CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Faraday’s
laws of electrolysis
Question Bank - Set 3
Liberty University
Question 1
Question
An electrolysis experiment was performed using a cell with a constant current
of 2.5 A for 30 minutes. During this time, a mass of 3.75 g of copper was
deposited at the cathode. Calculate the Faraday constant and the number of
moles of electrons that passed through the cell.
Solution
Step 1: Calculate the charge that passed through the cell. Given: Current,
I= 2.5 A Time, t= 30 minutes = 30 ×60 seconds = 1800 s
The total charge, Q, that passed through the cell can be calculated using
the formula:
Q=I×t
Q= 2.5 A ×1800 s
Q= 4500 C
Step 2: Calculate the number of moles of copper deposited. Given: Mass of
copper deposited, m= 3.75 g Molar mass of copper, M= 63.55 g/mol
The number of moles of copper deposited at the cathode can be calculated
using the formula:
Number of moles = Mass
Molar mass
Number of moles = 3.75 g
63.55 g/mol
Number of moles ≈0.0591 mol
Step 3: Calculate the Faraday constant. The Faraday constant, F, is the
charge per mole of electrons. It can be calculated using the formula:
F=Total charge
Number of moles of electrons
Given that 1 Faraday is equal to the charge of 1 mole of electrons (approxi-
mately 96485 C/mol), the Faraday constant can be calculated as:
F≈96485 C/mol
Step 4: Calculate the number of moles of electrons that passed through the
cell.
Number of moles of electrons = Total charge
F
Number of moles of electrons = 4500 C
96485 C/mol
Number of moles of electrons ≈0.0466 mol
Therefore, the Faraday constant is approximately 96485 C/mol and the num-
ber of moles of electrons that passed through the cell is approximately 0.0466
mol.
Question 2
Question
During electrolysis, a current of 2.5 A is passed through a solution of silver
nitrate for 15 minutes. If silver ions have a valency of +1, calculate the mass of
silver deposited on the cathode. (Given: Atomic mass of silver = 107.87 amu)
Solution
To find the mass of silver deposited on the cathode during the electrolysis of
silver nitrate, we can use Faraday’s laws of electrolysis.
Step 1: Calculate the total charge passed through the solution
Given:
Current (I) = 2.5 A
Time (t) = 15 minutes = 15 ×60 seconds = 900 s
Total charge (Q) passed through the solution can be calculated using the for-
mula:
Q=I×t
Q= 2.5 A ×900 s
2
Q= 2250 C
Step 2: Calculate the number of moles of electrons transferred
One Faraday of charge is equivalent to 96500 C, which corresponds to 1 mole of
electrons. Therefore:
1 F = 96500 C
1 mol e−= 1 F = 96500 C
Given that the total charge passed (Q) is 2250 C, we can find the number of
moles of electrons transferred:
Moles of electrons = Q
96500
Moles of electrons = 2250
96500
Moles of electrons = 0.023 mol
Step 3: Calculate the mass of silver deposited Since silver ions have a
valency of +1, the number of moles of silver deposited will be equal to the num-
ber of moles of electrons transferred. The molar mass of silver is 107.87 g/mol,
so the mass of silver deposited can be calculated as:
Mass of silver deposited = Moles of silver ×Molar mass of silver
Mass of silver deposited = 0.023 mol ×107.87 g/mol
Mass of silver deposited ≈2.48 g
Therefore, the mass of silver deposited on the cathode during the electrolysis
of silver nitrate is approximately 2.48 g.
Question 3
Question
Consider an electrolytic cell containing an aqueous solution of copper sulfate,
CuSO4. If a current of 2.5 A is passed through the cell for 2.0 hours, calculate
the mass of copper deposited at the cathode. (Given: 1 F = 96500 Coulombs/mol,
Molar mass of Cu = 63.5 g/mol)
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current (I)
= 2.5 A, Time (t) = 2.0 hours = 7200 s
Q=It = 2.5 A ×7200 s = 18000 Coulombs
3
Step 2: Calculate the number of moles of electrons that passed through the
cell.
1 F = 96500 Coulombs/mol
Number of moles of electrons = Q
96500 =18000
96500 = 0.186 mol
Step 3: Determine the number of moles of copper deposited at the cathode.
Since the reaction at the cathode is Cu2+ + 2e−→Cu, 1 mole of copper is
deposited by 2 moles of electrons.
Number of moles of copper = 0.186 mol
2= 0.093 mol
Step 4: Calculate the mass of copper deposited at the cathode. Given: Molar
mass of copper (Cu) = 63.5 g/mol
Mass of copper = 0.093 mol ×63.5 g/mol = 5.9055 g
Therefore, the mass of copper deposited at the cathode is 5.9055 grams.
Question 4
Question
A student is performing an electrolysis experiment with a solution of sodium
chloride (NaCl). If a current of 2.5 A is passed through the solution for 30
minutes, what mass of sodium metal is deposited at the cathode? (Given:
molar mass of sodium = 22.99 g/mol, Faraday’s constant = 96485 C/mol)
Solution
Step 1: Find the total charge passed through the solution. Given: Current,
I= 2.5ATime, t= 30 min = 30 ×60 sec = 1800 sec
The total charge passed, Q, can be calculated using the formula:
Q=I×t
Q= 2.5A×1800 sec
Q= 4500 C
Step 2: Calculate the number of moles of electrons passed through the so-
lution. Since 1 F (Faraday) is equal to 96485 C/mol, the number of moles of
electrons, n, can be calculated as:
n=Q
96485
n=4500
96485
4
n≈0.0466 mol
Step 3: Use the stoichiometry of the reaction to find the mass of sodium
deposited. From the balanced chemical equation for the electrolysis of sodium
chloride:
2 Na++ 2 Cl−→2 Na + Cl2
It is clear that 2 moles of electrons are required to deposit 2 moles of sodium
metal. So the number of moles of sodium metal deposited, nNa, is equal to half
of the number of moles of electrons, nNa = 0.5n.
nNa = 0.5×0.0466
nNa = 0.0233 mol
Step 4: Calculate the mass of sodium metal deposited. The molar mass of
sodium is given as 22.99 g/mol.
Mass of sodium = nNa ×Molar mass of sodium
Mass of sodium = 0.0233 mol ×22.99 g/mol
Mass of sodium ≈0.535 g
Therefore, approximately 0.535 grams of sodium metal is deposited at the
cathode after passing a current of 2.5 A for 30 minutes.
Question 5
Question
An electrochemical cell contains a solution of silver nitrate (AgNO3) with silver
electrodes. If a constant current of 2.5 A is passed through the cell for 30
minutes, what mass of silver is deposited on each electrode? (Given: Faraday
constant, F= 96500 C/mol; molar mass of silver, M= 107.87 g/mol)
Solution
Step 1: Calculate the total charge passing through the cell. The total charge
passing through the cell can be calculated using the formula:
Q=I×t
where Qis the total charge in Coulombs, Iis the current in Amperes, and tis
the time in seconds.
Given that I= 2.5 A and t= 30 minutes, we need to convert the time to
seconds:
t= 30 ×60 = 1800 s
Therefore,
Q= 2.5 A ×1800 s = 4500 C
5
Step 2: Calculate the number of moles of silver deposited. The amount of
silver deposited is related to the charge passed through the cell by Faraday’s
laws of electrolysis. The formula to calculate the number of moles of a substance
deposited is:
moles = Q
nF
where nis the number of electrons transferred during the reaction, and Fis the
Faraday constant.
Since the reaction involves the deposition of silver (Ag+receiving one elec-
tron), n= 1.
Substitute Q= 4500 C, n= 1, and F= 96500 C/mol into the formula:
moles = 4500 C
1×96500 C/mol ≈0.0468 mol
Step 3: Calculate the mass of silver deposited on each electrode. The mass
of silver can be calculated using the formula:
mass = moles ×M
where Mis the molar mass of silver.
Substitute moles = 0.0468 mol and M= 107.87 g/mol into the formula:
mass = 0.0468 mol ×107.87 g/mol ≈5.05 g
Therefore, approximately 5.05 g of silver is deposited on each electrode.
Question 6
Question
A solution of copper sulfate (CuSO4) was electrolyzed using platinum elec-
trodes. If a current of 2.5 A was passed through the solution for 3 hours,
what mass of copper would be deposited on the cathode? (Atomic masses:
Cu = 63.5g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution.
Given: Current, I= 2.5ATime, t= 3 hours = 3×3600 seconds = 10800 seconds
The total charge, Q, passed through the electrolyte solution can be calculated
using the formula:
Q=I×t
Substitute the given values:
Q= 2.5A×10800 s= 27000 C
6
Step 2: Determine the number of moles of copper deposited on the cathode.
The reaction at the cathode can be represented as:
Cu2+ + 2e−→Cu
From the reaction, it can be observed that 2 moles of electrons are required
to deposit 1 mole of copper.
Given: Total charge, Q= 27000 CCharge per mole of electrons, e= 1.6×
10−19 C(charge of 1 electron)
The number of moles of electrons, n, passed during electrolysis can be cal-
culated as:
n=Q
e
Substitute the given values:
n=27000 C
1.6×10−19 C≈1.69 ×1020 electrons
The number of moles of copper deposited can be calculated as:
Number of moles of copper = 1
2×n
NA
Where NAis Avogadro’s number (6.022 ×1023 mol−1).
Substitute the values and calculate:
Number of moles of copper = 1
2×1.69 ×1020
6.022 ×1023 ≈1.41 ×10−4mol
Step 3: Calculate the mass of copper deposited on the cathode. The mass
of copper deposited can be calculated using the formula:
Mass = Number of moles ×Molar mass
Substitute the values:
Mass = 1.41 ×10−4mol ×63.5g/mol ≈8.94 ×10−3g
Therefore, approximately 8.94 mg of copper would be deposited on the cath-
ode.
Question 7
Question
A student is performing an electrolysis experiment where a current of 2.5 A is
passed through an electrolytic cell containing a solution of copper(II) sulfate.
If the student was able to deposit 12.5 g of copper metal at the cathode after
20 minutes, what is the efficiency of the process? (Given: Faraday’s constant =
96500 C/mol, atomic mass of copper = 63.5 g/mol)
7
Solution
Step 1: Calculate the charge passing through the electrolytic cell. Given current
I= 2.5 A and time t= 20 min = 20
60 h = 1
3h. The charge Qpassing through the
cell can be calculated using the formula Q=It.
Q= 2.5 A ×1
3h = 2.5
3C = 5
6C
Step 2: Determine the number of moles of copper deposited. To calculate
the number of moles of copper deposited, we will use the formula:
moles = mass
molar mass =12.5 g
63.5 g/mol = 0.197 mol
Step 3: Calculate the efficiency of the process. The theoretical amount
of charge required to deposit 1 mole of copper is given by Faraday’s constant
F= 96500 C/mol. Therefore, the charge required to deposit 0.197 moles of
copper is:
Charge = 0.197 mol ×96500 C/mol = 19058.5 C
The efficiency of the process can be calculated using the formula:
Efficiency = actual charge
theoretical charge ×100% = 5/6
19058.5×100% ≈0.0026%
Therefore, the efficiency of the process is approximately 0.0026
Question 8
Question
Faraday’s laws of electrolysis state that the mass of a substance produced at an
electrode during electrolysis is directly proportional to the quantity of electricity
passed through the cell. In an electrolysis experiment, a current of 2.5 A is
passed through a cell containing a solution of copper (II) sulfate (CuSO4) for
30 minutes. If the atomic mass of copper is 63.5 g/mol, what mass of copper is
deposited at the cathode?
Solution
Step 1: Find the total charge passed through the cell. Given: Current, I= 2.5
A Time, t= 30 minutes
First convert the time to seconds:
t= 30 ×60 = 1800 seconds
Now, use the formula for charge:
Q=It
8
Q= 2.5×1800 = 4500 C
Step 2: Calculate the number of moles of copper deposition. The molar
charge of an electron is 9.65 ×10−4C/mol, so the total charge is equivalent to:
n=4500
9.65 ×10−4= 46714.85 mol
Step 3: Find the mass of copper deposited at the cathode. The molar mass
of copper is 63.5 g/mol, so the mass deposited is:
Mass = 46714.85 ×63.5 = 2964066.475 g ≈2.96 kg
Therefore, approximately 2.96 kg of copper is deposited at the cathode dur-
ing the electrolysis of a solution of copper (II) sulfate for 30 minutes with a
current of 2.5 A.
Question 9
Question
A student performed an experiment on electrolysis and found that when a cur-
rent of 2.5 A was passed through an electrolyte for 30 minutes, 0.45 g of sub-
stance was deposited at one of the electrodes. Calculate the equivalent weight
of the substance.
Given: Faraday constant = 96485 C/mol
Solution
Step 1: Find the total charge passed through the electrolyte. Given: Current,
I= 2.5 A Time, t= 30 minutes = 1800 seconds
The total charge passed, Q, is given by:
Q=I×t
Q= 2.5 A ×1800 seconds
Q= 4500 C
Step 2: Calculate the number of moles of substance deposited. Given: Mass
deposited, m= 0.45 g Molar mass of substance, M
Number of moles, n, is given by:
n=m
M
Step 3: Determine the equivalent weight of the substance. The equivalent
weight is the mass of a substance that reacts with one mole of electrons. It is
given by:
Equivalent weight = M
n=M
(m
M)=M2
m
9
Step 4: Use Faraday’s laws to find the equivalent weight. Faraday’s first
law states that the mass of substance deposited is directly proportional to the
quantity of electricity passed through the electrolyte.
We know that,
Charge of 1 mole of electrons = 1 Faraday
Charge of n moles of electrons = nFaradays
Given: Faraday constant, F= 96485 C/mol
Using Faraday’s laws:
n×F=Q
m
M×96485 = 4500
Solving for M:
M=4500 ×M
0.45 ×96485
Therefore, the equivalent weight of the substance can be calculated using
the above formula.
Question 10
Question
A copper electroplating cell operates using a copper(II) sulfate solution. In one
experiment, a current of 2.00 A is passed through the cell for 4.00 hours. During
this time, 6.35 grams of copper metal is deposited on the cathode. Calculate
the number of moles of electrons that passed through the cell.
Solution
Step 1: Calculate the number of moles of copper deposited on the cathode.
Given: Current (I) = 2.00 A Time (t) = 4.00 hours = 14400 seconds Mass of
copper deposited = 6.35 g Molar mass of copper (Cu) = 63.55 g/mol
We first convert the time to seconds:
t= 4.00 hours ×3600 seconds/hour = 14400 seconds
Next, we calculate the number of moles of copper deposited using the for-
mula:
Moles of copper = Mass of copper deposited
Molar mass of copper
Moles of copper = 6.35 g
63.55 g/mol = 0.100 mol
Step 2: Calculate the number of moles of electrons that passed through the
cell. The relationship between the number of moles of substance deposited in
10
an electroplating cell and the number of moles of electrons passed through the
cell is given by Faraday’s laws of electrolysis:
Moles of substance deposited = Q
nF
where: Q= charge passed through the cell (in coulombs) n= number of elec-
trons transferred in the reaction F= Faraday constant = 96485 C/mol
We rearrange the equation to solve for n:
n=Q
F
The total charge passed through the cell (Q) can be calculated using the
formula:
Q=It
Q= 2.00 A ×14400 s = 28800 C
Now, we can calculate the number of moles of electrons passed through the
cell:
n=28800 C
96485 C/mol = 0.298 mol
Therefore, the number of moles of electrons that passed through the cell is
0.298 mol.
Question 11
Question
A solution contains a mixture of copper (Cu2+) and silver (Ag+) ions. A current
of 2.50 A is passed through the solution for 90.0 minutes. If 0.200 g of copper is
deposited at the cathode, how many grams of silver are deposited at the cathode
during this time?
(Hint: Use Faraday’s laws of electrolysis to relate the amount of substance
deposited to the charge passed and the molar masses of copper and silver.)
Solution
Step 1: Calculate the total charge passed through the solution. Given: current,
I= 2.50 A and time, t= 90.0 minutes. We first convert time to seconds:
90.0 minutes = 90.0×60 seconds = 5400 seconds. The total charge passed,
Q=I×t= 2.50 A ×5400 s = 13500 C.
Step 2: Determine the number of moles of copper deposited. Given: mass
of copper deposited, mCu = 0.200 g and molar mass of copper, MCu = 63.55
g/mol. First, convert the mass of copper deposited to moles:
nCu =mCu
MCu
=0.200 g
63.55 g/mol ≈0.00315 mol
11
Step 3: Use Faraday’s laws to determine the number of coulombs required
to deposit 1 mole of copper. The charge required to deposit 1 mole of Cu is
given by the equation:
Q=nF
where Fis the Faraday constant (96485 C/mol). Hence, the charge required to
deposit 0.00315 moles of Cu is:
Q=nCu ×F= 0.00315 mol ×96485 C/mol ≈304 C
Step 4: Calculate the number of moles of Ag deposited using the charge
passed. Since the total charge passed is 13500 C and the charge required to
deposit 1 mole of Ag is also 96485 C, the number of moles of Ag deposited is
given by:
nAg =Q
F=13500 C
96485 C/mol ≈0.140 mol
Step 5: Determine the mass of silver deposited. Given: molar mass of silver,
MAg = 107.87 g/mol. The mass of silver deposited is:
mAg =nAg ×MAg = 0.140 mol ×107.87 g/mol ≈15.0 g
Therefore, approximately 15.0 grams of silver are deposited at the cathode
during this time.
Question 12
Question
A student is conducting an experiment to investigate Faraday’s laws of elec-
trolysis. The student passes a current of 2.50 A through a solution of silver
nitrate for 30 minutes. During this time, 2.80 grams of silver is deposited at the
cathode. Determine the molar mass of silver.
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current
(I) = 2.50 A Time (t) = 30 minutes = 1800 seconds We can use the formula
Q=It to calculate the total charge passed through the solution.
Q= 2.50 A ×1800 s = 4500 C
Step 2: Determine the number of moles of silver deposited. The total charge
passed through the circuit is used to calculate the number of moles of silver
deposited. 1 Faraday is equal to 96485 C/mol e. Convert the total charge (in
Coulombs) to moles of electrons using this relationship.
1 Faraday = 96485 C/mol e−
12
Moles of electrons = 4500 C
96485 C/mol e−≈0.0466 mol e−
Since 1 mol of silver ions (Ag) requires 1 mol of electrons to form 1 mol of silver
atoms (Ag), the number of moles of silver deposited is the same as the number
of moles of electrons.
Moles of silver = 0.0466 mol
Step 3: Calculate the molar mass of silver. Given that 2.80 grams of silver
is deposited, we can use the formula m=n×Mwhere mis the mass of silver,
nis the number of moles of silver, and Mis the molar mass of silver.
2.80 g = 0.0466 mol ×M
Solve for the molar mass of silver.
M=2.80 g
0.0466 mol ≈60.08 g/mol
Therefore, the molar mass of silver is approximately 60.08 g/mol.
Question 13
Question
A student wants to perform an electrolysis experiment using a solution of silver
nitrate (AgNO3). The student places two silver electrodes in the solution and
applies a current of 2.50 A for 30.0 minutes.
Given that the molar mass of silver is 107.87 g/mol and the Faraday constant
is 9.65×104C/mol, calculate: (i) The amount of silver deposited on the cathode
during the experiment. (ii) The amount of silver nitrate consumed during the
experiment.
Solution
Step 1: Calculate the total charge passed through the solution. The total charge
passed can be calculated using the formula:
Q=I·t
Where: - Qis the total charge (in Coulombs), - Iis the current (in Amperes),
and - tis the time (in seconds).
Given that I= 2.50 A and t= 30.0 min = 30.0×60 s = 1800.0 s, we can
substitute these values into the formula to find:
Q= 2.50 A ×1800.0 s = 4500.0 C
13
Step 2: Calculate the amount of silver deposited on the cathode. The amount
of substance deposited can be calculated using the formula:
Amount of substance = Q
n·F
Where: - Qis the total charge passed (in Coulombs) calculated in Step 1, - n
is the number of electrons involved in the reaction (1 for silver ions), and - Fis
the Faraday constant (9.65 ×104C/mol).
Since n= 1 for silver ions, we can substitute the values into the formula:
Amount of silver = 4500.0 C
1·(9.65 ×104) C/mol = 4.67 ×10−2mol
To find the mass of silver deposited on the cathode, we use the molar mass
of silver:
Mass of silver = Amount of silver ×Molar mass of silver
Mass of silver = 4.67 ×10−2mol ×107.87 g/mol ≈5.03 g
Step 3: Calculate the amount of silver nitrate consumed during the exper-
iment. The stoichiometry of the reaction shows that one mole of silver ion
requires one mole of silver nitrate. Therefore, the amount of silver nitrate con-
sumed will also be 4.67×10−2mol, which is equal to 4.67 g since the molar mass
of silver nitrate is 169.87 g/mol.
Question 14
Question
In an electrolytic cell, a current of 2.00 A is passed through a lead(II) iodide,
PbI2, solution for 3.00 hours. If 0.500 g of lead is deposited at the cathode,
determine the standard electrode potential of the lead half-cell.
Solution
Step 1: Determine the amount of charge passed through the cell. Given: Cur-
rent, I= 2.00 A Time, t= 3.00 hours = 10800 s Charge, Q=I×t
Q= 2.00 A ×10800 s = 21600 C
Step 2: Convert the amount of lead deposited to moles. Given: Mass of lead
deposited, m= 0.500 g Molar mass of lead = 207.2 g/mol
Number of moles of lead deposited, n=m
Molar mass of lead
n=0.500 g
207.2 g/mol = 0.002415 mol
14
Step 3: Determine the number of electrons involved in the reduction of
lead(II) ions to lead atoms. The overall reaction for the reduction of Pb2+ ions
to Pb atoms is:
Pb2+ + 2e−→Pb
From the reaction, it is clear that 2 moles of electrons are involved in the re-
duction of 1 mole of Pb2+ ions.
Step 4: Calculate the charge required to deposit the amount of lead. Since
2 moles of electrons are involved in the reduction of 1 mole of Pb2+ ions:
Q=n×Faraday constant ×2
Given: Faraday constant, F= 96485 C/mol
Substitute the values to find Q.
21600 C = 0.002415 mol ×96485 C/mol ×2
21600 C = 0.4651 mol ×96485 C/mol
Step 5: Calculate the standard electrode potential of the lead half-cell. The
standard electrode potential, E◦
Pb2+/Pb, can be calculated using the equation:
E◦
Pb2+/Pb =−Q
nF
Substitute the known values of Q,n, and Finto the equation to calculate
E◦
Pb2+/Pb.
E◦
Pb2+/Pb =−21600
0.002415 ×96485
E◦
Pb2+/Pb =−0.898 V
Therefore, the standard electrode potential of the lead half-cell is −0.898 V.
Question 15
Question
Consider an electrolytic cell where a current of 2.50 A is passed through a
solution of copper (II) chloride (CuCl2). If the cell operates for 2.00 hours,
what mass of copper is deposited at the cathode? (Given: Faraday’s constant,
F= 96,500 C/mol)
15
Solution
Step 1: Find the total charge passing through the cell. The total charge passed
through the cell can be calculated using the formula:
Q=I·t
where Q= total charge passed in Coulombs, I= current in Amperes, and t=
time in seconds.
Given I= 2.50 Aand t= 2.00 hours = 7200 s:
Q= 2.50 A·7200 s= 18,000 C
Step 2: Calculate the number of moles of electrons passed through the cell.
Since 1 Faraday of charge is equivalent to 1 mole of electrons (where 1 Faraday
is equal to 96,500 C/mol), the number of moles of electrons passed through the
cell can be calculated as:
moles of electrons = Q
F=18,000 C
96,500 C/mol = 0.186 mol
Step 3: Determine the mass of copper deposited at the cathode. The sto-
ichiometry of the reaction can be used to relate the moles of electrons to the
moles of copper deposited. From the balanced equation for the reduction of
copper (II) ions:
Cu2+ + 2e−→Cu
we see that 2 moles of electrons are required to reduce 1 mole of Cu2+ to copper.
Therefore, the moles of copper deposited will be half the moles of electrons
passed through the cell:
moles of copper = 0.186 mol
2= 0.093 mol
Step 4: Calculate the mass of copper deposited. The molar mass of copper
is approximately 63.5 g/mol. Therefore, the mass of copper deposited at the
cathode can be calculated as:
mass of copper = moles of copper×molar mass of copper = 0.093 mol×63.5g/mol = 5.91 g
Therefore, 5.91 grams of copper will be deposited at the cathode after 2
hours of electrolysis.
Question 16
Question
State Faraday’s laws of electrolysis and discuss how they can be applied to
determine the quantity of a substance deposited or liberated during electrolysis.
16
Solution
Faraday’s laws of electrolysis describe the fundamental relationships between the
amount of substance produced during electrolysis, the current passing through
the electrolyte, the time the current is applied, and the molar mass of the
substance being produced.
Faraday’s First Law: The amount of a substance produced at an electrode
during electrolysis is directly proportional to the quantity of electric charge (in
Coulombs) passed through the electrolyte.
Faraday’s Second Law: The amounts of different substances produced by
the same quantity of electric charge passing through the electrolyte are directly
proportional to their chemical equivalent weights.
To determine the quantity of a substance deposited or liberated during elec-
trolysis, the following steps can be followed:
Step 1: Identify the half-reaction occurring at each electrode and determine
the number of moles of electrons (n) involved in the reaction.
Step 2: Calculate the total charge passed through the electrolyte using the
formula: Q=I×t, where Qis the charge in Coulombs, Iis the current in
Amperes, and tis the time in seconds.
Step 3: Determine the Faraday constant (F) using the equation: F=
96,485 Coulombs per equivalent.
Step 4: Calculate the amount of substance produced using the formula:
n=Q
n×F, where nis the number of moles of electrons, Qis the charge passed
in Coulombs, and Fis the Faraday constant.
Step 5: Convert the number of moles to grams using the molar mass of the
substance.
By following these steps and applying Faraday’s laws of electrolysis, one can
accurately determine the quantity of a substance deposited or liberated during
electrolysis.
Question 17
Question
A student is conducting an experiment on electrolysis using a sulfuric acid so-
lution. The student passes a current of 2.5 A through the solution for 3 hours.
During this time, hydrogen gas is produced at one electrode and oxygen gas at
the other electrode.
What mass of hydrogen gas is produced during the electrolysis process?
(Assume standard conditions.)
Solution
To find the mass of hydrogen gas produced during electrolysis, we first need to
calculate the total charge passed through the solution using Faraday’s laws of
17
electrolysis. We can then use the ideal gas law to convert the charge passed to
the mass of hydrogen gas produced.
Step 1: Calculate the total charge passed Given: Current, I= 2.5 A
Time, t= 3 hours = 10800 s Faraday’s constant, F= 96500 C/mol
The total charge passed through the solution is given by the formula:
Q=I×t
Substitute the given values to find Q:
Q= 2.5 A ×10800 s = 27000 C
Step 2: Calculate the number of moles of hydrogen gas produced
The number of moles of electrons required to produce one mole of hydrogen gas
is 2 according to the balanced electrolysis equation for water.
Since 1 Faraday of charge liberates 1 mole of electrons, the number of moles
of electrons liberated is given by:
Moles of electrons = Q
F
Substitute Qand Finto the formula to find the moles of electrons:
Moles of electrons = 27000 C
96500 C/mol ≈0.28 mol
Since 2 moles of electrons are required to produce 1 mole of hydrogen gas,
the number of moles of hydrogen gas produced is half of the moles of electrons
liberated:
Moles of hydrogen gas = 1
2×0.28 mol = 0.14 mol
Step 3: Calculate the mass of hydrogen gas produced The ideal gas
law relates the number of moles of a gas to its mass:
P V =nRT
For hydrogen gas at standard conditions, P= 1 atm, V= 22.4 L/mol, R=
0.0821 L-atm/mol-K, and T= 273 K.
The mass of 1 mole of hydrogen gas at standard conditions can be calculated
using its molar mass:
Molar mass of hydrogen = 2×atomic mass of hydrogen = 2×1 g/mol = 2 g/mol
Now, we can calculate the mass of hydrogen gas produced:
Mass of hydrogen gas = Moles of hydrogen gas ×Molar mass of hydrogen
Mass of hydrogen gas = 0.14 mol ×2 g/mol = 0.28 g
Therefore, approximately 0.28 g of hydrogen gas is produced during the
electrolysis process.
18
Question 18
Question
In an electrolytic cell, a constant current of 3.00 A is passed through a solution
of molten lead (II) bromide (PbBr2) for 45.0 minutes. Calculate the mass of lead
deposited at the cathode. Given: molar mass of lead = 207.2 g/mol, Faraday’s
constant = 96,485 C/mol.
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current (I)
= 3.00 A, Time (t) = 45.0 minutes = 45.0 minutes ×60 seconds/min = 2700
s, Faraday’s constant (F) = 96,485 C/mol.
The total charge (Q) passed through the cell can be calculated using the
formula:
Q=I×t
Q= 3.00 A ×2700 s = 8100 C
Step 2: Calculate the number of moles of electrons (n) transferred. Since 1
mole of electrons is equivalent to 1 Faraday, we can use the formula:
n=Q
F
n=8100 C
96,485 C/mol ≈0.0839 mol
Step 3: Determine the number of moles of lead deposited. From the balanced
chemical equation for the electrolysis of lead (II) bromide:
Pb2+ + 2e−→Pb(s)
It is evident that 2 moles of electrons are required to deposit 1 mole of lead.
Therefore, the number of moles of lead deposited (nPb) is given by:
nPb =n
2
nPb =0.0839 mol
2= 0.0419 mol
Step 4: Calculate the mass of lead deposited at the cathode. Given: Molar
mass of lead (Pb) = 207.2 g/mol, Number of moles of lead deposited (nPb) =
0.0419 mol.
The mass of lead deposited (m) can be calculated using the formula:
m=nPb ×Molar mass of Pb
m= 0.0419 mol ×207.2 g/mol = 8.68 g
Therefore, the mass of lead deposited at the cathode is 8.68 g.
19
Question 19
Question
In an electrolysis experiment, a constant current of 2.5 A is passed through a
solution of copper sulfate (CuSO4) for 45 minutes. If a mass of 3.2 g of copper
is deposited at the cathode, determine the number of moles of electrons that
were transferred during the electrolysis.
Solution
Step 1: Calculate the total charge transferred during the electrolysis. Given
that the current = 2.5 A and time = 45 minutes = 2700 seconds, we can use
the formula Q=I×t.
Q= 2.5 A ×2700 s = 6750 C
Step 2: Calculate the number of moles of copper deposited at the cathode.
Given that the molar mass of copper (Cu) is 63.55 g/mol, we can use the formula
moles = mass
molar mass .
moles = 3.2 g
63.55 g/mol ≈0.05 mol
Step 3: Determine the number of moles of electrons that were transferred
during the electrolysis. From the balanced equation for the electrolysis of copper
sulfate: Cu2+ + 2e−→Cu we can see that 2 moles of electrons are required to
deposit 1 mol of copper. Therefore, the number of moles of electrons transferred
is twice the number of moles of copper deposited.
moles of electrons = 2 ×0.05 = 0.1 mol
Hence, the number of moles of electrons that were transferred during the
electrolysis is 0.1 mol.
Question 20
Question
State Faraday’s laws of electrolysis and explain how they are related to each
other.
Solution
Faraday’s laws of electrolysis are fundamental principles in electrochemistry
that describe the relationship between the amount of substance produced or
consumed during electrolysis and the amount of electric charge passed through
the electrolyte. There are two laws:
20
Faraday’s First Law: The amount of substance produced at an electrode
during electrolysis is directly proportional to the quantity of electricity passed
through the cell.
Faraday’s Second Law: The amounts of different substances produced or
consumed by the same quantity of electricity are in the ratio of their equivalent
weights.
Step 1: Faraday’s First Law states that the mass of a substance (m) pro-
duced at an electrode during electrolysis is directly proportional to the amount
of electric charge (Q) passed through the cell. Mathematically, this can be
expressed as:
m∝Q
m=k·Q
where k is a constant of proportionality known as the electrochemical equivalent.
Step 2: Faraday’s Second Law relates the amounts of different substances
produced or consumed during electrolysis to their equivalent weights. The equiv-
alent weight of a substance is the mass that reacts with or produces one mole
of either electrons or protons.
Step 3: By combining Faraday’s First and Second Laws, we get the relation-
ship between the amount of substance produced or consumed during electrolysis
and the quantity of electricity passed through the cell. This relationship is given
by:
m1/m2=E1/E2
where: - m1 and m2 are the masses of substances produced or consumed, - E1
and E2 are the equivalent weights of the substances.
Therefore, Faraday’s laws of electrolysis provide a comprehensive under-
standing of the relationship between the quantities of substances involved in
electrolysis and the electric charge passed through the cell.
Question 21
Question
A solution containing aqueous sodium chloride, NaCl(aq), is electrolyzed using
inert electrodes. During the electrolysis, 3.50 g of chlorine gas is produced at the
anode. Calculate the volume of hydrogen gas produced at the cathode (standard
conditions) by passing the same amount of electricity through the solution.
(Hint: Start by determining the number of moles of chlorine gas produced.)
Solution
Step 1: Calculate the number of moles of chlorine gas produced.
Given: Mass of chlorine gas produced = 3.50 g Molar mass of Cl2= 70.906
g/mol
Number of moles of Cl2=3.50 g
70.906 g/mol
21
Step 2: Determine the number of moles of electrons transferred during the
electrolysis of NaCl.
Each molecule of Cl2is produced by the transfer of 2 electrons. So, the number
of moles of electrons transferred can be calculated using the formula:
Number of moles of electrons = Number of moles of Cl2×2
2
Step 3: Calculate the number of moles of hydrogen gas produced.
From the number of moles of electrons transferred, we can find the number of
moles of H2produced since 2 moles of electrons are required to produce 1 mole
of H2.
Number of moles of H2= Number of moles of electrons ×1
2
Step 4: Convert the number of moles of hydrogen gas to volume at standard
conditions.
At standard conditions, 1 mole of any gas occupies 22.4 L. Hence, the volume
of H2gas produced can be calculated as follows:
Volume of H2gas (at STP) = Number of moles of H2×22.4 L/mol
Now, follow these steps to find the final answer.
Question 22
Question
A student performs an electrolysis experiment using a copper(II) sulfate solu-
tion. The student connects a copper electrode to the positive terminal of a
power supply and a platinum electrode to the negative terminal. During the ex-
periment, the student observes that the mass of the copper electrode decreases
over time.
Explain this observation using Faraday’s laws of electrolysis.
Solution
To explain the student’s observation using Faraday’s laws of electrolysis, we
need to consider the reactions occurring at the electrodes.
Step 1: Reactions at the electrodes At the anode (copper electrode
connected to the positive terminal), copper(II) ions will be reduced:
Cu2+ + 2e−→Cu
At the cathode (platinum electrode connected to the negative terminal),
water will be oxidized:
2H2O→O2+ 4H++ 4e−
Step 2: Application of Faraday’s First Law Faraday’s First Law states
that the mass of a substance deposited or liberated at an electrode during elec-
trolysis is directly proportional to the quantity of electricity passed through the
cell.
22
Since the copper electrode is losing mass over time, it indicates that more
copper ions are being reduced (deposited as solid copper) at the electrode than
the amount of copper dissolving into the solution.
Step 3: Explanation The reason for the decrease in mass of the copper
electrode is that the reduction of copper(II) ions at the anode is consuming
copper from the electrode to form solid copper. This results in the copper
electrode losing mass over time as more copper(II) ions are reduced to solid
copper than the amount of copper dissolving into the solution.
Therefore, the observation of the copper electrode losing mass during the
experiment can be explained by Faraday’s laws of electrolysis.
Question 23
Question
An electrolytic cell contains a solution of silver nitrate (AgNO3). If a current
of 5.00 A is passed through the cell for 2.00 hours, what mass of pure silver is
deposited on the cathode? (Given: Atomic masses: Ag = 107.87 g/mol, N =
14.01 g/mol, O = 16.00 g/mol)
Solution
Step 1: Calculate the total charge passed through the cell. Step 2: Determine
the number of moles of silver deposited. Step 3: Convert the moles of silver to
mass in grams.
Step 1: Calculate the total charge passed through the cell. The total charge
(Q) can be calculated using the formula:
Q=I×t
where Iis the current (5.00 A) and tis the time in seconds (2.00 hours = 7200
s).
Q= 5.00A×7200s= 36000C
Step 2: Determine the number of moles of silver deposited. The Faraday’s
constant, F, is 96485 C/mol. The number of moles of silver deposited can be
calculated by dividing the total charge by the Faraday’s constant:
n=Q
F=36000C
96485C/mol ≈0.373mol
Step 3: Convert the moles of silver to mass in grams. The molar mass
of silver is 107.87 g/mol. The mass of silver deposited can be calculated by
multiplying the number of moles by the molar mass:
Mass of silver = 0.373mol ×107.87g/mol ≈40.24g
Therefore, approximately 40.24 grams of pure silver is deposited on the cath-
ode.
23
Question 24
Question
An aqueous solution of silver nitrate is electrolyzed using a current of 2.50 A
for 4.00 hours. If 25.0 g of pure silver is deposited at the cathode, determine
the number of moles of silver ions reduced and calculate the Faraday constant.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given: Cur-
rent, I= 2.50 A Time, t= 4.00 hours
The total charge passed, Q, can be calculated using the formula:
Q=I×t
Let’s first convert the time from hours to seconds:
t= 4.00 ×3600 = 14400 seconds
Therefore,
Q= 2.50 ×14400 = 36000 C
Step 2: Calculate the number of moles of silver ions reduced. Given: Mass
of silver deposited, m= 25.0 g Molar mass of silver, MAg = 107.87 g/mol
The number of moles of silver deposited, n, can be calculated using the
formula:
n=m
MAg
Substitute the given values to find n:
n=25.0
107.87 ≈0.232 mol
Step 3: Calculate the number of moles of electrons transferred. From the
balanced half-reaction for the reduction of silver ions:
2e−+ Ag+→Ag
We can see that 2 moles of electrons are required to reduce 1 mole of Ag+.
Therefore, the number of moles of electrons transferred, ne, is given by:
ne= 2n
Substitute the value of ncalculated earlier to find ne:
ne= 2 ×0.232 = 0.464 mol
Step 4: Calculate the Faraday constant. The Faraday constant, F, represents
the charge of 1 mole of electrons, and can be calculated using the formula:
F=Q
ne×96485
24
Substitute the given values to find F:
F=36000
0.464 ×96485 ≈7.20 ×104C/mol
Therefore, the Faraday constant is approximately 7.20 ×104C/mol.
Question 25
Question
An aqueous solution of sodium chloride is electrolyzed using platinum electrodes.
If a current of 3.00 A is passed through the solution for 1 hour, calculate the
mass of sodium metal deposited on the cathode. (Given: Atomic masses - Na
= 23 g/mol, Cl = 35.5 g/mol, Faraday constant = 9.65 ×104C/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution.
Given: Current, I= 3.00 A Time, t= 1 hour = 3600 s Using the formula
Q=I·t, where Qis the total charge passed, Q= 3.00 A ×3600 s Q= 10800 C
Step 2: Calculate the number of moles of electrons passed through the so-
lution. Given: Faraday constant, F= 9.65 ×104C/mol Using the formula
n=Q
F, where nis the number of moles of electrons passed, n=10800 C
9.65×104C/mol
n= 0.112 mol
Step 3: Determine the stoichiometry of the reaction at the cathode. The
balanced half-reaction for the reduction of sodium ions to sodium metal is:
2 Na++ 2 e−→2 Na
Step 4: Calculate the mass of sodium metal deposited on the cathode. Given:
Molar mass of Na = 23 g/mol Using the formula m=n×Molar mass, where
mis the mass of sodium deposited, m= 0.112 mol ×23 g/mol m= 2.57 g
Therefore, the mass of sodium metal deposited on the cathode after passing
a current of 3.00 A for 1 hour is 2.57 g.
Question 26
Question
A copper sulfate solution is electrolyzed using copper electrodes. If a current
of 2.5 A is passed through the solution for 30 minutes, calculate the mass of
copper deposited on one electrode. The molar mass of copper is 63.5 g/mol.
Solution
Step 1: Find the charge passed through the solution using the formula:
Charge (C) = Current (A) ×Time (s)
25
Charge = 2.5 A ×(30 ×60) s = 4500 C
Step 2: Using Faraday’s first law of electrolysis, we know that 1 mole of
electrons carries a charge of 1 Faraday, which is 96500 C/mol. Therefore, the
number of moles of electrons passed through the solution is:
Moles of electrons = Total charge
Charge per mole of electrons
Moles of electrons = 4500 C
96500 C/mol ≈0.0467 mol
Step 3: Since the reaction is Cu2+ + 2e−→Cu, we need 2 moles of elec-
trons to deposit 1 mole of copper. Therefore, moles of copper deposited on one
electrode is:
Moles of copper = 0.0467 mol
2= 0.0234 mol
Step 4: Calculate the mass of copper deposited using the molar mass of
copper:
Mass of copper (g) = Moles of copper ×Molar mass of copper
Mass of copper = 0.0234 mol ×63.5 g/mol ≈1.49 g
Therefore, the mass of copper deposited on one electrode is approximately
1.49 grams.
Question 27
Question
In an electrolytic cell, a student passes a current of 2.50 A through a solution
of copper(II) sulfate, CuSO4, for 1.50 hours. During this time, the student
observes that 6.35 grams of copper is deposited at the cathode.
Calculate the number of moles of electrons that are required to reduce 1
mole of Cu2+ ions to copper metal.
Solution
Step 1: Find the charge required to reduce 1 mole of Cu2+ ions to copper metal.
Given:
Current, I= 2.50 A
Time, t= 1.50 hours = 5400 s
Atomic weight of copper, mCu = 63.55 g/mol
1 faraday, 1F= 96,485 C
26
Charge required to reduce 1 mole of Cu2+ ions to copper metal:
Q= Current ×Time
Q=I×t= 2.50 A ×5400 s = 13500 C
Step 2: Calculate the number of moles of electrons required to reduce 1
mole of Cu2+ ions. The number of moles of electrons can be calculated using
Faraday’s law:
Q=n×F
n=Q
F
Substitute the values:
n=13500 C
96485 C/mol ≈0.140 mol
So, approximately 0.140 moles of electrons are required to reduce 1 mole of
Cu2+ ions to copper metal.
Question 28
Question
During the electrolysis of copper(II) sulfate solution using copper electrodes, a
current of 2 A is passed for 30 minutes. If the Faraday constant is 96500 C/mol,
calculate the mass of copper deposited on the cathode.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given current
I= 2 Aand time t= 30 min = 30 ×60 s= 1800 s.
Q=I×t= 2 ×1800 = 3600 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday is equivalent to 96500 C/mol.
moles of electrons = Q
96500 =3600
96500 mol
Step 3: Calculate the mass of copper deposited. The balanced equation for
the electrolysis of copper(II) sulfate solution is:
Cu2+(aq)+2e−→Cu(s)
27
One mole of copper is deposited for every 2 moles of electrons.
moles of copper = 1
2×moles of electrons = 1
2×3600
96500 mol
Step 4: Calculate the mass of copper using the molar mass of copper. The
molar mass of copper is approximately 63.5g/mol.
mass of copper = moles of copper ×molar mass of copper = 1
2×3600
96500 ×63.5g
mass of copper ≈1
2×3600
96500 ×63.5≈0.746 g
Therefore, the mass of copper deposited on the cathode is approximately
0.746 grams.
Question 29
Question
A student is performing an electrolysis experiment using a solution of copper(II)
sulfate, CuSO4, using a current of 2.5 A. If the student deposits 1.25 g of copper
at the cathode in 30 minutes, calculate the Faraday constant, F.
Solution
Step 1: Find the number of moles of copper deposited at the cathode. Given
that the molar mass of copper is 63.55 g/mol, the number of moles of copper
deposited can be calculated as follows:
Moles of Cu = Mass of Cu
Molar mass of Cu =1.25 g
63.55 g/mol
Step 2: Calculate the charge transferred. Since 1 mole of copper corresponds to
2 moles of electrons in the electrolytic cell, the total charge transferred can be
calculated using Faraday’s laws of electrolysis:
Charge transferred = Moles of Cu ×2×F
where Fis the Faraday constant. Step 3: Determine the time of electrolysis.
Given that the current is 2.5 A and the time is 30 minutes, convert the time
into seconds:
Time = 30 ×60 s
Step 4: Calculate the charge transferred using the formula:
Charge transferred = Current ×Time
Step 5: Equate the two expressions for the charge transferred and solve for F.
Equate the two expressions for the charge transferred obtained in Step 2 and
Step 4, and solve for F. Remember to convert the time into seconds.
Moles of Cu ×2×F= Current ×Time
28
Question 30
Question
An electrolysis experiment is set up using a copper(II) sulfate solution. A cur-
rent of 2.5 A is passed through the solution for 30 minutes. Calculate the mass of
copper deposited on the cathode during this time. (Given: Faraday’s constant
= 96500 C mol−1, atomic mass of copper = 63.5 g mol−1)
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current,
I= 2.5 A Time, t= 30 minutes = 30 ×60 s = 1800 s
Total charge, Q=It Q = 2.5 A ×1800 s = 4500 C
Step 2: Calculate the number of moles of electrons passed. Given: Faraday’s
constant, F= 96500 C mol−1
Number of moles of electrons, n=Q
Fn=4500 C
96500 C mol−1≈0.0467 mol
Step 3: Calculate the mass of copper deposited. Given: Atomic mass of
copper, M= 63.5 g mol−1
Mass of copper deposited, m=n×M m = 0.0467 mol×63.5 g mol−1≈2.97 g
Therefore, the mass of copper deposited on the cathode during this time is
approximately 2.97 g.
Question 31
Question
In an electrolytic cell, a current of 2.50 A is passed through a solution of iron(II)
chloride for 3.00 hours. During this time, 1.20 g of iron is deposited at the
cathode. Determine the oxidation state of iron in the product and write the
balanced half-reaction occurring at the cathode.
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current,
I= 2.50 A Time, t= 3.00 hours
The total charge passed, Q, can be calculated as:
Q=I×t
Q= 2.50 A ×3.00 h ×3600 s/h
Q= 2.50 A ×10800 s
Q= 27000 C
Step 2: Calculate the number of moles of iron deposited at the cathode.
Given: Mass of iron deposited, m= 1.20 g Molar mass of iron, M= 55.85
g/mol
29
The number of moles of iron deposited, n, can be calculated as:
n=m
M
n=1.20 g
55.85 g/mol
n≈0.0215 mol
Step 3: Determine the oxidation state of iron in the product. Each mole
of deposited iron corresponds to 2 moles of electrons transferred in the half-
reaction. Therefore, the oxidation state of iron in the product is +2.
Step 4: Write the balanced half-reaction occurring at the cathode. The half-
reaction at the cathode involves the reduction of iron(II) ions to form elemental
iron:
Fe2+ + 2e−→Fe
Therefore, the balanced half-reaction at the cathode is:
Fe2+ + 2e−→Fe
Question 32
Question
A student is conducting an electrolysis experiment using a solution of copper(II)
sulfate. The student passes a current of 2.50 A through the solution for 3.00
hours. During this time, copper is deposited on one electrode. If the student
measures that 5.00 g of copper is deposited, determine the number of moles of
electrons that pass through the cell during the experiment. Assume that the
only reaction occurring is the reduction of Cu2+ to Cu.
Solution
Step 1: Calculate the charge passed through the cell. The charge passed through
the cell can be calculated using the formula:
Q=I·t
where Qis the charge in coulombs, Iis the current in amperes, and tis the time
in seconds. Given that I= 2.50 A and t= 3.00 hours, we first convert hours to
seconds:
t= 3.00 ×3600 = 10800 s
Now, we can calculate the charge passed through the cell:
Q= 2.50 A ×10800 s = 27000 C
30
Step 2: Calculate the number of moles of electrons. The number of moles of
electrons can be calculated using Faraday’s law of electrolysis:
mol e−=Q
nF
where Qis the charge passed through the cell in coulombs, nis the number of
moles of electrons, and Fis the Faraday constant, 96485 C/mol. Substitute the
values to find the number of moles of electrons:
n=27000 C
96485 C/mol = 0.280 mol e−
Therefore, the number of moles of electrons that passed through the cell during
the experiment is 0.280 mol.
Question 33
Question
Explain Faraday’s laws of electrolysis and how they are applied in electrochem-
ical processes.
Solution
Faraday’s First Law states that the amount of a substance produced at an
electrode during electrolysis is directly proportional to the quantity of electricity
passed through the electrolyte. This can be expressed as:
w=Zq
where: - wis the amount of substance produced, - Zis the electrochemical
equivalent of the substance (the mass of the substance produced by one coulomb
of charge), and - qis the total charge passed through the electrolyte.
Faraday’s Second Law states that the amounts of different substances
produced by the same quantity of electricity are proportional to the equivalent
weights of these substances. Mathematically, it can be represented as:
w
Z=m
E
where: - wis the amount of substance produced, - mis the mass of the sub-
stance, - Eis the equivalent weight of the substance, - Zis the electrochemical
equivalent, and - qis the total charge passed through the electrolyte.
These laws are crucial in understanding and predicting the outcomes of var-
ious electrochemical processes, such as electroplating, electrolysis of water, and
corrosion prevention, among others.
31
Question 34
Question
A student is performing an electrolysis experiment using a solution of silver
nitrate (AgNO3). The student passes a current of 2.5 A through the solution
for 30 minutes. During this time, silver metal is deposited on the cathode.
Given that the standard electrode potential of the silver/silver ion (Ag+/Ag)
half-cell is +0.80 V, calculate the mass of silver deposited on the cathode.
Solution
Step 1: Calculate the total charge passed through the electrolytic cell. The total
charge passed (Q) is given by the formula:
Q=I×t
where Iis the current (in amperes) and tis the time (in seconds). Given that
I= 2.5 A and t= 30 minutes = 30 ×60 s, we have:
Q= 2.5×30 ×60 = 4500 C
Step 2: Calculate the number of moles of electrons passed. Since silver
ions (Ag+) gain one electron to form silver metal (Ag) during electrolysis, the
number of moles of silver deposited is equal to the number of moles of electrons
passed through the cell. This can be calculated using Faraday’s first law of
electrolysis:
n=Q
F
where nis the number of moles of electrons, Qis the total charge passed, and F
is the Faraday constant (96485 C/mol). Substitute Q= 4500 C and F= 96485
C/mol into the formula to get:
n=4500
96485 ≈0.0467 mol
Step 3: Calculate the mass of silver deposited. The molar mass of silver is
107.87 g/mol. Therefore, the mass of silver deposited (m) can be calculated
using:
m=n×molar mass
Substitute n= 0.0467 mol into the formula:
m= 0.0467 ×107.87 ≈5.04 g
Therefore, the mass of silver deposited on the cathode is approximately 5.04
grams.
32
Step 2: Calculate the number of moles of electrons that passed through the
cell.
1 F = 96500 Coulombs/mol
Number of moles of electrons = Q
96500 =18000
96500 = 0.186 mol
Step 3: Determine the number of moles of copper deposited at the cathode.
Since the reaction at the cathode is Cu2+ + 2e−→Cu, 1 mole of copper is
deposited by 2 moles of electrons.
Number of moles of copper = 0.186 mol
2= 0.093 mol
Step 4: Calculate the mass of copper deposited at the cathode. Given: Molar
mass of copper (Cu) = 63.5 g/mol
Mass of copper = 0.093 mol ×63.5 g/mol = 5.9055 g
Therefore, the mass of copper deposited at the cathode is 5.9055 grams.
Question 4
Question
A student is performing an electrolysis experiment with a solution of sodium
chloride (NaCl). If a current of 2.5 A is passed through the solution for 30
minutes, what mass of sodium metal is deposited at the cathode? (Given:
molar mass of sodium = 22.99 g/mol, Faraday’s constant = 96485 C/mol)
Solution
Step 1: Find the total charge passed through the solution. Given: Current,
I= 2.5ATime, t= 30 min = 30 ×60 sec = 1800 sec
The total charge passed, Q, can be calculated using the formula:
Q=I×t
Q= 2.5A×1800 sec
Q= 4500 C
Step 2: Calculate the number of moles of electrons passed through the so-
lution. Since 1 F (Faraday) is equal to 96485 C/mol, the number of moles of
electrons, n, can be calculated as:
n=Q
96485
n=4500
96485
4
n≈0.0466 mol
Step 3: Use the stoichiometry of the reaction to find the mass of sodium
deposited. From the balanced chemical equation for the electrolysis of sodium
chloride:
2 Na++ 2 Cl−→2 Na + Cl2
It is clear that 2 moles of electrons are required to deposit 2 moles of sodium
metal. So the number of moles of sodium metal deposited, nNa, is equal to half
of the number of moles of electrons, nNa = 0.5n.
nNa = 0.5×0.0466
nNa = 0.0233 mol
Step 4: Calculate the mass of sodium metal deposited. The molar mass of
sodium is given as 22.99 g/mol.
Mass of sodium = nNa ×Molar mass of sodium
Mass of sodium = 0.0233 mol ×22.99 g/mol
Mass of sodium ≈0.535 g
Therefore, approximately 0.535 grams of sodium metal is deposited at the
cathode after passing a current of 2.5 A for 30 minutes.
Question 5
Question
An electrochemical cell contains a solution of silver nitrate (AgNO3) with silver
electrodes. If a constant current of 2.5 A is passed through the cell for 30
minutes, what mass of silver is deposited on each electrode? (Given: Faraday
constant, F= 96500 C/mol; molar mass of silver, M= 107.87 g/mol)
Solution
Step 1: Calculate the total charge passing through the cell. The total charge
passing through the cell can be calculated using the formula:
Q=I×t
where Qis the total charge in Coulombs, Iis the current in Amperes, and tis
the time in seconds.
Given that I= 2.5 A and t= 30 minutes, we need to convert the time to
seconds:
t= 30 ×60 = 1800 s
Therefore,
Q= 2.5 A ×1800 s = 4500 C
5
Step 2: Calculate the number of moles of silver deposited. The amount of
silver deposited is related to the charge passed through the cell by Faraday’s
laws of electrolysis. The formula to calculate the number of moles of a substance
deposited is:
moles = Q
nF
where nis the number of electrons transferred during the reaction, and Fis the
Faraday constant.
Since the reaction involves the deposition of silver (Ag+receiving one elec-
tron), n= 1.
Substitute Q= 4500 C, n= 1, and F= 96500 C/mol into the formula:
moles = 4500 C
1×96500 C/mol ≈0.0468 mol
Step 3: Calculate the mass of silver deposited on each electrode. The mass
of silver can be calculated using the formula:
mass = moles ×M
where Mis the molar mass of silver.
Substitute moles = 0.0468 mol and M= 107.87 g/mol into the formula:
mass = 0.0468 mol ×107.87 g/mol ≈5.05 g
Therefore, approximately 5.05 g of silver is deposited on each electrode.
Question 6
Question
A solution of copper sulfate (CuSO4) was electrolyzed using platinum elec-
trodes. If a current of 2.5 A was passed through the solution for 3 hours,
what mass of copper would be deposited on the cathode? (Atomic masses:
Cu = 63.5g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution.
Given: Current, I= 2.5ATime, t= 3 hours = 3×3600 seconds = 10800 seconds
The total charge, Q, passed through the electrolyte solution can be calculated
using the formula:
Q=I×t
Substitute the given values:
Q= 2.5A×10800 s= 27000 C
6
Step 2: Determine the number of moles of copper deposited on the cathode.
The reaction at the cathode can be represented as:
Cu2+ + 2e−→Cu
From the reaction, it can be observed that 2 moles of electrons are required
to deposit 1 mole of copper.
Given: Total charge, Q= 27000 CCharge per mole of electrons, e= 1.6×
10−19 C(charge of 1 electron)
The number of moles of electrons, n, passed during electrolysis can be cal-
culated as:
n=Q
e
Substitute the given values:
n=27000 C
1.6×10−19 C≈1.69 ×1020 electrons
The number of moles of copper deposited can be calculated as:
Number of moles of copper = 1
2×n
NA
Where NAis Avogadro’s number (6.022 ×1023 mol−1).
Substitute the values and calculate:
Number of moles of copper = 1
2×1.69 ×1020
6.022 ×1023 ≈1.41 ×10−4mol
Step 3: Calculate the mass of copper deposited on the cathode. The mass
of copper deposited can be calculated using the formula:
Mass = Number of moles ×Molar mass
Substitute the values:
Mass = 1.41 ×10−4mol ×63.5g/mol ≈8.94 ×10−3g
Therefore, approximately 8.94 mg of copper would be deposited on the cath-
ode.
Question 7
Question
A student is performing an electrolysis experiment where a current of 2.5 A is
passed through an electrolytic cell containing a solution of copper(II) sulfate.
If the student was able to deposit 12.5 g of copper metal at the cathode after
20 minutes, what is the efficiency of the process? (Given: Faraday’s constant =
96500 C/mol, atomic mass of copper = 63.5 g/mol)
7
Solution
Step 1: Calculate the charge passing through the electrolytic cell. Given current
I= 2.5 A and time t= 20 min = 20
60 h = 1
3h. The charge Qpassing through the
cell can be calculated using the formula Q=It.
Q= 2.5 A ×1
3h = 2.5
3C = 5
6C
Step 2: Determine the number of moles of copper deposited. To calculate
the number of moles of copper deposited, we will use the formula:
moles = mass
molar mass =12.5 g
63.5 g/mol = 0.197 mol
Step 3: Calculate the efficiency of the process. The theoretical amount
of charge required to deposit 1 mole of copper is given by Faraday’s constant
F= 96500 C/mol. Therefore, the charge required to deposit 0.197 moles of
copper is:
Charge = 0.197 mol ×96500 C/mol = 19058.5 C
The efficiency of the process can be calculated using the formula:
Efficiency = actual charge
theoretical charge ×100% = 5/6
19058.5×100% ≈0.0026%
Therefore, the efficiency of the process is approximately 0.0026
Question 8
Question
Faraday’s laws of electrolysis state that the mass of a substance produced at an
electrode during electrolysis is directly proportional to the quantity of electricity
passed through the cell. In an electrolysis experiment, a current of 2.5 A is
passed through a cell containing a solution of copper (II) sulfate (CuSO4) for
30 minutes. If the atomic mass of copper is 63.5 g/mol, what mass of copper is
deposited at the cathode?
Solution
Step 1: Find the total charge passed through the cell. Given: Current, I= 2.5
A Time, t= 30 minutes
First convert the time to seconds:
t= 30 ×60 = 1800 seconds
Now, use the formula for charge:
Q=It
8
Q= 2.5×1800 = 4500 C
Step 2: Calculate the number of moles of copper deposition. The molar
charge of an electron is 9.65 ×10−4C/mol, so the total charge is equivalent to:
n=4500
9.65 ×10−4= 46714.85 mol
Step 3: Find the mass of copper deposited at the cathode. The molar mass
of copper is 63.5 g/mol, so the mass deposited is:
Mass = 46714.85 ×63.5 = 2964066.475 g ≈2.96 kg
Therefore, approximately 2.96 kg of copper is deposited at the cathode dur-
ing the electrolysis of a solution of copper (II) sulfate for 30 minutes with a
current of 2.5 A.
Question 9
Question
A student performed an experiment on electrolysis and found that when a cur-
rent of 2.5 A was passed through an electrolyte for 30 minutes, 0.45 g of sub-
stance was deposited at one of the electrodes. Calculate the equivalent weight
of the substance.
Given: Faraday constant = 96485 C/mol
Solution
Step 1: Find the total charge passed through the electrolyte. Given: Current,
I= 2.5 A Time, t= 30 minutes = 1800 seconds
The total charge passed, Q, is given by:
Q=I×t
Q= 2.5 A ×1800 seconds
Q= 4500 C
Step 2: Calculate the number of moles of substance deposited. Given: Mass
deposited, m= 0.45 g Molar mass of substance, M
Number of moles, n, is given by:
n=m
M
Step 3: Determine the equivalent weight of the substance. The equivalent
weight is the mass of a substance that reacts with one mole of electrons. It is
given by:
Equivalent weight = M
n=M
(m
M)=M2
m
9
Step 4: Use Faraday’s laws to find the equivalent weight. Faraday’s first
law states that the mass of substance deposited is directly proportional to the
quantity of electricity passed through the electrolyte.
We know that,
Charge of 1 mole of electrons = 1 Faraday
Charge of n moles of electrons = nFaradays
Given: Faraday constant, F= 96485 C/mol
Using Faraday’s laws:
n×F=Q
m
M×96485 = 4500
Solving for M:
M=4500 ×M
0.45 ×96485
Therefore, the equivalent weight of the substance can be calculated using
the above formula.
Question 10
Question
A copper electroplating cell operates using a copper(II) sulfate solution. In one
experiment, a current of 2.00 A is passed through the cell for 4.00 hours. During
this time, 6.35 grams of copper metal is deposited on the cathode. Calculate
the number of moles of electrons that passed through the cell.
Solution
Step 1: Calculate the number of moles of copper deposited on the cathode.
Given: Current (I) = 2.00 A Time (t) = 4.00 hours = 14400 seconds Mass of
copper deposited = 6.35 g Molar mass of copper (Cu) = 63.55 g/mol
We first convert the time to seconds:
t= 4.00 hours ×3600 seconds/hour = 14400 seconds
Next, we calculate the number of moles of copper deposited using the for-
mula:
Moles of copper = Mass of copper deposited
Molar mass of copper
Moles of copper = 6.35 g
63.55 g/mol = 0.100 mol
Step 2: Calculate the number of moles of electrons that passed through the
cell. The relationship between the number of moles of substance deposited in
10
an electroplating cell and the number of moles of electrons passed through the
cell is given by Faraday’s laws of electrolysis:
Moles of substance deposited = Q
nF
where: Q= charge passed through the cell (in coulombs) n= number of elec-
trons transferred in the reaction F= Faraday constant = 96485 C/mol
We rearrange the equation to solve for n:
n=Q
F
The total charge passed through the cell (Q) can be calculated using the
formula:
Q=It
Q= 2.00 A ×14400 s = 28800 C
Now, we can calculate the number of moles of electrons passed through the
cell:
n=28800 C
96485 C/mol = 0.298 mol
Therefore, the number of moles of electrons that passed through the cell is
0.298 mol.
Question 11
Question
A solution contains a mixture of copper (Cu2+) and silver (Ag+) ions. A current
of 2.50 A is passed through the solution for 90.0 minutes. If 0.200 g of copper is
deposited at the cathode, how many grams of silver are deposited at the cathode
during this time?
(Hint: Use Faraday’s laws of electrolysis to relate the amount of substance
deposited to the charge passed and the molar masses of copper and silver.)
Solution
Step 1: Calculate the total charge passed through the solution. Given: current,
I= 2.50 A and time, t= 90.0 minutes. We first convert time to seconds:
90.0 minutes = 90.0×60 seconds = 5400 seconds. The total charge passed,
Q=I×t= 2.50 A ×5400 s = 13500 C.
Step 2: Determine the number of moles of copper deposited. Given: mass
of copper deposited, mCu = 0.200 g and molar mass of copper, MCu = 63.55
g/mol. First, convert the mass of copper deposited to moles:
nCu =mCu
MCu
=0.200 g
63.55 g/mol ≈0.00315 mol
11
Step 3: Use Faraday’s laws to determine the number of coulombs required
to deposit 1 mole of copper. The charge required to deposit 1 mole of Cu is
given by the equation:
Q=nF
where Fis the Faraday constant (96485 C/mol). Hence, the charge required to
deposit 0.00315 moles of Cu is:
Q=nCu ×F= 0.00315 mol ×96485 C/mol ≈304 C
Step 4: Calculate the number of moles of Ag deposited using the charge
passed. Since the total charge passed is 13500 C and the charge required to
deposit 1 mole of Ag is also 96485 C, the number of moles of Ag deposited is
given by:
nAg =Q
F=13500 C
96485 C/mol ≈0.140 mol
Step 5: Determine the mass of silver deposited. Given: molar mass of silver,
MAg = 107.87 g/mol. The mass of silver deposited is:
mAg =nAg ×MAg = 0.140 mol ×107.87 g/mol ≈15.0 g
Therefore, approximately 15.0 grams of silver are deposited at the cathode
during this time.
Question 12
Question
A student is conducting an experiment to investigate Faraday’s laws of elec-
trolysis. The student passes a current of 2.50 A through a solution of silver
nitrate for 30 minutes. During this time, 2.80 grams of silver is deposited at the
cathode. Determine the molar mass of silver.
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current
(I) = 2.50 A Time (t) = 30 minutes = 1800 seconds We can use the formula
Q=It to calculate the total charge passed through the solution.
Q= 2.50 A ×1800 s = 4500 C
Step 2: Determine the number of moles of silver deposited. The total charge
passed through the circuit is used to calculate the number of moles of silver
deposited. 1 Faraday is equal to 96485 C/mol e. Convert the total charge (in
Coulombs) to moles of electrons using this relationship.
1 Faraday = 96485 C/mol e−
12
Moles of electrons = 4500 C
96485 C/mol e−≈0.0466 mol e−
Since 1 mol of silver ions (Ag) requires 1 mol of electrons to form 1 mol of silver
atoms (Ag), the number of moles of silver deposited is the same as the number
of moles of electrons.
Moles of silver = 0.0466 mol
Step 3: Calculate the molar mass of silver. Given that 2.80 grams of silver
is deposited, we can use the formula m=n×Mwhere mis the mass of silver,
nis the number of moles of silver, and Mis the molar mass of silver.
2.80 g = 0.0466 mol ×M
Solve for the molar mass of silver.
M=2.80 g
0.0466 mol ≈60.08 g/mol
Therefore, the molar mass of silver is approximately 60.08 g/mol.
Question 13
Question
A student wants to perform an electrolysis experiment using a solution of silver
nitrate (AgNO3). The student places two silver electrodes in the solution and
applies a current of 2.50 A for 30.0 minutes.
Given that the molar mass of silver is 107.87 g/mol and the Faraday constant
is 9.65×104C/mol, calculate: (i) The amount of silver deposited on the cathode
during the experiment. (ii) The amount of silver nitrate consumed during the
experiment.
Solution
Step 1: Calculate the total charge passed through the solution. The total charge
passed can be calculated using the formula:
Q=I·t
Where: - Qis the total charge (in Coulombs), - Iis the current (in Amperes),
and - tis the time (in seconds).
Given that I= 2.50 A and t= 30.0 min = 30.0×60 s = 1800.0 s, we can
substitute these values into the formula to find:
Q= 2.50 A ×1800.0 s = 4500.0 C
13
Step 2: Calculate the amount of silver deposited on the cathode. The amount
of substance deposited can be calculated using the formula:
Amount of substance = Q
n·F
Where: - Qis the total charge passed (in Coulombs) calculated in Step 1, - n
is the number of electrons involved in the reaction (1 for silver ions), and - Fis
the Faraday constant (9.65 ×104C/mol).
Since n= 1 for silver ions, we can substitute the values into the formula:
Amount of silver = 4500.0 C
1·(9.65 ×104) C/mol = 4.67 ×10−2mol
To find the mass of silver deposited on the cathode, we use the molar mass
of silver:
Mass of silver = Amount of silver ×Molar mass of silver
Mass of silver = 4.67 ×10−2mol ×107.87 g/mol ≈5.03 g
Step 3: Calculate the amount of silver nitrate consumed during the exper-
iment. The stoichiometry of the reaction shows that one mole of silver ion
requires one mole of silver nitrate. Therefore, the amount of silver nitrate con-
sumed will also be 4.67×10−2mol, which is equal to 4.67 g since the molar mass
of silver nitrate is 169.87 g/mol.
Question 14
Question
In an electrolytic cell, a current of 2.00 A is passed through a lead(II) iodide,
PbI2, solution for 3.00 hours. If 0.500 g of lead is deposited at the cathode,
determine the standard electrode potential of the lead half-cell.
Solution
Step 1: Determine the amount of charge passed through the cell. Given: Cur-
rent, I= 2.00 A Time, t= 3.00 hours = 10800 s Charge, Q=I×t
Q= 2.00 A ×10800 s = 21600 C
Step 2: Convert the amount of lead deposited to moles. Given: Mass of lead
deposited, m= 0.500 g Molar mass of lead = 207.2 g/mol
Number of moles of lead deposited, n=m
Molar mass of lead
n=0.500 g
207.2 g/mol = 0.002415 mol
14
Step 3: Determine the number of electrons involved in the reduction of
lead(II) ions to lead atoms. The overall reaction for the reduction of Pb2+ ions
to Pb atoms is:
Pb2+ + 2e−→Pb
From the reaction, it is clear that 2 moles of electrons are involved in the re-
duction of 1 mole of Pb2+ ions.
Step 4: Calculate the charge required to deposit the amount of lead. Since
2 moles of electrons are involved in the reduction of 1 mole of Pb2+ ions:
Q=n×Faraday constant ×2
Given: Faraday constant, F= 96485 C/mol
Substitute the values to find Q.
21600 C = 0.002415 mol ×96485 C/mol ×2
21600 C = 0.4651 mol ×96485 C/mol
Step 5: Calculate the standard electrode potential of the lead half-cell. The
standard electrode potential, E◦
Pb2+/Pb, can be calculated using the equation:
E◦
Pb2+/Pb =−Q
nF
Substitute the known values of Q,n, and Finto the equation to calculate
E◦
Pb2+/Pb.
E◦
Pb2+/Pb =−21600
0.002415 ×96485
E◦
Pb2+/Pb =−0.898 V
Therefore, the standard electrode potential of the lead half-cell is −0.898 V.
Question 15
Question
Consider an electrolytic cell where a current of 2.50 A is passed through a
solution of copper (II) chloride (CuCl2). If the cell operates for 2.00 hours,
what mass of copper is deposited at the cathode? (Given: Faraday’s constant,
F= 96,500 C/mol)
15
Solution
Step 1: Find the total charge passing through the cell. The total charge passed
through the cell can be calculated using the formula:
Q=I·t
where Q= total charge passed in Coulombs, I= current in Amperes, and t=
time in seconds.
Given I= 2.50 Aand t= 2.00 hours = 7200 s:
Q= 2.50 A·7200 s= 18,000 C
Step 2: Calculate the number of moles of electrons passed through the cell.
Since 1 Faraday of charge is equivalent to 1 mole of electrons (where 1 Faraday
is equal to 96,500 C/mol), the number of moles of electrons passed through the
cell can be calculated as:
moles of electrons = Q
F=18,000 C
96,500 C/mol = 0.186 mol
Step 3: Determine the mass of copper deposited at the cathode. The sto-
ichiometry of the reaction can be used to relate the moles of electrons to the
moles of copper deposited. From the balanced equation for the reduction of
copper (II) ions:
Cu2+ + 2e−→Cu
we see that 2 moles of electrons are required to reduce 1 mole of Cu2+ to copper.
Therefore, the moles of copper deposited will be half the moles of electrons
passed through the cell:
moles of copper = 0.186 mol
2= 0.093 mol
Step 4: Calculate the mass of copper deposited. The molar mass of copper
is approximately 63.5 g/mol. Therefore, the mass of copper deposited at the
cathode can be calculated as:
mass of copper = moles of copper×molar mass of copper = 0.093 mol×63.5g/mol = 5.91 g
Therefore, 5.91 grams of copper will be deposited at the cathode after 2
hours of electrolysis.
Question 16
Question
State Faraday’s laws of electrolysis and discuss how they can be applied to
determine the quantity of a substance deposited or liberated during electrolysis.
16
Solution
Faraday’s laws of electrolysis describe the fundamental relationships between the
amount of substance produced during electrolysis, the current passing through
the electrolyte, the time the current is applied, and the molar mass of the
substance being produced.
Faraday’s First Law: The amount of a substance produced at an electrode
during electrolysis is directly proportional to the quantity of electric charge (in
Coulombs) passed through the electrolyte.
Faraday’s Second Law: The amounts of different substances produced by
the same quantity of electric charge passing through the electrolyte are directly
proportional to their chemical equivalent weights.
To determine the quantity of a substance deposited or liberated during elec-
trolysis, the following steps can be followed:
Step 1: Identify the half-reaction occurring at each electrode and determine
the number of moles of electrons (n) involved in the reaction.
Step 2: Calculate the total charge passed through the electrolyte using the
formula: Q=I×t, where Qis the charge in Coulombs, Iis the current in
Amperes, and tis the time in seconds.
Step 3: Determine the Faraday constant (F) using the equation: F=
96,485 Coulombs per equivalent.
Step 4: Calculate the amount of substance produced using the formula:
n=Q
n×F, where nis the number of moles of electrons, Qis the charge passed
in Coulombs, and Fis the Faraday constant.
Step 5: Convert the number of moles to grams using the molar mass of the
substance.
By following these steps and applying Faraday’s laws of electrolysis, one can
accurately determine the quantity of a substance deposited or liberated during
electrolysis.
Question 17
Question
A student is conducting an experiment on electrolysis using a sulfuric acid so-
lution. The student passes a current of 2.5 A through the solution for 3 hours.
During this time, hydrogen gas is produced at one electrode and oxygen gas at
the other electrode.
What mass of hydrogen gas is produced during the electrolysis process?
(Assume standard conditions.)
Solution
To find the mass of hydrogen gas produced during electrolysis, we first need to
calculate the total charge passed through the solution using Faraday’s laws of
17
electrolysis. We can then use the ideal gas law to convert the charge passed to
the mass of hydrogen gas produced.
Step 1: Calculate the total charge passed Given: Current, I= 2.5 A
Time, t= 3 hours = 10800 s Faraday’s constant, F= 96500 C/mol
The total charge passed through the solution is given by the formula:
Q=I×t
Substitute the given values to find Q:
Q= 2.5 A ×10800 s = 27000 C
Step 2: Calculate the number of moles of hydrogen gas produced
The number of moles of electrons required to produce one mole of hydrogen gas
is 2 according to the balanced electrolysis equation for water.
Since 1 Faraday of charge liberates 1 mole of electrons, the number of moles
of electrons liberated is given by:
Moles of electrons = Q
F
Substitute Qand Finto the formula to find the moles of electrons:
Moles of electrons = 27000 C
96500 C/mol ≈0.28 mol
Since 2 moles of electrons are required to produce 1 mole of hydrogen gas,
the number of moles of hydrogen gas produced is half of the moles of electrons
liberated:
Moles of hydrogen gas = 1
2×0.28 mol = 0.14 mol
Step 3: Calculate the mass of hydrogen gas produced The ideal gas
law relates the number of moles of a gas to its mass:
P V =nRT
For hydrogen gas at standard conditions, P= 1 atm, V= 22.4 L/mol, R=
0.0821 L-atm/mol-K, and T= 273 K.
The mass of 1 mole of hydrogen gas at standard conditions can be calculated
using its molar mass:
Molar mass of hydrogen = 2×atomic mass of hydrogen = 2×1 g/mol = 2 g/mol
Now, we can calculate the mass of hydrogen gas produced:
Mass of hydrogen gas = Moles of hydrogen gas ×Molar mass of hydrogen
Mass of hydrogen gas = 0.14 mol ×2 g/mol = 0.28 g
Therefore, approximately 0.28 g of hydrogen gas is produced during the
electrolysis process.
18
Question 18
Question
In an electrolytic cell, a constant current of 3.00 A is passed through a solution
of molten lead (II) bromide (PbBr2) for 45.0 minutes. Calculate the mass of lead
deposited at the cathode. Given: molar mass of lead = 207.2 g/mol, Faraday’s
constant = 96,485 C/mol.
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current (I)
= 3.00 A, Time (t) = 45.0 minutes = 45.0 minutes ×60 seconds/min = 2700
s, Faraday’s constant (F) = 96,485 C/mol.
The total charge (Q) passed through the cell can be calculated using the
formula:
Q=I×t
Q= 3.00 A ×2700 s = 8100 C
Step 2: Calculate the number of moles of electrons (n) transferred. Since 1
mole of electrons is equivalent to 1 Faraday, we can use the formula:
n=Q
F
n=8100 C
96,485 C/mol ≈0.0839 mol
Step 3: Determine the number of moles of lead deposited. From the balanced
chemical equation for the electrolysis of lead (II) bromide:
Pb2+ + 2e−→Pb(s)
It is evident that 2 moles of electrons are required to deposit 1 mole of lead.
Therefore, the number of moles of lead deposited (nPb) is given by:
nPb =n
2
nPb =0.0839 mol
2= 0.0419 mol
Step 4: Calculate the mass of lead deposited at the cathode. Given: Molar
mass of lead (Pb) = 207.2 g/mol, Number of moles of lead deposited (nPb) =
0.0419 mol.
The mass of lead deposited (m) can be calculated using the formula:
m=nPb ×Molar mass of Pb
m= 0.0419 mol ×207.2 g/mol = 8.68 g
Therefore, the mass of lead deposited at the cathode is 8.68 g.
19
Question 19
Question
In an electrolysis experiment, a constant current of 2.5 A is passed through a
solution of copper sulfate (CuSO4) for 45 minutes. If a mass of 3.2 g of copper
is deposited at the cathode, determine the number of moles of electrons that
were transferred during the electrolysis.
Solution
Step 1: Calculate the total charge transferred during the electrolysis. Given
that the current = 2.5 A and time = 45 minutes = 2700 seconds, we can use
the formula Q=I×t.
Q= 2.5 A ×2700 s = 6750 C
Step 2: Calculate the number of moles of copper deposited at the cathode.
Given that the molar mass of copper (Cu) is 63.55 g/mol, we can use the formula
moles = mass
molar mass .
moles = 3.2 g
63.55 g/mol ≈0.05 mol
Step 3: Determine the number of moles of electrons that were transferred
during the electrolysis. From the balanced equation for the electrolysis of copper
sulfate: Cu2+ + 2e−→Cu we can see that 2 moles of electrons are required to
deposit 1 mol of copper. Therefore, the number of moles of electrons transferred
is twice the number of moles of copper deposited.
moles of electrons = 2 ×0.05 = 0.1 mol
Hence, the number of moles of electrons that were transferred during the
electrolysis is 0.1 mol.
Question 20
Question
State Faraday’s laws of electrolysis and explain how they are related to each
other.
Solution
Faraday’s laws of electrolysis are fundamental principles in electrochemistry
that describe the relationship between the amount of substance produced or
consumed during electrolysis and the amount of electric charge passed through
the electrolyte. There are two laws:
20
Faraday’s First Law: The amount of substance produced at an electrode
during electrolysis is directly proportional to the quantity of electricity passed
through the cell.
Faraday’s Second Law: The amounts of different substances produced or
consumed by the same quantity of electricity are in the ratio of their equivalent
weights.
Step 1: Faraday’s First Law states that the mass of a substance (m) pro-
duced at an electrode during electrolysis is directly proportional to the amount
of electric charge (Q) passed through the cell. Mathematically, this can be
expressed as:
m∝Q
m=k·Q
where k is a constant of proportionality known as the electrochemical equivalent.
Step 2: Faraday’s Second Law relates the amounts of different substances
produced or consumed during electrolysis to their equivalent weights. The equiv-
alent weight of a substance is the mass that reacts with or produces one mole
of either electrons or protons.
Step 3: By combining Faraday’s First and Second Laws, we get the relation-
ship between the amount of substance produced or consumed during electrolysis
and the quantity of electricity passed through the cell. This relationship is given
by:
m1/m2=E1/E2
where: - m1 and m2 are the masses of substances produced or consumed, - E1
and E2 are the equivalent weights of the substances.
Therefore, Faraday’s laws of electrolysis provide a comprehensive under-
standing of the relationship between the quantities of substances involved in
electrolysis and the electric charge passed through the cell.
Question 21
Question
A solution containing aqueous sodium chloride, NaCl(aq), is electrolyzed using
inert electrodes. During the electrolysis, 3.50 g of chlorine gas is produced at the
anode. Calculate the volume of hydrogen gas produced at the cathode (standard
conditions) by passing the same amount of electricity through the solution.
(Hint: Start by determining the number of moles of chlorine gas produced.)
Solution
Step 1: Calculate the number of moles of chlorine gas produced.
Given: Mass of chlorine gas produced = 3.50 g Molar mass of Cl2= 70.906
g/mol
Number of moles of Cl2=3.50 g
70.906 g/mol
21
Step 2: Determine the number of moles of electrons transferred during the
electrolysis of NaCl.
Each molecule of Cl2is produced by the transfer of 2 electrons. So, the number
of moles of electrons transferred can be calculated using the formula:
Number of moles of electrons = Number of moles of Cl2×2
2
Step 3: Calculate the number of moles of hydrogen gas produced.
From the number of moles of electrons transferred, we can find the number of
moles of H2produced since 2 moles of electrons are required to produce 1 mole
of H2.
Number of moles of H2= Number of moles of electrons ×1
2
Step 4: Convert the number of moles of hydrogen gas to volume at standard
conditions.
At standard conditions, 1 mole of any gas occupies 22.4 L. Hence, the volume
of H2gas produced can be calculated as follows:
Volume of H2gas (at STP) = Number of moles of H2×22.4 L/mol
Now, follow these steps to find the final answer.
Question 22
Question
A student performs an electrolysis experiment using a copper(II) sulfate solu-
tion. The student connects a copper electrode to the positive terminal of a
power supply and a platinum electrode to the negative terminal. During the ex-
periment, the student observes that the mass of the copper electrode decreases
over time.
Explain this observation using Faraday’s laws of electrolysis.
Solution
To explain the student’s observation using Faraday’s laws of electrolysis, we
need to consider the reactions occurring at the electrodes.
Step 1: Reactions at the electrodes At the anode (copper electrode
connected to the positive terminal), copper(II) ions will be reduced:
Cu2+ + 2e−→Cu
At the cathode (platinum electrode connected to the negative terminal),
water will be oxidized:
2H2O→O2+ 4H++ 4e−
Step 2: Application of Faraday’s First Law Faraday’s First Law states
that the mass of a substance deposited or liberated at an electrode during elec-
trolysis is directly proportional to the quantity of electricity passed through the
cell.
22
Since the copper electrode is losing mass over time, it indicates that more
copper ions are being reduced (deposited as solid copper) at the electrode than
the amount of copper dissolving into the solution.
Step 3: Explanation The reason for the decrease in mass of the copper
electrode is that the reduction of copper(II) ions at the anode is consuming
copper from the electrode to form solid copper. This results in the copper
electrode losing mass over time as more copper(II) ions are reduced to solid
copper than the amount of copper dissolving into the solution.
Therefore, the observation of the copper electrode losing mass during the
experiment can be explained by Faraday’s laws of electrolysis.
Question 23
Question
An electrolytic cell contains a solution of silver nitrate (AgNO3). If a current
of 5.00 A is passed through the cell for 2.00 hours, what mass of pure silver is
deposited on the cathode? (Given: Atomic masses: Ag = 107.87 g/mol, N =
14.01 g/mol, O = 16.00 g/mol)
Solution
Step 1: Calculate the total charge passed through the cell. Step 2: Determine
the number of moles of silver deposited. Step 3: Convert the moles of silver to
mass in grams.
Step 1: Calculate the total charge passed through the cell. The total charge
(Q) can be calculated using the formula:
Q=I×t
where Iis the current (5.00 A) and tis the time in seconds (2.00 hours = 7200
s).
Q= 5.00A×7200s= 36000C
Step 2: Determine the number of moles of silver deposited. The Faraday’s
constant, F, is 96485 C/mol. The number of moles of silver deposited can be
calculated by dividing the total charge by the Faraday’s constant:
n=Q
F=36000C
96485C/mol ≈0.373mol
Step 3: Convert the moles of silver to mass in grams. The molar mass
of silver is 107.87 g/mol. The mass of silver deposited can be calculated by
multiplying the number of moles by the molar mass:
Mass of silver = 0.373mol ×107.87g/mol ≈40.24g
Therefore, approximately 40.24 grams of pure silver is deposited on the cath-
ode.
23
Question 24
Question
An aqueous solution of silver nitrate is electrolyzed using a current of 2.50 A
for 4.00 hours. If 25.0 g of pure silver is deposited at the cathode, determine
the number of moles of silver ions reduced and calculate the Faraday constant.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given: Cur-
rent, I= 2.50 A Time, t= 4.00 hours
The total charge passed, Q, can be calculated using the formula:
Q=I×t
Let’s first convert the time from hours to seconds:
t= 4.00 ×3600 = 14400 seconds
Therefore,
Q= 2.50 ×14400 = 36000 C
Step 2: Calculate the number of moles of silver ions reduced. Given: Mass
of silver deposited, m= 25.0 g Molar mass of silver, MAg = 107.87 g/mol
The number of moles of silver deposited, n, can be calculated using the
formula:
n=m
MAg
Substitute the given values to find n:
n=25.0
107.87 ≈0.232 mol
Step 3: Calculate the number of moles of electrons transferred. From the
balanced half-reaction for the reduction of silver ions:
2e−+ Ag+→Ag
We can see that 2 moles of electrons are required to reduce 1 mole of Ag+.
Therefore, the number of moles of electrons transferred, ne, is given by:
ne= 2n
Substitute the value of ncalculated earlier to find ne:
ne= 2 ×0.232 = 0.464 mol
Step 4: Calculate the Faraday constant. The Faraday constant, F, represents
the charge of 1 mole of electrons, and can be calculated using the formula:
F=Q
ne×96485
24
Substitute the given values to find F:
F=36000
0.464 ×96485 ≈7.20 ×104C/mol
Therefore, the Faraday constant is approximately 7.20 ×104C/mol.
Question 25
Question
An aqueous solution of sodium chloride is electrolyzed using platinum electrodes.
If a current of 3.00 A is passed through the solution for 1 hour, calculate the
mass of sodium metal deposited on the cathode. (Given: Atomic masses - Na
= 23 g/mol, Cl = 35.5 g/mol, Faraday constant = 9.65 ×104C/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution.
Given: Current, I= 3.00 A Time, t= 1 hour = 3600 s Using the formula
Q=I·t, where Qis the total charge passed, Q= 3.00 A ×3600 s Q= 10800 C
Step 2: Calculate the number of moles of electrons passed through the so-
lution. Given: Faraday constant, F= 9.65 ×104C/mol Using the formula
n=Q
F, where nis the number of moles of electrons passed, n=10800 C
9.65×104C/mol
n= 0.112 mol
Step 3: Determine the stoichiometry of the reaction at the cathode. The
balanced half-reaction for the reduction of sodium ions to sodium metal is:
2 Na++ 2 e−→2 Na
Step 4: Calculate the mass of sodium metal deposited on the cathode. Given:
Molar mass of Na = 23 g/mol Using the formula m=n×Molar mass, where
mis the mass of sodium deposited, m= 0.112 mol ×23 g/mol m= 2.57 g
Therefore, the mass of sodium metal deposited on the cathode after passing
a current of 3.00 A for 1 hour is 2.57 g.
Question 26
Question
A copper sulfate solution is electrolyzed using copper electrodes. If a current
of 2.5 A is passed through the solution for 30 minutes, calculate the mass of
copper deposited on one electrode. The molar mass of copper is 63.5 g/mol.
Solution
Step 1: Find the charge passed through the solution using the formula:
Charge (C) = Current (A) ×Time (s)
25
Charge = 2.5 A ×(30 ×60) s = 4500 C
Step 2: Using Faraday’s first law of electrolysis, we know that 1 mole of
electrons carries a charge of 1 Faraday, which is 96500 C/mol. Therefore, the
number of moles of electrons passed through the solution is:
Moles of electrons = Total charge
Charge per mole of electrons
Moles of electrons = 4500 C
96500 C/mol ≈0.0467 mol
Step 3: Since the reaction is Cu2+ + 2e−→Cu, we need 2 moles of elec-
trons to deposit 1 mole of copper. Therefore, moles of copper deposited on one
electrode is:
Moles of copper = 0.0467 mol
2= 0.0234 mol
Step 4: Calculate the mass of copper deposited using the molar mass of
copper:
Mass of copper (g) = Moles of copper ×Molar mass of copper
Mass of copper = 0.0234 mol ×63.5 g/mol ≈1.49 g
Therefore, the mass of copper deposited on one electrode is approximately
1.49 grams.
Question 27
Question
In an electrolytic cell, a student passes a current of 2.50 A through a solution
of copper(II) sulfate, CuSO4, for 1.50 hours. During this time, the student
observes that 6.35 grams of copper is deposited at the cathode.
Calculate the number of moles of electrons that are required to reduce 1
mole of Cu2+ ions to copper metal.
Solution
Step 1: Find the charge required to reduce 1 mole of Cu2+ ions to copper metal.
Given:
Current, I= 2.50 A
Time, t= 1.50 hours = 5400 s
Atomic weight of copper, mCu = 63.55 g/mol
1 faraday, 1F= 96,485 C
26
Charge required to reduce 1 mole of Cu2+ ions to copper metal:
Q= Current ×Time
Q=I×t= 2.50 A ×5400 s = 13500 C
Step 2: Calculate the number of moles of electrons required to reduce 1
mole of Cu2+ ions. The number of moles of electrons can be calculated using
Faraday’s law:
Q=n×F
n=Q
F
Substitute the values:
n=13500 C
96485 C/mol ≈0.140 mol
So, approximately 0.140 moles of electrons are required to reduce 1 mole of
Cu2+ ions to copper metal.
Question 28
Question
During the electrolysis of copper(II) sulfate solution using copper electrodes, a
current of 2 A is passed for 30 minutes. If the Faraday constant is 96500 C/mol,
calculate the mass of copper deposited on the cathode.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given current
I= 2 Aand time t= 30 min = 30 ×60 s= 1800 s.
Q=I×t= 2 ×1800 = 3600 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday is equivalent to 96500 C/mol.
moles of electrons = Q
96500 =3600
96500 mol
Step 3: Calculate the mass of copper deposited. The balanced equation for
the electrolysis of copper(II) sulfate solution is:
Cu2+(aq)+2e−→Cu(s)
27
One mole of copper is deposited for every 2 moles of electrons.
moles of copper = 1
2×moles of electrons = 1
2×3600
96500 mol
Step 4: Calculate the mass of copper using the molar mass of copper. The
molar mass of copper is approximately 63.5g/mol.
mass of copper = moles of copper ×molar mass of copper = 1
2×3600
96500 ×63.5g
mass of copper ≈1
2×3600
96500 ×63.5≈0.746 g
Therefore, the mass of copper deposited on the cathode is approximately
0.746 grams.
Question 29
Question
A student is performing an electrolysis experiment using a solution of copper(II)
sulfate, CuSO4, using a current of 2.5 A. If the student deposits 1.25 g of copper
at the cathode in 30 minutes, calculate the Faraday constant, F.
Solution
Step 1: Find the number of moles of copper deposited at the cathode. Given
that the molar mass of copper is 63.55 g/mol, the number of moles of copper
deposited can be calculated as follows:
Moles of Cu = Mass of Cu
Molar mass of Cu =1.25 g
63.55 g/mol
Step 2: Calculate the charge transferred. Since 1 mole of copper corresponds to
2 moles of electrons in the electrolytic cell, the total charge transferred can be
calculated using Faraday’s laws of electrolysis:
Charge transferred = Moles of Cu ×2×F
where Fis the Faraday constant. Step 3: Determine the time of electrolysis.
Given that the current is 2.5 A and the time is 30 minutes, convert the time
into seconds:
Time = 30 ×60 s
Step 4: Calculate the charge transferred using the formula:
Charge transferred = Current ×Time
Step 5: Equate the two expressions for the charge transferred and solve for F.
Equate the two expressions for the charge transferred obtained in Step 2 and
Step 4, and solve for F. Remember to convert the time into seconds.
Moles of Cu ×2×F= Current ×Time
28
Question 30
Question
An electrolysis experiment is set up using a copper(II) sulfate solution. A cur-
rent of 2.5 A is passed through the solution for 30 minutes. Calculate the mass of
copper deposited on the cathode during this time. (Given: Faraday’s constant
= 96500 C mol−1, atomic mass of copper = 63.5 g mol−1)
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current,
I= 2.5 A Time, t= 30 minutes = 30 ×60 s = 1800 s
Total charge, Q=It Q = 2.5 A ×1800 s = 4500 C
Step 2: Calculate the number of moles of electrons passed. Given: Faraday’s
constant, F= 96500 C mol−1
Number of moles of electrons, n=Q
Fn=4500 C
96500 C mol−1≈0.0467 mol
Step 3: Calculate the mass of copper deposited. Given: Atomic mass of
copper, M= 63.5 g mol−1
Mass of copper deposited, m=n×M m = 0.0467 mol×63.5 g mol−1≈2.97 g
Therefore, the mass of copper deposited on the cathode during this time is
approximately 2.97 g.
Question 31
Question
In an electrolytic cell, a current of 2.50 A is passed through a solution of iron(II)
chloride for 3.00 hours. During this time, 1.20 g of iron is deposited at the
cathode. Determine the oxidation state of iron in the product and write the
balanced half-reaction occurring at the cathode.
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current,
I= 2.50 A Time, t= 3.00 hours
The total charge passed, Q, can be calculated as:
Q=I×t
Q= 2.50 A ×3.00 h ×3600 s/h
Q= 2.50 A ×10800 s
Q= 27000 C
Step 2: Calculate the number of moles of iron deposited at the cathode.
Given: Mass of iron deposited, m= 1.20 g Molar mass of iron, M= 55.85
g/mol
29
The number of moles of iron deposited, n, can be calculated as:
n=m
M
n=1.20 g
55.85 g/mol
n≈0.0215 mol
Step 3: Determine the oxidation state of iron in the product. Each mole
of deposited iron corresponds to 2 moles of electrons transferred in the half-
reaction. Therefore, the oxidation state of iron in the product is +2.
Step 4: Write the balanced half-reaction occurring at the cathode. The half-
reaction at the cathode involves the reduction of iron(II) ions to form elemental
iron:
Fe2+ + 2e−→Fe
Therefore, the balanced half-reaction at the cathode is:
Fe2+ + 2e−→Fe
Question 32
Question
A student is conducting an electrolysis experiment using a solution of copper(II)
sulfate. The student passes a current of 2.50 A through the solution for 3.00
hours. During this time, copper is deposited on one electrode. If the student
measures that 5.00 g of copper is deposited, determine the number of moles of
electrons that pass through the cell during the experiment. Assume that the
only reaction occurring is the reduction of Cu2+ to Cu.
Solution
Step 1: Calculate the charge passed through the cell. The charge passed through
the cell can be calculated using the formula:
Q=I·t
where Qis the charge in coulombs, Iis the current in amperes, and tis the time
in seconds. Given that I= 2.50 A and t= 3.00 hours, we first convert hours to
seconds:
t= 3.00 ×3600 = 10800 s
Now, we can calculate the charge passed through the cell:
Q= 2.50 A ×10800 s = 27000 C
30
Step 2: Calculate the number of moles of electrons. The number of moles of
electrons can be calculated using Faraday’s law of electrolysis:
mol e−=Q
nF
where Qis the charge passed through the cell in coulombs, nis the number of
moles of electrons, and Fis the Faraday constant, 96485 C/mol. Substitute the
values to find the number of moles of electrons:
n=27000 C
96485 C/mol = 0.280 mol e−
Therefore, the number of moles of electrons that passed through the cell during
the experiment is 0.280 mol.
Question 33
Question
Explain Faraday’s laws of electrolysis and how they are applied in electrochem-
ical processes.
Solution
Faraday’s First Law states that the amount of a substance produced at an
electrode during electrolysis is directly proportional to the quantity of electricity
passed through the electrolyte. This can be expressed as:
w=Zq
where: - wis the amount of substance produced, - Zis the electrochemical
equivalent of the substance (the mass of the substance produced by one coulomb
of charge), and - qis the total charge passed through the electrolyte.
Faraday’s Second Law states that the amounts of different substances
produced by the same quantity of electricity are proportional to the equivalent
weights of these substances. Mathematically, it can be represented as:
w
Z=m
E
where: - wis the amount of substance produced, - mis the mass of the sub-
stance, - Eis the equivalent weight of the substance, - Zis the electrochemical
equivalent, and - qis the total charge passed through the electrolyte.
These laws are crucial in understanding and predicting the outcomes of var-
ious electrochemical processes, such as electroplating, electrolysis of water, and
corrosion prevention, among others.
31
Question 34
Question
A student is performing an electrolysis experiment using a solution of silver
nitrate (AgNO3). The student passes a current of 2.5 A through the solution
for 30 minutes. During this time, silver metal is deposited on the cathode.
Given that the standard electrode potential of the silver/silver ion (Ag+/Ag)
half-cell is +0.80 V, calculate the mass of silver deposited on the cathode.
Solution
Step 1: Calculate the total charge passed through the electrolytic cell. The total
charge passed (Q) is given by the formula:
Q=I×t
where Iis the current (in amperes) and tis the time (in seconds). Given that
I= 2.5 A and t= 30 minutes = 30 ×60 s, we have:
Q= 2.5×30 ×60 = 4500 C
Step 2: Calculate the number of moles of electrons passed. Since silver
ions (Ag+) gain one electron to form silver metal (Ag) during electrolysis, the
number of moles of silver deposited is equal to the number of moles of electrons
passed through the cell. This can be calculated using Faraday’s first law of
electrolysis:
n=Q
F
where nis the number of moles of electrons, Qis the total charge passed, and F
is the Faraday constant (96485 C/mol). Substitute Q= 4500 C and F= 96485
C/mol into the formula to get:
n=4500
96485 ≈0.0467 mol
Step 3: Calculate the mass of silver deposited. The molar mass of silver is
107.87 g/mol. Therefore, the mass of silver deposited (m) can be calculated
using:
m=n×molar mass
Substitute n= 0.0467 mol into the formula:
m= 0.0467 ×107.87 ≈5.04 g
Therefore, the mass of silver deposited on the cathode is approximately 5.04
grams.
32
Question 35
Question
In an electrolytic cell, a current of 2.5 A is passed through a solution of copper
(II) sulfate for 30 minutes. Calculate the mass of copper deposited at the
cathode. (Molar mass of copper = 63.5 g/mol)
Solution
Step 1: Calculate the total charge passed through the cell. Given: Current,
I= 2.5 A Time, t= 30 minutes = 30 ×60 seconds = 1800 seconds
Using the formula Q=I×t, where Qis the total charge,
Q= 2.5 A ×1800 s = 4500 C
Step 2: Determine the number of moles of electrons passed. 1 mole of
electrons carries a charge of 1 Faraday, which is equivalent to 96485 C.
Thus, the number of moles of electrons,
moles = 4500 C
96485 C/mol ≈0.0466 mol
Step 3: Calculate the mass of copper deposited at the cathode. For the
copper (II) sulfate electrolysis, one mole of copper deposits 1 mole of copper.
Molar mass of copper, M= 63.5 g/mol
Mass of copper deposited,
mass = moles ×M= 0.0466 mol ×63.5 g/mol ≈2.96 g
Therefore, the mass of copper deposited at the cathode is approximately
2.96 g.
33
