CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Faraday’s
laws of electrolysis
Question Bank - Set 4
Liberty University
Question 1
Question
A copper electrode in a solution of copper (II) sulfate (CuSO4) is connected to
a battery. If a current of 2.5 A is passed through the solution for 2.5 hours,
what mass of copper is deposited on the electrode? (Given: Atomic mass of
copper = 63.5 g/mol, Faraday’s constant = 9.65 ×104C/mol)
Solution
Step 1: Calculate the total charge passed through the solution using the formula
Q=I×t, where Qis the charge in Coulombs, Iis the current in Amperes, and
tis the time in seconds.
Q= 2.5 A ×(2.5×3600) s = 22500 C
Step 2: Determine the number of moles of electrons involved in the electro-
chemical reaction using Faraday’s first law, which states that the charge passed
is proportional to the number of moles of electrons involved. The formula is
given by moles of electrons = Q
96500 .
moles of electrons = 22500 C
96500 C/mol = 0.233 mol
Step 3: Since 2 moles of electrons are involved in the reduction of Cu2+ to
Cu, the moles of copper deposited on the electrode is equal to the number of
moles of electrons. Therefore, moles of copper = 0.233 mol.
Step 4: Calculate the mass of copper deposited using the formula mass =
moles ×molar mass with the atomic mass of copper given as 63.5 g/mol.
mass of copper = 0.233 mol ×63.5 g/mol = 14.8055 g
Therefore, the mass of copper deposited on the electrode is 14.81 g.
Question 2
Question
A student is performing an electrolysis experiment using a copper sulfate solu-
tion. Initially, there is no current passing through the solution. The student
then turns on the power supply and allows a current of 2.00 A to pass through
the solution for 30 minutes. During this time, copper ions are reduced at the
cathode to form copper metal.
If the student observes 1.00 g of copper deposited at the cathode, calculate
the Faraday’s constant and determine the number of electrons required for the
reduction of 1 mole of copper ions.
Given: Atomic mass of copper = 63.5 g/mol, Faraday’s constant = 9.65×104
C/mol.
Solution
Step 1: Find the total charge passed through the solution. The total charge
passed through the solution can be calculated using the formula:
Q=It
where Qis the total charge passed (in coulombs), Iis the current (in amperes),
and tis the time (in seconds). Given I= 2.00 A and t= 30 minutes = 30 ×60
s = 1800 s, we have:
Q= 2.00 ×1800 = 3600 C
Step 2: Calculate the moles of copper deposited at the cathode. To calculate
the moles of copper deposited, we use the formula:
moles = mass
molar mass
Given that the mass of copper deposited is 1.00 g and the molar mass of copper
is 63.5 g/mol, we have:
moles = 1.00
63.5= 0.0157 moles
Step 3: Determine the number of electrons required for the reduction of
1 mole of copper ions. We know that the Faraday’s constant represents the
charge of 1 mole of electrons. Therefore, the number of electrons required for
the reduction of 1 mole of copper ions is given by:
number of electrons = total charge passed (C)
Faraday’s constant (C/mol) ×moles of copper ions
Substitute the values:
number of electrons = 3600
9.65 ×104×0.0157
number of electrons = 0.5918 moles of electrons
2
Question 3
Question
In an electrolysis experiment, a current of 2.5 A is passed through a solution of
copper(II) sulfate for 30 minutes. During this time, 3.75 g of copper metal is
deposited on the cathode. Calculate the Faraday constant and the number of
moles of electrons transferred during the experiment.
Given: Atomic mass of copper = 63.5 g/mol, and the charge of an electron
= 1.602 ×10−19 C.
Solution
Step 1: Calculate the charge required to deposit 3.75 g of copper. The number
of moles of copper deposited can be calculated using the atomic mass of copper.
Let the charge required be Q.
Q= charge needed to deposit 3.75 g of copper
Q=mass deposited
molar mass of copper ×charge of one mole of electrons
Q=3.75 g
63.5 g/mol ×1 mol ×6.022 ×1023 molecules/mol ×1.602 ×10−19 C
Step 2: Calculate the Faraday constant (F). The Faraday constant can be
calculated using the charge on one mole of electrons. Let Fbe the Faraday
constant.
F= Charge on one mole of electrons
Step 3: Calculate the number of moles of electrons (n) transferred. The
number of moles of electrons can be calculated using the charge required and
the Faraday constant. Let nbe the number of moles of electrons transferred.
n=Q
F
Step 4: Perform the calculations. Substitute the given values to calculate
Q,F, and n.
Step 5: Analyze the results. Discuss the significance of the calculated Fara-
day constant and the number of moles of electrons transferred in the context of
the electrolysis experiment.
Question 4
Question
A student is conducting an electrolysis experiment using a solution of copper
sulfate (CuSO4). If the student passes a current of 2.5 A through the solution
3
for 30 minutes, what mass of copper will be deposited at the cathode? (Given:
atomic mass of copper = 63.55 g/mol, Faraday’s constant = 9.65 ×104C/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. The
total charge can be calculated using the formula:
Q=I×t
where Q= total charge passed (coulombs), I= current (amperes), t= time
(seconds).
Given that I= 2.5Aand t= 30 ×60 seconds, we have:
Q= 2.5×30 ×60 = 4500 C
Step 2: Determine the number of moles of electrons transferred. Each mole
of electrons corresponds to a charge of 1 Faraday, which is equal to the charge
of 1 mole of electrons. The number of moles of electrons can be calculated using
Faraday’s constant:
Moles of electrons = Q
F
where F= 9.65 ×104C/mol.
Substitute the values to find moles of electrons:
Moles of electrons = 4500
9.65 ×104= 0.0467 mol
Step 3: Determine the moles of copper deposited. From the balanced chem-
ical equation for electrolysis of copper sulfate (CuSO4):
Cu2+ + 2e−→Cu
we know that 2 moles of electrons are required to deposit 1 mole of copper.
Therefore, the moles of copper deposited (nCu) is half of the moles of electrons:
nCu =0.0467
2= 0.0234 mol
Step 4: Calculate the mass of copper deposited. The mass of copper de-
posited can be calculated using the atomic mass of copper:
Mass of copper = nCu ×Molar mass of copper
Mass of copper = 0.0234 ×63.55 = 1.49 g
Therefore, the student will deposit 1.49 g of copper at the cathode during
the electrolysis of copper sulfate solution.
4
Question 5
Question
Faraday’s laws of electrolysis state that the amount of a substance produced at
an electrode is directly proportional to the quantity of electricity passed through
the electrolyte. A student wants to electroplate a metal with a mass of 7.5 g
onto an electrode using a current of 2.5 A for 20 minutes. If the atomic mass
of the metal is 63.55 g/mol, determine the efficiency of the electrolysis process.
Assume 100
Solution
Step 1: Determine the number of moles of metal deposited on the electrode.
Given: - Mass of metal deposited, m= 7.5 g - Atomic mass of the metal,
M= 63.55 g/mol
The number of moles of metal deposited is given by:
moles = m
M
moles = 7.5
63.55
moles ≈0.118 mol
Step 2: Calculate the total charge passed through the electrolyte.
Given: - Current, I= 2.5 A - Time, t= 20 minutes
Firstly, convert the time to seconds:
t= 20 ×60 = 1200 s
The total charge passed through the electrolyte is given by:
Q=It
Q= 2.5×1200
Q= 3000 C
Step 3: Determine the number of electrons involved in the reaction.
One Faraday of charge is equivalent to Avogadro’s number of electrons:
1 Faraday = 96,485 C/mol
The number of electrons involved in the reaction is given by:
n=Q
1 Faraday
n=3000
96485
5
n≈0.0311 mol
Step 4: Calculate the efficiency of the electrolysis process.
The efficiency of the electrolysis process is given by:
Efficiency = moles deposited
moles of electrons passed
Efficiency = 0.118
0.0311 ×100%
Efficiency ≈379.74%
Therefore, the efficiency of the electrolysis process is approximately 379.74
Question 6
Question
A solution of potassium iodide, KI, is electrolyzed using platinum electrodes. If
a current of 5.00 A is passed through the solution for 30.0 minutes, what mass
of iodine, I2, is produced? (Assume 100
Solution
Step 1: Write the balanced chemical equation for the electrolysis of potassium
iodide.
The balanced half-reactions are:
At the anode: 2I−→I2+ 2e−
At the cathode: 2H++ 2e−→H2
The overall reaction is:
2KI →I2+H2
Step 2: Calculate the total charge that passed through the solution.
Given: current, I= 5.00 A
Time, t= 30.0 minutes = 30.0 minutes ×1 hour
60 minutes = 0.50 hours
The total charge, Q, is given by Q=I×t.
Q= 5.00 A ×0.50 hours = 2.50 C
Step 3: Determine the number of moles of electrons that reacted.
From the balanced reaction, 2 moles of electrons are involved in the production
of 1 mole of I2.
Hence, the number of moles of electrons, n, that passed through the solution is
given by:
n=Q
96,485 C/mol
6
n=2.50 C
96,485 C/mol ≈2.59 ×10−5mol
Step 4: Calculate the mass of iodine produced.
From the balanced reaction, 2 moles of I−produce 1 mole of I2.
Therefore, the number of moles of I2produced is half the number of moles of
electrons:
moles of I2=n
2=2.59 ×10−5mol
2= 1.30 ×10−5mol
The molar mass of I2is 2×Atomic mass of I = 2×126.90 g/mol = 253.80 g/mol.
The mass of I2produced is:
Mass of I2= moles of I2×Molar mass of I2
Mass of I2= 1.30 ×10−5mol ×253.80 g/mol
Mass of I2≈3.29 ×10−3g
Therefore, approximately 3.29 ×10−3g of iodine is produced during the
electrolysis of potassium iodide.
Question 7
Question
A student performs an electrolysis experiment using a copper(II) sulfate solu-
tion and two copper electrodes. The student passes a constant current of 2.50
A through the solution for 2.00 hours. During this time, 5.00 g of copper is
deposited on the cathode.
Calculate the charge passed through the solution.
Solution
Step 1: Determine the number of moles of copper deposited on the cathode. To
calculate the number of moles of copper deposited, we first convert the mass
of copper to moles using the molar mass of copper (Cu). The molar mass of
copper is 63.55 g/mol.
Number of moles of Cu = Mass of Cu
Molar mass of Cu =5.00 g
63.55 g/mol
Step 2: Calculate the charge passed through the solution. The charge passed
through the solution can be calculated using Faraday’s first law of electrolysis:
Q=n×F
where: - Qis the charge in coulombs, - nis the number of moles of elec-
trons transferred (moles of copper deposited/mol), - Fis the Faraday constant
(96500 C/mol).
7
Now, we need to determine the number of moles of electrons transferred
from the number of moles of copper deposited. Copper(II) ions (Cu
²
) gain two
electrons to form solid copper. Therefore, the number of moles of electrons
transferred is twice the moles of copper deposited.
Step 3: Calculate the charge passed through the solution.
Q= 2 ×Number of moles of Cu ×F
Substitute the values into the equation to find the charge passed through
the solution.
Question 8
Question
A student is conducting an electrolysis experiment using a copper (Cu) electrode
and a silver (Ag) electrode in a solution of copper(II) sulfate (CuSO4) to deposit
silver onto the copper electrode. If the student applies a constant current of 2.0
A for 1 hour, what mass of silver will be deposited on the copper electrode?
(Given: atomic masses - Cu: 63.5 g/mol, Ag: 107.9 g/mol)
Solution
Step 1: Calculate the charge transferred during electrolysis. Given: Current
(I) = 2.0 A, Time (t) = 1 hour = 3600 seconds, Charge (Q) = ? The charge
transferred during electrolysis can be calculated using the formula:
Q=I×t
Q= 2.0 A ×3600 s = 7200 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday (F) = 96500 C (charge of 1 mole of electrons), we can calculate the
number of moles of electrons transferred using the formula:
Moles of electrons = Q
96500
Moles of electrons = 7200
96500
Moles of electrons ≈0.075 mol
Step 3: Determine the mass of silver deposited on the copper electrode. The
molar ratio of electrons to silver ions in the electrolysis of silver ions (Ag) is 2:1.
This means that for every 2 moles of electrons transferred, 1 mole of silver is
deposited. So, the number of moles of silver deposited can be calculated as:
Moles of silver = 1
2×Moles of electrons
8
Moles of silver = 1
2×0.075
Moles of silver = 0.0375 mol
Step 4: Calculate the mass of silver deposited on the copper electrode. Using
the molar mass of silver (Ag):
Mass of silver = Moles of silver ×Molar mass of Ag
Mass of silver = 0.0375 mol ×107.9 g/mol
Mass of silver ≈4.04 g
Therefore, approximately 4.04 grams of silver will be deposited on the copper
electrode after electrolysis.
Question 9
Question
An electrolytic cell is set up with a copper(II) sulfate solution and platinum
electrodes. If a current of 2.50 A is passed through the cell for 2.00 hours, what
mass of copper will be deposited at the cathode? (Given: 1 F = 96,485 C/mol,
M(Cu) = 63.55 g/mol)
Solution
Step 1: Find the total charge passed through the cell. Given: Current, I=
2.50 A Time, t= 2.00 hours = 2.00 ×3600 s = 7200 s
Total charge, Q=I×t
Plugging in the values: Q= 2.50 A ×7200 s = 18,000 C
Step 2: Convert the total charge to moles of electrons. Given: 1 F =
96,485 C/mol
Number of moles of electrons, n=Q
96485
Plugging in the values: n=18,000 C
96485 C/mol ≈0.1867 mol
Step 3: Determine the mass of copper deposited at the cathode. Given:
Molar mass of copper, M(Cu) = 63.55 g/mol
Mass of copper, m(Cu) = n×M(Cu)
Plugging in the values: m(Cu) = 0.1867 mol ×63.55 g/mol ≈11.86 g
Therefore, approximately 11.86 grams of copper will be deposited at the
cathode.
Question 10
Question
An aqueous solution of silver nitrate (AgNO3) is electrolyzed using a current of
1.5 A for 2 hours. If the reaction produces solid silver and oxygen gas, calculate:
9
1. the mass of silver deposited on the cathode,
2. the volume of oxygen gas produced at STP.
(Assume 100
Solution
Let’s first determine the total charge passed through the electrolyte solution:
Total charge = Current ×Time = 1.5 A ×2 hours
Total charge = 1.5 C/s ×7200 s = 10800 C
Step 1: Calculate the mass of silver deposited on the cathode. The molar
quantity of electric charge required to deposit one mole of silver is equal to the
Faraday constant (F), which is 96500 C/mol.
1 mol of Ag requires 96500 C
1 C will deposit 1
96500 mol of Ag
Mass of Ag deposited = Molar mass of Ag ×moles of Ag
Mass of Ag deposited = 108 g/mol ×10800
96500mol
Step 2: Calculate the volume of oxygen gas produced at STP. Since 2
moles of electrons are required to produce 1 mole of oxygen gas when water is
electrolyzed, and the molar volume of gas at STP is 22.4 L:
2 mol of e−produces 1 mol of O2
1 C will produce 1
2×1
96500 mol of O2
Volume of O2produced = 22.4 L/mol ×1
2×1
96500 ×10800mol
Question 11
Question
In an electrolytic cell, a solution of copper sulfate (CuSO4) is electrolyzed using
a current of 3.00 A for 30.0 minutes. If the mass of copper deposited on the
cathode is 4.50 grams, determine the molar mass of copper.
10
Solution
Step 1: Find the total charge passed through the cell. Given that the current
is 3.00 A and the time is 30.0 minutes, we can calculate the total charge using
the formula:
Q=I×t
Q= 3.00 A ×30.0 min ×60 s
1 min
Q= 5400 C
Step 2: Convert the charge to moles of electrons. Since 1 mole of electrons
has a charge of 1 Faraday (F = 96500 C), we can find the moles of electrons
using:
moles of electrons = Q
96500
moles of electrons = 5400
96500
moles of electrons = 0.0561 mol
Step 3: Find the number of moles of copper (Cu) deposited. From the bal-
anced equation for the electrolysis of CuSO4, we know that 2 moles of electrons
are required to deposit 1 mole of copper. Therefore:
moles of Cu = 0.0561 mol ×1 mol Cu
2 mol e−
moles of Cu = 0.0281 mol
Step 4: Calculate the molar mass of copper. The mass of copper deposited
is given as 4.50 grams. Using the formula for molar mass:
Molar mass of Cu = mass of Cu
moles of Cu
Molar mass of Cu = 4.50 g
0.0281 mol
Molar mass of Cu = 160.14 g/mol
Therefore, the molar mass of copper is 160.14 g/mol.
Question 12
Question
Describe Faraday’s laws of electrolysis and explain the relationship between the
amount of substance deposited during electrolysis, the amount of charge passed
through the electrolyte, and the Faraday constant.
11
Solution
Faraday’s laws of electrolysis describe the relationship between the amount of
substance deposited during electrolysis, the amount of charge passed through
the electrolyte, and the Faraday constant.
Faraday’s First Law: The amount of substance deposited at an electrode
during electrolysis is directly proportional to the quantity of charge passing
through the electrolyte.
Faraday’s Second Law: When the same quantity of charge passes through
different electrolytes, the amounts of different substances liberated or deposited
at the electrodes are proportional to their chemical equivalent weights.
Relationship between the Amount of Substance Deposited, Charge
Passed, and Faraday Constant: Let’s consider the relationship between the
amount of substance deposited (m), the quantity of charge passed through the
electrolyte (Q), and the Faraday constant (F).
Step 1: We can express Faraday’s laws mathematically as:
m=k·Q
where kis a proportionality constant.
Step 2: The charge passed (Q) can be related to the current (I) and time
(t) using the equation:
Q=I·t
Step 3: Substituting Q=I·tinto the equation m=k·Q, we get:
m=k·I·t
Step 4: The proportionality constant kcan be expressed in terms of the
Faraday constant (F) as:
k=F
z
where zis the number of moles of electrons exchanged per mole of substance
undergoing electrolysis.
Step 5: Substituting k=F
zinto the equation m=k·I·t, we get:
m=F
z·I·t
Step 6: Therefore, the relationship between the amount of substance de-
posited, the charge passed, and the Faraday constant is given by:
m=F
z·I·t
Question 13
Question
According to Faraday’s laws of electrolysis, what is the relationship between the
amount of substance deposited during electrolysis and the quantity of electricity
passed through the electrolyte solution?
12
Solution
To answer this question, we need to understand Faraday’s laws of electrolysis
and the relationships they describe.
Faraday’s First Law: The amount of substance deposited in electrolysis is
directly proportional to the quantity of electricity passed through the electrolyte
solution.
Faraday’s Second Law: The amount of substance deposited by the same
quantity of electricity is directly proportional to the equivalent weight of the
substance.
Therefore, the relationship between the amount of substance deposited dur-
ing electrolysis and the quantity of electricity passed through the electrolyte
solution can be summarized as follows:
Step 1: Let Qbe the quantity of electricity passed through the electrolyte
solution in coulombs, mbe the mass of the substance deposited in grams, and
zbe the electrochemical equivalent of the substance in grams per coulomb.
According to Faraday’s laws, we have the following relationship:
m=zQ
This equation shows that the mass of substance deposited is directly pro-
portional to the quantity of electricity passed through the electrolyte solution.
Step 2: If the current Ipassing through the electrolyte for a time tis
known, we can relate the quantity of electricity Qto the current and time using
the formula:
Q=It
Step 3: Substituting Q=It into the equation m=zQ, we get:
m=zIt
Step 4: This final equation shows the direct proportionality between the
amount of substance deposited during electrolysis, the quantity of electricity
passed through the electrolyte solution, and the time of electrolysis.
Therefore, the relationship between the amount of substance deposited dur-
ing electrolysis and the quantity of electricity passed through the electrolyte
solution is given by the equation m=zIt.
Question 14
Question
An electrolytic cell is set up with a solution containing silver ions, Ag+, and
copper ions, Cu2+. A constant current of 2.50 A is passed through the cell for
50.0 minutes. If the electrolysis of the solution results in the deposition of 1.25
g of silver, calculate the mass of copper that is deposited. The molar masses of
Ag and Cu are 107.87 g/mol and 63.55 g/mol, respectively.
13
Solution
Step 1: Calculate the total charge that has passed through the electrolyte. Let’s
use Faraday’s first law of electrolysis to find the total charge passed:
Q=I·t
where: - Qis the total charge passed (in Coulombs), - Iis the current (in
Amperes), and - tis the time in seconds.
Given that the current is 2.50 A and the time is 50.0 minutes (or 3000
seconds), we can calculate Q:
Q= 2.50 A ×3000 s = 7500 C
Step 2: Determine the moles of silver deposited. Since 1 F (Faraday) is
equivalent to 1 mole of electrons (96485 C), we can find the moles of silver
deposited using Faraday’s second law:
molAg =massAg
molar massAg
molAg =1.25 g
107.87 g/mol ≈0.0116 mol
Step 3: Calculate the moles of copper deposited. The ratio of moles of
silver to moles of copper deposited is 2:1 according to the stoichiometry of the
reaction. Therefore, the moles of copper deposited is half the moles of silver:
molCu = 0.5×molAg = 0.5×0.0116 mol = 0.0058 mol
Step 4: Determine the mass of copper deposited.
massCu = molCu ×molar massCu
massCu = 0.0058 mol ×63.55 g/mol ≈0.37 g
Therefore, approximately 0.37 g of copper is deposited during the electrolysis
process.
Question 15
Question
A certain electrolysis cell operates using a current of 1.5 A for 45 minutes.
During this time, 0.25 g of substance is deposited at one of the electrodes.
Assume that the substance is a divalent metal. Determine the molar mass of
the metal.
14
Solution
Step 1: Calculate the total charge passed through the cell. Given that the
current is 1.5 A and the time of operation is 45 minutes, first convert the time
to seconds:
45 minutes ×60 seconds/minute = 2700 seconds
Using the formula Q=I×t, where Qis the charge, Iis the current, and t
is the time, we have:
Q= 1.5 A ×2700 s = 4050 C
Step 2: Calculate the number of moles of metal deposited. The faraday
constant, F, is 96500 C/mol. The charge passed through the cell (4050 C) can
be related to the number of moles of metal deposited (0.25 g) using the equation
Q=nF :
n=Q
F=4050 C
96500 C/mol
Step 3: Calculate the molar mass of the metal. Since the substance is a
divalent metal, the number of moles calculated above represents half the molar
quantity of the substance (M/2). Therefore, the molar mass of the metal (M)
can be calculated as follows:
M= 2n×molar mass of one mole of substance
Substitute the value of ninto the above equation to find the molar mass of
the metal.
Question 16
Question
In an electrolysis experiment, a current of 2.5 A was passed through a solution
of copper(II) sulfate for 30 minutes. If the mass of copper deposited at the
cathode was 1.8 g, calculate the atomic mass of copper. (Assume 100
Solution
Step 1: Calculate the charge passed through the electrolyte. Given: Current (I)
= 2.5 A, Time (t) = 30 minutes = 1800 seconds Using the formula Q = I ×t,
we can find the charge.
Q= 2.5 A ×1800 s
Q= 4500 C
15
Step 2: Calculate the number of moles of copper deposited. Given: Mass
of copper deposited = 1.8 g, Atomic mass of copper = 63.5 g/mol Using the
formula moles = mass / atomic mass, we can find the number of moles deposited.
Moles of copper = 1.8 g
63.5 g/mol
Moles of copper ≈0.02835 mol
Step 3: Determine the number of electrons involved in the reaction. Since
each mole of copper requires 2 moles of electrons for reduction, the number of
electrons involved can be calculated as:
Number of electrons = 2 ×Moles of copper
Number of electrons = 2 ×0.02835 mol
Number of electrons ≈0.0567 mol
Step 4: Calculate the Faraday constant. The Faraday constant, F, is the
charge of 1 mole of electrons, which is 96,485 C/mol.
Faraday constant (F) = 96485 C/mol
Step 5: Determine the atomic mass of copper. The atomic mass of copper
can be calculated using the formula:
Faraday constant (F) = Total charge passed (Q)
Number of electrons = Atomic mass of copper
96485 = 4500
0.0567 = Atomic mass of copper
96485 ≈79323.529 g/mol
Therefore, the calculated atomic mass of copper is approximately 79.3235
g/mol.
Question 17
Question
A student conducts an electrolysis experiment using a solution of molten sodium
chloride. Initially, the student passes a current of 2.5 A through the solution for
15 minutes. Later, the student passes a current of 4.0 A through the solution
for another 20 minutes. Calculate the total charge passed through the solution
in coulombs during the entire experiment.
16
Solution
Step 1: Calculate the charge passed during the first part of the experiment.
Given: Current, I1= 2.5 A Time, t1= 15 minutes
The total charge passed during the first part of the experiment is given by:
Q1=I1×t1
Substitute I1= 2.5 A and t1= 15 minutes into the formula:
Q1= 2.5×15
Calculate Q1:
Q1= 37.5 C
Step 2: Calculate the charge passed during the second part of the experi-
ment. Given: Current, I2= 4.0 A Time, t2= 20 minutes
The total charge passed during the second part of the experiment is given
by:
Q2=I2×t2
Substitute I2= 4.0 A and t2= 20 minutes into the formula:
Q2= 4.0×20
Calculate Q2:
Q2= 80 C
Step 3: Calculate the total charge passed during the entire experiment. The
total charge passed during the entire experiment is the sum of Q1and Q2:
Qtotal =Q1+Q2
Substitute Q1= 37.5 C and Q2= 80 C into the formula:
Qtotal = 37.5 + 80
Calculate Qtotal:
Qtotal = 117.5 C
Therefore, the total charge passed through the solution in coulombs during
the entire experiment is 117.5 C.
Question 18
Question
An electric current is passed through a solution of silver nitrate, AgNO3, using
silver electrodes. If 1.50 g of silver is deposited on the cathode in 15 minutes,
determine the current passing through the cell. The molar mass of silver is
107.87 g/mol.
17
Solution
Step 1: Find the number of moles of deposited silver.
Given that the molar mass of silver is 107.87 g/mol, we can calculate the number
of moles of silver deposited on the cathode using the formula:
moles of silver = mass of silver
molar mass of silver
moles of silver = 1.50 g
107.87 g/mol
moles of silver ≈0.0139 mol
Step 2: Find the charge passed through the cell.
Since 1 mole of silver corresponds to 1 mole of electrons (Ag++ 1e−→Ag), we
can determine the charge passed through the cell in coulombs using Faraday’s
constant, F= 96485 C/mol, and the formula:
charge = moles of silver ×F
charge = 0.0139 mol ×96485 C/mol
charge ≈1343.0 C
Step 3: Find the current passing through the cell.
Since the time taken for the deposition of silver is 15 minutes (or 900 seconds),
we can calculate the current passing through the cell using the formula:
current = charge
time
current = 1343.0 C
900 s
current ≈1.49 A
Therefore, the current passing through the cell is approximately 1.49 A.
Question 19
Question
Explain Faraday’s laws of electrolysis and how they are applied in electrochem-
istry.
18
Solution
To understand Faraday’s laws of electrolysis, one must first recall that electrol-
ysis is the process of using electricity to bring about a chemical change in an
electrolyte. Faraday’s laws describe the quantitative relationships between the
amount of substance produced or consumed at an electrode during electrolysis
and the quantity of electricity that flows through the cell.
Faraday’s First Law: The amount of a substance produced or consumed
during electrolysis is directly proportional to the quantity of electricity passed
through the cell.
1. This law can be expressed mathematically as:
Amount of substance = Current ×Time ×Equivalent weight
Faraday’s Second Law: The amounts of different substances produced or
consumed by the same quantity of electricity are in the ratio of their chemical
equivalent weights.
2. Mathematically, this law can be written as:
n1
n2
=E1
E2
where n1and n2are the amounts of two substances obtained at the electrodes,
and E1and E2are their chemical equivalent weights.
Applications of Faraday’s laws in electrochemistry include calculating the
amount of product formed during electrolysis, determining the valency of an
ion in an electrolyte solution, and predicting the products of electrolysis.
In conclusion, Faraday’s laws of electrolysis are fundamental principles in
electrochemistry that relate the amount of substance produced or consumed
during electrolysis to the quantity of electricity passed through the cell. These
laws are essential for understanding and predicting the outcomes of electrolytic
processes.
Question 20
Question
A student is conducting an electrolysis experiment using a copper(II) sulfate
solution with copper electrodes. If a current of 0.5 A is passed through the
solution for 30 minutes, calculate the mass of copper deposited on the cathode.
(Given: Atomic mass of copper = 63.5 g/mol, Faraday constant = 96485 C/mol)
Solution
Step 1: Calculate the total charge passed through the solution.
Total charge = Current ×Time
19
Total charge = 0.5 A ×(30 min) ×60 s
1 min
Total charge = 0.5×30 ×60 C
Total charge = 900 C
Step 2: Calculate the number of moles of electrons passed through the solu-
tion using Faraday’s law.
1 mol of electrons = 1 Faraday = 96485 C
Number of moles of electrons = Total charge
Faraday constant
Number of moles of electrons = 900 C
96485 C/mol
Number of moles of electrons ≈0.00933 mol
Step 3: Since 1 mole of electrons deposits 1 mol of copper, calculate the
mass of copper deposited on the cathode.
Mass of copper = Number of moles of electrons ×Molar mass of copper
Mass of copper = 0.00933 mol ×63.5 g/mol
Mass of copper ≈0.592 g
Therefore, the mass of copper deposited on the cathode is approximately
0.592 g.
Question 21
Question
In an electrolysis experiment, a current of 2.5 A is passed through a solution of
molten lead (II) bromide for 15 minutes. Calculate the mass of lead produced
during this time. (Molar mass of Pb = 207.2 g/mol, Faraday’s constant =
96485 C/mol)
Solution
Step 1: Determine the number of moles of lead produced. Given: Current (I) =
2.5 A, Time (t) = 15 minutes = 900 seconds, Molar mass of lead (Pb) = 207.2
g/mol, Faraday’s constant (F) = 96485 C/mol.
We know that the charge passed (Q) during electrolysis is given by:
Q=I×t
Substitute the given values:
Q= 2.5A×900 s= 2250 C
20
Step 2: Calculate the number of moles of lead produced. The number of
moles of lead produced can be determined by dividing the total charge passed
by the Faraday’s constant:
Moles of Pb = Q
F
Substitute the values:
Moles of Pb = 2250 C
96485 C/mol ≈0.0233 mol
Step 3: Calculate the mass of lead produced. The mass of lead produced
can be calculated using the number of moles and the molar mass of lead:
Mass of Pb = Moles of Pb ×Molar mass of Pb
Substitute the values:
Mass of Pb = 0.0233 mol ×207.2g/mol ≈4.82 g
Therefore, the mass of lead produced during the electrolysis of molten lead
(II) bromide for 15 minutes is approximately 4.82 grams.
Question 22
Question
An aqueous solution of sodium chloride (NaCl) is electrolyzed using inert elec-
trodes. If a current of 2.5 A is passed through the solution for 3 hours, calculate
the amount of chlorine gas (Cl2) produced during the electrolysis. Assume 100
Solution
Step 1: Write the balanced half-reactions for the electrolysis of aqueous sodium
chloride:
Anode: 2Cl−→Cl2+ 2e−
Cathode: 2H2O + 2e−→H2+ 2OH−
Step 2: Calculate the total charge passed through the cell using the formula
Q=I×t, where Qis the charge in Coulombs (C), Iis the current in Amperes
(A), and tis the time in seconds:
Q= 2.5 A ×3 hours ×3600 s/hour = 27,000 C
Step 3: Determine the moles of electrons passed through the circuit using
Faraday’s law, where 1 Faraday (F) is equivalent to 96,485 C/mol electrons:
Moles of electrons = 27,000 C
96500 C/mol ≈0.28 mol
21
Step 4: Calculate the moles of chlorine gas produced based on the stoichiom-
etry of the reaction (1 mol Cl2is produced per 2 mol e−passed):
Moles of Cl2= 0.28 mol ×1 mol Cl2
2 mol e−= 0.14 mol
Step 5: Convert moles of chlorine gas to volume using the ideal gas law at
standard temperature and pressure (STP) conditions (22.4 L/mol):
Volume of Cl2= 0.14 mol ×22.4 L/mol = 3.14 L
Therefore, approximately 3.14 L of chlorine gas will be produced during the
electrolysis.
Question 23
Question
An aqueous solution of silver nitrate is subjected to electrolysis using an external
electromotive force of 2.0 V. If 5.0 grams of silver is deposited in 30 minutes,
determine the amount of current passing through the solution during this time.
Solution
Step 1: Convert the given mass of silver to moles using the molar mass of silver.
Given: Mass of silver, m= 5.0 g Molar mass of silver, M= 107.87 g/mol
Number of moles of silver, n=m
M=5.0 g
107.87 g/mol
Step 2: Calculate the charge passed in coulombs using Faraday’s laws of
electrolysis.
Given: Faraday constant, F= 96,485 C/mol
Charge passed, Q=n×F
Step 3: Determine the time period in seconds.
Given: Time, t= 30 minutes = 30 ×60 s
Step 4: Calculate the current passing through the solution.
Given: Voltage, V= 2.0 V
Current, I=Q
t=n×F
t
22
Question 24
Question
In an electrolytic cell, a current of 2.5 A is passed through a solution of silver
nitrate (AgNO3) for 2 hours. If 3.5 g of silver is deposited at the cathode,
determine the oxidation state of silver in silver nitrate and the mass of silver
nitrate that decomposed.
Solution
Step 1: Find the equivalent weight of silver. Given that 3.5 g of silver is de-
posited, we can calculate the equivalent weight of silver using the formula:
Equivalent weight (E) = Atomic weight
Valency ×Mass deposited
Since the atomic weight of silver is 107.87 g/mol and the valency of silver is 1,
we have:
E=107.87
1×3.5=107.87
3.5≈30.82 g/equiv
Step 2: Determine the oxidation state of silver in silver nitrate (AgN O3).
From the equivalent weight of silver calculated in Step 1, we know that 1 equiv
of silver is deposited when 3.5 g is deposited. The molar mass of silver nitrate
is 169.87 g/mol. Thus, the number of equivalents of silver is given by:
Equivalents of silver = Mass of silver deposited
Equivalent weight of silver =3.5
30.82 ≈0.1137 equiv
Since 1 mole of AgNO3produces 1 mole of silver, the number of moles of AgNO3
decomposed is equal to the number of equivalents of silver deposited. Therefore,
the mass of AgNO3decomposed is:
Mass of AgNO3decomposed = Molar mass of AgNO3×Equivalents of silver
= 169.87 ×0.1137 ≈19.33 g
Therefore, the oxidation state of silver in silver nitrate is +1, and the mass
of silver nitrate that decomposed is approximately 19.33 g.
Question 25
Question
A student sets up an electrolysis experiment using a copper(II) sulfate solution.
The student uses a current of 2.0 A to electrolyze the solution for 30 minutes.
During this time, the student measures that 0.20 g of copper is deposited onto
one of the electrodes. Calculate the Faraday constant.
Given: Atomic mass of copper = 63.55 g/mol
23
Solution
Step 1: Calculate the moles of copper deposited. Step 2: Use the moles of
copper to find the amount of charge passed through the circuit. Step 3: Apply
Faraday’s law of electrolysis to find the Faraday constant.
Step 1: To find the moles of copper deposited, we use the formula:
moles = mass
molar mass
Substitute the given values:
moles = 0.20 g
63.55 g/mol = 0.00315 mol
Step 2: The amount of charge passed through the circuit can be found using
Faraday’s law, which states:
Charge (C) = moles ×Faraday constant (C/mol)
Since the current is given in Amperes and time is given in seconds, we need to
convert the time to seconds:
Time (s) = 30 minutes ×60 s/min = 1800 s
Then, we can find the charge passed:
Charge (C) = 2.0 A ×1800 s = 3600 C
Step 3: Now, we can find the Faraday constant using the charge passed and
moles of copper. Rearranging the formula from Step 2 gives:
Faraday constant (C/mol) = Charge (C)
moles
Substitute the calculated values:
Faraday constant = 3600 C
0.00315 mol ≈1.14 ×104C/mol
Therefore, the Faraday constant is approximately 1.14 ×104C/mol.
Question 26
Question
A student is conducting an electrolysis experiment using a solution of copper
sulfate (CuSO4). They connect a power supply to two electrodes: one made
of copper and one made of platinum. During the experiment, they observe
that 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.
24
Solution
To explain the observed phenomenon, we need to consider the implications of
Faraday’s laws of electrolysis.
Step 1: Electrolysis of Copper(II) Sulfate In the electrolysis of copper
sulfate solution, copper ions (Cu2+) will be attracted to the negative electrode
(cathode) where they will gain electrons and be deposited as solid copper. The
copper electrode will therefore increase in mass due to the deposition of copper.
Step 2: Faraday’s First Law Faraday’s First Law states that the amount
of substance deposited at an electrode is directly proportional to the amount
of charge passed through the electrolyte. Therefore, the increase in mass of
the copper electrode can be attributed to the transfer of copper ions from the
solution to the electrode by the passage of electric current.
Step 3: Electrolysis of Water At the positive electrode (anode) made of
platinum, water molecules will undergo electrolysis to release oxygen gas. Since
platinum is inert and does not react with water or the products of electrolysis,
the mass of the platinum electrode remains constant.
Step 4: Faraday’s Second Law Faraday’s Second Law states that the
amounts of different substances deposited by a given quantity of electricity are
proportional to their chemical equivalent weights. In this case, copper ions
are reduced to solid copper at the cathode, while water is oxidized to release
oxygen gas at the anode. The mass changes at each electrode reflect the different
chemical reactions occurring due to the applied electric current.
In conclusion, the decrease in mass of the copper electrode and the constant
mass of the platinum electrode can be explained by the principles of Faraday’s
laws of electrolysis, which govern the transfer of material during the electrolysis
process.
Question 27
Question
A copper sulfate solution contains 0.1 moles of copper ions. If a current of 2A
is passed through the solution for 1 hour, calculate:
1. The mass of copper deposited at the cathode.
2. The volume of oxygen produced at the anode at STP.
3. The volume of hydrogen produced at the cathode at STP.
Given: Atomic mass of copper = 63.5 g/mol, 1 Faraday = 96500 C/mol, 1 mol
of any gas at STP occupies 22.4 L.
Solution
1. Step 1: Calculate the equivalent weight of copper.
25
The equivalent weight of copper is the atomic mass divided by the valency.
Valency of copper = 2 (since Cu2+ ions are produced in the solution during
electrolysis)
Equivalent weight of copper = 63.5
2g/mol = 31.75 g/mol
Step 2: Calculate the mass of copper deposited at the cathode.
The mass of copper deposited can be calculated using the formula: Mass
= (Equivalent weight ×Current ×Time) / (96500)
Mass = 31.75×2×3600
96500 g Mass = 57240
96500 g Mass ≈0.593 g
Therefore, the mass of copper deposited at the cathode is 0.593 g.
2. Step 3: Calculate the moles of oxygen produced at the anode.
Since copper ions are being discharged at the cathode, oxygen gas is pro-
duced at the anode during electrolysis. 4 moles of electrons are required
to produce 1 mole of oxygen gas.
Moles of electrons = Current ×Time / 96500 Moles of oxygen = Moles
of electrons / 4
Moles of oxygen = 2×3600
96500 /4 mol Moles of oxygen ≈0.0187 mol
Step 4: Calculate the volume of oxygen produced at STP.
Volume of oxygen = Moles of oxygen ×22.4 L Volume of oxygen = 0.0187
×22.4 L Volume of oxygen ≈0.419 L
Therefore, the volume of oxygen produced at the anode at STP is 0.419
L.
3. Step 5: Calculate the moles of hydrogen produced at the cathode.
For every mole of copper deposited at the cathode, 2 moles of electrons
are consumed which are utilized to generate hydrogen gas.
Moles of electrons = 2 ×current ×time / 96500 Moles of hydrogen =
Moles of electrons / 2
Moles of hydrogen = 2×2×3600
96500 /2 mol Moles of hydrogen ≈0.075 mol
Step 6: Calculate the volume of hydrogen produced at STP.
Volume of hydrogen = Moles of hydrogen ×22.4 L Volume of hydrogen
= 0.075 ×22.4 L Volume of hydrogen ≈1.68 L
Therefore, the volume of hydrogen produced at the cathode at STP is 1.68
L.
Question 28
Question
Consider an electrolytic cell containing a solution of sodium chloride (NaCl).
When a current of 2.00 A is passed through the cell for 1.00 hour, 0.190 g of
26
sodium metal is deposited at the cathode. Calculate the standard electrode
potential of sodium.
Solution
Step 1: First, we need to convert the amount of sodium deposited from grams
to moles. The molar mass of sodium (Na) is 22.99 g/mol.
Given: Amount of sodium deposited, m= 0.190 g Molar mass of sodium,
M= 22.99 g/mol
Number of moles of sodium:
n=m
M=0.190
22.99 = 0.00826 mol
Step 2: Next, we need to determine the number of electrons transferred dur-
ing the electrolysis based on the reaction at the cathode, which is the reduction
of sodium ions (Na+) to form sodium metal (Na):
2Na++ 2e−→2Na
Since 2 moles of electrons are required to deposit 2 moles of sodium, the
number of electrons transferred (ne) is equal to the number of moles of sodium
deposited:
ne= 2 ×n= 2 ×0.00826 = 0.0165 mol
Step 3: Using Faraday’s laws of electrolysis, we know that the quantity of
electricity transferred is directly proportional to the number of moles of electrons
transferred. The constant of proportionality is the Faraday constant (F), which
is equal to 9.6485 ×104C/mol.
Quantity of electricity transferred:
Q=ne×F= 0.0165 ×9.6485 ×104= 1590 C
Step 4: The standard electrode potential of sodium, denoted as E◦(Na+/Na),
can be calculated using the formula:
E◦=Q
I×t
Given: Current, I= 2.00 A Time, t= 1.00 hour = 3600 s
Substitute the known values:
E◦=1590
2.00 ×3600 =1590
7200 = 0.221 V
Therefore, the standard electrode potential of sodium is 0.221 V .
27
Question 29
Question
A solution of silver nitrate (AgNO3) is electrolyzed using a current of 4.5 A for
2.5 hours. If 0.15 g of silver is deposited at the cathode, determine the number
of moles of electrons involved in the reaction. Assume 100
Solution
Step 1: Calculate the total charge passed
Total charge (Q) = Current (I) ×Time (t)
Given current, I= 4.5 A and time, t= 2.5 hours, we convert time to seconds:
t= 2.5×60 ×60 = 9000 s
Now, calculating total charge:
Q= 4.5 A ×9000 s = 40500 C
Step 2: Determine the number of moles of electrons involved To find the
number of moles of electrons, we need to first calculate the moles of silver
deposited using the given mass. Given mass of silver, m= 0.15 g, and molar
mass of silver, MAg = 107.87 g/mol.
Converting mass to moles:
nAg =m
MAg
=0.15 g
107.87 g/mol ≈0.00139 mol
Each mole of silver deposited corresponds to 1 mole of electrons. Therefore,
the number of moles of electrons involved in the reaction is:
ne−= 0.00139 mol
So, approximately 0.00139 moles of electrons are involved in the electrolysis
reaction.
Question 30
Question
Discuss Faraday’s laws of electrolysis and explain how they can be used to
determine the amount of substance deposited or liberated during electrolysis.
28
Solution
Faraday’s laws of electrolysis are a set of two laws formulated by Michael Fara-
day in 1834. These laws describe the quantitative relationship between the
amount of substance produced or consumed during electrolysis and the amount
of electric charge passed through the electrolyte.
Faraday’s First Law: The mass of a substance deposited or liberated
during electrolysis is directly proportional to the amount of electrical charge
passed through the electrolyte.
Faraday’s Second Law: The mass of different substances deposited or
liberated by the same quantity of electricity is directly proportional to their
chemical equivalents (equivalent weight).
Using Faraday’s laws, we can determine the amount of substance deposited
or liberated during electrolysis using the following steps:
Step 1: Calculate the quantity of electrical charge passed through the elec-
trolyte using the formula
Q=I×t,
where Qis the electric charge (in Coulombs), Iis the current (in Amperes), and
tis the time (in seconds).
Step 2: Determine the number of moles of electrons transferred during the
electrolysis by dividing the electric charge by the Faraday constant F:
n=Q
F,
where nis the number of moles of electrons, and F≈96485 C/mol is Faraday’s
constant.
Step 3: Identify the balanced chemical equation for the electrolysis process
and determine the mole ratio between the substance of interest and electrons.
Step 4: Calculate the number of moles of the substance deposited or liber-
ated by multiplying the number of moles of electrons by the appropriate mole
ratio.
Step 5: Convert the number of moles of the substance to mass using its
molar mass.
By following these steps and applying Faraday’s laws, one can accurately
determine the amount of substance deposited or liberated during electrolysis
based on the amount of electric charge passed through the system.
Question 31
Question
In an electrolytic cell, a metal M is deposited at the cathode when a current
of 2.00 A is passed through a solution of MCl2for 30.0 minutes. Calculate the
mass of M deposited. Given: - Atomic mass of M = 63.5 g/mol - Faraday’s
constant F= 96500 C/mol
29
Solution
Step 1: Find the charge passed through the cell.
Charge (Q) = Current (I) ×Time (t)
Q= 2.00 A ×(30.0 min ×60 s/min)
Q= 2.00 A ×1800 s = 3600 C
Step 2: Calculate the number of moles of M deposited.
Number of moles of M = Charge (Q)
Faraday’s constant (F)
Number of moles of M = 3600 C
96500 C/mol
Number of moles of M ≈0.0373 mol
Step 3: Determine the mass of M deposited.
Mass of M = Number of moles ×Atomic mass
Mass of M = 0.0373 mol ×63.5 g/mol
Mass of M ≈2.37 g
Therefore, the mass of metal M deposited in the given electrolytic cell is
approximately 2.37 grams.
Question 32
Question
An electroplating apparatus contains a solution of silver nitrate (AgNO3). If
a current of 2.5 A is passed through the solution for 1 hour, calculate the
mass of silver deposited on the cathode. Given that the molar mass of silver is
107.87 g/mol and Faraday’s constant is 96500 C/mol.
Solution
Step 1: Calculate the charge passed through the solution.
Given: Current (I) = 2.5 A, Time (t) = 1 hour = 3600 s
The charge (Q) passed through the solution can be calculated using the formula:
Q=I×t
Q= 2.5 A ×3600 s = 9000 C
Step 2: Calculate the number of moles of electrons that passed through the
solution.
30
Since Faraday’s constant is the charge consumed by 1 mole of electrons, the
number of moles (n) can be calculated as:
n=Q
96500
n=9000 C
96500 C/mol ≈0.093 mol
Step 3: Calculate the mass of silver deposited on the cathode.
The chemical reaction at the cathode is:
Ag++e−−→ Ag
From the balanced equation, we can see that 1 mole of electrons corresponds to
the deposition of 1 mole of silver. Since the molar mass of silver is 107.87 g/mol,
the mass of silver deposited is:
Mass of silver = n×Molar mass of silver
Mass of silver = 0.093 mol ×107.87 g/mol ≈10.02 g
Therefore, the mass of silver deposited on the cathode is approximately
10.02 g after passing a current of 2.5 A for 1 hour.
Question 33
Question
State Faraday’s laws of electrolysis. Explain how these laws are derived from
the laws of conservation of mass and charge.
Solution
Faraday’s laws of electrolysis describe the quantitative relationship between
the amount of substance produced at an electrode during electrolysis and the
amount of charge passed through the electrolyte. These laws can be derived
from the laws of conservation of mass and charge.
Faraday’s First Law: The amount of substance deposited on each elec-
trode during electrolysis is directly proportional to the quantity of electricity
passing through the cell.
Faraday’s Second Law: The amounts of different substances deposited
by the same quantity of electricity are proportional to their chemical equivalent
weights.
Derivation from Conservation of Mass and Charge:
Conservation of Mass: This law states that mass can neither be created
nor destroyed in a chemical reaction. During electrolysis, the mass of the
substance deposited on each electrode is proportional to the quantity of
electricity passed.
31
Conservation of Charge: This law states that electric charge is con-
served in any process. The charge passed during electrolysis is directly
proportional to the number of electrons transferred in the redox reactions
at the electrodes.
Step 1: Let Qbe the total charge passed through the electrolyte during
electrolysis, and nbe the number of moles of electrons transferred. The charge
Qis related to nby Faraday’s constant F:
Q=n×F
Step 2: The quantity of electricity needed to deposit one mole of a substance
can be determined using its chemical equivalent weight (M). According to
Faraday’s laws, if mgrams of a substance are deposited, then:
m=M×n
Step 3: By combining the two equations from Step 1 and Step 2, we get:
Q=M×F×m
This equation represents Faraday’s laws of electrolysis, relating the amount
of substance deposited (m) to the quantity of electricity passed (Q), with the
constant of proportionality being a product of the chemical equivalent weight
and Faraday’s constant.
Question 34
Question
A copper(II) sulfate solution is electrolyzed using two inert electrodes. If a
current of 2.50 A is passed through the solution for 3.00 hours, what mass of
copper is deposited on the cathode? The molar mass of copper is 63.55 g/mol.
Solution
Step 1: Calculate the total charge passed through the circuit Given values:
Current, I= 2.50 A Time, t= 3.00 hours
The total charge, Q, passed through the circuit can be calculated using the
formula:
Q=I×t
Substitute the given values:
Q= 2.50 A ×3.00 hours = 7.50 C
Step 2: Determine the number of moles of electrons Each mole of electrons
(1 Faraday) is equivalent to 6.022 ×1023 electrons and carries a charge of 1
32
C. Therefore, the number of moles of electrons, n, can be calculated using the
formula:
n=Q
Faraday constant =Q
96485
Substitute the value for Q:
n=7.50 C
96485 C/mol = 7.78 ×10−5mol
Step 3: Determine the mass of copper deposited The balanced half-reaction
for the deposition of copper:
Cu2+(aq)+2e−→Cu(s)
From the balanced reaction, it is observed that 2 moles of electrons are
needed to deposit 1 mole of copper. Therefore, the mole ratio between electrons
and copper is 2:1.
Since 2 moles of electrons correspond to the deposition of 1 mole of copper,
the number of moles of copper deposited is half the number of moles of electrons:
nCu =ne−
2=7.78 ×10−5
2= 3.89 ×10−5mol
Step 4: Calculate the mass of copper deposited The mass of copper, m,
deposited can be calculated using the formula:
m=n×Molar mass of copper
Substitute the values for nand the molar mass of copper:
m= 3.89 ×10−5mol ×63.55 g/mol = 0.00247 g
Therefore, the mass of copper deposited on the cathode is 0.00247 g.
Question 35
Question
Consider an electrolytic cell containing a solution of copper(II) sulfate, CuSO4
with copper electrodes. During the electrolysis process, a current of 2.50 A is
passed through the cell for 3.00 hours. If the mass of copper deposited on the
cathode is 6.72 g, calculate the number of moles of electrons that passed through
the electrolytic cell.
33
Question 2
Question
A student is performing an electrolysis experiment using a copper sulfate solu-
tion. Initially, there is no current passing through the solution. The student
then turns on the power supply and allows a current of 2.00 A to pass through
the solution for 30 minutes. During this time, copper ions are reduced at the
cathode to form copper metal.
If the student observes 1.00 g of copper deposited at the cathode, calculate
the Faraday’s constant and determine the number of electrons required for the
reduction of 1 mole of copper ions.
Given: Atomic mass of copper = 63.5 g/mol, Faraday’s constant = 9.65×104
C/mol.
Solution
Step 1: Find the total charge passed through the solution. The total charge
passed through the solution can be calculated using the formula:
Q=It
where Qis the total charge passed (in coulombs), Iis the current (in amperes),
and tis the time (in seconds). Given I= 2.00 A and t= 30 minutes = 30 ×60
s = 1800 s, we have:
Q= 2.00 ×1800 = 3600 C
Step 2: Calculate the moles of copper deposited at the cathode. To calculate
the moles of copper deposited, we use the formula:
moles = mass
molar mass
Given that the mass of copper deposited is 1.00 g and the molar mass of copper
is 63.5 g/mol, we have:
moles = 1.00
63.5= 0.0157 moles
Step 3: Determine the number of electrons required for the reduction of
1 mole of copper ions. We know that the Faraday’s constant represents the
charge of 1 mole of electrons. Therefore, the number of electrons required for
the reduction of 1 mole of copper ions is given by:
number of electrons = total charge passed (C)
Faraday’s constant (C/mol) ×moles of copper ions
Substitute the values:
number of electrons = 3600
9.65 ×104×0.0157
number of electrons = 0.5918 moles of electrons
2
Question 3
Question
In an electrolysis experiment, a current of 2.5 A is passed through a solution of
copper(II) sulfate for 30 minutes. During this time, 3.75 g of copper metal is
deposited on the cathode. Calculate the Faraday constant and the number of
moles of electrons transferred during the experiment.
Given: Atomic mass of copper = 63.5 g/mol, and the charge of an electron
= 1.602 ×10−19 C.
Solution
Step 1: Calculate the charge required to deposit 3.75 g of copper. The number
of moles of copper deposited can be calculated using the atomic mass of copper.
Let the charge required be Q.
Q= charge needed to deposit 3.75 g of copper
Q=mass deposited
molar mass of copper ×charge of one mole of electrons
Q=3.75 g
63.5 g/mol ×1 mol ×6.022 ×1023 molecules/mol ×1.602 ×10−19 C
Step 2: Calculate the Faraday constant (F). The Faraday constant can be
calculated using the charge on one mole of electrons. Let Fbe the Faraday
constant.
F= Charge on one mole of electrons
Step 3: Calculate the number of moles of electrons (n) transferred. The
number of moles of electrons can be calculated using the charge required and
the Faraday constant. Let nbe the number of moles of electrons transferred.
n=Q
F
Step 4: Perform the calculations. Substitute the given values to calculate
Q,F, and n.
Step 5: Analyze the results. Discuss the significance of the calculated Fara-
day constant and the number of moles of electrons transferred in the context of
the electrolysis experiment.
Question 4
Question
A student is conducting an electrolysis experiment using a solution of copper
sulfate (CuSO4). If the student passes a current of 2.5 A through the solution
3
for 30 minutes, what mass of copper will be deposited at the cathode? (Given:
atomic mass of copper = 63.55 g/mol, Faraday’s constant = 9.65 ×104C/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. The
total charge can be calculated using the formula:
Q=I×t
where Q= total charge passed (coulombs), I= current (amperes), t= time
(seconds).
Given that I= 2.5Aand t= 30 ×60 seconds, we have:
Q= 2.5×30 ×60 = 4500 C
Step 2: Determine the number of moles of electrons transferred. Each mole
of electrons corresponds to a charge of 1 Faraday, which is equal to the charge
of 1 mole of electrons. The number of moles of electrons can be calculated using
Faraday’s constant:
Moles of electrons = Q
F
where F= 9.65 ×104C/mol.
Substitute the values to find moles of electrons:
Moles of electrons = 4500
9.65 ×104= 0.0467 mol
Step 3: Determine the moles of copper deposited. From the balanced chem-
ical equation for electrolysis of copper sulfate (CuSO4):
Cu2+ + 2e−→Cu
we know that 2 moles of electrons are required to deposit 1 mole of copper.
Therefore, the moles of copper deposited (nCu) is half of the moles of electrons:
nCu =0.0467
2= 0.0234 mol
Step 4: Calculate the mass of copper deposited. The mass of copper de-
posited can be calculated using the atomic mass of copper:
Mass of copper = nCu ×Molar mass of copper
Mass of copper = 0.0234 ×63.55 = 1.49 g
Therefore, the student will deposit 1.49 g of copper at the cathode during
the electrolysis of copper sulfate solution.
4
Question 5
Question
Faraday’s laws of electrolysis state that the amount of a substance produced at
an electrode is directly proportional to the quantity of electricity passed through
the electrolyte. A student wants to electroplate a metal with a mass of 7.5 g
onto an electrode using a current of 2.5 A for 20 minutes. If the atomic mass
of the metal is 63.55 g/mol, determine the efficiency of the electrolysis process.
Assume 100
Solution
Step 1: Determine the number of moles of metal deposited on the electrode.
Given: - Mass of metal deposited, m= 7.5 g - Atomic mass of the metal,
M= 63.55 g/mol
The number of moles of metal deposited is given by:
moles = m
M
moles = 7.5
63.55
moles ≈0.118 mol
Step 2: Calculate the total charge passed through the electrolyte.
Given: - Current, I= 2.5 A - Time, t= 20 minutes
Firstly, convert the time to seconds:
t= 20 ×60 = 1200 s
The total charge passed through the electrolyte is given by:
Q=It
Q= 2.5×1200
Q= 3000 C
Step 3: Determine the number of electrons involved in the reaction.
One Faraday of charge is equivalent to Avogadro’s number of electrons:
1 Faraday = 96,485 C/mol
The number of electrons involved in the reaction is given by:
n=Q
1 Faraday
n=3000
96485
5
n≈0.0311 mol
Step 4: Calculate the efficiency of the electrolysis process.
The efficiency of the electrolysis process is given by:
Efficiency = moles deposited
moles of electrons passed
Efficiency = 0.118
0.0311 ×100%
Efficiency ≈379.74%
Therefore, the efficiency of the electrolysis process is approximately 379.74
Question 6
Question
A solution of potassium iodide, KI, is electrolyzed using platinum electrodes. If
a current of 5.00 A is passed through the solution for 30.0 minutes, what mass
of iodine, I2, is produced? (Assume 100
Solution
Step 1: Write the balanced chemical equation for the electrolysis of potassium
iodide.
The balanced half-reactions are:
At the anode: 2I−→I2+ 2e−
At the cathode: 2H++ 2e−→H2
The overall reaction is:
2KI →I2+H2
Step 2: Calculate the total charge that passed through the solution.
Given: current, I= 5.00 A
Time, t= 30.0 minutes = 30.0 minutes ×1 hour
60 minutes = 0.50 hours
The total charge, Q, is given by Q=I×t.
Q= 5.00 A ×0.50 hours = 2.50 C
Step 3: Determine the number of moles of electrons that reacted.
From the balanced reaction, 2 moles of electrons are involved in the production
of 1 mole of I2.
Hence, the number of moles of electrons, n, that passed through the solution is
given by:
n=Q
96,485 C/mol
6
n=2.50 C
96,485 C/mol ≈2.59 ×10−5mol
Step 4: Calculate the mass of iodine produced.
From the balanced reaction, 2 moles of I−produce 1 mole of I2.
Therefore, the number of moles of I2produced is half the number of moles of
electrons:
moles of I2=n
2=2.59 ×10−5mol
2= 1.30 ×10−5mol
The molar mass of I2is 2×Atomic mass of I = 2×126.90 g/mol = 253.80 g/mol.
The mass of I2produced is:
Mass of I2= moles of I2×Molar mass of I2
Mass of I2= 1.30 ×10−5mol ×253.80 g/mol
Mass of I2≈3.29 ×10−3g
Therefore, approximately 3.29 ×10−3g of iodine is produced during the
electrolysis of potassium iodide.
Question 7
Question
A student performs an electrolysis experiment using a copper(II) sulfate solu-
tion and two copper electrodes. The student passes a constant current of 2.50
A through the solution for 2.00 hours. During this time, 5.00 g of copper is
deposited on the cathode.
Calculate the charge passed through the solution.
Solution
Step 1: Determine the number of moles of copper deposited on the cathode. To
calculate the number of moles of copper deposited, we first convert the mass
of copper to moles using the molar mass of copper (Cu). The molar mass of
copper is 63.55 g/mol.
Number of moles of Cu = Mass of Cu
Molar mass of Cu =5.00 g
63.55 g/mol
Step 2: Calculate the charge passed through the solution. The charge passed
through the solution can be calculated using Faraday’s first law of electrolysis:
Q=n×F
where: - Qis the charge in coulombs, - nis the number of moles of elec-
trons transferred (moles of copper deposited/mol), - Fis the Faraday constant
(96500 C/mol).
7
Now, we need to determine the number of moles of electrons transferred
from the number of moles of copper deposited. Copper(II) ions (Cu
²
) gain two
electrons to form solid copper. Therefore, the number of moles of electrons
transferred is twice the moles of copper deposited.
Step 3: Calculate the charge passed through the solution.
Q= 2 ×Number of moles of Cu ×F
Substitute the values into the equation to find the charge passed through
the solution.
Question 8
Question
A student is conducting an electrolysis experiment using a copper (Cu) electrode
and a silver (Ag) electrode in a solution of copper(II) sulfate (CuSO4) to deposit
silver onto the copper electrode. If the student applies a constant current of 2.0
A for 1 hour, what mass of silver will be deposited on the copper electrode?
(Given: atomic masses - Cu: 63.5 g/mol, Ag: 107.9 g/mol)
Solution
Step 1: Calculate the charge transferred during electrolysis. Given: Current
(I) = 2.0 A, Time (t) = 1 hour = 3600 seconds, Charge (Q) = ? The charge
transferred during electrolysis can be calculated using the formula:
Q=I×t
Q= 2.0 A ×3600 s = 7200 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday (F) = 96500 C (charge of 1 mole of electrons), we can calculate the
number of moles of electrons transferred using the formula:
Moles of electrons = Q
96500
Moles of electrons = 7200
96500
Moles of electrons ≈0.075 mol
Step 3: Determine the mass of silver deposited on the copper electrode. The
molar ratio of electrons to silver ions in the electrolysis of silver ions (Ag) is 2:1.
This means that for every 2 moles of electrons transferred, 1 mole of silver is
deposited. So, the number of moles of silver deposited can be calculated as:
Moles of silver = 1
2×Moles of electrons
8
Moles of silver = 1
2×0.075
Moles of silver = 0.0375 mol
Step 4: Calculate the mass of silver deposited on the copper electrode. Using
the molar mass of silver (Ag):
Mass of silver = Moles of silver ×Molar mass of Ag
Mass of silver = 0.0375 mol ×107.9 g/mol
Mass of silver ≈4.04 g
Therefore, approximately 4.04 grams of silver will be deposited on the copper
electrode after electrolysis.
Question 9
Question
An electrolytic cell is set up with a copper(II) sulfate solution and platinum
electrodes. If a current of 2.50 A is passed through the cell for 2.00 hours, what
mass of copper will be deposited at the cathode? (Given: 1 F = 96,485 C/mol,
M(Cu) = 63.55 g/mol)
Solution
Step 1: Find the total charge passed through the cell. Given: Current, I=
2.50 A Time, t= 2.00 hours = 2.00 ×3600 s = 7200 s
Total charge, Q=I×t
Plugging in the values: Q= 2.50 A ×7200 s = 18,000 C
Step 2: Convert the total charge to moles of electrons. Given: 1 F =
96,485 C/mol
Number of moles of electrons, n=Q
96485
Plugging in the values: n=18,000 C
96485 C/mol ≈0.1867 mol
Step 3: Determine the mass of copper deposited at the cathode. Given:
Molar mass of copper, M(Cu) = 63.55 g/mol
Mass of copper, m(Cu) = n×M(Cu)
Plugging in the values: m(Cu) = 0.1867 mol ×63.55 g/mol ≈11.86 g
Therefore, approximately 11.86 grams of copper will be deposited at the
cathode.
Question 10
Question
An aqueous solution of silver nitrate (AgNO3) is electrolyzed using a current of
1.5 A for 2 hours. If the reaction produces solid silver and oxygen gas, calculate:
9
1. the mass of silver deposited on the cathode,
2. the volume of oxygen gas produced at STP.
(Assume 100
Solution
Let’s first determine the total charge passed through the electrolyte solution:
Total charge = Current ×Time = 1.5 A ×2 hours
Total charge = 1.5 C/s ×7200 s = 10800 C
Step 1: Calculate the mass of silver deposited on the cathode. The molar
quantity of electric charge required to deposit one mole of silver is equal to the
Faraday constant (F), which is 96500 C/mol.
1 mol of Ag requires 96500 C
1 C will deposit 1
96500 mol of Ag
Mass of Ag deposited = Molar mass of Ag ×moles of Ag
Mass of Ag deposited = 108 g/mol ×10800
96500mol
Step 2: Calculate the volume of oxygen gas produced at STP. Since 2
moles of electrons are required to produce 1 mole of oxygen gas when water is
electrolyzed, and the molar volume of gas at STP is 22.4 L:
2 mol of e−produces 1 mol of O2
1 C will produce 1
2×1
96500 mol of O2
Volume of O2produced = 22.4 L/mol ×1
2×1
96500 ×10800mol
Question 11
Question
In an electrolytic cell, a solution of copper sulfate (CuSO4) is electrolyzed using
a current of 3.00 A for 30.0 minutes. If the mass of copper deposited on the
cathode is 4.50 grams, determine the molar mass of copper.
10
Solution
Step 1: Find the total charge passed through the cell. Given that the current
is 3.00 A and the time is 30.0 minutes, we can calculate the total charge using
the formula:
Q=I×t
Q= 3.00 A ×30.0 min ×60 s
1 min
Q= 5400 C
Step 2: Convert the charge to moles of electrons. Since 1 mole of electrons
has a charge of 1 Faraday (F = 96500 C), we can find the moles of electrons
using:
moles of electrons = Q
96500
moles of electrons = 5400
96500
moles of electrons = 0.0561 mol
Step 3: Find the number of moles of copper (Cu) deposited. From the bal-
anced equation for the electrolysis of CuSO4, we know that 2 moles of electrons
are required to deposit 1 mole of copper. Therefore:
moles of Cu = 0.0561 mol ×1 mol Cu
2 mol e−
moles of Cu = 0.0281 mol
Step 4: Calculate the molar mass of copper. The mass of copper deposited
is given as 4.50 grams. Using the formula for molar mass:
Molar mass of Cu = mass of Cu
moles of Cu
Molar mass of Cu = 4.50 g
0.0281 mol
Molar mass of Cu = 160.14 g/mol
Therefore, the molar mass of copper is 160.14 g/mol.
Question 12
Question
Describe Faraday’s laws of electrolysis and explain the relationship between the
amount of substance deposited during electrolysis, the amount of charge passed
through the electrolyte, and the Faraday constant.
11
Solution
Faraday’s laws of electrolysis describe the relationship between the amount of
substance deposited during electrolysis, the amount of charge passed through
the electrolyte, and the Faraday constant.
Faraday’s First Law: The amount of substance deposited at an electrode
during electrolysis is directly proportional to the quantity of charge passing
through the electrolyte.
Faraday’s Second Law: When the same quantity of charge passes through
different electrolytes, the amounts of different substances liberated or deposited
at the electrodes are proportional to their chemical equivalent weights.
Relationship between the Amount of Substance Deposited, Charge
Passed, and Faraday Constant: Let’s consider the relationship between the
amount of substance deposited (m), the quantity of charge passed through the
electrolyte (Q), and the Faraday constant (F).
Step 1: We can express Faraday’s laws mathematically as:
m=k·Q
where kis a proportionality constant.
Step 2: The charge passed (Q) can be related to the current (I) and time
(t) using the equation:
Q=I·t
Step 3: Substituting Q=I·tinto the equation m=k·Q, we get:
m=k·I·t
Step 4: The proportionality constant kcan be expressed in terms of the
Faraday constant (F) as:
k=F
z
where zis the number of moles of electrons exchanged per mole of substance
undergoing electrolysis.
Step 5: Substituting k=F
zinto the equation m=k·I·t, we get:
m=F
z·I·t
Step 6: Therefore, the relationship between the amount of substance de-
posited, the charge passed, and the Faraday constant is given by:
m=F
z·I·t
Question 13
Question
According to Faraday’s laws of electrolysis, what is the relationship between the
amount of substance deposited during electrolysis and the quantity of electricity
passed through the electrolyte solution?
12
Solution
To answer this question, we need to understand Faraday’s laws of electrolysis
and the relationships they describe.
Faraday’s First Law: The amount of substance deposited in electrolysis is
directly proportional to the quantity of electricity passed through the electrolyte
solution.
Faraday’s Second Law: The amount of substance deposited by the same
quantity of electricity is directly proportional to the equivalent weight of the
substance.
Therefore, the relationship between the amount of substance deposited dur-
ing electrolysis and the quantity of electricity passed through the electrolyte
solution can be summarized as follows:
Step 1: Let Qbe the quantity of electricity passed through the electrolyte
solution in coulombs, mbe the mass of the substance deposited in grams, and
zbe the electrochemical equivalent of the substance in grams per coulomb.
According to Faraday’s laws, we have the following relationship:
m=zQ
This equation shows that the mass of substance deposited is directly pro-
portional to the quantity of electricity passed through the electrolyte solution.
Step 2: If the current Ipassing through the electrolyte for a time tis
known, we can relate the quantity of electricity Qto the current and time using
the formula:
Q=It
Step 3: Substituting Q=It into the equation m=zQ, we get:
m=zIt
Step 4: This final equation shows the direct proportionality between the
amount of substance deposited during electrolysis, the quantity of electricity
passed through the electrolyte solution, and the time of electrolysis.
Therefore, the relationship between the amount of substance deposited dur-
ing electrolysis and the quantity of electricity passed through the electrolyte
solution is given by the equation m=zIt.
Question 14
Question
An electrolytic cell is set up with a solution containing silver ions, Ag+, and
copper ions, Cu2+. A constant current of 2.50 A is passed through the cell for
50.0 minutes. If the electrolysis of the solution results in the deposition of 1.25
g of silver, calculate the mass of copper that is deposited. The molar masses of
Ag and Cu are 107.87 g/mol and 63.55 g/mol, respectively.
13
Solution
Step 1: Calculate the total charge that has passed through the electrolyte. Let’s
use Faraday’s first law of electrolysis to find the total charge passed:
Q=I·t
where: - Qis the total charge passed (in Coulombs), - Iis the current (in
Amperes), and - tis the time in seconds.
Given that the current is 2.50 A and the time is 50.0 minutes (or 3000
seconds), we can calculate Q:
Q= 2.50 A ×3000 s = 7500 C
Step 2: Determine the moles of silver deposited. Since 1 F (Faraday) is
equivalent to 1 mole of electrons (96485 C), we can find the moles of silver
deposited using Faraday’s second law:
molAg =massAg
molar massAg
molAg =1.25 g
107.87 g/mol ≈0.0116 mol
Step 3: Calculate the moles of copper deposited. The ratio of moles of
silver to moles of copper deposited is 2:1 according to the stoichiometry of the
reaction. Therefore, the moles of copper deposited is half the moles of silver:
molCu = 0.5×molAg = 0.5×0.0116 mol = 0.0058 mol
Step 4: Determine the mass of copper deposited.
massCu = molCu ×molar massCu
massCu = 0.0058 mol ×63.55 g/mol ≈0.37 g
Therefore, approximately 0.37 g of copper is deposited during the electrolysis
process.
Question 15
Question
A certain electrolysis cell operates using a current of 1.5 A for 45 minutes.
During this time, 0.25 g of substance is deposited at one of the electrodes.
Assume that the substance is a divalent metal. Determine the molar mass of
the metal.
14
Solution
Step 1: Calculate the total charge passed through the cell. Given that the
current is 1.5 A and the time of operation is 45 minutes, first convert the time
to seconds:
45 minutes ×60 seconds/minute = 2700 seconds
Using the formula Q=I×t, where Qis the charge, Iis the current, and t
is the time, we have:
Q= 1.5 A ×2700 s = 4050 C
Step 2: Calculate the number of moles of metal deposited. The faraday
constant, F, is 96500 C/mol. The charge passed through the cell (4050 C) can
be related to the number of moles of metal deposited (0.25 g) using the equation
Q=nF :
n=Q
F=4050 C
96500 C/mol
Step 3: Calculate the molar mass of the metal. Since the substance is a
divalent metal, the number of moles calculated above represents half the molar
quantity of the substance (M/2). Therefore, the molar mass of the metal (M)
can be calculated as follows:
M= 2n×molar mass of one mole of substance
Substitute the value of ninto the above equation to find the molar mass of
the metal.
Question 16
Question
In an electrolysis experiment, a current of 2.5 A was passed through a solution
of copper(II) sulfate for 30 minutes. If the mass of copper deposited at the
cathode was 1.8 g, calculate the atomic mass of copper. (Assume 100
Solution
Step 1: Calculate the charge passed through the electrolyte. Given: Current (I)
= 2.5 A, Time (t) = 30 minutes = 1800 seconds Using the formula Q = I ×t,
we can find the charge.
Q= 2.5 A ×1800 s
Q= 4500 C
15
Step 2: Calculate the number of moles of copper deposited. Given: Mass
of copper deposited = 1.8 g, Atomic mass of copper = 63.5 g/mol Using the
formula moles = mass / atomic mass, we can find the number of moles deposited.
Moles of copper = 1.8 g
63.5 g/mol
Moles of copper ≈0.02835 mol
Step 3: Determine the number of electrons involved in the reaction. Since
each mole of copper requires 2 moles of electrons for reduction, the number of
electrons involved can be calculated as:
Number of electrons = 2 ×Moles of copper
Number of electrons = 2 ×0.02835 mol
Number of electrons ≈0.0567 mol
Step 4: Calculate the Faraday constant. The Faraday constant, F, is the
charge of 1 mole of electrons, which is 96,485 C/mol.
Faraday constant (F) = 96485 C/mol
Step 5: Determine the atomic mass of copper. The atomic mass of copper
can be calculated using the formula:
Faraday constant (F) = Total charge passed (Q)
Number of electrons = Atomic mass of copper
96485 = 4500
0.0567 = Atomic mass of copper
96485 ≈79323.529 g/mol
Therefore, the calculated atomic mass of copper is approximately 79.3235
g/mol.
Question 17
Question
A student conducts an electrolysis experiment using a solution of molten sodium
chloride. Initially, the student passes a current of 2.5 A through the solution for
15 minutes. Later, the student passes a current of 4.0 A through the solution
for another 20 minutes. Calculate the total charge passed through the solution
in coulombs during the entire experiment.
16
Solution
Step 1: Calculate the charge passed during the first part of the experiment.
Given: Current, I1= 2.5 A Time, t1= 15 minutes
The total charge passed during the first part of the experiment is given by:
Q1=I1×t1
Substitute I1= 2.5 A and t1= 15 minutes into the formula:
Q1= 2.5×15
Calculate Q1:
Q1= 37.5 C
Step 2: Calculate the charge passed during the second part of the experi-
ment. Given: Current, I2= 4.0 A Time, t2= 20 minutes
The total charge passed during the second part of the experiment is given
by:
Q2=I2×t2
Substitute I2= 4.0 A and t2= 20 minutes into the formula:
Q2= 4.0×20
Calculate Q2:
Q2= 80 C
Step 3: Calculate the total charge passed during the entire experiment. The
total charge passed during the entire experiment is the sum of Q1and Q2:
Qtotal =Q1+Q2
Substitute Q1= 37.5 C and Q2= 80 C into the formula:
Qtotal = 37.5 + 80
Calculate Qtotal:
Qtotal = 117.5 C
Therefore, the total charge passed through the solution in coulombs during
the entire experiment is 117.5 C.
Question 18
Question
An electric current is passed through a solution of silver nitrate, AgNO3, using
silver electrodes. If 1.50 g of silver is deposited on the cathode in 15 minutes,
determine the current passing through the cell. The molar mass of silver is
107.87 g/mol.
17
Solution
Step 1: Find the number of moles of deposited silver.
Given that the molar mass of silver is 107.87 g/mol, we can calculate the number
of moles of silver deposited on the cathode using the formula:
moles of silver = mass of silver
molar mass of silver
moles of silver = 1.50 g
107.87 g/mol
moles of silver ≈0.0139 mol
Step 2: Find the charge passed through the cell.
Since 1 mole of silver corresponds to 1 mole of electrons (Ag++ 1e−→Ag), we
can determine the charge passed through the cell in coulombs using Faraday’s
constant, F= 96485 C/mol, and the formula:
charge = moles of silver ×F
charge = 0.0139 mol ×96485 C/mol
charge ≈1343.0 C
Step 3: Find the current passing through the cell.
Since the time taken for the deposition of silver is 15 minutes (or 900 seconds),
we can calculate the current passing through the cell using the formula:
current = charge
time
current = 1343.0 C
900 s
current ≈1.49 A
Therefore, the current passing through the cell is approximately 1.49 A.
Question 19
Question
Explain Faraday’s laws of electrolysis and how they are applied in electrochem-
istry.
18
Solution
To understand Faraday’s laws of electrolysis, one must first recall that electrol-
ysis is the process of using electricity to bring about a chemical change in an
electrolyte. Faraday’s laws describe the quantitative relationships between the
amount of substance produced or consumed at an electrode during electrolysis
and the quantity of electricity that flows through the cell.
Faraday’s First Law: The amount of a substance produced or consumed
during electrolysis is directly proportional to the quantity of electricity passed
through the cell.
1. This law can be expressed mathematically as:
Amount of substance = Current ×Time ×Equivalent weight
Faraday’s Second Law: The amounts of different substances produced or
consumed by the same quantity of electricity are in the ratio of their chemical
equivalent weights.
2. Mathematically, this law can be written as:
n1
n2
=E1
E2
where n1and n2are the amounts of two substances obtained at the electrodes,
and E1and E2are their chemical equivalent weights.
Applications of Faraday’s laws in electrochemistry include calculating the
amount of product formed during electrolysis, determining the valency of an
ion in an electrolyte solution, and predicting the products of electrolysis.
In conclusion, Faraday’s laws of electrolysis are fundamental principles in
electrochemistry that relate the amount of substance produced or consumed
during electrolysis to the quantity of electricity passed through the cell. These
laws are essential for understanding and predicting the outcomes of electrolytic
processes.
Question 20
Question
A student is conducting an electrolysis experiment using a copper(II) sulfate
solution with copper electrodes. If a current of 0.5 A is passed through the
solution for 30 minutes, calculate the mass of copper deposited on the cathode.
(Given: Atomic mass of copper = 63.5 g/mol, Faraday constant = 96485 C/mol)
Solution
Step 1: Calculate the total charge passed through the solution.
Total charge = Current ×Time
19
Total charge = 0.5 A ×(30 min) ×60 s
1 min
Total charge = 0.5×30 ×60 C
Total charge = 900 C
Step 2: Calculate the number of moles of electrons passed through the solu-
tion using Faraday’s law.
1 mol of electrons = 1 Faraday = 96485 C
Number of moles of electrons = Total charge
Faraday constant
Number of moles of electrons = 900 C
96485 C/mol
Number of moles of electrons ≈0.00933 mol
Step 3: Since 1 mole of electrons deposits 1 mol of copper, calculate the
mass of copper deposited on the cathode.
Mass of copper = Number of moles of electrons ×Molar mass of copper
Mass of copper = 0.00933 mol ×63.5 g/mol
Mass of copper ≈0.592 g
Therefore, the mass of copper deposited on the cathode is approximately
0.592 g.
Question 21
Question
In an electrolysis experiment, a current of 2.5 A is passed through a solution of
molten lead (II) bromide for 15 minutes. Calculate the mass of lead produced
during this time. (Molar mass of Pb = 207.2 g/mol, Faraday’s constant =
96485 C/mol)
Solution
Step 1: Determine the number of moles of lead produced. Given: Current (I) =
2.5 A, Time (t) = 15 minutes = 900 seconds, Molar mass of lead (Pb) = 207.2
g/mol, Faraday’s constant (F) = 96485 C/mol.
We know that the charge passed (Q) during electrolysis is given by:
Q=I×t
Substitute the given values:
Q= 2.5A×900 s= 2250 C
20
Step 2: Calculate the number of moles of lead produced. The number of
moles of lead produced can be determined by dividing the total charge passed
by the Faraday’s constant:
Moles of Pb = Q
F
Substitute the values:
Moles of Pb = 2250 C
96485 C/mol ≈0.0233 mol
Step 3: Calculate the mass of lead produced. The mass of lead produced
can be calculated using the number of moles and the molar mass of lead:
Mass of Pb = Moles of Pb ×Molar mass of Pb
Substitute the values:
Mass of Pb = 0.0233 mol ×207.2g/mol ≈4.82 g
Therefore, the mass of lead produced during the electrolysis of molten lead
(II) bromide for 15 minutes is approximately 4.82 grams.
Question 22
Question
An aqueous solution of sodium chloride (NaCl) is electrolyzed using inert elec-
trodes. If a current of 2.5 A is passed through the solution for 3 hours, calculate
the amount of chlorine gas (Cl2) produced during the electrolysis. Assume 100
Solution
Step 1: Write the balanced half-reactions for the electrolysis of aqueous sodium
chloride:
Anode: 2Cl−→Cl2+ 2e−
Cathode: 2H2O + 2e−→H2+ 2OH−
Step 2: Calculate the total charge passed through the cell using the formula
Q=I×t, where Qis the charge in Coulombs (C), Iis the current in Amperes
(A), and tis the time in seconds:
Q= 2.5 A ×3 hours ×3600 s/hour = 27,000 C
Step 3: Determine the moles of electrons passed through the circuit using
Faraday’s law, where 1 Faraday (F) is equivalent to 96,485 C/mol electrons:
Moles of electrons = 27,000 C
96500 C/mol ≈0.28 mol
21
Step 4: Calculate the moles of chlorine gas produced based on the stoichiom-
etry of the reaction (1 mol Cl2is produced per 2 mol e−passed):
Moles of Cl2= 0.28 mol ×1 mol Cl2
2 mol e−= 0.14 mol
Step 5: Convert moles of chlorine gas to volume using the ideal gas law at
standard temperature and pressure (STP) conditions (22.4 L/mol):
Volume of Cl2= 0.14 mol ×22.4 L/mol = 3.14 L
Therefore, approximately 3.14 L of chlorine gas will be produced during the
electrolysis.
Question 23
Question
An aqueous solution of silver nitrate is subjected to electrolysis using an external
electromotive force of 2.0 V. If 5.0 grams of silver is deposited in 30 minutes,
determine the amount of current passing through the solution during this time.
Solution
Step 1: Convert the given mass of silver to moles using the molar mass of silver.
Given: Mass of silver, m= 5.0 g Molar mass of silver, M= 107.87 g/mol
Number of moles of silver, n=m
M=5.0 g
107.87 g/mol
Step 2: Calculate the charge passed in coulombs using Faraday’s laws of
electrolysis.
Given: Faraday constant, F= 96,485 C/mol
Charge passed, Q=n×F
Step 3: Determine the time period in seconds.
Given: Time, t= 30 minutes = 30 ×60 s
Step 4: Calculate the current passing through the solution.
Given: Voltage, V= 2.0 V
Current, I=Q
t=n×F
t
22
Question 24
Question
In an electrolytic cell, a current of 2.5 A is passed through a solution of silver
nitrate (AgNO3) for 2 hours. If 3.5 g of silver is deposited at the cathode,
determine the oxidation state of silver in silver nitrate and the mass of silver
nitrate that decomposed.
Solution
Step 1: Find the equivalent weight of silver. Given that 3.5 g of silver is de-
posited, we can calculate the equivalent weight of silver using the formula:
Equivalent weight (E) = Atomic weight
Valency ×Mass deposited
Since the atomic weight of silver is 107.87 g/mol and the valency of silver is 1,
we have:
E=107.87
1×3.5=107.87
3.5≈30.82 g/equiv
Step 2: Determine the oxidation state of silver in silver nitrate (AgN O3).
From the equivalent weight of silver calculated in Step 1, we know that 1 equiv
of silver is deposited when 3.5 g is deposited. The molar mass of silver nitrate
is 169.87 g/mol. Thus, the number of equivalents of silver is given by:
Equivalents of silver = Mass of silver deposited
Equivalent weight of silver =3.5
30.82 ≈0.1137 equiv
Since 1 mole of AgNO3produces 1 mole of silver, the number of moles of AgNO3
decomposed is equal to the number of equivalents of silver deposited. Therefore,
the mass of AgNO3decomposed is:
Mass of AgNO3decomposed = Molar mass of AgNO3×Equivalents of silver
= 169.87 ×0.1137 ≈19.33 g
Therefore, the oxidation state of silver in silver nitrate is +1, and the mass
of silver nitrate that decomposed is approximately 19.33 g.
Question 25
Question
A student sets up an electrolysis experiment using a copper(II) sulfate solution.
The student uses a current of 2.0 A to electrolyze the solution for 30 minutes.
During this time, the student measures that 0.20 g of copper is deposited onto
one of the electrodes. Calculate the Faraday constant.
Given: Atomic mass of copper = 63.55 g/mol
23
Solution
Step 1: Calculate the moles of copper deposited. Step 2: Use the moles of
copper to find the amount of charge passed through the circuit. Step 3: Apply
Faraday’s law of electrolysis to find the Faraday constant.
Step 1: To find the moles of copper deposited, we use the formula:
moles = mass
molar mass
Substitute the given values:
moles = 0.20 g
63.55 g/mol = 0.00315 mol
Step 2: The amount of charge passed through the circuit can be found using
Faraday’s law, which states:
Charge (C) = moles ×Faraday constant (C/mol)
Since the current is given in Amperes and time is given in seconds, we need to
convert the time to seconds:
Time (s) = 30 minutes ×60 s/min = 1800 s
Then, we can find the charge passed:
Charge (C) = 2.0 A ×1800 s = 3600 C
Step 3: Now, we can find the Faraday constant using the charge passed and
moles of copper. Rearranging the formula from Step 2 gives:
Faraday constant (C/mol) = Charge (C)
moles
Substitute the calculated values:
Faraday constant = 3600 C
0.00315 mol ≈1.14 ×104C/mol
Therefore, the Faraday constant is approximately 1.14 ×104C/mol.
Question 26
Question
A student is conducting an electrolysis experiment using a solution of copper
sulfate (CuSO4). They connect a power supply to two electrodes: one made
of copper and one made of platinum. During the experiment, they observe
that 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.
24
Solution
To explain the observed phenomenon, we need to consider the implications of
Faraday’s laws of electrolysis.
Step 1: Electrolysis of Copper(II) Sulfate In the electrolysis of copper
sulfate solution, copper ions (Cu2+) will be attracted to the negative electrode
(cathode) where they will gain electrons and be deposited as solid copper. The
copper electrode will therefore increase in mass due to the deposition of copper.
Step 2: Faraday’s First Law Faraday’s First Law states that the amount
of substance deposited at an electrode is directly proportional to the amount
of charge passed through the electrolyte. Therefore, the increase in mass of
the copper electrode can be attributed to the transfer of copper ions from the
solution to the electrode by the passage of electric current.
Step 3: Electrolysis of Water At the positive electrode (anode) made of
platinum, water molecules will undergo electrolysis to release oxygen gas. Since
platinum is inert and does not react with water or the products of electrolysis,
the mass of the platinum electrode remains constant.
Step 4: Faraday’s Second Law Faraday’s Second Law states that the
amounts of different substances deposited by a given quantity of electricity are
proportional to their chemical equivalent weights. In this case, copper ions
are reduced to solid copper at the cathode, while water is oxidized to release
oxygen gas at the anode. The mass changes at each electrode reflect the different
chemical reactions occurring due to the applied electric current.
In conclusion, the decrease in mass of the copper electrode and the constant
mass of the platinum electrode can be explained by the principles of Faraday’s
laws of electrolysis, which govern the transfer of material during the electrolysis
process.
Question 27
Question
A copper sulfate solution contains 0.1 moles of copper ions. If a current of 2A
is passed through the solution for 1 hour, calculate:
1. The mass of copper deposited at the cathode.
2. The volume of oxygen produced at the anode at STP.
3. The volume of hydrogen produced at the cathode at STP.
Given: Atomic mass of copper = 63.5 g/mol, 1 Faraday = 96500 C/mol, 1 mol
of any gas at STP occupies 22.4 L.
Solution
1. Step 1: Calculate the equivalent weight of copper.
25
The equivalent weight of copper is the atomic mass divided by the valency.
Valency of copper = 2 (since Cu2+ ions are produced in the solution during
electrolysis)
Equivalent weight of copper = 63.5
2g/mol = 31.75 g/mol
Step 2: Calculate the mass of copper deposited at the cathode.
The mass of copper deposited can be calculated using the formula: Mass
= (Equivalent weight ×Current ×Time) / (96500)
Mass = 31.75×2×3600
96500 g Mass = 57240
96500 g Mass ≈0.593 g
Therefore, the mass of copper deposited at the cathode is 0.593 g.
2. Step 3: Calculate the moles of oxygen produced at the anode.
Since copper ions are being discharged at the cathode, oxygen gas is pro-
duced at the anode during electrolysis. 4 moles of electrons are required
to produce 1 mole of oxygen gas.
Moles of electrons = Current ×Time / 96500 Moles of oxygen = Moles
of electrons / 4
Moles of oxygen = 2×3600
96500 /4 mol Moles of oxygen ≈0.0187 mol
Step 4: Calculate the volume of oxygen produced at STP.
Volume of oxygen = Moles of oxygen ×22.4 L Volume of oxygen = 0.0187
×22.4 L Volume of oxygen ≈0.419 L
Therefore, the volume of oxygen produced at the anode at STP is 0.419
L.
3. Step 5: Calculate the moles of hydrogen produced at the cathode.
For every mole of copper deposited at the cathode, 2 moles of electrons
are consumed which are utilized to generate hydrogen gas.
Moles of electrons = 2 ×current ×time / 96500 Moles of hydrogen =
Moles of electrons / 2
Moles of hydrogen = 2×2×3600
96500 /2 mol Moles of hydrogen ≈0.075 mol
Step 6: Calculate the volume of hydrogen produced at STP.
Volume of hydrogen = Moles of hydrogen ×22.4 L Volume of hydrogen
= 0.075 ×22.4 L Volume of hydrogen ≈1.68 L
Therefore, the volume of hydrogen produced at the cathode at STP is 1.68
L.
Question 28
Question
Consider an electrolytic cell containing a solution of sodium chloride (NaCl).
When a current of 2.00 A is passed through the cell for 1.00 hour, 0.190 g of
26
sodium metal is deposited at the cathode. Calculate the standard electrode
potential of sodium.
Solution
Step 1: First, we need to convert the amount of sodium deposited from grams
to moles. The molar mass of sodium (Na) is 22.99 g/mol.
Given: Amount of sodium deposited, m= 0.190 g Molar mass of sodium,
M= 22.99 g/mol
Number of moles of sodium:
n=m
M=0.190
22.99 = 0.00826 mol
Step 2: Next, we need to determine the number of electrons transferred dur-
ing the electrolysis based on the reaction at the cathode, which is the reduction
of sodium ions (Na+) to form sodium metal (Na):
2Na++ 2e−→2Na
Since 2 moles of electrons are required to deposit 2 moles of sodium, the
number of electrons transferred (ne) is equal to the number of moles of sodium
deposited:
ne= 2 ×n= 2 ×0.00826 = 0.0165 mol
Step 3: Using Faraday’s laws of electrolysis, we know that the quantity of
electricity transferred is directly proportional to the number of moles of electrons
transferred. The constant of proportionality is the Faraday constant (F), which
is equal to 9.6485 ×104C/mol.
Quantity of electricity transferred:
Q=ne×F= 0.0165 ×9.6485 ×104= 1590 C
Step 4: The standard electrode potential of sodium, denoted as E◦(Na+/Na),
can be calculated using the formula:
E◦=Q
I×t
Given: Current, I= 2.00 A Time, t= 1.00 hour = 3600 s
Substitute the known values:
E◦=1590
2.00 ×3600 =1590
7200 = 0.221 V
Therefore, the standard electrode potential of sodium is 0.221 V .
27
Question 29
Question
A solution of silver nitrate (AgNO3) is electrolyzed using a current of 4.5 A for
2.5 hours. If 0.15 g of silver is deposited at the cathode, determine the number
of moles of electrons involved in the reaction. Assume 100
Solution
Step 1: Calculate the total charge passed
Total charge (Q) = Current (I) ×Time (t)
Given current, I= 4.5 A and time, t= 2.5 hours, we convert time to seconds:
t= 2.5×60 ×60 = 9000 s
Now, calculating total charge:
Q= 4.5 A ×9000 s = 40500 C
Step 2: Determine the number of moles of electrons involved To find the
number of moles of electrons, we need to first calculate the moles of silver
deposited using the given mass. Given mass of silver, m= 0.15 g, and molar
mass of silver, MAg = 107.87 g/mol.
Converting mass to moles:
nAg =m
MAg
=0.15 g
107.87 g/mol ≈0.00139 mol
Each mole of silver deposited corresponds to 1 mole of electrons. Therefore,
the number of moles of electrons involved in the reaction is:
ne−= 0.00139 mol
So, approximately 0.00139 moles of electrons are involved in the electrolysis
reaction.
Question 30
Question
Discuss Faraday’s laws of electrolysis and explain how they can be used to
determine the amount of substance deposited or liberated during electrolysis.
28
Solution
Faraday’s laws of electrolysis are a set of two laws formulated by Michael Fara-
day in 1834. These laws describe the quantitative relationship between the
amount of substance produced or consumed during electrolysis and the amount
of electric charge passed through the electrolyte.
Faraday’s First Law: The mass of a substance deposited or liberated
during electrolysis is directly proportional to the amount of electrical charge
passed through the electrolyte.
Faraday’s Second Law: The mass of different substances deposited or
liberated by the same quantity of electricity is directly proportional to their
chemical equivalents (equivalent weight).
Using Faraday’s laws, we can determine the amount of substance deposited
or liberated during electrolysis using the following steps:
Step 1: Calculate the quantity of electrical charge passed through the elec-
trolyte using the formula
Q=I×t,
where Qis the electric charge (in Coulombs), Iis the current (in Amperes), and
tis the time (in seconds).
Step 2: Determine the number of moles of electrons transferred during the
electrolysis by dividing the electric charge by the Faraday constant F:
n=Q
F,
where nis the number of moles of electrons, and F≈96485 C/mol is Faraday’s
constant.
Step 3: Identify the balanced chemical equation for the electrolysis process
and determine the mole ratio between the substance of interest and electrons.
Step 4: Calculate the number of moles of the substance deposited or liber-
ated by multiplying the number of moles of electrons by the appropriate mole
ratio.
Step 5: Convert the number of moles of the substance to mass using its
molar mass.
By following these steps and applying Faraday’s laws, one can accurately
determine the amount of substance deposited or liberated during electrolysis
based on the amount of electric charge passed through the system.
Question 31
Question
In an electrolytic cell, a metal M is deposited at the cathode when a current
of 2.00 A is passed through a solution of MCl2for 30.0 minutes. Calculate the
mass of M deposited. Given: - Atomic mass of M = 63.5 g/mol - Faraday’s
constant F= 96500 C/mol
29
Solution
Step 1: Find the charge passed through the cell.
Charge (Q) = Current (I) ×Time (t)
Q= 2.00 A ×(30.0 min ×60 s/min)
Q= 2.00 A ×1800 s = 3600 C
Step 2: Calculate the number of moles of M deposited.
Number of moles of M = Charge (Q)
Faraday’s constant (F)
Number of moles of M = 3600 C
96500 C/mol
Number of moles of M ≈0.0373 mol
Step 3: Determine the mass of M deposited.
Mass of M = Number of moles ×Atomic mass
Mass of M = 0.0373 mol ×63.5 g/mol
Mass of M ≈2.37 g
Therefore, the mass of metal M deposited in the given electrolytic cell is
approximately 2.37 grams.
Question 32
Question
An electroplating apparatus contains a solution of silver nitrate (AgNO3). If
a current of 2.5 A is passed through the solution for 1 hour, calculate the
mass of silver deposited on the cathode. Given that the molar mass of silver is
107.87 g/mol and Faraday’s constant is 96500 C/mol.
Solution
Step 1: Calculate the charge passed through the solution.
Given: Current (I) = 2.5 A, Time (t) = 1 hour = 3600 s
The charge (Q) passed through the solution can be calculated using the formula:
Q=I×t
Q= 2.5 A ×3600 s = 9000 C
Step 2: Calculate the number of moles of electrons that passed through the
solution.
30
Since Faraday’s constant is the charge consumed by 1 mole of electrons, the
number of moles (n) can be calculated as:
n=Q
96500
n=9000 C
96500 C/mol ≈0.093 mol
Step 3: Calculate the mass of silver deposited on the cathode.
The chemical reaction at the cathode is:
Ag++e−−→ Ag
From the balanced equation, we can see that 1 mole of electrons corresponds to
the deposition of 1 mole of silver. Since the molar mass of silver is 107.87 g/mol,
the mass of silver deposited is:
Mass of silver = n×Molar mass of silver
Mass of silver = 0.093 mol ×107.87 g/mol ≈10.02 g
Therefore, the mass of silver deposited on the cathode is approximately
10.02 g after passing a current of 2.5 A for 1 hour.
Question 33
Question
State Faraday’s laws of electrolysis. Explain how these laws are derived from
the laws of conservation of mass and charge.
Solution
Faraday’s laws of electrolysis describe the quantitative relationship between
the amount of substance produced at an electrode during electrolysis and the
amount of charge passed through the electrolyte. These laws can be derived
from the laws of conservation of mass and charge.
Faraday’s First Law: The amount of substance deposited on each elec-
trode during electrolysis is directly proportional to the quantity of electricity
passing through the cell.
Faraday’s Second Law: The amounts of different substances deposited
by the same quantity of electricity are proportional to their chemical equivalent
weights.
Derivation from Conservation of Mass and Charge:
Conservation of Mass: This law states that mass can neither be created
nor destroyed in a chemical reaction. During electrolysis, the mass of the
substance deposited on each electrode is proportional to the quantity of
electricity passed.
31
Conservation of Charge: This law states that electric charge is con-
served in any process. The charge passed during electrolysis is directly
proportional to the number of electrons transferred in the redox reactions
at the electrodes.
Step 1: Let Qbe the total charge passed through the electrolyte during
electrolysis, and nbe the number of moles of electrons transferred. The charge
Qis related to nby Faraday’s constant F:
Q=n×F
Step 2: The quantity of electricity needed to deposit one mole of a substance
can be determined using its chemical equivalent weight (M). According to
Faraday’s laws, if mgrams of a substance are deposited, then:
m=M×n
Step 3: By combining the two equations from Step 1 and Step 2, we get:
Q=M×F×m
This equation represents Faraday’s laws of electrolysis, relating the amount
of substance deposited (m) to the quantity of electricity passed (Q), with the
constant of proportionality being a product of the chemical equivalent weight
and Faraday’s constant.
Question 34
Question
A copper(II) sulfate solution is electrolyzed using two inert electrodes. If a
current of 2.50 A is passed through the solution for 3.00 hours, what mass of
copper is deposited on the cathode? The molar mass of copper is 63.55 g/mol.
Solution
Step 1: Calculate the total charge passed through the circuit Given values:
Current, I= 2.50 A Time, t= 3.00 hours
The total charge, Q, passed through the circuit can be calculated using the
formula:
Q=I×t
Substitute the given values:
Q= 2.50 A ×3.00 hours = 7.50 C
Step 2: Determine the number of moles of electrons Each mole of electrons
(1 Faraday) is equivalent to 6.022 ×1023 electrons and carries a charge of 1
32
C. Therefore, the number of moles of electrons, n, can be calculated using the
formula:
n=Q
Faraday constant =Q
96485
Substitute the value for Q:
n=7.50 C
96485 C/mol = 7.78 ×10−5mol
Step 3: Determine the mass of copper deposited The balanced half-reaction
for the deposition of copper:
Cu2+(aq)+2e−→Cu(s)
From the balanced reaction, it is observed that 2 moles of electrons are
needed to deposit 1 mole of copper. Therefore, the mole ratio between electrons
and copper is 2:1.
Since 2 moles of electrons correspond to the deposition of 1 mole of copper,
the number of moles of copper deposited is half the number of moles of electrons:
nCu =ne−
2=7.78 ×10−5
2= 3.89 ×10−5mol
Step 4: Calculate the mass of copper deposited The mass of copper, m,
deposited can be calculated using the formula:
m=n×Molar mass of copper
Substitute the values for nand the molar mass of copper:
m= 3.89 ×10−5mol ×63.55 g/mol = 0.00247 g
Therefore, the mass of copper deposited on the cathode is 0.00247 g.
Question 35
Question
Consider an electrolytic cell containing a solution of copper(II) sulfate, CuSO4
with copper electrodes. During the electrolysis process, a current of 2.50 A is
passed through the cell for 3.00 hours. If the mass of copper deposited on the
cathode is 6.72 g, calculate the number of moles of electrons that passed through
the electrolytic cell.
33
Solution
Step 1: Find the total charge that passed through the cell. The total charge
can be calculated using the formula Q=It, where Qis the charge in coulombs,
Iis the current in amperes, and tis the time in seconds. Given: I= 2.50 Aand
t= 3.00 hours = 3.00 ×3600 s= 10800 s.
So,
Q= 2.50 A×10800 s= 27000 C
Step 2: Calculate the number of moles of electrons. The number of moles of
electrons can be determined using the equation n=Q
F, where nis the number
of moles of electrons, Qis the charge in coulombs (already calculated), and F
is the Faraday constant (96485 C/mol).
Substitute the values:
n=27000 C
96485 C/mol
Calculating,
n≈0.280 mol
Therefore, approximately 0.280 moles of electrons passed through the elec-
trolytic cell during the electrolysis process.
34
