution

Step 1: Find the total charge that passed through the cell. The total charge Q

passed 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 that

I= 2.50 A and t= 3.00 hours, we first convert time to seconds:

t= 3.00 ×3600 = 10800 s

Plugging in the values, we find:

Q= 2.50 ×10800 = 27000 C

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

of electrons has a charge of 1 Faraday, which is equal to 96485 C. Hence, the

number of moles of electrons transferred, n, can be calculated as:

n=Q

96485

Substitute Q= 27000 C into the formula to find:

n=27000

96485 ≈0.280 mol

Step 3: Determine the number of moles of sodium being reduced. The

balanced half-reaction for the reduction of Na+ions to Na metal is:

Na+(aq) + e−

→Na(s)

From the half-reaction, we see that 1 mole of electrons corresponds to 1 mole of

Na being reduced. Therefore, the number of moles of sodium reduced will also

be approximately 0.280 mol.

Step 4: Calculate the mass of sodium deposited. The molar mass of sodium

(Na) is approximately 23.0 g/mol. The mass of sodium deposited, m, can be

calculated using the formula:

m=n×molar mass of Na

Substitute the values to find:

m≈0.280 ×23.0≈6.44 g

Therefore, approximately 6.44 grams of sodium metal will be deposited at

the cathode.

Question 24

Question

Faraday’s laws of electrolysis state that the amount of a substance produced

at an electrode during electrolysis is directly proportional to the quantity of

electricity passed through the electrode.

Given a problem in which a copper sulfate solution is electrolyzed using

a current of 2.5 A for 3 hours, calculate the mass of copper deposited at the

cathode. The molar mass of copper is 63.5 g/mol, and the charge on an electron

is 1.6×10−19 C.

Solution

Step 1: Calculate the total charge passed through the circuit using the formula:

Total charge = Current ×Time

Total charge = 2.5 A ×3 hours ×3600 s/hour

Total charge = 2.5 A ×10800 s

Total charge = 27000 C

Step 2: Calculate the number of electrons transferred during the electrolysis

using Faraday’s constant:

Number of electrons = Total charge

Charge on one electron

Number of electrons = 27000 C

1.6×10−19 C

Number of electrons ≈1.6875 ×1020

Step 3: Calculate the number of moles of copper deposited using the equa-

tion:

Number of moles = Number of electrons

2×Valency

Since copper has a valency of 2, we have:

Number of moles = 1.6875 ×1020

2×2

Number of moles = 1.6875 ×1020

Step 4: Calculate the mass of copper deposited using the equation:

Mass of copper = Number of moles ×Molar mass of copper

Mass of copper = 1.6875 ×1020 ×63.5 g/mol

Mass of copper = 1.0706 ×1022 g

Therefore, the mass of copper deposited at the cathode during the electrol-

ysis is approximately 1.0706 ×1022 g.

Question 25

Question

An electrolysis cell is set up with a copper sulfate solution. If a current of 2.0

A is passed through the cell for 2.5 hours, what mass of copper is deposited on

the cathode? (Given: Faraday’s constant = 9.65 ×104C/mol, atomic mass of

copper = 63.5 g/mol)

Solution

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

Current(I)=2.0 A

Time(t)=2.5 hours = 2.5×3600 s = 9000 s

The total charge passed through the cell can be calculated using the formula

Q=I×t.

Q= 2.0 A ×9000 s

= 18000 C

Step 2: Calculate the number of moles of electrons that have passed through

the cell. Since 1 Faraday (F) is equal to 9.65 ×104C/mol, the number of moles

of electrons (n) is given by n=Q

n=18000 C

9.65 ×104C/mol

≈0.1866 mol

Step 3: Calculate the mass of copper deposited on the cathode. Since the

atomic mass of copper is 63.5 g/mol, the mass of copper can be calculated using

the formula m=n×atomic mass.

m= 0.1866 mol ×63.5 g/mol

≈11.85 g

Therefore, the mass of copper deposited on the cathode is approximately

11.85 g.

Question 26

Question

Two electrolytic cells are set up in series. In the first cell, a current of 2.5

A is passed through a solution of sodium chloride for 2 hours. In the second

cell, a current of 3.5 A is passed through a solution of copper(II) sulfate for

1 hour. Calculate the mass of sodium metal and copper metal produced in

each cell. (Given: Faraday’s constant F= 96500 C/mol; molar mass of sodium

= 22.99 g/mol; molar mass of copper = 63.55 g/mol)

Solution

Step 1: Calculate the charge passed through each cell For the first cell: Given

current I1= 2.5 A and time t1= 2 hours = 7200 seconds

Charge Q1=I1×t1= 2.5 A ×7200 s = 18000 C

For the second cell: Given current I2= 3.5 A and time t2= 1 hour =

3600 seconds

Charge Q2=I2×t2= 3.5 A ×3600 s = 12600 C

Step 2: Calculate the number of moles of each metal deposited For the first

cell (sodium deposition): Number of moles of sodium n1=Q1

F=18000 C

96500 C/mol ≈

0.1865 mol

Mass of sodium produced = n1×molar mass of sodium = 0.1865 mol×22.99 g/mol ≈

4.28 g

For the second cell (copper deposition): Number of moles of copper n2=

F=12600 C

96500 C/mol ≈0.1306 mol

Mass of copper produced = n2×molar mass of copper = 0.1306 mol×63.55 g/mol ≈

8.30 g

Therefore, the mass of sodium metal produced in the first cell is approx-

imately 4.28 g, and the mass of copper metal produced in the second cell is

approximately 8.30 g.

Question 27

Question

In a laboratory experiment, a researcher passes a current of 2.5 A through a

solution of a copper salt for 15 minutes. During this electrolysis process, the

researcher observes that a mass of 1.8 g of copper is deposited on the cathode.

Calculate the molar mass of the copper salt. (Assume 100

Solution

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

current is 2.5 A and the time is 15 minutes, we first convert the time into

seconds:

15 minutes = 15 ×60 = 900 seconds

The total charge passed through the solution can be found using the formula:

Q=I×t

Q= 2.5 A ×900 s = 2250 C

Step 2: Determine the number of moles of electrons and the number of moles

of copper deposited on the cathode. Since the electrolysis process is 1001 mole

of electrons = 1 Faraday = 9.65 ×104C

nelectrons =Q/Faraday constant

nelectrons = 2250 C/9.65 ×104C/mol = 0.0234 mol

Step 3: Calculate the molar mass of the copper salt. The molar mass of

the copper salt can be calculated using the mass of copper deposited and the

number of moles of copper:

mcopper = 1.8 g

ncopper =mcopper/molar mass of copper

0.0234 mol = 1.8 g/molar mass of copper

Solving for the molar mass of copper:

Molar mass of copper = 1.8 g/0.0234 mol = 76.92 g/mol

Therefore, the molar mass of the copper salt is 76.92 g/mol.

Question 28

Question

A student conducts an experiment to study the electrolysis of a certain salt

solution using a constant current of 2.0 A for 30 minutes. The student discovers

that 2.40 g of the salt is deposited at one electrode. Determine the charge on

each ion in the molecular formula of the salt.

Solution

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

current is 2.0 A and the time is 30 minutes, we first convert the time to seconds:

Time (s) = 30 minutes ×60 s/min = 1800 s

Then, we calculate the total charge passed through the solution using the

formula:

Charge = Current ×Time

Charge = 2.0 A ×1800 s = 3600 Coulombs

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

mass of the salt deposited is 2.40 g. To find the number of moles, we divide the

mass by the molar mass of the salt.

Step 3: Find the molar mass of the salt. To find the molar mass of the salt,

we need to know its formula. Let the molecular formula of the salt be AxBy,

where xand yare the charges on A and B ions, respectively.

Step 4: Calculate the charge on each ion. From the charge passed through

the solution and the number of moles of salt deposited, we can determine the

charge on each ion using the formula:

Charge = Number of moles ×1×(Charge on A + Charge on B)

Solving this equation will give us the charges on ions A and B in the molec-

ular formula of the salt.

Question 29

Question

In an electrolytic cell, a current of 5.0 A is passed through a solution of cop-

per(II) sulfate for 20 minutes. During this time, 1.0 g of copper is deposited at

the cathode.

Assuming 100

Solution

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

I= 5.0 A

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

The total charge passing through the cell, Q=I·t

Q= 5.0 A ×1200 s = 6000 C

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

copper, m= 1.0 g Molar mass of copper, M= 63.55 g/mol

Number of moles of copper deposited, n=m

n=1.0 g

63.55 g/mol ≈0.0157 mol

Step 3: Determine the number of electrons passed. The number of moles of

electrons passed is equal to the number of moles of copper deposited (since in

the redox reaction, 1 mole of Cu2+ gains 2 moles of electrons to form 1 mole of

copper). Number of moles of electrons, ne= 2n

ne= 2 ×0.0157 mol = 0.0314 mol

Step 4: Calculate the charge transferred in coulombs. Since 1 mole of

electrons corresponds to the charge of 1 F = 96485 C, Charge transferred,

Q=ne·96485 C/mol

Q= 0.0314 mol ×96485 C/mol ≈3033 C

Step 5: Calculate the standard electrode potential of the copper half-cell.

Using Faraday’s laws of electrolysis:

∆G=−nF E

where ∆Gis the change in Gibbs free energy, nis the number of moles of

electrons, Fis the Faraday constant (96485 C/mol), and Eis the standard

electrode potential.

We know that ∆G=−QV , where Vis the potential difference in volts.

Thus,

E=QV

nF =3033 C ×V

0.0314 mol ×96485 C/mol

Since the question assumes 100

3033 C = 1 mol ×96485 C/mol

Thus, V=E= 1 V is the standard electrode potential of the copper half-

cell.

Question 30

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution of

CuCl2for 20 minutes. If 1.2 g of copper is deposited at the cathode, what is

the efficiency of the electrolysis process?

Solution

Step 1: Find the total charge passing through the solution. Given that the

current passing through the solution is 2.5 A and the time is 20 minutes, we

first convert the time to seconds:

20 minutes = 20 ×60 seconds = 1200 seconds

The total charge passing through the solution is given by:

Q=I×t

Q= 2.5 A ×1200 s = 3000 C

Step 2: Calculate the amount of charge required to deposit 1.2 g of copper.

The number of moles of copper deposited can be found using the molar mass of

copper (Cu) as 63.55 g/mol. The charge required to deposit 1 mole of copper

is the same as the charge on 1 mole of electrons (1 F = 96485 C/mol). First,

calculate the number of moles of copper:

n=mass

molar mass =1.2 g

63.55 g/mol ≈0.0189 mol

Then, find the charge required to deposit this amount of copper:

QCu =n×96485 C/mol = 0.0189 mol ×96485 C/mol = 1822.46 C

Step 3: Calculate the efficiency of the electrolysis process. The efficiency of

the electrolysis process can be calculated using the formula:

Efficiency = Actual charge

Theoretical charge ×100%

where the actual charge passing through the solution is 3000 C and the theo-

retical charge required to deposit 1.2 g of copper is 1822.46 C. Calculating the

efficiency:

Efficiency = 3000

1822.46 ×100% ≈164.6%

The efficiency of the electrolysis process is approximately 164.6

Question 31

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of CuSO4

for 45 minutes. If the electrodeposition of copper occurs, calculate the mass of

copper deposited. (Atomic masses: Cu = 63.5, O = 16, S = 32)

Solution

Step 1: Find the total charge passed through the cell. Given current, I= 2.5

A and time, t= 45 minutes, we can convert time to seconds:

t= 45 minutes ×60 = 2700 seconds

The total charge passed through the cell is given by:

Q=I×t= 2.5 A ×2700 s = 6750 C

Step 2: Convert charge to moles of electrons. Since 1 faraday (F) is equiv-

alent to 1 mol e−, we can calculate the moles of electrons transferred using

Faraday’s constant:

1 F = 96500 C/mol

Moles of electrons = 6750 C

96500 C/mol = 0.07 mol

Step 3: Determine the moles of copper deposited. From the balanced equa-

tion for the electrodeposition of copper:

2 mol e−+Cu2+ →Cu

we know that 2 mol of electrons are needed to deposit 1 mol of copper. So,

moles of copper deposited = 0.07 mol

2= 0.035 mol

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

is 63.5 g/mol. The mass of copper deposited can be calculated as follows:

Mass of Cu deposited = moles of Cu deposited ×atomic mass of Cu

= 0.035 mol ×63.5 g/mol = 2.225 g

Therefore, the mass of copper deposited is 2.225 g.

Question 32

Question

Two electrolytic cells are set up in series using a power supply with a constant

voltage. Cell A contains a solution of silver nitrate (AgNO3), while cell B

contains a solution of sodium chloride (NaCl). If 0.5 grams of silver is deposited

at the cathode in cell A after 20 minutes, how much sodium metal will be

deposited at the cathode in cell B after 30 minutes? (Assume the same current

passes through both cells.)

Solution

Step 1: Calculate the amount of charge passed through cell A. The molar mass

of silver is 107.87 g/mol. Since 1 mole of electrons is required to deposit 1 mole

of silver, we can calculate the charge passed through the cell using the formula:

Q=Mass

Molar mass ×Faraday constant

Q=0.5 g

107.87 g/mol ×96485 C/mol

Q=0.5

107.87 ×96485

Q≈447.12 C

Step 2: Determine the current passing through cell B. Since the same volt-

age is applied to both cells, the current flowing through both cells is the same.

Therefore, the current passing through cell B can be calculated using the for-

mula:

I=Q

where tis the time in seconds.

I=447.12

20 ×60

I≈0.3726 A

Step 3: Calculate the amount of charge passed through cell B in 30 minutes.

QB=I×tB

QB= 0.3726 ×30 ×60

QB= 672.72 C

Step 4: Determine the amount of sodium metal deposited at the cathode

in cell B. The molar mass of sodium is 22.99 g/mol. Since 1 mole of electrons

is required to deposit 1 mole of sodium, we can calculate the mass of sodium

deposited using:

Mass of sodium = QB

Faraday constant ×Molar mass of sodium

Mass of sodium = 672.72

96485 ×22.99

Mass of sodium ≈0.1600 g

Therefore, approximately 0.1600 grams of sodium metal will be deposited at

the cathode in cell B after 30 minutes.

Question 33

Question

An aqueous solution of copper sulfate, CuSO4, is electrolyzed using inert elec-

trodes. If 2.50 amperes of current is passed through the solution for 2.00 hours,

what mass of copper will be deposited at the cathode? (Given: Faraday constant

= 96500 C/mol, atomic mass of copper = 63.5 g/mol)

Solution

Step 1: Calculate the total charge passed through the solution. The total charge

(Q) passed can be calculated using the formula:

Q=I×t

where I= 2.50 A is the current and t= 2.00 h is the time in hours.

Q= 2.50 A ×2.00 h ×3600 s/h = 18000 C

Step 2: Calculate the number of moles of electrons transferred. One mole of

electrons corresponds to one Faraday (96500 C/mol). Therefore, the number of

moles of electrons (n) can be calculated as:

n=Q

96500

n=18000 C

96500 C/mol ≈0.1866 mol

Step 3: Determine the number of moles of copper deposited. By the stoi-

chiometry of the reaction, 2 moles of electrons are required to deposit 1 mole of

copper. Therefore, the number of moles of copper deposited (nCu) is:

nCu =1

2×n

nCu =1

2×0.1866 mol ≈0.0933 mol

Step 4: Calculate the mass of copper deposited. To find the mass of copper

deposited, we use the atomic mass of copper (63.5 g/mol). The mass of copper

deposited (m) can be calculated as:

m=nCu ×Atomic mass of copper

m= 0.0933 mol ×63.5 g/mol ≈5.93 g

Therefore, approximately 5.93 grams of copper will be deposited at the cath-

ode.

Question 34

Question

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

sulfate. The student connects a copper electrode to the positive terminal of a

battery and a platinum electrode to the negative terminal. The student observes

that after some time, the mass of the copper electrode decreases while the mass

of the platinum electrode remains constant. Explain this observation in terms

of Faraday’s laws of electrolysis.

Solution

To explain this observation in terms of Faraday’s laws of electrolysis, we need

to consider the reactions that occur at each electrode and the Faraday’s laws

associated with the amount of substance deposited or dissolved during electrol-

ysis.

Step 1: Reactions at the Electrodes At the copper electrode (cathode):

The copper(II) ions in the solution gain electrons and are reduced to copper

atoms:

Cu2+ + 2e−

→Cu

At the platinum electrode (anode): Water molecules in the solution are oxi-

dized to produce oxygen gas and protons:

2H2O→O2+ 4H++ 4e−

Step 2: Faraday’s Laws of Electrolysis Faraday’s first law states that

the amount of substance deposited or dissolved at an electrode during elec-

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

electrolyte. Faraday’s second law states that the amounts of different substances

deposited or dissolved by the same quantity of electricity are proportional to

their chemical equivalent weights.

Step 3: Explanation of the Observation Since copper ions are reduced

at the copper electrode, copper atoms are deposited on the electrode. This

explains why the mass of the copper electrode decreases.

On the other hand, at the platinum electrode, water molecules are oxidized

to produce oxygen gas and protons. Since the platinum electrode does not

participate in the redox reaction involving copper ions, its mass remains constant

throughout the electrolysis.

Therefore, the observation that the mass of the copper electrode decreases

while the mass of the platinum electrode remains constant can be explained by

the different reactions occurring at each electrode as governed by Faraday’s laws

of electrolysis.

Question 35

Question

State Faraday’s laws of electrolysis and explain the significance of these laws in

the context of electrochemistry.

Solution

Faraday’s Laws of Electrolysis: Faraday’s laws of electrolysis are two quan-

titative laws that describe the amount of chemical change that occurs during

electrolysis. These laws were formulated by Michael Faraday in the 1830s. The

two laws are:

1. The amount of chemical change (in mass) produced by a current passing

through an electrolyte is proportional to the quantity of electricity that

passes through the electrolyte.

2. The amounts of different substances produced by passing the same quan-

tity of electricity through different electrolytes are proportional to their

equivalent weights.

Significance:



Quantitative Relationship: Faraday’s laws provide a quantitative re-

lationship between the amount of current passed through an electrolyte

and the chemical changes that occur. This allows for precise control and

prediction of outcomes in electrochemical processes.



Equivalent Weights: By considering the equivalent weights of sub-

stances, Faraday’s laws provide a way to compare the amounts of different

substances produced in electrolysis. This is essential for understanding

and designing electrochemical processes.



Fundamental Principles: Faraday’s laws are fundamental to the field of

electrochemistry, providing a basis for understanding the transformation

of chemical energy into electrical energy and vice versa.



Experimental Verification: The laws have been extensively verified

through experiments, reinforcing their importance in the field of electro-

chemistry.

where the Faraday constant is 96500 C/mol. Therefore, the total charge passed

through the cell is:

0.0118 mol ×31.75 g/C ×96500 C/mol = 35.11 C

Step 3: Calculate the current passing through the cell. Given that the

current passed through the cell in 30 minutes (0.5 hours) is:

35.11 C

0.5 h = 70.22 C/h = 70.22 A

Therefore, the current passing through the cell is 70.22 Amperes.

Question 2

Question

A certain electrochemical cell consists of a copper electrode immersed in a cop-

per(II) sulfate solution and a platinum electrode immersed in a potassium iodide

solution. If a constant current of 2.50 A flows through the cell for 2.00 hours, cal-

culate the mass of copper deposited on the copper electrode. (Given: Faraday’s

constant F= 96,485 C/mol, molar mass of copper MCu = 63.55 g/mol)

Solution

Step 1: Find the total charge passed through the cell. Given that the current is

2.50 A and the time is 2.00 hours, we can calculate the total charge using the

formula:

Q=I×t

Q= 2.50 A ×2.00 hours ×3600 s/hour (to convert hours to seconds)

Q= 18,000 C

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

We know that 1 Faraday (F) is equal to 1 mole of electrons. Therefore, the

number of moles of electrons passed through the cell is:

moles of electrons = Q

moles of electrons = 18,000 C

96,485 C/mol

moles of electrons ≈0.186 mol

Step 3: Calculate the mass of copper deposited on the electrode. Since the

reaction at the copper electrode is Cu(II) ions gaining electrons to form solid

copper, the balanced chemical equation is:

Cu2+ + 2e−

→Cu

From the equation, we see that 2 moles of electrons are required to deposit

1 mole of copper. Therefore, the number of moles of copper deposited is equal

to half the number of moles of electrons passed through the cell. Calculate the

mass of copper deposited using the molar mass of copper (MCu):

mass of copper = 1

2×0.186 mol ×63.55 g/mol

mass of copper ≈5.93 g

Therefore, the mass of copper deposited on the copper electrode is approxi-

mately 5.93 grams.

Question 3

Question

A solution of silver nitrate (AgNO3) is electrolyzed using a current of 2.00 A

for 3.00 hours. If the reduction of Ag+ions occurs at the cathode, determine

the mass of silver deposited. (Hint: The molar mass of silver is 107.87 g/mol)

Solution

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

I= 2.00 A Time, t= 3.00 hours = 10800 seconds

The total charge passed, Q=I×t Q = 2.00 A ×10800 s Q= 21600 C

Step 2: Determine the moles of Ag+ions reduced. From the balanced chem-

ical equation for the reduction of Ag+ions: Ag++e−

→Ag

It is a 1:1 ratio, so the moles of Ag+ions reduced is equal to the total charge

passed, since 1 mole of electrons carries a charge of 1 F (Faraday): Moles of

Ag+ions reduced = Q

96500 mol Moles of Ag+ions reduced = 21600

96500 mol Moles

of Ag+ions reduced = 0.224 mol

Step 3: Calculate the mass of silver deposited. Using the molar mass of

silver (Ag): Mass of Ag deposited = moles ×molar mass Mass of Ag deposited

= 0.224 mol ×107.87 g/mol Mass of Ag deposited = 24.16 g

Therefore, the mass of silver deposited during the electrolysis of the solution

of silver nitrate is 24.16 g.

Question 4

Question

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

trodes. The electrolysis of this solution leads to the deposition of silver metal

on the cathode. If a current of 2.50 A is passed through the solution for 45.0

minutes, calculate the mass of silver deposited on the cathode.

Given: Atomic mass of silver (Ag) = 107.87 g/mol. Faraday’s constant (F)

= 96485 C/mol.

Solution

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

I= 2.50 A Time, t= 45.0 minutes = 2700 seconds

The total charge passing through the solution can be calculated using the

formula:

Q=I×t

Q= 2.50 A ×2700 s

Q= 6750 C

Step 2: Determine the number of moles of silver deposited. The amount

of charge required to deposit one mole of silver is equal to the charge of one

mole of electrons (Faraday’s constant). Therefore, the number of moles of silver

deposited is given by:

moles of Ag = Q

moles of Ag = 6750 C

96485 C/mol

moles of Ag ≈0.07 mol

Step 3: Calculate the mass of silver deposited on the cathode. Given: Atomic

mass of silver (Ag) = 107.87 g/mol

The mass of silver deposited is given by:

Mass of Ag = moles of Ag ×atomic mass of Ag

Mass of Ag = 0.07 mol ×107.87 g/mol

Mass of Ag ≈7.55 g

Therefore, the mass of silver deposited on the cathode is approximately 7.55

grams.

Question 5

Question

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

amount of substance liberated at an electrode during electrolysis and the quan-

tity of electricity passed through the electrolyte. A student performs an electrol-

ysis experiment where a current of 2.5 A is passed through a solution of silver

nitrate for 45 minutes. Calculate the mass of silver deposited on the cathode

during this time. Given: Atomic mass of silver = 107.87 g/mol, and Faraday’s

constant = 96500 C/mol.

Solution

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

Charge = Current ×Time

Charge = 2.5 A ×45 min ×60 s/min

Charge = 2.5 A ×2700 s

Charge = 6750 C

Step 2: Determine the number of moles of silver ions (Ag+) reduced. One

mole of electrons corresponds to the reduction of one mole of silver ions.

Moles of electrons = Charge

Faraday’s constant

Moles of electrons = 6750 C

96500 C/mol

Moles of electrons ≈0.07003 mol

Step 3: Calculate the mass of silver deposited.

Mass of silver = Moles of electrons ×Atomic mass of silver

Mass of silver = 0.07003 mol ×107.87 g/mol

Mass of silver ≈7.56 g

Therefore, the mass of silver deposited on the cathode during this time is

approximately 7.56 g.

Question 6

Question

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

sulfate for 30 minutes. Calculate the mass of copper deposited at the cathode.

Given: Atomic mass of copper = 63.5 u, Faraday’s constant = 96500 C mol−1.

Solution

Let’s start by calculating the total charge passed through the cell, which can be

used to determine the amount of copper deposited.

Step 1: Calculate the total charge passed through the cell The formula

relating charge, current, and time is:

Q=I×t

where: - Qis the total charge passed through the cell, - Iis the current (2.5

A), and - tis the time (30 minutes).

Substitute the given values:

Q= 2.5A×30 min

Convert the time from minutes to seconds (1 min = 60 s):

t= 30 min ×60 s/min = 1800 s

Calculate the total charge:

Q= 2.5A×1800 s

Q= 4500 C

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

Since 1 Faraday is equal to 96500 Cand corresponds to 1 mole of electrons, we

can determine the number of moles of electrons passed through the cell.

1F= 96500 C/mol

Number of moles of electrons = Q

96500

Substitute the value of Q:

Number of moles of electrons = 4500 C

96500

Number of moles of electrons ≈0.0466 mol

Step 3: Calculate the mass of copper deposited From the balanced chemical

equation for the electrolysis of copper sulfate with copper deposited at the cath-

ode, we know that for every mole of electrons, one mole of copper is deposited.

Given the atomic mass of copper as 63.5 u, we can calculate the mass of

copper deposited.

Mass of copper deposited = Number of moles of electrons×Atomic mass of copper

Mass of copper deposited = 0.0466 mol ×63.5g/mol

Mass of copper deposited ≈2.96 g

Therefore, the mass of copper deposited at the cathode is approximately

2.96 grams.

Question 7

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution of

copper(II) sulfate, CuSO4, for 30 minutes. If the atomic mass of copper is 63.5

g/mol, calculate the mass of copper deposited on the cathode during this time.

Solution

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

(I) = 2.5 A, Time (t) = 30 minutes = 1800 seconds. The total charge (Q) passed

is given by Q=I·t. Substituting the given values, we get

Q= 2.5 A ×1800 s = 4500 C.

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

copper(II) sulfate, the mole ratio of electrons to copper ions is 2:1. Therefore,

the number of moles of electrons passed is

Moles of electrons = Q

where Faraday’s constant F= 96485 C/mol. Substituting the value of Q, we

get

Moles of electrons = 4500 C

96485 C/mol ≈0.0466 mol.

Step 3: Calculate the mass of copper deposited. Since the mole ratio of

electrons to copper ions is 2:1, the moles of copper deposited will be half of the

moles of electrons passed.

Moles of copper = 0.5×Moles of electrons = 0.5×0.0466 mol = 0.0233 mol.

The mass of copper deposited can be calculated as follows:

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

Mass of copper = 0.0233 mol ×63.5 g/mol ≈1.48 g.

Therefore, the mass of copper deposited on the cathode during this electrol-

ysis experiment is approximately 1.48 g.

Question 8

Question

A student is conducting an electrolysis experiment where they are passing a

current through a solution of CuSO4. If they observe the deposition of 3.50 g

of copper in 45 minutes, calculate the current passing through the electrolyte.

Assume the Faraday constant is 96,500 C/mol and the molar mass of copper is

63.5 g/mol.

Solution

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

Moles of Cu = Mass of Cu

Molar mass of Cu =3.50 g

63.5 g/mol

Step 2: Calculate the charge required to deposit this amount of copper using

Faraday’s laws of electrolysis.

Q= Moles of Cu ×Faraday constant = 3.50 g

63.5 g/mol×96500 C/mol

Step 3: Determine the current passing through the electrolyte using the

definition of current.

I=Q

Step 4: Substitute the values and solve for the current.

I=3.50 g

63.5 g/mol ×96500 C/mol

45 ×60 s

Question 9

Question

An electrolysis cell contains a solution of silver nitrate, AgNO3, and is connected

to a source of electrical current. If a current of 2.50 A is passed through the cell

for 3.00 hours, calculate the mass of silver (Ag) deposited. (Atomic masses: Ag

= 107.87 g/mol)

Solution

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

I= 2.50 A Time, t= 3.00 hours

The total charge passed through the cell can be calculated using the formula:

Q=I×t

Substitute the given values:

Q= 2.50 A ×3.00 hours

Step 2: Convert the time from hours to seconds for consistency in units.

Since 1 hour is equal to 3600 seconds, we convert 3.00 hours to seconds:

3.00 hours ×3600 s/h = 10800 s

Step 3: Calculate the total charge passed through the cell.

Q= 2.50 A ×10800 s

Step 4: Calculate the moles of electrons transferred using Faraday’s laws.

We know that 1 Faraday (F) is equal to 96485 C/mol.

First, calculate the total charge in coulombs (C):

Q= 2.50 A ×10800 s

Now, convert the charge to moles of electrons:

n=Q

Step 5: Calculate the moles of silver deposited. In the electrolysis of silver ni-

trate (AgNO3), each mole of electrons transferred corresponds to the deposition

of 1 mole of silver (Ag).

Given that the molar mass of silver (Ag) is 107.87 g/mol, we can calculate

the mass of silver deposited.

Step 6: Calculate the mass of silver deposited. Using the relationship:

1 mol Ag = 107.87 g

Therefore, the mass of silver deposited is:

Mass of Ag = n×Molar mass of Ag

Substitute the calculated value of n into the formula.

Step 7: Calculate the final answer. Substitute the calculated values into

the formula to find the mass of silver deposited. Calculate the final answer in

grams, with appropriate units.

Question 10

Question

Consider an electrolytic cell where a constant current of 2.5 A is passed through

a solution of copper sulfate for 30 minutes. If copper metal is deposited at the

cathode, determine the mass of copper deposited. Given: atomic mass of copper

= 63.5 g/mol, Faraday’s constant = 96,485 C/mol.

Solution

Step 1: Determine the charge transferred The charge transferred can be found

using the formula:

Charge = Current ×Time

Charge = 2.5 A ×30 minutes ×60 s/min

Charge = 2.5 A ×1800 s

Charge = 4500 C

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

moles of copper deposited can be determined by dividing the charge by Faraday’s

constant:

Moles of Cu = Charge

Faraday’s constant

Moles of Cu = 4500 C

96485 C/mol

Moles of Cu ≈0.0467 mol

Step 3: Find the mass of copper deposited The mass of copper deposited

can be calculated using the formula:

Mass = Moles ×Molar mass

Mass = 0.0467 mol ×63.5 g/mol

Mass ≈2.97 g

Therefore, approximately 2.97 grams of copper will be deposited at the cath-

ode after 30 minutes with a constant current of 2.5 A flowing through the solu-

tion.

Question 11

Question

An aqueous solution of sodium chloride (NaCl) is electrolyzed using inert elec-

trodes. If a current of 2.50 A is passed through the solution, what mass of

chlorine gas will be produced in 1 hour?

Given: Faraday’s constant = 96485 C/mol Molar mass of chlorine = 35.45 g/mol

Solution

Step 1: Calculate the charge passed through the circuit in 1 hour. Given:

Current, I= 2.50 A Time, t= 1 hour = 3600 s

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

Q=I·t

Substitute the given values:

Q= 2.50 A ×3600 s = 9000 C

Step 2: Determine the number of moles of electrons involved in the reac-

tion. Each mole of electrons corresponds to 1 Faraday of charge (96485 C/mol).

Therefore, the number of moles of electrons (n) can be calculated by dividing

the total charge passed (Q) by Faraday’s constant:

n=Q

Faraday’s constant

Substitute the values:

n=9000 C

96485 C/mol ≈0.0933 mol

Step 3: Determine the moles of chlorine gas produced. Since the stoichiom-

etry of the reaction is 2 moles of electrons per mole of chlorine, the moles of

chlorine gas produced will be half of the moles of electrons:

Moles of chlorine gas = n

2=0.0933 mol

2= 0.0466 mol

Step 4: Calculate the mass of chlorine gas produced. The mass of chlorine

gas produced can be calculated using the molar mass of chlorine:

Mass of chlorine gas = Moles ×Molar mass

Substitute the values:

Mass of chlorine gas = 0.0466 mol ×35.45 g/mol ≈1.65 g

Therefore, approximately 1.65 grams of chlorine gas will be produced in 1

hour.

Question 12

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution

of silver nitrate (AgNO3). If 0.5 g of silver is deposited at the cathode in 10

minutes, determine the number of moles of silver ions reduced.

Solution

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

current is 2.5 A and the time is 10 minutes, we first convert the time to seconds:

10 minutes = 10 ×60 seconds = 600 seconds

The total charge Qpassed is given by:

Q=I×t

Q= 2.5 A ×600 s

Q= 1500 C

Step 2: Calculate the number of moles of silver deposited. 1. Determine the

molar mass of silver, Ag:

Ag = 107.87 g/mol

2. Calculate the number of moles using the given mass of silver:

0.5 g of silver = 0.5 g ×1 mol

107.87 g

0.5 g of silver = 0.00464 mol

Step 3: Determine the number of moles of silver ions reduced. From Fara-

day’s laws of electrolysis, the charge required to deposit one mole of any ion

is equal to its equivalent mass in grams. For silver, the equivalent mass is its

molar mass which is 107.87 g/mol.

Therefore, the number of moles of silver ions reduced is:

Moles of silver ions reduced = Q

Charge per mole

Moles of silver ions reduced = 1500 C

107.87 C/mol

Moles of silver ions reduced ≈13.90 mol

Thus, the number of moles of silver ions reduced is approximately 13.90 mol.

Question 13

Question

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

solution. The student passes a current of 2.50 A through the solution for 30.0

minutes. If the experiment resulted in the deposition of 5.00 grams of copper

onto the cathode, calculate the Faraday constant (F).

Solution

Step 1: Determine the amount of charge passed through the solution. Given

that the current is 2.50 A and the time is 30.0 minutes, we need to convert the

time to seconds:

30.0 minutes = 30.0×60 seconds = 1800 seconds

The amount of charge passed through the solution can be calculated using

the formula:

Q=I×t

where Qis the charge in Coulombs, Iis the current in amperes, and tis the

time in seconds. Substitute the given values into the formula:

Q= 2.50 A ×1800 s = 4500 C

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

tion. One equivalent (or one mole) of electrons has a charge of 1 Faraday, which

is equal to F= 96485 C/mol.

To find the number of moles of electrons, we divide the total charge by the

charge of one mole of electrons:

Moles of electrons = 4500 C

96485 C/mol

Step 3: Calculate the amount of copper deposited using the given mass. The

molar mass of copper is 63.55 g/mol. Use this to find the number of moles of

copper deposited:

Moles of copper = 5.00 g

63.55 g/mol

Step 4: Using Faraday’s laws of electrolysis, relate the number of moles of

electrons to the number of moles of copper. The balanced equation for the

electrolysis of copper(II) sulfate is:

Cu2+ + 2e−

→Cu

For each mole of copper deposited, 2 moles of electrons are required. There-

fore, the moles of electrons should be twice the moles of copper deposited.

Step 5: Calculate the Faraday constant. Since 2 moles of electrons are

required for each mole of copper deposited, we have:

Moles of electrons = 2 ×Moles of copper

Equating this to the ratio calculated in Step 2, we can solve for the Faraday

constant:

2×5.00 g

63.55 g/mol =4500 C

96485 C/mol

Solving for Fgives:

F=2×5.00 g ×96485 C/mol

4500 C ×63.55 g/mol

Question 14

Question

An aqueous solution of copper(II) sulfate (CuSO4) is electrolyzed using graphite

electrodes. If a current of 2.5 A is passed through the solution for 1 hour, what

mass of copper is deposited at the cathode? (Given: atomic masses - Cu = 63.5

g/mol, S= 32.1 g/mol, O= 16.0 g/mol, H= 1.0 g/mol; Faraday constant =

96500 C/mol; 1 hour = 3600 s)

Solution

Step 1: Calculate the total charge passed through the solution. The total charge

can be calculated using the formula:

Q=I×t

where Qis the total charge in Coulombs, Iis the current in Amperes, and tis

the time in seconds. Given I= 2.5Aand t= 1 hour = 3600 s, we have:

Q= 2.5A×3600 s= 9000 C

Step 2: Determine the number of moles of electrons transferred. Since 1 F

(Faraday) of charge is equivalent to 96500 Cof charge, we can calculate the

number of moles of electrons transferred using the formula:

n=Q

where nis the number of moles of electrons transferred, Qis the total charge

in Coulombs, and Fis the Faraday constant. Substitute Q= 9000 Cand

F= 96500 C/mol to get:

n=9000 C

96500 C/mol = 0.0935 mol

Step 3: Determine the number of moles of copper deposited. Since each

mole of Cu2+ ion is reduced to one mole of copper metal, the number of moles

of copper deposited is the same as the number of moles of electrons transferred.

Thus, 0.0935 mol of copper is deposited.

Step 4: Calculate the mass of copper deposited. To find the mass of copper

deposited, we can use the formula:

Mass = Number of moles ×Molar mass

Given the molar mass of copper as 63.5g/mol, we have:

Mass = 0.0935 mol ×63.5g/mol = 5.92 g

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

Question 15

Question

A certain electrolytic cell operates under a constant current of 2.5 A for 3.5

hours. During this time, 1.25 g of aluminum is deposited on the cathode. Cal-

culate the faraday constant F. Given that the molar mass of aluminum is

27 g/mol.

Solution

Step 1: Calculate the charge (in coulombs) passed through the electrolytic cell.

We know that the charge can be calculated using the formula:

Charge (C) = Current (A) ×Time (s)

Given that the current is 2.5 A and the time is 3.5 hours = 3.5×3600 s, we

have:

Charge (C) = 2.5 A ×3.5×3600 s

Charge (C) = 2.5×3.5×3600 C

Step 2: Calculate the number of moles of aluminum deposited. Using the

formula:

Moles = Mass (g)

Molar Mass (g/mol)

Given that the mass of aluminum deposited is 1.25 g and the molar mass is

27 g/mol, we have:

Moles = 1.25 g

27 g/mol

Step 3: Calculate the faraday constant. The faraday constant is the charge

required to deposit one mole of a substance. Since the charge is directly pro-

portional to the moles of substance deposited, we have:

F=Charge (C)

Moles

Substitute the values calculated in Step 1 and Step 2 into the formula above

to find the faraday constant F.

Question 16

Question

An electrolytic cell containing a molten salt with a molar mass of 80.0 g mol−1

is connected to a 6.0 V battery. If the mass of the metal deposited during

electrolysis is 1.6 g, calculate the time taken for the deposition.

Solution

Step 1: Calculate the number of moles of the deposited metal. Given that the

molar mass of the metal is 80.0 g mol−1and the mass of the metal deposited is

1.6 g, we can calculate the number of moles (n) using the formula:

n=Mass

Molar Mass

n=1.6 g

80.0 g mol−1

n= 0.02 mol

Step 2: Determine the charge required to deposit the given amount of metal.

From Faraday’s first law of electrolysis, we know that the amount of substance

deposited at an electrode is directly proportional to the quantity of electricity

passed through it. The formula for charge (Q) in terms of moles of substance

(n) is:

Q=n×F

where Fis the Faraday constant which is ≈96500 C mol−1.

Substitute n= 0.02 mol and F= 96500 C mol−1into the formula:

Q= 0.02 ×96500

Q= 1930 C

Step 3: Calculate the current and time taken for deposition. Given that the

potential difference across the cell is 6.0 V, we can use the formula Q=I×t

to find the current (I) and time (t).

I=Q

t=Q

Substitute Q= 1930 C and I=6.0 V

Rinto the formula:

t=1930 C

6.0 V

t=1930 C ×R

6.0 V

Therefore, the time taken for the deposition will depend on the resistance of

the electrolytic cell.

Question 17

Question

Consider an electrolytic cell containing a solution of silver nitrate, where silver

is deposited on the cathode and the current flowing through the cell is 2.0 A. If

1.0 g of silver is deposited in 30 minutes, determine the Faraday constant.

Solution

Step 1: Find the number of moles of silver deposited. Given: Current, I= 2.0 A

Time, t= 30 min = 1800 s Mass of silver deposited, m= 1.0 g

We know that the charge deposited (Q) is given by:

Q=It

Substitute the given values:

Q= 2.0 A ×1800 s = 3600 C

Step 2: Find the number of moles of silver deposited. The charge on one

mole of electrons is given by the Faraday constant F, which is equal to 96485 C

mol−1. The number of moles of electrons can be calculated using the formula:

n=Q

Where eis the elementary charge (1.6×10−19 C). Therefore,

n=3600 C

1.6×10−19 C

Step 3: Calculate the Faraday constant. The Faraday constant is defined as

the charge on one mole of electrons, so:

F=Q/n

Substitute the calculated values:

F=3600 C

Finally, simplify the expression to find the Faraday constant.

Question 18

Question

When a current of 2.5 A is passed through a solution of a compound of silver

for 2 hours, 10.8 g of silver is deposited at the cathode. Calculate the atomic

weight of silver.

Solution

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

I= 2.5 A Time, t= 2 hours = 2×3600 seconds We know that charge, Q=I×t

Q= 2.5×2×3600

Q= 2.5×7200

Q= 18000 C

Step 2: Calculate the number of moles of silver deposited. Given: Mass of

silver deposited, m= 10.8 g Atomic weight of silver, MNumber of moles of

silver, n=m

MSince 1 F (Faraday) is the charge needed to deposit 1 mole of Ag,

we know that charge for 1 mole of Ag is Q= 1 ×96500 C So, 18000 C of charge

is needed to deposit 1 mole of Ag.

18000

96500 =m/M

M=10.8×96500

18000 = 58.17 g/mol

Therefore, the atomic weight of silver is 107.87 g/mol.

Question 19

Question

In an electrolytic cell, a certain amount of current is passed through a solution

of copper sulfate using copper electrodes. If 2.75 grams of copper is deposited

on the cathode in 25 minutes, determine the current passing through the cell in

amperes.

(Given: Atomic mass of copper = 63.5 g/mol, Faraday constant = 9.65×104

C/mol)

Solution

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

Moles of copper = Mass

Molar mass

Moles of copper = 2.75 g

63.5 g/mol

Step 2: Use Faraday’s laws to determine the charge passed through the cell.

Charge = Moles ×Faraday constant

Step 3: Calculate the current passing through the cell.

Current = Charge

Time

Now, let’s calculate the current passing through the cell in amperes.

Question 20

Question

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

sulfate, CuSO4. If a current of 2.50 A is passed through the solution for 1.50

hours, how many grams of copper will be deposited at the cathode? (Given:

atomic mass of copper = 63.5 g/mol, Faraday constant = 96,500 C/mol)

Solution

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

I= 2.50 A Time, t= 1.50 hours = 1.50 ×3600 s = 5400 s

The total charge passed, Q=I×t

Calculate Q:

Q= 2.50 A ×5400 s = 13,500 C

Step 2: Find the number of moles of copper deposited. Using Faraday’s first

law of electrolysis, 1 mole of electrons produces 1 mole of copper:

1 mol of electrons = 1 Faraday = 96485 C

Number of moles of electrons, n=Q

96500 =13,500 C

96500 C/mol

Calculate n:

n=13,500 C

96500 C/mol ≈0.140 mol

Step 3: Calculate the mass of copper deposited at the cathode. Given:

Atomic mass of copper, MCu = 63.5 g/mol

Mass of copper deposited, m=n×MCu

Calculate m:

m= 0.140 mol ×63.5 g/mol ≈8.89 g

Therefore, approximately 8.89 grams of copper will be deposited at the cath-

ode.

Question 21

Question

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

tion. The student collects the data below:

- Mass of copper deposited on the cathode: 6.25 g - Current passing through

the electrolyte: 2.50 A - Time taken: 45 minutes

Determine the following: a) The charge passed through the electrolyte b)

The number of moles of electrons passed c) The number of moles of copper

deposited d) Calculate the experimental Faraday’s constant value using the

data provided

Given: - Atomic mass of copper = 63.5 g/mol - Faraday’s constant =

96500 C/mol

Solution

a) To determine the charge passed through the electrolyte, we use the formula:

Q=I·t

where: - Qis the charge passed through the electrolyte - Iis the current passing

through the electrolyte - tis the time taken

Step 1: Substitute the given values into the formula.

Q= 2.50 A×(45 min ×60 s/min)

Step 2: Calculate the charge passed through the electrolyte.

Q= 2.50 A×2700 s= 6750 C

Therefore, the charge passed through the electrolyte is 6750 C.

b) To find the number of moles of electrons passed, we use the formula:

Moles of electrons = Q

where: - Qis the charge passed through the electrolyte - Fis Faraday’s constant

Step 1: Substitute the given values into the formula.

Moles of electrons = 6750 C

96500 C/mol

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

Moles of electrons = 6750

96500 mol ≈0.0701 mol

Therefore, the number of moles of electrons passed is approximately 0.0701

mol.

c) The number of moles of copper deposited can be determined by stoichiom-

etry. Since each copper ion (Cu2+) requires 2 moles of electrons to be reduced

to copper, the number of moles of copper deposited is half the number of moles

of electrons passed.

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

Moles of copper deposited = 0.5×Moles of electrons = 0.5×0.0701 mol

Step 2: Simplify the expression.

Moles of copper deposited = 0.0350 mol

Therefore, the number of moles of copper deposited is 0.0350 mol.

d) The experimental Faraday’s constant value can be calculated using the

mass of copper deposited and the number of moles of copper deposited.

Step 1: Calculate the experimental Faraday’s constant value.

Experimental Faraday’s constant = Mass of copper deposited

Moles of copper deposited

Step 2: Substitute the given values into the formula.

Experimental Faraday’s constant = 6.25 g

0.0350 mol

Step 3: Calculate the experimental value of Faraday’s constant.

Experimental Faraday’s constant = 6.25

0.0350 mol/g ≈178.57 mol/g

Therefore, the experimental Faraday’s constant value is approximately 178.57

mol/g.

Question 22

Question

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

tion. The student passes a constant current of 0.5 A through the solution for

30 minutes. During this time, copper(II) ions are reduced at the cathode and

copper metal is deposited. If the student measures that 2 grams of copper is

deposited, calculate the number of moles of electrons that have passed through

the cell.

Solution

Step 1: Find the molar mass of copper. Given that the molar mass of copper is

63.55 g/mol.

Step 2: Calculate the number of moles of copper deposited. Given that 2

grams of copper is deposited.

Number of moles = 2 g

63.55 g/mol = 0.0315 mol

Step 3: Use Faraday’s laws of electrolysis to relate the number of moles of

copper to the number of moles of electrons. The balanced half-reaction for the

reduction of copper(II) ions to copper is:

Cu2+ + 2e−

→Cu

From the half-reaction, 2 moles of electrons are required to deposit 1 mole of

copper.

Step 4: Calculate the number of moles of electrons. Given that 2 moles of

electrons are required to deposit 1 mole of copper.

Number of moles of electrons = 2 ×0.0315 mol = 0.0630 mol

Therefore, the number of moles of electrons that have passed through the

cell is 0.0630 mol.

Question 23

Question

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

chloride (NaCl) for 3.00 hours. If the standard electrode potential for the re-

duction of Na+ions to Na metal is -2.71 V, calculate the mass of sodium metal

deposited at the cathode.

Solution

Step 1: Find the total charge that passed through the cell. The total charge Q

passed 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 that

I= 2.50 A and t= 3.00 hours, we first convert time to seconds:

t= 3.00 ×3600 = 10800 s

Plugging in the values, we find:

Q= 2.50 ×10800 = 27000 C

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

of electrons has a charge of 1 Faraday, which is equal to 96485 C. Hence, the

number of moles of electrons transferred, n, can be calculated as:

n=Q

96485

Substitute Q= 27000 C into the formula to find:

n=27000

96485 ≈0.280 mol

Step 3: Determine the number of moles of sodium being reduced. The

balanced half-reaction for the reduction of Na+ions to Na metal is:

Na+(aq) + e−

→Na(s)

From the half-reaction, we see that 1 mole of electrons corresponds to 1 mole of

Na being reduced. Therefore, the number of moles of sodium reduced will also

be approximately 0.280 mol.

Step 4: Calculate the mass of sodium deposited. The molar mass of sodium

(Na) is approximately 23.0 g/mol. The mass of sodium deposited, m, can be

calculated using the formula:

m=n×molar mass of Na

Substitute the values to find:

m≈0.280 ×23.0≈6.44 g

Therefore, approximately 6.44 grams of sodium metal will be deposited at

the cathode.

Question 24

Question

Faraday’s laws of electrolysis state that the amount of a substance produced

at an electrode during electrolysis is directly proportional to the quantity of

electricity passed through the electrode.

Given a problem in which a copper sulfate solution is electrolyzed using

a current of 2.5 A for 3 hours, calculate the mass of copper deposited at the

cathode. The molar mass of copper is 63.5 g/mol, and the charge on an electron

is 1.6×10−19 C.

Solution

Step 1: Calculate the total charge passed through the circuit using the formula:

Total charge = Current ×Time

Total charge = 2.5 A ×3 hours ×3600 s/hour

Total charge = 2.5 A ×10800 s

Total charge = 27000 C

Step 2: Calculate the number of electrons transferred during the electrolysis

using Faraday’s constant:

Number of electrons = Total charge

Charge on one electron

Number of electrons = 27000 C

1.6×10−19 C

Number of electrons ≈1.6875 ×1020

Step 3: Calculate the number of moles of copper deposited using the equa-

tion:

Number of moles = Number of electrons

2×Valency

Since copper has a valency of 2, we have:

Number of moles = 1.6875 ×1020

2×2

Number of moles = 1.6875 ×1020

Step 4: Calculate the mass of copper deposited using the equation:

Mass of copper = Number of moles ×Molar mass of copper

Mass of copper = 1.6875 ×1020 ×63.5 g/mol

Mass of copper = 1.0706 ×1022 g

Therefore, the mass of copper deposited at the cathode during the electrol-

ysis is approximately 1.0706 ×1022 g.

Question 25

Question

An electrolysis cell is set up with a copper sulfate solution. If a current of 2.0

A is passed through the cell for 2.5 hours, what mass of copper is deposited on

the cathode? (Given: Faraday’s constant = 9.65 ×104C/mol, atomic mass of

copper = 63.5 g/mol)

Solution

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

Current(I)=2.0 A

Time(t)=2.5 hours = 2.5×3600 s = 9000 s

The total charge passed through the cell can be calculated using the formula

Q=I×t.

Q= 2.0 A ×9000 s

= 18000 C

Step 2: Calculate the number of moles of electrons that have passed through

the cell. Since 1 Faraday (F) is equal to 9.65 ×104C/mol, the number of moles

of electrons (n) is given by n=Q

n=18000 C

9.65 ×104C/mol

≈0.1866 mol

Step 3: Calculate the mass of copper deposited on the cathode. Since the

atomic mass of copper is 63.5 g/mol, the mass of copper can be calculated using

the formula m=n×atomic mass.

m= 0.1866 mol ×63.5 g/mol

≈11.85 g

Therefore, the mass of copper deposited on the cathode is approximately

11.85 g.

Question 26

Question

Two electrolytic cells are set up in series. In the first cell, a current of 2.5

A is passed through a solution of sodium chloride for 2 hours. In the second

cell, a current of 3.5 A is passed through a solution of copper(II) sulfate for

1 hour. Calculate the mass of sodium metal and copper metal produced in

each cell. (Given: Faraday’s constant F= 96500 C/mol; molar mass of sodium

= 22.99 g/mol; molar mass of copper = 63.55 g/mol)

Solution

Step 1: Calculate the charge passed through each cell For the first cell: Given

current I1= 2.5 A and time t1= 2 hours = 7200 seconds

Charge Q1=I1×t1= 2.5 A ×7200 s = 18000 C

For the second cell: Given current I2= 3.5 A and time t2= 1 hour =

3600 seconds

Charge Q2=I2×t2= 3.5 A ×3600 s = 12600 C

Step 2: Calculate the number of moles of each metal deposited For the first

cell (sodium deposition): Number of moles of sodium n1=Q1

F=18000 C

96500 C/mol ≈

0.1865 mol

Mass of sodium produced = n1×molar mass of sodium = 0.1865 mol×22.99 g/mol ≈

4.28 g

For the second cell (copper deposition): Number of moles of copper n2=

F=12600 C

96500 C/mol ≈0.1306 mol

Mass of copper produced = n2×molar mass of copper = 0.1306 mol×63.55 g/mol ≈

8.30 g

Therefore, the mass of sodium metal produced in the first cell is approx-

imately 4.28 g, and the mass of copper metal produced in the second cell is

approximately 8.30 g.

Question 27

Question

In a laboratory experiment, a researcher passes a current of 2.5 A through a

solution of a copper salt for 15 minutes. During this electrolysis process, the

researcher observes that a mass of 1.8 g of copper is deposited on the cathode.

Calculate the molar mass of the copper salt. (Assume 100

Solution

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

current is 2.5 A and the time is 15 minutes, we first convert the time into

seconds:

15 minutes = 15 ×60 = 900 seconds

The total charge passed through the solution can be found using the formula:

Q=I×t

Q= 2.5 A ×900 s = 2250 C

Step 2: Determine the number of moles of electrons and the number of moles

of copper deposited on the cathode. Since the electrolysis process is 1001 mole

of electrons = 1 Faraday = 9.65 ×104C

nelectrons =Q/Faraday constant

nelectrons = 2250 C/9.65 ×104C/mol = 0.0234 mol

Step 3: Calculate the molar mass of the copper salt. The molar mass of

the copper salt can be calculated using the mass of copper deposited and the

number of moles of copper:

mcopper = 1.8 g

ncopper =mcopper/molar mass of copper

0.0234 mol = 1.8 g/molar mass of copper

Solving for the molar mass of copper:

Molar mass of copper = 1.8 g/0.0234 mol = 76.92 g/mol

Therefore, the molar mass of the copper salt is 76.92 g/mol.

Question 28

Question

A student conducts an experiment to study the electrolysis of a certain salt

solution using a constant current of 2.0 A for 30 minutes. The student discovers

that 2.40 g of the salt is deposited at one electrode. Determine the charge on

each ion in the molecular formula of the salt.

Solution

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

current is 2.0 A and the time is 30 minutes, we first convert the time to seconds:

Time (s) = 30 minutes ×60 s/min = 1800 s

Then, we calculate the total charge passed through the solution using the

formula:

Charge = Current ×Time

Charge = 2.0 A ×1800 s = 3600 Coulombs

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

mass of the salt deposited is 2.40 g. To find the number of moles, we divide the

mass by the molar mass of the salt.

Step 3: Find the molar mass of the salt. To find the molar mass of the salt,

we need to know its formula. Let the molecular formula of the salt be AxBy,

where xand yare the charges on A and B ions, respectively.

Step 4: Calculate the charge on each ion. From the charge passed through

the solution and the number of moles of salt deposited, we can determine the

charge on each ion using the formula:

Charge = Number of moles ×1×(Charge on A + Charge on B)

Solving this equation will give us the charges on ions A and B in the molec-

ular formula of the salt.

Question 29

Question

In an electrolytic cell, a current of 5.0 A is passed through a solution of cop-

per(II) sulfate for 20 minutes. During this time, 1.0 g of copper is deposited at

the cathode.

Assuming 100

Solution

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

I= 5.0 A

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

The total charge passing through the cell, Q=I·t

Q= 5.0 A ×1200 s = 6000 C

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

copper, m= 1.0 g Molar mass of copper, M= 63.55 g/mol

Number of moles of copper deposited, n=m

n=1.0 g

63.55 g/mol ≈0.0157 mol

Step 3: Determine the number of electrons passed. The number of moles of

electrons passed is equal to the number of moles of copper deposited (since in

the redox reaction, 1 mole of Cu2+ gains 2 moles of electrons to form 1 mole of

copper). Number of moles of electrons, ne= 2n

ne= 2 ×0.0157 mol = 0.0314 mol

Step 4: Calculate the charge transferred in coulombs. Since 1 mole of

electrons corresponds to the charge of 1 F = 96485 C, Charge transferred,

Q=ne·96485 C/mol

Q= 0.0314 mol ×96485 C/mol ≈3033 C

Step 5: Calculate the standard electrode potential of the copper half-cell.

Using Faraday’s laws of electrolysis:

∆G=−nF E

where ∆Gis the change in Gibbs free energy, nis the number of moles of

electrons, Fis the Faraday constant (96485 C/mol), and Eis the standard

electrode potential.

We know that ∆G=−QV , where Vis the potential difference in volts.

Thus,

E=QV

nF =3033 C ×V

0.0314 mol ×96485 C/mol

Since the question assumes 100

3033 C = 1 mol ×96485 C/mol

Thus, V=E= 1 V is the standard electrode potential of the copper half-

cell.

Question 30

Question

In an electrolysis experiment, a current of 2.5 A is passed through a solution of

CuCl2for 20 minutes. If 1.2 g of copper is deposited at the cathode, what is

the efficiency of the electrolysis process?

Solution

Step 1: Find the total charge passing through the solution. Given that the

current passing through the solution is 2.5 A and the time is 20 minutes, we

first convert the time to seconds:

20 minutes = 20 ×60 seconds = 1200 seconds

The total charge passing through the solution is given by:

Q=I×t

Q= 2.5 A ×1200 s = 3000 C

Step 2: Calculate the amount of charge required to deposit 1.2 g of copper.

The number of moles of copper deposited can be found using the molar mass of

copper (Cu) as 63.55 g/mol. The charge required to deposit 1 mole of copper

is the same as the charge on 1 mole of electrons (1 F = 96485 C/mol). First,

calculate the number of moles of copper:

n=mass

molar mass =1.2 g

63.55 g/mol ≈0.0189 mol

Then, find the charge required to deposit this amount of copper:

QCu =n×96485 C/mol = 0.0189 mol ×96485 C/mol = 1822.46 C

Step 3: Calculate the efficiency of the electrolysis process. The efficiency of

the electrolysis process can be calculated using the formula:

Efficiency = Actual charge

Theoretical charge ×100%

where the actual charge passing through the solution is 3000 C and the theo-

retical charge required to deposit 1.2 g of copper is 1822.46 C. Calculating the

efficiency:

Efficiency = 3000

1822.46 ×100% ≈164.6%

The efficiency of the electrolysis process is approximately 164.6

Question 31

Question

In an electrolytic cell, a current of 2.5 A is passed through a solution of CuSO4

for 45 minutes. If the electrodeposition of copper occurs, calculate the mass of

copper deposited. (Atomic masses: Cu = 63.5, O = 16, S = 32)

Solution

Step 1: Find the total charge passed through the cell. Given current, I= 2.5

A and time, t= 45 minutes, we can convert time to seconds:

t= 45 minutes ×60 = 2700 seconds

The total charge passed through the cell is given by:

Q=I×t= 2.5 A ×2700 s = 6750 C

Step 2: Convert charge to moles of electrons. Since 1 faraday (F) is equiv-

alent to 1 mol e−, we can calculate the moles of electrons transferred using

Faraday’s constant:

1 F = 96500 C/mol

Moles of electrons = 6750 C

96500 C/mol = 0.07 mol

Step 3: Determine the moles of copper deposited. From the balanced equa-

tion for the electrodeposition of copper:

2 mol e−+Cu2+ →Cu

we know that 2 mol of electrons are needed to deposit 1 mol of copper. So,

moles of copper deposited = 0.07 mol

2= 0.035 mol

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

is 63.5 g/mol. The mass of copper deposited can be calculated as follows:

Mass of Cu deposited = moles of Cu deposited ×atomic mass of Cu

= 0.035 mol ×63.5 g/mol = 2.225 g

Therefore, the mass of copper deposited is 2.225 g.

Question 32

Question

Two electrolytic cells are set up in series using a power supply with a constant

voltage. Cell A contains a solution of silver nitrate (AgNO3), while cell B

contains a solution of sodium chloride (NaCl). If 0.5 grams of silver is deposited

at the cathode in cell A after 20 minutes, how much sodium metal will be

deposited at the cathode in cell B after 30 minutes? (Assume the same current

passes through both cells.)

Solution

Step 1: Calculate the amount of charge passed through cell A. The molar mass

of silver is 107.87 g/mol. Since 1 mole of electrons is required to deposit 1 mole

of silver, we can calculate the charge passed through the cell using the formula:

Q=Mass

Molar mass ×Faraday constant

Q=0.5 g

107.87 g/mol ×96485 C/mol

Q=0.5

107.87 ×96485

Q≈447.12 C

Step 2: Determine the current passing through cell B. Since the same volt-

age is applied to both cells, the current flowing through both cells is the same.

Therefore, the current passing through cell B can be calculated using the for-

mula:

I=Q

where tis the time in seconds.

I=447.12

20 ×60

I≈0.3726 A

Step 3: Calculate the amount of charge passed through cell B in 30 minutes.

QB=I×tB

QB= 0.3726 ×30 ×60

QB= 672.72 C

Step 4: Determine the amount of sodium metal deposited at the cathode

in cell B. The molar mass of sodium is 22.99 g/mol. Since 1 mole of electrons

is required to deposit 1 mole of sodium, we can calculate the mass of sodium

deposited using:

Mass of sodium = QB

Faraday constant ×Molar mass of sodium

Mass of sodium = 672.72

96485 ×22.99

Mass of sodium ≈0.1600 g

Therefore, approximately 0.1600 grams of sodium metal will be deposited at

the cathode in cell B after 30 minutes.

Question 33

Question

An aqueous solution of copper sulfate, CuSO4, is electrolyzed using inert elec-

trodes. If 2.50 amperes of current is passed through the solution for 2.00 hours,

what mass of copper will be deposited at the cathode? (Given: Faraday constant

= 96500 C/mol, atomic mass of copper = 63.5 g/mol)

Solution

Step 1: Calculate the total charge passed through the solution. The total charge

(Q) passed can be calculated using the formula:

Q=I×t

where I= 2.50 A is the current and t= 2.00 h is the time in hours.

Q= 2.50 A ×2.00 h ×3600 s/h = 18000 C

Step 2: Calculate the number of moles of electrons transferred. One mole of

electrons corresponds to one Faraday (96500 C/mol). Therefore, the number of

moles of electrons (n) can be calculated as:

n=Q

96500

n=18000 C

96500 C/mol ≈0.1866 mol

Step 3: Determine the number of moles of copper deposited. By the stoi-

chiometry of the reaction, 2 moles of electrons are required to deposit 1 mole of

copper. Therefore, the number of moles of copper deposited (nCu) is:

nCu =1

2×n

nCu =1

2×0.1866 mol ≈0.0933 mol

Step 4: Calculate the mass of copper deposited. To find the mass of copper

deposited, we use the atomic mass of copper (63.5 g/mol). The mass of copper

deposited (m) can be calculated as:

m=nCu ×Atomic mass of copper

m= 0.0933 mol ×63.5 g/mol ≈5.93 g

Therefore, approximately 5.93 grams of copper will be deposited at the cath-

ode.

Question 34

Question

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

sulfate. The student connects a copper electrode to the positive terminal of a

battery and a platinum electrode to the negative terminal. The student observes

that after some time, the mass of the copper electrode decreases while the mass

of the platinum electrode remains constant. Explain this observation in terms

of Faraday’s laws of electrolysis.

Solution

To explain this observation in terms of Faraday’s laws of electrolysis, we need

to consider the reactions that occur at each electrode and the Faraday’s laws

associated with the amount of substance deposited or dissolved during electrol-

ysis.

Step 1: Reactions at the Electrodes At the copper electrode (cathode):

The copper(II) ions in the solution gain electrons and are reduced to copper

atoms:

Cu2+ + 2e−

→Cu

At the platinum electrode (anode): Water molecules in the solution are oxi-

dized to produce oxygen gas and protons:

2H2O→O2+ 4H++ 4e−

Step 2: Faraday’s Laws of Electrolysis Faraday’s first law states that

the amount of substance deposited or dissolved at an electrode during elec-

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

electrolyte. Faraday’s second law states that the amounts of different substances

deposited or dissolved by the same quantity of electricity are proportional to

their chemical equivalent weights.

Step 3: Explanation of the Observation Since copper ions are reduced

at the copper electrode, copper atoms are deposited on the electrode. This

explains why the mass of the copper electrode decreases.

On the other hand, at the platinum electrode, water molecules are oxidized

to produce oxygen gas and protons. Since the platinum electrode does not

participate in the redox reaction involving copper ions, its mass remains constant

throughout the electrolysis.

Therefore, the observation that the mass of the copper electrode decreases

while the mass of the platinum electrode remains constant can be explained by

the different reactions occurring at each electrode as governed by Faraday’s laws

of electrolysis.

Question 35

Question

State Faraday’s laws of electrolysis and explain the significance of these laws in

the context of electrochemistry.

Solution

Faraday’s Laws of Electrolysis: Faraday’s laws of electrolysis are two quan-

titative laws that describe the amount of chemical change that occurs during

electrolysis. These laws were formulated by Michael Faraday in the 1830s. The

two laws are:

1. The amount of chemical change (in mass) produced by a current passing

through an electrolyte is proportional to the quantity of electricity that

passes through the electrolyte.

2. The amounts of different substances produced by passing the same quan-

tity of electricity through different electrolytes are proportional to their

equivalent weights.

Significance:



Quantitative Relationship: Faraday’s laws provide a quantitative re-

lationship between the amount of current passed through an electrolyte

and the chemical changes that occur. This allows for precise control and

prediction of outcomes in electrochemical processes.



Equivalent Weights: By considering the equivalent weights of sub-

stances, Faraday’s laws provide a way to compare the amounts of different

substances produced in electrolysis. This is essential for understanding

and designing electrochemical processes.



Fundamental Principles: Faraday’s laws are fundamental to the field of

electrochemistry, providing a basis for understanding the transformation

of chemical energy into electrical energy and vice versa.



Experimental Verification: The laws have been extensively verified

through experiments, reinforcing their importance in the field of electro-

chemistry.

CHEM 105 - ELEMENTS OF GENERAL CHEMISTRY Faraday’s laws of electrolysis Question Bank Set 1

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