CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Faraday’s
laws of electrolysis
Question Bank - Set 5
Liberty University
Question 1
Question
An electrolysis experiment was conducted using a copper (Cu) electrode and
a silver (Ag) electrode connected to a power supply with a current of 2.5 A
passing through the circuit for 20 minutes. If the standard electrode potentials
are E◦
Cu2+/Cu = 0.34 V and E◦
Ag+/Ag = 0.80 V, calculate the following: (a) The
mass of copper deposited on the electrode. (b) The mass of silver deposited on
the electrode.
Solution
(a) To find the mass of copper deposited on the electrode, you can use Faraday’s
laws of electrolysis. The formula to calculate the mass of a substance deposited
during electrolysis is given by:
Mass = Q×M
n×F
where: - Qis the total charge passing through the circuit (in Coulombs), -
Mis the molar mass of the substance (in g/mol), - nis the number of moles of
electrons transferred in the reaction, - Fis the Faraday constant (96485 C/mol).
First, calculate the total charge passing through the circuit:
Q=I×t
Q= 2.5 A ×20 min ×60 s
1 min
Q= 3000 C
Step 1: Calculate the number of moles of electrons transferred in the reaction
for copper. Given the balanced half-reaction: Cu2+ + 2e−→Cu, one mole of
copper requires 2 moles of electrons. As the total charge passing through the
circuit is 3000 C, we can calculate the number of moles transferred:
n=Q
F
n=3000 C
96485 C/mol
n≈0.0311 mol
Step 2: Calculate the mass of copper deposited on the electrode: Given that
the molar mass of copper is 63.55 g/mol:
Mass of Cu = Q×M
n×F
Mass of Cu = 3000 C ×63.55 g/mol
0.0311 mol ×96485 C/mol
Mass of Cu ≈6.41 g
Therefore, the mass of copper deposited on the electrode is approximately
6.41 g.
(b) The mass of silver deposited on the electrode can be calculated using the
same formula with the appropriate values for silver.
Step 1: Calculate the number of moles of electrons transferred in the reaction
for silver. Given the balanced half-reaction: Ag++e−→Ag, one mole of silver
requires 1 mole of electrons. As the total charge passing through the circuit is
3000 C, we can calculate the number of moles transferred:
n=Q
F
n=3000 C
96485 C/mol
n≈0.0311 mol
Step 2: Calculate the mass of silver deposited on the electrode: Given that
the molar mass of silver is 107.87 g/mol:
Mass of Ag = Q×M
n×F
Mass of Ag = 3000 C ×107.87 g/mol
0.0311 mol ×96485 C/mol
Mass of Ag ≈10.88 g
Therefore, the mass of silver deposited on the electrode is approximately
10.88 g.
2
Question 2
Question
A copper sulfate solution is electrolyzed using a current of 2.50 A. If 1.00 g of
copper is deposited at the cathode in 30 minutes, determine the half-reaction at
the cathode and the mass of copper sulfate consumed during the electrolysis.
Solution
Step 1: Write the half-reaction at the cathode. The half-reaction for the depo-
sition of copper at the cathode is:
Cu2+ + 2e−→Cu
Step 2: Calculate the number of moles of copper deposited at the cathode.
Given that the current is 2.50 A and the time is 30 minutes, we first convert
the time to seconds:
30 minutes ×60 seconds/minute = 1800 seconds
Next, we calculate the charge passed through the circuit:
Charge = Current ×Time = 2.50 A ×1800 s = 4500 C
Since 1 mole of electrons is equivalent to 1 Faraday (96485 C), we can cal-
culate the number of moles of copper deposited:
Number of moles of Cu = Charge
2×96485 =4500
192970
Step 3: Calculate the mass of copper deposited. Given that the molar mass
of copper is 63.55 g/mol, we can find the mass of copper deposited:
Mass of Cu = Number of moles of Cu×Molar mass of Cu = Number of moles of Cu×63.55 g/mol
Step 4: Calculate the mass of copper sulfate consumed. Since the stoi-
chiometry between copper and copper sulfate is 1:1, the mass of copper sulfate
consumed is equal to the mass of copper deposited.
Therefore, the mass of copper sulfate consumed during the electrolysis is the
same as the mass of copper deposited at the cathode.
Question 3
Question
A solution of copper sulfate (CuSO4) is electrolyzed using a current of 2.50 A for
2.50 hours. If copper is deposited at the cathode, determine the mass of copper
deposited. (Given: Atomic mass of copper = 63.5 g/mol, Faraday’s constant =
96,500 C/mol)
3
Solution
Step 1: Calculate the total charge passed through the solution.
Total charge = Current ×Time
Total charge = 2.50 A ×2.50 hours ×3600 s/hour
Total charge = 22,500 C
Step 2: Calculate the moles of copper deposited using Faraday’s law.
Moles of copper = Total charge
1 F ×1 mol
96,500 C
Moles of copper = 22,500 C
96,500 C/mol
Moles of copper = 0.2337 mol
Step 3: Calculate the mass of copper deposited.
Mass of copper = Moles of copper ×Atomic mass of copper
Mass of copper = 0.2337 mol ×63.5 g/mol
Mass of copper = 14.84 g
Therefore, the mass of copper deposited is 14.84 g.
Question 4
Question
In an electrolysis experiment, a current of 2.5 A was passed through a solution
of copper sulfate for 30 minutes. If 1.5 g of copper was deposited at the cathode,
calculate the molar mass of copper.
Given: Faraday constant, F = 96485 C mol−1.
Solution
Step 1: Calculate the total charge passed through the electrolyte using the
equation Q=I×t, where Qis the total charge, Iis the current, and tis the
time.
Total charge, Q= 2.5A×30 min = 2.5A×30 min
60 s/min = 75 C
Step 2: Calculate the number of moles of copper deposited at the cathode
using the equation n=m
M, where nis the number of moles, mis the mass, and
Mis the molar mass.
Number of moles of copper, n=1.5g
M
4
Step 3: Calculate the number of electrons required for the deposition of 1
mole of copper ions using the half-reaction of copper deposition: Cu2+ + 2e−→
Cu.
Number of electrons, = 2 (per mole of Cu)
Step 4: Calculate the number of coulombs required for the deposition of 1
mole of copper ions using Faraday’s laws: Q=n×F×z, where zis the number
of electrons per ion in the half-reaction.
Q=n×F×z
75 C=1.5g
M×96485 C mol−1×2
Step 5: Solve for the molar mass of copper, M.
1.5×96485 ×2
75 =M
M≈7726 g/mol
Therefore, the molar mass of copper is approximately 7726 g/mol.
Question 5
Question
A student is conducting an experiment on Faraday’s laws of electrolysis. In one
trial, the student passes a current of 2.5 A through an electrolytic cell containing
a copper (Cu) sulfate solution. If the student observes the deposition of 12.0
g of copper metal in 20 minutes, determine the electrochemical equivalent of
copper in the electrolyte. Assume 100
Solution
Step 1: Find the total charge passed through the cell. Given that the current
is 2.5 A and the time is 20 minutes, we first convert the time to seconds:
Time (s) = 20 ×60 = 1200 s
The total charge passed through the cell can be calculated using the formula:
Q=I×t
Q= 2.5 A ×1200 s = 3000 C
Step 2: Calculate the amount of charge required to deposit 1 mole of copper.
The molar mass of copper (Cu) is approximately 63.5 g/mol. Therefore, 1 mole
of copper requires 1 mole of electrons which is equivalent to a charge of:
1 mol of Cu ×2×96,485 C/mol = 2 ×96,485 C
5
= 192,970 C
Step 3: Determine the number of moles of copper deposited. Using the
deposited mass of copper (12.0 g) and the molar mass of copper (63.5 g/mol),
we calculate the number of moles of copper:
Moles of Cu = 12.0 g
63.5 g/mol ≈0.189 mol
Step 4: Compute the electrochemical equivalent of copper. Comparing the
charge passed through the cell with the charge required to deposit the observed
amount of copper: 3000 C
0.189 mol ≈15,873 C/mol
Therefore, the electrochemical equivalent of copper in the electrolyte is ap-
proximately 15,873 C/mol.
Question 6
Question
Faraday’s laws of electrolysis describe the relationship between the amount of
substance produced in an electrolytic cell and the electric charge passed through
it. A student is performing an electrolysis experiment using a solution of cop-
per(II) sulfate with copper electrodes. Calculate the mass of copper deposited
at the cathode after passing a current of 2.50 A for 2.00 hours. Given: Atomic
mass of copper = 63.5 g/mol, Faraday’s constant = 96,485 C/mol, and assume
100
Solution
Step 1: Calculate the total charge passed through the electrolytic cell.
Charge (C) = Current (A) ×Time (s)
Convert the time from hours to seconds (1 hour = 3600 s).
Time = 2.00 ×3600 = 7200 s
Charge = 2.50 A ×7200 s = 18,000 C
Step 2: Determine the number of moles of electrons transferred.
Number of moles of electrons = Charge (C)
F
where Fis Faraday’s constant.
Number of moles of electrons = 18,000 C
96,485 C/mol ≈0.187 mol
6
Step 3: Calculate the mass of copper deposited at the cathode using the
mole ratio and the atomic mass of copper.
Mass of copper (g) = 0.187 mol ×63.5 g/mol = 11.85 g
Therefore, the mass of copper deposited at the cathode after passing a cur-
rent of 2.50 A for 2.00 hours is 11.85 grams.
Question 7
Question
For an electrolytic cell, if a current of 2.5 A is passed through a solution of
molten lead(II) bromide, how long (in seconds) would it take to deposit 50.0 g
of lead metal?
(Assume 1 F is required to deposit 107.9 g of Ag.)
Solution
Step 1: Calculate the amount of charge required to deposit 50.0 g of lead metal.
Given that 1 Faraday (F) is required to deposit 107.9 g of Ag, the amount of
charge required to deposit 50.0 g of Pb is:
Charge = 50.0 g
107.9 g/F×1 F
Charge = 0.463F
Step 2: Use the formula Q = It to find the time (t) taken to deposit this
amount of charge. Given current (I) = 2.5 A, charge (Q) = 0.463 F, we have:
t=Q
I
t=0.463 C
2.5 A
t= 0.1852 s
Therefore, it would take 0.1852 seconds to deposit 50.0 g of lead metal.
Question 8
Question
A student performs an electrolysis experiment using a solution of copper(II)
sulfate. The student applies a current of 2.5 A for 30 minutes to a copper(II)
sulfate solution using platinum electrodes. If the student measures that 0.5 g
of copper is deposited on the cathode during this time, calculate the number of
moles of electrons transferred during the electrolysis.
Given: Faraday’s constant F= 96500 C/mol
7
Solution
Step 1: Calculate the charge passed through the circuit. The charge passed
through the circuit 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. Convert the time to seconds:
t= 30 minutes ×60 s/min = 1800 s
Now, substitute the given values to find Q:
Q= 2.5 A ×1800 s = 4500 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday (F) is equivalent to the charge of one mole of electrons (Avogadro’s
number), we can calculate the number of moles of electrons transferred using
the formula:
Moles of electrons transferred = Q
F
Substitute the values to find the number of moles of electrons transferred:
Moles of electrons transferred = 4500 C
96500 C/mol
Moles of electrons transferred = 0.0468 mol
Therefore, the number of moles of electrons transferred during the electrol-
ysis of copper(II) sulfate solution is 0.0468 mol.
Question 9
Question
An electrolytic cell is set up where a current of 3.00 A is passed through a
solution of CuSO4. If 0.500 g of copper is deposited at the cathode in 30
minutes, determine the faraday constant.
Solution
Step 1: Determine the moles of copper deposited. Given that the molar mass of
copper is 63.55 g/mol, the number of moles of copper deposited can be calculated
as:
Moles of Cu = Mass of Cu deposited
Molar mass of Cu =0.500 g
63.55 g/mol
8
Step 2: Calculate the charge passed through the circuit. The charge passed
can be calculated using the formula: Q=It, where Iis the current in amperes
(3.00 A) and tis the time in seconds (30 minutes = 1800 seconds).
Q= 3.00 A ×1800 s
Step 3: Calculate the Faraday constant. Since 1 Faraday is equivalent to the
charge required to deposit 1 mole of a monovalent ion (e.g., Cu2+ in this case),
the Faraday constant can be calculated as:
Faraday constant = Q
n×F
where nis the moles of Cu deposited and Fis Faraday constant.
Step 4: Substitute the values and calculate the Faraday constant. Substitute
the values calculated in steps 1 and 2 into the formula from step 3, and solve
for F. Remember to convert the time from minutes to seconds.
Faraday constant = 3.00 A ×1800 s
0.500 g/63.55 g/mol
Question 10
Question
In an electrolysis experiment, a constant current of 2.5 A is passed through
an aqueous solution of copper(II) sulfate for 20 minutes. If copper metal is
deposited at the cathode, calculate the mass of copper deposited. (Atomic mass
of copper = 63.5 g/mol, Faraday’s constant = 96,485 C/mol)
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current,
I= 2.5 A Time, t= 20 minutes = 1200 s
The total charge, Q, passed through the solution can be calculated using the
formula:
Q=I×t
Q= 2.5 A ×1200 s = 3000 C
Step 2: Calculate the moles of electrons (n) that have passed through the
solution. Since 1 Faraday is the charge of 1 mole of electrons, we can use the
relation:
1 mol e−= 1 F
Given Faraday’s constant, F= 96,485 C/mol, we can calculate:
9
n=Q
F=3000 C
96,485 C/mol ≈0.0311 mol
Step 3: Calculate the mass of copper deposited. The balanced half-reaction
for the reduction of copper(II) ions to copper metal is:
Cu2+(aq)+2e−→Cu(s)
From the reaction, we can see that 2 moles of electrons reduce 1 mole of
Cu2+ ions. Hence, the moles of copper deposited will be equal to the moles of
electrons passed.
Thus, the mass of copper deposited can be calculated using the formula:
Mass of Cu = n×Molar mass of Cu
Mass of Cu = 0.0311 mol ×63.5 g/mol ≈1.97 g
Therefore, approximately 1.97 g of copper will be deposited at the cathode.
Question 11
Question
A student is conducting an electrolysis experiment using a copper(II) sulfate
solution with copper electrodes. The student passes a current of 2.00 A through
the solution for 90.0 minutes. During this time, copper is deposited on one
electrode. If the student measures that 0.350 g of copper is deposited, determine
the yield of the process in percent.
Given: Atomic mass of copper = 63.5 g/mol
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current
(I) = 2.00 A Time (t) = 90.0 minutes
To calculate the total charge (Q) passed through the solution, use the for-
mula:
Q=I×t
Plugging in the values:
Q= 2.00 A×90.0min ×1min
60 s×1C
1A·s
Q= 2.00 A×5400 s×1C
Q= 10800 C
Therefore, the total charge passed through the solution is 10800 C.
10
Step 2: Calculate the number of moles of copper deposited. Given: Mass of
copper deposited (m) = 0.350 g Atomic mass of copper = 63.5 g/mol
To calculate the number of moles of copper deposited, use the formula:
Number of moles = mass
molar mass
Plugging in the values:
Number of moles = 0.350 g
63.5g/mol
Number of moles = 0.00551 mol
Therefore, the number of moles of copper deposited is 0.00551 mol.
Step 3: Calculate the yield of the process in percent. The theoretical number
of moles of copper that should have been deposited can be calculated from
the Faraday’s laws of electrolysis. The molar equivalent of one Faraday of
charge (1 F) is equal to the number of moles of electrons in one mole of copper
(1 F = 2 mol).
The theoretical number of moles of copper (ntheoretical) that should have
been deposited can be calculated using:
ntheoretical =Q
F
Where: Q= Total charge passed through the solution (10800 C) F= Fara-
day constant = 96485 C/mol
Plugging in the values:
ntheoretical =10800 C
96485 C/mol
ntheoretical = 0.112mol
Now, the yield of the process can be calculated using the formula:
Yield (%) = actual yield
theoretical yield ×100%
Plugging in the values:
Yield (%) = 0.00551 mol
0.112 mol ×100%
Yield (%) = 0.0491
0.112 ×100%
Yield (%) = 43.8%
Therefore, the yield of the process is 43.8
11
Question 12
Question
Faraday’s laws of electrolysis state that the amount of chemical reaction that
occurs during electrolysis is proportional to the quantity of electricity that flows
through the cell.
Suppose a solution of silver chloride is electrolyzed using a current of 2.50
A for 2.00 hours. If 0.500 g of silver is deposited at the cathode, determine the
Faraday constant (96485 C/mol).
Solution
Step 1: Find the total charge passing through the cell. Given: Current, I =
2.50 A Time, t = 2.00 hours = 7200 s
The total charge passing through the cell can be calculated using the formula:
Q=It
Substitute the given values:
Q= 2.50 A ×7200 s = 18000 C
Step 2: Determine the number of moles of silver deposited. The amount of
charge required to deposit 1 mole of silver is given by the Faraday constant:
F= 96485 C/mol
The number of moles of silver deposited can be calculated using the formula:
moles of Ag = mass of Ag
molar mass of Ag =0.500 g
107.87 g/mol
Step 3: Calculate the Faraday constant. The Faraday constant relates the
total charge passed through the cell to the number of moles of substance de-
posited:
F=Q
n
Substitute the values:
F=18000 C
0.500 g
107.87 g/mol
= 21404 C/mol
Therefore, the Faraday constant is 21404 C/mol.
12
Question 13
Question
An aqueous solution of NaCl (sodium chloride) is electrolyzed using platinum
electrodes. If a current of 2.50 A is passed through the solution for 2.00 hours,
calculate the mass of sodium metal deposited at the cathode. (Given: Faraday
constant F= 96,485 C/mol, molar mass of Na is 23.0 g/mol and the ionic
charge of Na is +1.)
Solution
Step 1: Calculate the total charge passed through the electrolytic cell. The total
charge passed is given by the formula:
Q=I×t
where Qis the charge in coulombs, Iis the current in amperes, and tis the
time in seconds.
Given I= 2.50 A and t= 2.00 hours = 2.00 ×3600 s, we have:
Q= 2.50 A ×2.00 ×3600 s
Q= 2.50 C/s ×7200 s
Q= 18,000 C
Step 2: Calculate the number of moles of electrons passed through the elec-
trolytic cell. Since 1 faraday (F) is equivalent to 1 mole of electrons, we can
calculate the number of moles of electrons passed using the Faraday constant:
Moles of electrons (n) = Q
F
Moles of electrons (n) = 18,000 C
96,485 C/mol
Moles of electrons (n) ≈0.1865 mol
Step 3: Calculate the moles of sodium that will be deposited. Since each
sodium ion (Na+) gains one electron to form a sodium atom, the moles of sodium
deposited will be equal to the moles of electrons passed:
Moles of Na deposited = 0.1865 mol
Step 4: Calculate the mass of sodium deposited. Using the molar mass of
sodium (Na = 23.0 g/mol), we can calculate the mass of sodium deposited:
13
Mass of Na deposited = Moles of Na deposited ×Molar mass of Na
Mass of Na deposited = 0.1865 mol ×23.0 g/mol
Mass of Na deposited ≈4.29 g
Therefore, the mass of sodium metal deposited at the cathode is approxi-
mately 4.29 g.
Question 14
Question
A copper sulfate solution (CuSO4) is electrolyzed using platinum electrodes.
If a current of 2.5 A is passed through the solution for 2 hours, what mass of
copper is deposited on the cathode? (Given: 1 F araday = 96,500 C/mol and
Molar mass of Cu = 63.5g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given that
the current is 2.5 A and the time is 2 hours, we first convert the time from hours
to seconds:
2hours = 2 ×3600 seconds = 7200 seconds
Now, calculate the total charge passed:
Q=I×t= 2.5A×7200 s= 18000 C
Step 2: Calculate the number of moles of electrons passed. Since 1 Faraday
is equivalent to 96,500 C/mol, we can calculate the number of moles of electrons
passed:
n=Q
F=18000 C
96500 C/mol ≈0.1865 mol
Step 3: Calculate the mass of copper deposited. The reaction at the cathode
is:
Cu2+ + 2e−→Cu
From the balanced equation, we see that one mole of copper is deposited by 2
moles of electrons. Therefore, the moles of copper deposited will be half the
moles of electrons passed:
nCu =n
2=0.1865 mol
2= 0.09325 mol
14
Finally, calculate the mass of copper deposited on the cathode:
Mass = nCu ×Molar mass of Cu = 0.09325 mol ×63.5g/mol ≈5.92 g
Therefore, approximately 5.92 g of copper is deposited on the cathode after
2 hours of electrolysis.
Question 15
Question
A student is conducting an electrolysis experiment using two electrodes con-
nected to a power supply. The student observes that the mass of one of the
electrodes decreases over time while the mass of the other electrode increases.
Explain this observation in terms of Faraday’s laws of electrolysis.
Solution
To explain the observation of the changing masses of the electrodes during
electrolysis, we can refer to Faraday’s laws of electrolysis. These laws are based
on the principles of conservation of mass and conservation of charge.
Step 1: Oxidation and Reduction Reactions
During electrolysis, oxidation and reduction reactions occur at the respective
electrodes. The anode undergoes oxidation, where electrons are lost, while the
cathode undergoes reduction, where electrons are gained.
Step 2: Faraday’s First Law
Faraday’s first law states that the amount of chemical change produced by
an electric current is proportional to the quantity of electricity passed through
the electrolyte. This can be represented by the equation:
Chemical change ∝Electric charge passed
Step 3: Electrolysis of Electrodes
In this case, as the anode undergoes oxidation, metal ions from the anode
dissolve into the electrolyte, leading to a decrease in mass of the anode. At
the same time, at the cathode, metal ions from the electrolyte get reduced and
deposited onto the cathode, causing an increase in mass.
Therefore, the observed changes in the masses of the electrodes can be ex-
plained by Faraday’s laws of electrolysis, specifically the relationship between
the amount of electricity passed, the chemical changes occurring at the elec-
trodes, and the resulting changes in mass.
15
Question 16
Question
A solution contains both iron(II) sulfate and copper(II) sulfate. When a cur-
rent is passed through this solution for a certain amount of time, the electrode
connected to the negative terminal gained 2.8 grams. If the faradic (electro-
chemical) efficiency is 80
Given: - Mass of iron deposited at the negative electrode = 2.8 grams -
Faradic efficiency = 80
Solution
Step 1: Determine the molar mass of Iron and Copper Iron (Fe) has a molar
mass of 55.85 g/mol. Copper (Cu) has a molar mass of 63.55 g/mol.
Step 2: Calculate the number of moles of Iron deposited Given that the
mass of iron deposited is 2.8 grams, we can calculate the number of moles of
iron deposited: Number of moles of iron = Mass of iron deposited / Molar mass
of iron Number of moles of iron = 2.8 g / 55.85 g/mol
Step 3: Calculate the theoretical amount of Copper deposited Since iron
has a valency of 2 and the faradic efficiency is 802 moles of Fe : 1 mole of Cu
Number of moles of Cu = 0.8 * Number of moles of Fe
Step 4: Calculate the mass of Copper deposited Mass of Cu deposited =
Number of moles of Cu * Molar mass of Cu
Step 5: Substitute the values into the formula and calculate the mass of
Copper deposited.
Question 17
Question
A student performs an electrolysis experiment using a solution of silver nitrate,
AgNO3. During the experiment, the student observes that the mass of silver
deposited at the cathode is 1.68 g. Calculate the total charge that passed
through the electrolyte.
Given:
Atomic mass of silver, m(Ag) = 107.87 g/mol
Number of electrons needed to reduce one mole of silver ions, n= 1
Solution
Step 1: Determine the number of moles of silver deposited at the cathode.
Mass of silver deposited = 1.68 g
Molar mass of silver = 107.87 g/mol
16
Number of moles of silver = Mass of silver deposited
Molar mass of silver
Number of moles of silver = 1.68
107.87 ≈0.0156 mol
Step 2: Calculate the total charge that passed through the electrolyte.
Q=n×F×N
where Q= total charge (C)
n= number of moles of silver deposited
F= Faraday constant = 96500 C/mol
N= number of electrons needed to reduce one mole of silver = 1
Q= 0.0156 ×96500 ×1≈1505.4 C
Therefore, the total charge that passed through the electrolyte is approxi-
mately 1505.4 C.
Question 18
Question
In an electrolytic cell, a current of 2.5 A is passed for 2 hours through a solution
of silver nitrate AgNO3. If 1.5 g of silver is deposited at the cathode, determine
the faraday constant.
Solution
Let’s start by calculating the amount of electric charge passed through the
circuit.
Step 1: Calculate the total charge passed. The total charge passed can be
calculated using the formula:
Total charge (C) = Current (A) ×Time (s)
Given that current = 2.5 A and time = 2 hours = 7200 s, we have:
Total charge = 2.5×7200 = 18000 C
Step 2: Calculate the number of moles of silver deposited. Using the formula
for calculating the number of moles:
moles = mass (g)
molar mass (g/mol)
17
Given that the mass of silver deposited = 1.5 g, and the molar mass of silver
= 107.87 g/mol, we have:
moles of silver = 1.5
107.87 ≈0.014 mol
Step 3: Calculate the number of electrons passed. Since each mole of silver
requires 1 mole of electrons to be deposited, the number of electrons passed will
be equal to the number of moles of silver deposited. Therefore, the number of
electrons passed = 0.014 mol.
Step 4: Calculate the Faraday constant. One Faraday is equal to the charge
of one mole of electrons, which is approximately equal to 96,485 C/mol. There-
fore, the Faraday constant is:
F=Total charge (C)
Number of electrons =18000
0.014 ≈1,285,714 C/mol
Question 19
Question
During the electrolysis of a molten compound, it was found that 0.5 grams
of the compound was decomposed by a current of 0.5 amperes flowing for 2
hours. If the compound contains only one type of cation and one type of anion,
determine:
1. The equivalent weight of the cation.
2. The valency of the cation.
Solution
1. We can find the equivalent weight of the cation using Faraday’s laws of
electrolysis. The mass of the substance decomposed is related to the equivalent
weight by the formula:
Equivalent weight = Atomic weight
n×F
where nis the valency of the ion being discharged and Fis the Faraday constant
(96500 C/mol). Given that 0.5 grams of the compound was decomposed by a
current of 0.5 amperes flowing for 2 hours, we can calculate the equivalent weight
as follows:
Step 1: Determine the total charge passed through the electrolyte.
The total charge Qcan be calculated using the formula:
Q=I×t
18
where Iis the current (0.5 A) and tis the time in seconds (2 hours = 7200
seconds).
Q= 0.5 A ×7200 s = 3600 C
Step 2: Calculate the equivalent weight of the cation. Given that 0.5
grams of the compound was decomposed, we first need to determine the number
of moles of the compound decomposed using its atomic weight. Let’s assume the
formula of the compound is MmXx, whereM representsthecationandXrepresentstheanion.Moles of M =
Mass
Atomic weight of M =0.5 g
Atomic weight of M Since the total charge passed through the
electrolyte is equivalent to the charge on 1 mole of M (as only ions of M are
involved in electrolysis), we have:
Charge = n×F= Moles of M ×F
Substitute the values of charge and moles of M:
3600 = 0.5
Atomic weight of M ×96500
Solve for the atomic weight of M to find the equivalent weight.
2. Now that we have the equivalent weight of the cation, we can determine
the valency of the cation.
Step 3: Calculate the valency of the cation. The valency of the cation,
denoted by n, can be found using the formula:
n=Atomic weight
Equivalent weight ×F
Given that we have found the equivalent weight in the previous step, we can
substitute its value into the formula along with the atomic weight of the cation
to obtain the valency.
Question 20
Question
Explain Faraday’s first and second laws of electrolysis. Use mathematical ex-
pressions to define each law and provide an example to illustrate each law.
Solution
Faraday’s First Law of Electrolysis: This law states that the amount of
chemical change produced by the passage of current through an electrolyte is
proportional to the quantity of electricity passed through it.
Mathematical expression: If mis the mass of a substance deposited or
liberated at an electrode, Qis the quantity of electricity passed through the
19
electrolyte, and Mis the atomic or molecular weight of the substance, then
according to Faraday’s First Law of Electrolysis,
m=kQ,
where kis a constant of proportionality.
Example: Consider the electrolysis of molten lead bromide (PbBr2) using
a current of 2.50 A for 1 hour. Calculate the mass of lead liberated at the
cathode. The atomic masses are Pb = 207, Br = 80.
Step 1: Calculate the quantity of electricity passed.
Q=I×t= 2.50 A ×1 hour = 2.50 ×3600 C = 9000 C
Step 2: Calculate the equivalent weight of lead (M) in grams.
M= 207 g/mol = 207 g/equiv
Step 3: Using Faraday’s First Law of Electrolysis:
m=kQ
Substitute k=M
96500 (since 1 Faraday = 96500 C) and Q= 9000 C.
m=M
96500 ×Q=207
96500 ×9000 = 19.37 g
Faraday’s Second Law of Electrolysis: This law states that if the same
quantity of electricity is passed through different electrolytes connected in series,
the mass of the substances liberated or deposited at the electrodes is directly
proportional to their chemical equivalent weights.
Mathematical expression: If m1and m2are the masses of two substances
liberated or deposited by the same quantity of electricity, and M1and M2are
their respective atomic or molecular weights, then according to Faraday’s Second
Law of Electrolysis, m1
m2
=M1
M2
Example: When 0.20 g of silver is plated out by the passage of a current,
what mass of copper will be deposited by the same quantity of electricity (atomic
masses: Ag = 108, Cu = 64)?
Step 4: Given m1= 0.20 g, M1= 108 g/mol, and M2= 64 g/mol. Use
Faraday’s Second Law of Electrolysis:
m1
m2
=M1
M2
0.20
m2
=108
64
m2=0.20 ×64
108 = 0.118 g
Therefore, 0.118 g of copper will be deposited by the same quantity of elec-
tricity.
20
Question 21
Question
Explain Faraday’s laws of electrolysis and how they can be used to calculate the
amount of substance deposited or liberated during an electrolytic reaction.
Solution
Faraday’s laws of electrolysis describe the relationship between the amount of
substance produced or consumed during electrolysis and the amount of electric-
ity passed through the electrolyte. They can be summarized as: 1. The amount
of chemical change produced by a current is proportional to the quantity of
electricity that passes through the electrolyte. 2. The quantities of different
substances liberated by the same quantity of electricity are proportional to their
equivalent weights.
Step 1: Explanation of Faraday’s Laws Let Qbe the total charge
passing through the electrolyte in coulombs (Q=I×t, where Iis the current in
amperes and tis the time in seconds), nbe the number of moles of the substance
liberated or consumed, zbe the valence of the substance, and Fbe the Faraday
constant (F= 96,485 C/mol). Faraday’s laws can be mathematically expressed
as: 1. Amount of substance liberated or consumed: n=Q
zF 2. Relationship
between the amounts of different substances: n1
z1=n2
z2
Step 2: Calculation using Faraday’s Laws Suppose we electrolyze an
aqueous solution of copper(II) sulfate using a current of 2.5 A for 30 minutes.
Calculate the mass of copper deposited on the electrode.
Given: I= 2.5 A, t= 30 minutes = 1800 s, z= 2 (valence of copper),
atomic mass of copper = 63.5 g/mol.
First, calculate the charge passed through the electrolyte: Q=I×t=
2.5×1800 = 4500 C
Now, use Faraday’s law to find the number of moles of copper deposited:
n=Q
zF =4500
2×96485 ≈0.0234 moles
Finally, calculate the mass of copper: Mass = n×Atomic mass = 0.0234 ×
63.5≈1.487 g
Therefore, approximately 1.487 grams of copper will be deposited on the
electrode.
Question 22
Question
Consider an electrolysis cell where a copper(I) sulfate solution is electrolyzed
using copper electrodes. The electrolysis cell operates for 1 hour at a constant
current of 2.5 A. Calculate the mass of copper deposited on the cathode during
this time.
21
Given: - The molar mass of copper is 63.55 g/mol. - 1 Faraday (F) is
equivalent to 96,485 C/mol.
Solution
Step 1: Calculate the total charge passed through the cell. Given current I= 2.5
A and time t= 1 hour = 3600 s. Using the formula Q=It, where Qis the
charge passed through the cell.
Charge passed (Q) = 2.5 A ×3600 s = 9000 C
Step 2: Determine the number of moles of copper deposited. Since 1 Fara-
day is equivalent to 96,485 C/mol, we can find the number of moles of copper
deposited using the formula:
Moles of copper = Charge passed (C)
Faraday constant (C/mol)
Moles of copper = 9000 C
96485 C/mol = 0.0934 mol
Step 3: Calculate the mass of copper deposited. Given the molar mass
of copper is 63.55 g/mol, we can find the mass of copper deposited using the
formula:
Mass of copper = Moles of copper ×Molar mass of copper
Mass of copper = 0.0934 mol ×63.55 g/mol = 5.928 g
Therefore, the mass of copper deposited on the cathode during 1 hour of
electrolysis is 5.928 g.
Question 23
Question
A solution containing NaCl is electrolyzed using a current of 3.20 A for 45.0
minutes. If 2.00 g of Na is deposited at the cathode, determine the half-reaction
occurring at the cathode and calculate the number of moles of electrons trans-
ferred during the electrolysis.
Solution
Step 1: Calculate the amount of charge passed through the electrolyte. Given:
Current, I= 3.20 A Time, t= 45.0 minutes
First, we convert the time to seconds:
t= 45.0×60 = 2700 s
22
The amount of charge, Q, passed through the electrolyte is given by:
Q=I×t= 3.20 ×2700 = 8640 C
Step 2: Determine the half-reaction occurring at the cathode. From the
amount of deposited Na, we can determine that the half-reaction at the cathode
is:
Na++e−→Na
Step 3: Calculate the number of moles of electrons transferred during the
electrolysis. To calculate the number of moles of electrons transferred, we need
to find the number of moles of Na deposited at the cathode first.
Given: Mass of Na deposited, m= 2.00 g Molar mass of Na, M= 22.99 g/mol
The number of moles of Na deposited is:
n=m
M=2.00
22.99 = 0.0870 mol
Since 1 mole of Na requires 1 mole of electrons, the number of moles of
electrons transferred is also 0.0870 mol.
Therefore, the number of moles of electrons transferred during the electrol-
ysis is 0.0870 mol.
Question 24
Question
A solution of copper(II) sulfate CuSO4is electrolyzed using a current of 2.0
A for 10 minutes. If the mass of copper deposited at the cathode is 2.45 g,
calculate the molar mass of copper(II) sulfate.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given that
the current is 2.0 A and the time is 10 minutes, which is equal to 600 seconds,
the total charge passed can be calculated using the formula:
Q=It
Q= 2.0 A ×600 s
Q= 1200 C
Step 2: Determine the number of moles of electrons passed. Since each mole
of copper(II) ions requires 2 moles of electrons for reduction, we can calculate
the number of moles of electrons passed using the formula:
moles of electrons = Q
F
23
where Fis the Faraday’s constant (96500 C mol−1).
moles of electrons = 1200 C
96500 C mol−1
moles of electrons ≈0.0124 mol
Step 3: Calculate the number of moles of copper deposited at the cathode.
From the balanced equation for the electrolysis of copper(II) sulfate (Cu2+ +
2e−→Cu), we see that 1 mole of copper corresponds to 2 moles of electrons.
Therefore, the number of moles of copper deposited can be calculated as follows:
moles of copper = 0.5×moles of electrons
moles of copper = 0.5×0.0124
moles of copper ≈0.0062 mol
Step 4: Calculate the molar mass of copper(II) sulfate. We are given that
the mass of copper deposited is 2.45 g. Therefore, the molar mass of copper(II)
sulfate can be calculated using the formula:
Molar mass of CuSO4=Mass of copper
moles of copper
Molar mass of CuSO4=2.45 g
0.0062 mol
Molar mass of CuSO4≈395 g/mol
Therefore, the molar mass of copper(II) sulfate is approximately 395 g/mol.
Question 25
Question
A solution of copper sulfate (CuSO4) is electrolyzed using platinum electrodes.
If a current of 2.50 A is passed through the solution for 30.0 minutes, what
mass of copper is deposited at the cathode? (Given: 1 F = 96500 C/mol,
M(Cu) = 63.5 g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. Given
that the current is I= 2.50 A and the time is t= 30.0 min, first convert the
time to seconds:
t= 30.0×60 = 1800 s.
Then, calculate the total charge passed using:
Q=I×t= 2.50 A ×1800 s.
24
Step 2: Determine the number of moles of electrons passed. Since 1 Faraday
(F) represents 96500 C:
1 F = 96500 C ⇒Q=1
96500 F.
Thus, the number of moles of electrons passed is:
n=Q
96500 mol.
Step 3: Use the stoichiometry to find the moles of copper deposited. Since
each mole of copper requires 2 moles of electrons to be deposited:
2 mol Cu : 2 mol e−⇒1 mol Cu : 1 mol e−.
Therefore, the moles of copper deposited at the cathode is equal to the moles
of electrons passed in Step 2.
Step 4: Calculate the mass of copper deposited. Finally, calculate the mass
of copper deposited using the molar mass of copper:
m=n×M(Cu) = n×63.5 g/mol.
Question 26
Question
Faraday’s laws of electrolysis state that the amount of a substance produced
during electrolysis is proportional to the quantity of electricity passed through
the electrolyte solution.
Consider an electrolysis experiment where a current of 3.2 A is passed
through a copper(II) sulfate solution for 45 minutes. If the mass of copper
deposited on the cathode is found to be 0.72 g, determine the half-reaction that
occurs at the cathode and calculate the Faraday constant (F) based on this
experiment.
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. The
total charge (Q) can be calculated using the formula:
Q=I×t
Given: Current, I= 3.2 A, Time, t= 45 minutes = 2700 seconds
Substitute the given values into the formula:
Q= 3.2 A ×2700 s = 8640 C
25
Step 2: Determine the half-reaction at the cathode. The half-reaction at
the cathode involves the reduction of copper(II) ions to form solid copper. The
balanced half-reaction is:
Cu2+ + 2e−→Cu
So, 2 electrons are required to reduce 1 copper(II) ion.
Step 3: Calculate the number of moles of copper deposited. The number of
moles of copper deposited can be calculated using the formula:
moles = mass
molar mass
Given: Mass of copper deposited, 0.72 g
The molar mass of copper (Cu) is 63.55 g/mol.
Substitute the given values into the formula:
moles = 0.72 g
63.55 g/mol ≈0.01133 mol
Step 4: Calculate the number of electrons transferred. Since 2 electrons are
required to reduce 1 copper(II) ion, the total number of electrons transferred
can be calculated as:
Total electrons = 2 ×moles of copper deposited
Total electrons = 2 ×0.01133 mol = 0.02266 mol
Step 5: Determine the Faraday constant. The Faraday constant (F) repre-
sents the charge of 1 mole of electrons and is equal to the charge of 1 mole of
electrons = 1 Faraday. It can be calculated using the formula:
F=Q
Total electrons
Substitute the calculated values into the formula:
F=8640 C
0.02266 mol ≈381103 C/mol
Therefore, the Faraday constant based on this experiment is approximately
381103 C/mol.
Question 27
Question
An electroplating experiment is set up using a solution of silver nitrate (AgNO3)
with two electrodes connected by a battery. The current passing through the
circuit is 0.5 A for 30 minutes. Calculate the amount of silver deposited on one
of the electrodes during this time.
(Assume the molar mass of silver is 107.87 g/mol, and the Faraday constant
is 96485 C/mol.)
26
Solution
Step 1: Calculate the total charge passed through the circuit. Given that the
current is 0.5 A and the time is 30 minutes, we first convert the time to seconds:
30 minutes = 30 ×60 seconds = 1800 seconds.
The total charge passed through the circuit can be calculated using the
formula Q=I×t, where Iis the current and tis the time: Q= 0.5 A×1800 s =
900 C.
Step 2: Convert the charge to moles of electrons. From Faraday’s laws
of electrolysis, we know that one mole of electrons has a charge of 96485 C.
Therefore, the number of moles of electrons passed through the circuit is given
by: Moles of electrons = 900 C
96485 C/mol = 0.00933 mol.
Step 3: Calculate the amount of silver deposited. In the electroplating pro-
cess of silver, each mole of silver deposited corresponds to one mole of electrons.
Given that the molar mass of silver is 107.87 g/mol, we can calculate the mass
of silver deposited using the formula: Mass of silver = Moles of electrons ×
Molar mass of silver.
Substitute the values to find the mass of silver deposited: Mass of silver =
0.00933 mol ×107.87 g/mol = 1.008 g.
Therefore, during the electrolysis process, 1.008 g of silver will be deposited
on one of the electrodes.
Question 28
Question
Faraday’s laws of electrolysis state that the mass of a substance deposited or lib-
erated at an electrode during electrolysis is directly proportional to the quantity
of electricity passed through the electrolyte.
Consider the electrolysis of molten lead(II) bromide using two inert elec-
trodes. If a current of 2.5 A is passed through the molten lead(II) bromide for
1 hour, calculate:
1. The mass of lead deposited at the cathode.
2. The volume of bromine gas evolved at the anode (at STP).
Given:
Atomic masses: Pb = 207 g/mol; Br = 80 g/mol
1 F (Faraday) of charge = 96500 C
Solution
Let’s start by calculating the number of coulombs of charge passed through the
electrolyte:
Total charge = Current ×Time
27
Total charge = 2.5 A ×3600 s = 9000 C
Step 1: Calculate the amount of substance deposited at the cathode (Pb):
First, find the number of moles of electrons passed:
Moles of electrons = Total charge
96500 C/mol
Moles of electrons = 9000 C
96500 C/mol = 0.09326 mol
Since the molten lead(II) bromide will be reduced at the cathode:
Pb2+ + 2e−→Pb
We can see that 1 mole of Pb requires 2 moles of electrons.
Therefore, the number of moles of Pb deposited at the cathode will be:
Moles of Pb = 0.09326 mol∇ · 2 = 0.04663 mol
Next, calculate the mass of lead deposited at the cathode:
Mass of Pb = Moles of Pb ×Molar mass of Pb
Mass of Pb = 0.04663 mol ×207 g/mol = 9.65 g
Therefore, the mass of lead deposited at the cathode is 9.65 g.
Step 2: Calculate the volume of bromine gas evolved at the anode:
The half-reaction at the anode will be the oxidation of bromide ions:
2Br−→Br2+ 2e−
The moles of electrons passed through the electrolyte will produce half as
many moles of bromine gas:
Moles of Br2=Moles of electrons
2= 0.09326 mol
Since 1 mole of gas occupies 24 L at STP:
Volume of Br2= Moles of Br2×24 L/mol = 0.09326 mol ×24 L/mol = 2.23 L
Therefore, the volume of bromine gas evolved at the anode is 2.23 L at STP.
Question 29
Question
A solution contains two electrolytes, copper sulfate and sodium chloride. If a
current of 2.5 A is passed through the solution for 1 hour, calculate the mass of
copper deposited. Given: atomic masses - Cu = 63.5 g/mol, Na = 23 g/mol, Cl
= 35.5 g/mol; Faraday constant = 96500 C/mol, and 1F = 96500 C. (Assume
100
28
Solution
Step 1: Calculate the total charge passed through the solution. Given current
I= 2.5Aand time t= 1 hour = 3600 seconds, we have
Q=I×t= 2.5A×3600 s= 9000 C
Step 2: Determine the number of moles of electrons passed. Since the charge
on 1 mole of electrons is equal to 1 Faraday, the number of moles of electrons
passed is given by
n=Q
96500 C/mol =9000 C
96500 C/mol ≈0.0933 mol
Step 3: Calculate the number of moles of copper deposited. From the bal-
anced chemical equation for the electrolysis of copper sulfate,
2e−+Cu2+ →Cu
we see that 2 moles of electrons deposited 1 mole of copper. Therefore,
nCu = 0.5×n= 0.5×0.0933 = 0.0467 mol
Step 4: Find the mass of copper deposited. The mass of copper deposited
can be calculated using the formula:
Mass = n×Atomic Mass of Cu = 0.0467 mol ×63.5g/mol = 2.971 g
Therefore, the mass of copper deposited is approximately 2.971 grams.
Question 30
Question
A student conducted an electrolysis experiment using a solution of copper(II)
sulfate, CuSO4, and two copper electrodes. During the experiment, they ob-
served that a current of 2.50 A flowed through the cell for 2.00 hours. The
student also measured the mass of the copper electrode connected to the posi-
tive terminal before and after the experiment and found that it had decreased
by 0.145 g.
Determine the following:
1. The total charge passed through the cell.
2. The number of moles of electrons that flowed through the cell.
3. The mass of copper deposited on the negative electrode during the 2-hour
period.
Given:
Faraday’s constant, F= 96,485 C/mol
Atomic mass of copper, 63.55 g/mol
29
Solution
1. First, let’s find the total charge passed through the cell using the formula:
Total charge = Current ×Time
Total charge = 2.50 A ×2.00 hours ×3600 s/hour
Step 1:
Total charge = 2.50 A ×2.00 hours ×3600 s/hour = 18,000 C
2. Next, to find the number of moles of electrons that flowed through the
cell, we use Faraday’s laws of electrolysis that relate the total charge passed to
the number of moles of electrons:
Total charge = Number of moles of electrons ×F araday′s constant
Number of moles of electrons = Total charge
F araday′s constant
Step 2:
Number of moles of electrons = 18,000 C
96,485 C/mol
Number of moles of electrons ≈0.1866 mol
3. Finally, to find the mass of copper deposited on the negative electrode
during the 2-hour period, we use the stoichiometry of the reaction:
2 moles of electrons + Cu2+ →Cu
Since the molar ratio between moles of electrons and moles of copper is 2:1,
we have:
Moles of copper deposited = 1
2×Number of moles of electrons
Step 3:
Moles of copper deposited = 1
2×0.1866 mol = 0.0933 mol
Now, we can find the mass of copper deposited using the molar mass of
copper:
Mass of copper deposited = Moles of copper deposited ×Molar mass of copper
Mass of copper deposited = 0.0933 mol ×63.55 g/mol
Mass of copper deposited ≈5.93 g
Therefore, the mass of copper deposited on the negative electrode during
the 2-hour period is approximately 5.93 g.
30
Question 31
Question
A solution of copper(II) ions is electrolyzed using a current of 2.00 A for 4.00
hours. If copper is deposited on one of the electrodes, calculate the mass of
copper deposited. Given that the molar mass of copper is 63.55 g/mol and the
Faraday constant is 9.65 ×104C/mol.
Solution
Step 1: Calculate the total charge passed through the solution.
Given: Current, I= 2.00 A
Time, t= 4.00 hours
The total charge, Q, passed through the solution can be calculated using the
formula:
Q=I×t
Q= 2.00 A ×4.00 hours ×3600 s/hour
Q= 2.00 ×4.00 ×3600 C
Q= 28800 C
Step 2: Calculate the number of moles of copper deposited.
The number of moles of electrons involved in the reduction of 1 mole of cop-
per(II) ions is 2. This is because each copper(II) ion gains 2 electrons to form
copper.
The number of moles of copper deposited can be calculated using the formula:
Moles of copper = Q
n×F
where: Q= 28800 C (total charge passed through the solution)
n= 2 (number of moles of electrons involved)
F= 9.65 ×104C/mol (Faraday constant)
Moles of copper = 28800
2×9.65 ×104
Moles of copper = 28800
1.93 ×105
Moles of copper ≈0.1493 mol
Step 3: Calculate the mass of copper deposited.
The mass of copper deposited can be calculated using the formula:
Mass = Moles ×Molar mass
Mass = 0.1493 mol ×63.55 g/mol
Mass = 9.48 g
Therefore, approximately 9.48 g of copper is deposited on one of the elec-
trodes during the electrolysis process.
31
Question 32
Question
Explain Faraday’s laws of electrolysis. An experiment was conducted using a
copper chloride solution with a current of 2.5 A passed through it for 2 hours.
If the mass of copper deposited was 6.78 g, calculate the Faraday constant.
Solution
Faraday’s laws of electrolysis state the following: 1. The mass of a substance
discharged at an electrode during electrolysis is directly proportional to the
quantity of electricity passed through the electrolyte. 2. The masses of different
substances discharged by the same quantity of electricity are proportional to
their chemical equivalent weights.
To calculate the Faraday constant, we will use the formula:
Faraday constant (F) = m×Z
n×F
where: - mis the mass of the substance deposited (in grams), - Zis the valency
of the ions, - nis the number of electrons exchanged in the reaction, - Fis the
Faraday constant (charge on one mole of electrons is 9.65 ×104C/mol).
Given: - Current, I= 2.5 A, - Time, t= 2 hours = 7200 s, - Mass of copper
deposited, m= 6.78 g, - Valency of copper ions, Z= 2.
First, let’s calculate the charge passed through the electrolyte using the
formula Q=It:
Q= 2.5×7200
Q= 18000 C
Next, calculate the number of moles of copper deposited:
Moles of Cu = Mass
Molar mass
The molar mass of copper is 63.55 g/mol, so:
Moles of Cu = 6.78
63.55
Now, we can calculate the Faraday constant:
F=m×Z
n×Q=6.78 ×2
6.78
63.55 ×18000 =13.56
6.78
63.55 ×18000
F=13.56
6.78
63.55 ×18000
F=13.56 ×63.55
6.78 ×18000
F=861.438
122040
F= 0.00706 C/mol
32
Question 33
Question
In an electrolytic cell, a metal plate is connected to the positive terminal of a
battery while a copper plate is connected to the negative terminal. The metal
plate loses mass at a rate of 0.2 grams per minute. If the Faraday constant is
9.65 ×104C/mol, calculate the current passing through the cell.
Solution
Step 1: Determine the number of moles of metal being deposited per minute.
Given that the metal plate loses mass at a rate of 0.2 grams per minute, we can
calculate the number of moles using the molar mass of the metal. Let’s assume
the molar mass of the metal is Mg/mol.
Molar mass of metal = Mg/mol
Number of moles of metal deposited per minute = 0.2 g/min
Mmol/min
Step 2: Determine the charge passing through the cell per minute. The
charge required to deposit one mole of the metal can be calculated using Fara-
day’s constant.
1 mole of metal requires 1 ×Faraday constant = 1 ×9.65 ×104C/mol
Therefore, the charge required to deposit 0.2
Mmoles of metal is:
Charge passing through the cell per minute = 0.2
M×9.65 ×104C/min
Step 3: Calculate the current passing through the cell. The current passing
through the cell can be calculated using the formula:
I=Q
t
where Iis the current, Qis the charge passing through the cell per minute, and
tis the time in minutes.
I=0.2×9.65 ×104
tM A
Question 34
Question
A copper sulfate solution is electrolyzed using inert electrodes. If 0.2 Faraday
of electricity is passed through the solution, what mass of copper is deposited
at the cathode? Assume 100
33
Solution
Faraday’s laws of electrolysis state that the amount of chemical change produced
by a current is proportional to the quantity of electricity passed through the
electrolyte. The amount of substance deposited or reacted can be calculated by
the equation:
Amount of substance = Charge passed
Faraday constant ×Valency
In the case of copper sulfate, the valency of copper is 2 and the Faraday
constant is 96500 C/mol.
Step 1: Calculate the amount of substance deposited at the cathode.
Amount of substance = 0.2 Faraday
96500 C/mol ×2=0.2
193000 mol
Step 2: Calculate the mass of copper deposited using the molar mass of
copper (Cu is 63.5 g/mol).
Mass of copper = Amount of substance ×Molar mass = 0.2
193000 ×63.5 g
Therefore, 0.000328 of copper will be deposited at the cathode.
Question 35
Question
An electric current of 5.0 A is passed through a solution of copper(II) sulfate
for 30 minutes. If the current causes the reduction of copper(II) ions to copper
metal, calculate:
1. The amount of copper deposited.
2. The mass of copper deposited, given that the molar mass of copper is
63.55 g/mol.
Solution
Let’s start by calculating the amount of charge passed through the solution us-
ing Faraday’s laws of electrolysis:
Step 1: Find the amount of charge passed The formula relating charge,
current, and time is given by:
Charge (Q) = Current (I) ×Time (t)
34
Question 2
Question
A copper sulfate solution is electrolyzed using a current of 2.50 A. If 1.00 g of
copper is deposited at the cathode in 30 minutes, determine the half-reaction at
the cathode and the mass of copper sulfate consumed during the electrolysis.
Solution
Step 1: Write the half-reaction at the cathode. The half-reaction for the depo-
sition of copper at the cathode is:
Cu2+ + 2e−→Cu
Step 2: Calculate the number of moles of copper deposited at the cathode.
Given that the current is 2.50 A and the time is 30 minutes, we first convert
the time to seconds:
30 minutes ×60 seconds/minute = 1800 seconds
Next, we calculate the charge passed through the circuit:
Charge = Current ×Time = 2.50 A ×1800 s = 4500 C
Since 1 mole of electrons is equivalent to 1 Faraday (96485 C), we can cal-
culate the number of moles of copper deposited:
Number of moles of Cu = Charge
2×96485 =4500
192970
Step 3: Calculate the mass of copper deposited. Given that the molar mass
of copper is 63.55 g/mol, we can find the mass of copper deposited:
Mass of Cu = Number of moles of Cu×Molar mass of Cu = Number of moles of Cu×63.55 g/mol
Step 4: Calculate the mass of copper sulfate consumed. Since the stoi-
chiometry between copper and copper sulfate is 1:1, the mass of copper sulfate
consumed is equal to the mass of copper deposited.
Therefore, the mass of copper sulfate consumed during the electrolysis is the
same as the mass of copper deposited at the cathode.
Question 3
Question
A solution of copper sulfate (CuSO4) is electrolyzed using a current of 2.50 A for
2.50 hours. If copper is deposited at the cathode, determine the mass of copper
deposited. (Given: Atomic mass of copper = 63.5 g/mol, Faraday’s constant =
96,500 C/mol)
3
Solution
Step 1: Calculate the total charge passed through the solution.
Total charge = Current ×Time
Total charge = 2.50 A ×2.50 hours ×3600 s/hour
Total charge = 22,500 C
Step 2: Calculate the moles of copper deposited using Faraday’s law.
Moles of copper = Total charge
1 F ×1 mol
96,500 C
Moles of copper = 22,500 C
96,500 C/mol
Moles of copper = 0.2337 mol
Step 3: Calculate the mass of copper deposited.
Mass of copper = Moles of copper ×Atomic mass of copper
Mass of copper = 0.2337 mol ×63.5 g/mol
Mass of copper = 14.84 g
Therefore, the mass of copper deposited is 14.84 g.
Question 4
Question
In an electrolysis experiment, a current of 2.5 A was passed through a solution
of copper sulfate for 30 minutes. If 1.5 g of copper was deposited at the cathode,
calculate the molar mass of copper.
Given: Faraday constant, F = 96485 C mol−1.
Solution
Step 1: Calculate the total charge passed through the electrolyte using the
equation Q=I×t, where Qis the total charge, Iis the current, and tis the
time.
Total charge, Q= 2.5A×30 min = 2.5A×30 min
60 s/min = 75 C
Step 2: Calculate the number of moles of copper deposited at the cathode
using the equation n=m
M, where nis the number of moles, mis the mass, and
Mis the molar mass.
Number of moles of copper, n=1.5g
M
4
Step 3: Calculate the number of electrons required for the deposition of 1
mole of copper ions using the half-reaction of copper deposition: Cu2+ + 2e−→
Cu.
Number of electrons, = 2 (per mole of Cu)
Step 4: Calculate the number of coulombs required for the deposition of 1
mole of copper ions using Faraday’s laws: Q=n×F×z, where zis the number
of electrons per ion in the half-reaction.
Q=n×F×z
75 C=1.5g
M×96485 C mol−1×2
Step 5: Solve for the molar mass of copper, M.
1.5×96485 ×2
75 =M
M≈7726 g/mol
Therefore, the molar mass of copper is approximately 7726 g/mol.
Question 5
Question
A student is conducting an experiment on Faraday’s laws of electrolysis. In one
trial, the student passes a current of 2.5 A through an electrolytic cell containing
a copper (Cu) sulfate solution. If the student observes the deposition of 12.0
g of copper metal in 20 minutes, determine the electrochemical equivalent of
copper in the electrolyte. Assume 100
Solution
Step 1: Find the total charge passed through the cell. Given that the current
is 2.5 A and the time is 20 minutes, we first convert the time to seconds:
Time (s) = 20 ×60 = 1200 s
The total charge passed through the cell can be calculated using the formula:
Q=I×t
Q= 2.5 A ×1200 s = 3000 C
Step 2: Calculate the amount of charge required to deposit 1 mole of copper.
The molar mass of copper (Cu) is approximately 63.5 g/mol. Therefore, 1 mole
of copper requires 1 mole of electrons which is equivalent to a charge of:
1 mol of Cu ×2×96,485 C/mol = 2 ×96,485 C
5
= 192,970 C
Step 3: Determine the number of moles of copper deposited. Using the
deposited mass of copper (12.0 g) and the molar mass of copper (63.5 g/mol),
we calculate the number of moles of copper:
Moles of Cu = 12.0 g
63.5 g/mol ≈0.189 mol
Step 4: Compute the electrochemical equivalent of copper. Comparing the
charge passed through the cell with the charge required to deposit the observed
amount of copper: 3000 C
0.189 mol ≈15,873 C/mol
Therefore, the electrochemical equivalent of copper in the electrolyte is ap-
proximately 15,873 C/mol.
Question 6
Question
Faraday’s laws of electrolysis describe the relationship between the amount of
substance produced in an electrolytic cell and the electric charge passed through
it. A student is performing an electrolysis experiment using a solution of cop-
per(II) sulfate with copper electrodes. Calculate the mass of copper deposited
at the cathode after passing a current of 2.50 A for 2.00 hours. Given: Atomic
mass of copper = 63.5 g/mol, Faraday’s constant = 96,485 C/mol, and assume
100
Solution
Step 1: Calculate the total charge passed through the electrolytic cell.
Charge (C) = Current (A) ×Time (s)
Convert the time from hours to seconds (1 hour = 3600 s).
Time = 2.00 ×3600 = 7200 s
Charge = 2.50 A ×7200 s = 18,000 C
Step 2: Determine the number of moles of electrons transferred.
Number of moles of electrons = Charge (C)
F
where Fis Faraday’s constant.
Number of moles of electrons = 18,000 C
96,485 C/mol ≈0.187 mol
6
Step 3: Calculate the mass of copper deposited at the cathode using the
mole ratio and the atomic mass of copper.
Mass of copper (g) = 0.187 mol ×63.5 g/mol = 11.85 g
Therefore, the mass of copper deposited at the cathode after passing a cur-
rent of 2.50 A for 2.00 hours is 11.85 grams.
Question 7
Question
For an electrolytic cell, if a current of 2.5 A is passed through a solution of
molten lead(II) bromide, how long (in seconds) would it take to deposit 50.0 g
of lead metal?
(Assume 1 F is required to deposit 107.9 g of Ag.)
Solution
Step 1: Calculate the amount of charge required to deposit 50.0 g of lead metal.
Given that 1 Faraday (F) is required to deposit 107.9 g of Ag, the amount of
charge required to deposit 50.0 g of Pb is:
Charge = 50.0 g
107.9 g/F×1 F
Charge = 0.463F
Step 2: Use the formula Q = It to find the time (t) taken to deposit this
amount of charge. Given current (I) = 2.5 A, charge (Q) = 0.463 F, we have:
t=Q
I
t=0.463 C
2.5 A
t= 0.1852 s
Therefore, it would take 0.1852 seconds to deposit 50.0 g of lead metal.
Question 8
Question
A student performs an electrolysis experiment using a solution of copper(II)
sulfate. The student applies a current of 2.5 A for 30 minutes to a copper(II)
sulfate solution using platinum electrodes. If the student measures that 0.5 g
of copper is deposited on the cathode during this time, calculate the number of
moles of electrons transferred during the electrolysis.
Given: Faraday’s constant F= 96500 C/mol
7
Solution
Step 1: Calculate the charge passed through the circuit. The charge passed
through the circuit 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. Convert the time to seconds:
t= 30 minutes ×60 s/min = 1800 s
Now, substitute the given values to find Q:
Q= 2.5 A ×1800 s = 4500 C
Step 2: Determine the number of moles of electrons transferred. Since 1
Faraday (F) is equivalent to the charge of one mole of electrons (Avogadro’s
number), we can calculate the number of moles of electrons transferred using
the formula:
Moles of electrons transferred = Q
F
Substitute the values to find the number of moles of electrons transferred:
Moles of electrons transferred = 4500 C
96500 C/mol
Moles of electrons transferred = 0.0468 mol
Therefore, the number of moles of electrons transferred during the electrol-
ysis of copper(II) sulfate solution is 0.0468 mol.
Question 9
Question
An electrolytic cell is set up where a current of 3.00 A is passed through a
solution of CuSO4. If 0.500 g of copper is deposited at the cathode in 30
minutes, determine the faraday constant.
Solution
Step 1: Determine the moles of copper deposited. Given that the molar mass of
copper is 63.55 g/mol, the number of moles of copper deposited can be calculated
as:
Moles of Cu = Mass of Cu deposited
Molar mass of Cu =0.500 g
63.55 g/mol
8
Step 2: Calculate the charge passed through the circuit. The charge passed
can be calculated using the formula: Q=It, where Iis the current in amperes
(3.00 A) and tis the time in seconds (30 minutes = 1800 seconds).
Q= 3.00 A ×1800 s
Step 3: Calculate the Faraday constant. Since 1 Faraday is equivalent to the
charge required to deposit 1 mole of a monovalent ion (e.g., Cu2+ in this case),
the Faraday constant can be calculated as:
Faraday constant = Q
n×F
where nis the moles of Cu deposited and Fis Faraday constant.
Step 4: Substitute the values and calculate the Faraday constant. Substitute
the values calculated in steps 1 and 2 into the formula from step 3, and solve
for F. Remember to convert the time from minutes to seconds.
Faraday constant = 3.00 A ×1800 s
0.500 g/63.55 g/mol
Question 10
Question
In an electrolysis experiment, a constant current of 2.5 A is passed through
an aqueous solution of copper(II) sulfate for 20 minutes. If copper metal is
deposited at the cathode, calculate the mass of copper deposited. (Atomic mass
of copper = 63.5 g/mol, Faraday’s constant = 96,485 C/mol)
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current,
I= 2.5 A Time, t= 20 minutes = 1200 s
The total charge, Q, passed through the solution can be calculated using the
formula:
Q=I×t
Q= 2.5 A ×1200 s = 3000 C
Step 2: Calculate the moles of electrons (n) that have passed through the
solution. Since 1 Faraday is the charge of 1 mole of electrons, we can use the
relation:
1 mol e−= 1 F
Given Faraday’s constant, F= 96,485 C/mol, we can calculate:
9
n=Q
F=3000 C
96,485 C/mol ≈0.0311 mol
Step 3: Calculate the mass of copper deposited. The balanced half-reaction
for the reduction of copper(II) ions to copper metal is:
Cu2+(aq)+2e−→Cu(s)
From the reaction, we can see that 2 moles of electrons reduce 1 mole of
Cu2+ ions. Hence, the moles of copper deposited will be equal to the moles of
electrons passed.
Thus, the mass of copper deposited can be calculated using the formula:
Mass of Cu = n×Molar mass of Cu
Mass of Cu = 0.0311 mol ×63.5 g/mol ≈1.97 g
Therefore, approximately 1.97 g of copper will be deposited at the cathode.
Question 11
Question
A student is conducting an electrolysis experiment using a copper(II) sulfate
solution with copper electrodes. The student passes a current of 2.00 A through
the solution for 90.0 minutes. During this time, copper is deposited on one
electrode. If the student measures that 0.350 g of copper is deposited, determine
the yield of the process in percent.
Given: Atomic mass of copper = 63.5 g/mol
Solution
Step 1: Calculate the total charge passed through the solution. Given: Current
(I) = 2.00 A Time (t) = 90.0 minutes
To calculate the total charge (Q) passed through the solution, use the for-
mula:
Q=I×t
Plugging in the values:
Q= 2.00 A×90.0min ×1min
60 s×1C
1A·s
Q= 2.00 A×5400 s×1C
Q= 10800 C
Therefore, the total charge passed through the solution is 10800 C.
10
Step 2: Calculate the number of moles of copper deposited. Given: Mass of
copper deposited (m) = 0.350 g Atomic mass of copper = 63.5 g/mol
To calculate the number of moles of copper deposited, use the formula:
Number of moles = mass
molar mass
Plugging in the values:
Number of moles = 0.350 g
63.5g/mol
Number of moles = 0.00551 mol
Therefore, the number of moles of copper deposited is 0.00551 mol.
Step 3: Calculate the yield of the process in percent. The theoretical number
of moles of copper that should have been deposited can be calculated from
the Faraday’s laws of electrolysis. The molar equivalent of one Faraday of
charge (1 F) is equal to the number of moles of electrons in one mole of copper
(1 F = 2 mol).
The theoretical number of moles of copper (ntheoretical) that should have
been deposited can be calculated using:
ntheoretical =Q
F
Where: Q= Total charge passed through the solution (10800 C) F= Fara-
day constant = 96485 C/mol
Plugging in the values:
ntheoretical =10800 C
96485 C/mol
ntheoretical = 0.112mol
Now, the yield of the process can be calculated using the formula:
Yield (%) = actual yield
theoretical yield ×100%
Plugging in the values:
Yield (%) = 0.00551 mol
0.112 mol ×100%
Yield (%) = 0.0491
0.112 ×100%
Yield (%) = 43.8%
Therefore, the yield of the process is 43.8
11
Question 12
Question
Faraday’s laws of electrolysis state that the amount of chemical reaction that
occurs during electrolysis is proportional to the quantity of electricity that flows
through the cell.
Suppose a solution of silver chloride is electrolyzed using a current of 2.50
A for 2.00 hours. If 0.500 g of silver is deposited at the cathode, determine the
Faraday constant (96485 C/mol).
Solution
Step 1: Find the total charge passing through the cell. Given: Current, I =
2.50 A Time, t = 2.00 hours = 7200 s
The total charge passing through the cell can be calculated using the formula:
Q=It
Substitute the given values:
Q= 2.50 A ×7200 s = 18000 C
Step 2: Determine the number of moles of silver deposited. The amount of
charge required to deposit 1 mole of silver is given by the Faraday constant:
F= 96485 C/mol
The number of moles of silver deposited can be calculated using the formula:
moles of Ag = mass of Ag
molar mass of Ag =0.500 g
107.87 g/mol
Step 3: Calculate the Faraday constant. The Faraday constant relates the
total charge passed through the cell to the number of moles of substance de-
posited:
F=Q
n
Substitute the values:
F=18000 C
0.500 g
107.87 g/mol
= 21404 C/mol
Therefore, the Faraday constant is 21404 C/mol.
12
Question 13
Question
An aqueous solution of NaCl (sodium chloride) is electrolyzed using platinum
electrodes. If a current of 2.50 A is passed through the solution for 2.00 hours,
calculate the mass of sodium metal deposited at the cathode. (Given: Faraday
constant F= 96,485 C/mol, molar mass of Na is 23.0 g/mol and the ionic
charge of Na is +1.)
Solution
Step 1: Calculate the total charge passed through the electrolytic cell. The total
charge passed is given by the formula:
Q=I×t
where Qis the charge in coulombs, Iis the current in amperes, and tis the
time in seconds.
Given I= 2.50 A and t= 2.00 hours = 2.00 ×3600 s, we have:
Q= 2.50 A ×2.00 ×3600 s
Q= 2.50 C/s ×7200 s
Q= 18,000 C
Step 2: Calculate the number of moles of electrons passed through the elec-
trolytic cell. Since 1 faraday (F) is equivalent to 1 mole of electrons, we can
calculate the number of moles of electrons passed using the Faraday constant:
Moles of electrons (n) = Q
F
Moles of electrons (n) = 18,000 C
96,485 C/mol
Moles of electrons (n) ≈0.1865 mol
Step 3: Calculate the moles of sodium that will be deposited. Since each
sodium ion (Na+) gains one electron to form a sodium atom, the moles of sodium
deposited will be equal to the moles of electrons passed:
Moles of Na deposited = 0.1865 mol
Step 4: Calculate the mass of sodium deposited. Using the molar mass of
sodium (Na = 23.0 g/mol), we can calculate the mass of sodium deposited:
13
Mass of Na deposited = Moles of Na deposited ×Molar mass of Na
Mass of Na deposited = 0.1865 mol ×23.0 g/mol
Mass of Na deposited ≈4.29 g
Therefore, the mass of sodium metal deposited at the cathode is approxi-
mately 4.29 g.
Question 14
Question
A copper sulfate solution (CuSO4) is electrolyzed using platinum electrodes.
If a current of 2.5 A is passed through the solution for 2 hours, what mass of
copper is deposited on the cathode? (Given: 1 F araday = 96,500 C/mol and
Molar mass of Cu = 63.5g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given that
the current is 2.5 A and the time is 2 hours, we first convert the time from hours
to seconds:
2hours = 2 ×3600 seconds = 7200 seconds
Now, calculate the total charge passed:
Q=I×t= 2.5A×7200 s= 18000 C
Step 2: Calculate the number of moles of electrons passed. Since 1 Faraday
is equivalent to 96,500 C/mol, we can calculate the number of moles of electrons
passed:
n=Q
F=18000 C
96500 C/mol ≈0.1865 mol
Step 3: Calculate the mass of copper deposited. The reaction at the cathode
is:
Cu2+ + 2e−→Cu
From the balanced equation, we see that one mole of copper is deposited by 2
moles of electrons. Therefore, the moles of copper deposited will be half the
moles of electrons passed:
nCu =n
2=0.1865 mol
2= 0.09325 mol
14
Finally, calculate the mass of copper deposited on the cathode:
Mass = nCu ×Molar mass of Cu = 0.09325 mol ×63.5g/mol ≈5.92 g
Therefore, approximately 5.92 g of copper is deposited on the cathode after
2 hours of electrolysis.
Question 15
Question
A student is conducting an electrolysis experiment using two electrodes con-
nected to a power supply. The student observes that the mass of one of the
electrodes decreases over time while the mass of the other electrode increases.
Explain this observation in terms of Faraday’s laws of electrolysis.
Solution
To explain the observation of the changing masses of the electrodes during
electrolysis, we can refer to Faraday’s laws of electrolysis. These laws are based
on the principles of conservation of mass and conservation of charge.
Step 1: Oxidation and Reduction Reactions
During electrolysis, oxidation and reduction reactions occur at the respective
electrodes. The anode undergoes oxidation, where electrons are lost, while the
cathode undergoes reduction, where electrons are gained.
Step 2: Faraday’s First Law
Faraday’s first law states that the amount of chemical change produced by
an electric current is proportional to the quantity of electricity passed through
the electrolyte. This can be represented by the equation:
Chemical change ∝Electric charge passed
Step 3: Electrolysis of Electrodes
In this case, as the anode undergoes oxidation, metal ions from the anode
dissolve into the electrolyte, leading to a decrease in mass of the anode. At
the same time, at the cathode, metal ions from the electrolyte get reduced and
deposited onto the cathode, causing an increase in mass.
Therefore, the observed changes in the masses of the electrodes can be ex-
plained by Faraday’s laws of electrolysis, specifically the relationship between
the amount of electricity passed, the chemical changes occurring at the elec-
trodes, and the resulting changes in mass.
15
Question 16
Question
A solution contains both iron(II) sulfate and copper(II) sulfate. When a cur-
rent is passed through this solution for a certain amount of time, the electrode
connected to the negative terminal gained 2.8 grams. If the faradic (electro-
chemical) efficiency is 80
Given: - Mass of iron deposited at the negative electrode = 2.8 grams -
Faradic efficiency = 80
Solution
Step 1: Determine the molar mass of Iron and Copper Iron (Fe) has a molar
mass of 55.85 g/mol. Copper (Cu) has a molar mass of 63.55 g/mol.
Step 2: Calculate the number of moles of Iron deposited Given that the
mass of iron deposited is 2.8 grams, we can calculate the number of moles of
iron deposited: Number of moles of iron = Mass of iron deposited / Molar mass
of iron Number of moles of iron = 2.8 g / 55.85 g/mol
Step 3: Calculate the theoretical amount of Copper deposited Since iron
has a valency of 2 and the faradic efficiency is 802 moles of Fe : 1 mole of Cu
Number of moles of Cu = 0.8 * Number of moles of Fe
Step 4: Calculate the mass of Copper deposited Mass of Cu deposited =
Number of moles of Cu * Molar mass of Cu
Step 5: Substitute the values into the formula and calculate the mass of
Copper deposited.
Question 17
Question
A student performs an electrolysis experiment using a solution of silver nitrate,
AgNO3. During the experiment, the student observes that the mass of silver
deposited at the cathode is 1.68 g. Calculate the total charge that passed
through the electrolyte.
Given:
Atomic mass of silver, m(Ag) = 107.87 g/mol
Number of electrons needed to reduce one mole of silver ions, n= 1
Solution
Step 1: Determine the number of moles of silver deposited at the cathode.
Mass of silver deposited = 1.68 g
Molar mass of silver = 107.87 g/mol
16
Number of moles of silver = Mass of silver deposited
Molar mass of silver
Number of moles of silver = 1.68
107.87 ≈0.0156 mol
Step 2: Calculate the total charge that passed through the electrolyte.
Q=n×F×N
where Q= total charge (C)
n= number of moles of silver deposited
F= Faraday constant = 96500 C/mol
N= number of electrons needed to reduce one mole of silver = 1
Q= 0.0156 ×96500 ×1≈1505.4 C
Therefore, the total charge that passed through the electrolyte is approxi-
mately 1505.4 C.
Question 18
Question
In an electrolytic cell, a current of 2.5 A is passed for 2 hours through a solution
of silver nitrate AgNO3. If 1.5 g of silver is deposited at the cathode, determine
the faraday constant.
Solution
Let’s start by calculating the amount of electric charge passed through the
circuit.
Step 1: Calculate the total charge passed. The total charge passed can be
calculated using the formula:
Total charge (C) = Current (A) ×Time (s)
Given that current = 2.5 A and time = 2 hours = 7200 s, we have:
Total charge = 2.5×7200 = 18000 C
Step 2: Calculate the number of moles of silver deposited. Using the formula
for calculating the number of moles:
moles = mass (g)
molar mass (g/mol)
17
Given that the mass of silver deposited = 1.5 g, and the molar mass of silver
= 107.87 g/mol, we have:
moles of silver = 1.5
107.87 ≈0.014 mol
Step 3: Calculate the number of electrons passed. Since each mole of silver
requires 1 mole of electrons to be deposited, the number of electrons passed will
be equal to the number of moles of silver deposited. Therefore, the number of
electrons passed = 0.014 mol.
Step 4: Calculate the Faraday constant. One Faraday is equal to the charge
of one mole of electrons, which is approximately equal to 96,485 C/mol. There-
fore, the Faraday constant is:
F=Total charge (C)
Number of electrons =18000
0.014 ≈1,285,714 C/mol
Question 19
Question
During the electrolysis of a molten compound, it was found that 0.5 grams
of the compound was decomposed by a current of 0.5 amperes flowing for 2
hours. If the compound contains only one type of cation and one type of anion,
determine:
1. The equivalent weight of the cation.
2. The valency of the cation.
Solution
1. We can find the equivalent weight of the cation using Faraday’s laws of
electrolysis. The mass of the substance decomposed is related to the equivalent
weight by the formula:
Equivalent weight = Atomic weight
n×F
where nis the valency of the ion being discharged and Fis the Faraday constant
(96500 C/mol). Given that 0.5 grams of the compound was decomposed by a
current of 0.5 amperes flowing for 2 hours, we can calculate the equivalent weight
as follows:
Step 1: Determine the total charge passed through the electrolyte.
The total charge Qcan be calculated using the formula:
Q=I×t
18
where Iis the current (0.5 A) and tis the time in seconds (2 hours = 7200
seconds).
Q= 0.5 A ×7200 s = 3600 C
Step 2: Calculate the equivalent weight of the cation. Given that 0.5
grams of the compound was decomposed, we first need to determine the number
of moles of the compound decomposed using its atomic weight. Let’s assume the
formula of the compound is MmXx, whereM representsthecationandXrepresentstheanion.Moles of M =
Mass
Atomic weight of M =0.5 g
Atomic weight of M Since the total charge passed through the
electrolyte is equivalent to the charge on 1 mole of M (as only ions of M are
involved in electrolysis), we have:
Charge = n×F= Moles of M ×F
Substitute the values of charge and moles of M:
3600 = 0.5
Atomic weight of M ×96500
Solve for the atomic weight of M to find the equivalent weight.
2. Now that we have the equivalent weight of the cation, we can determine
the valency of the cation.
Step 3: Calculate the valency of the cation. The valency of the cation,
denoted by n, can be found using the formula:
n=Atomic weight
Equivalent weight ×F
Given that we have found the equivalent weight in the previous step, we can
substitute its value into the formula along with the atomic weight of the cation
to obtain the valency.
Question 20
Question
Explain Faraday’s first and second laws of electrolysis. Use mathematical ex-
pressions to define each law and provide an example to illustrate each law.
Solution
Faraday’s First Law of Electrolysis: This law states that the amount of
chemical change produced by the passage of current through an electrolyte is
proportional to the quantity of electricity passed through it.
Mathematical expression: If mis the mass of a substance deposited or
liberated at an electrode, Qis the quantity of electricity passed through the
19
electrolyte, and Mis the atomic or molecular weight of the substance, then
according to Faraday’s First Law of Electrolysis,
m=kQ,
where kis a constant of proportionality.
Example: Consider the electrolysis of molten lead bromide (PbBr2) using
a current of 2.50 A for 1 hour. Calculate the mass of lead liberated at the
cathode. The atomic masses are Pb = 207, Br = 80.
Step 1: Calculate the quantity of electricity passed.
Q=I×t= 2.50 A ×1 hour = 2.50 ×3600 C = 9000 C
Step 2: Calculate the equivalent weight of lead (M) in grams.
M= 207 g/mol = 207 g/equiv
Step 3: Using Faraday’s First Law of Electrolysis:
m=kQ
Substitute k=M
96500 (since 1 Faraday = 96500 C) and Q= 9000 C.
m=M
96500 ×Q=207
96500 ×9000 = 19.37 g
Faraday’s Second Law of Electrolysis: This law states that if the same
quantity of electricity is passed through different electrolytes connected in series,
the mass of the substances liberated or deposited at the electrodes is directly
proportional to their chemical equivalent weights.
Mathematical expression: If m1and m2are the masses of two substances
liberated or deposited by the same quantity of electricity, and M1and M2are
their respective atomic or molecular weights, then according to Faraday’s Second
Law of Electrolysis, m1
m2
=M1
M2
Example: When 0.20 g of silver is plated out by the passage of a current,
what mass of copper will be deposited by the same quantity of electricity (atomic
masses: Ag = 108, Cu = 64)?
Step 4: Given m1= 0.20 g, M1= 108 g/mol, and M2= 64 g/mol. Use
Faraday’s Second Law of Electrolysis:
m1
m2
=M1
M2
0.20
m2
=108
64
m2=0.20 ×64
108 = 0.118 g
Therefore, 0.118 g of copper will be deposited by the same quantity of elec-
tricity.
20
Question 21
Question
Explain Faraday’s laws of electrolysis and how they can be used to calculate the
amount of substance deposited or liberated during an electrolytic reaction.
Solution
Faraday’s laws of electrolysis describe the relationship between the amount of
substance produced or consumed during electrolysis and the amount of electric-
ity passed through the electrolyte. They can be summarized as: 1. The amount
of chemical change produced by a current is proportional to the quantity of
electricity that passes through the electrolyte. 2. The quantities of different
substances liberated by the same quantity of electricity are proportional to their
equivalent weights.
Step 1: Explanation of Faraday’s Laws Let Qbe the total charge
passing through the electrolyte in coulombs (Q=I×t, where Iis the current in
amperes and tis the time in seconds), nbe the number of moles of the substance
liberated or consumed, zbe the valence of the substance, and Fbe the Faraday
constant (F= 96,485 C/mol). Faraday’s laws can be mathematically expressed
as: 1. Amount of substance liberated or consumed: n=Q
zF 2. Relationship
between the amounts of different substances: n1
z1=n2
z2
Step 2: Calculation using Faraday’s Laws Suppose we electrolyze an
aqueous solution of copper(II) sulfate using a current of 2.5 A for 30 minutes.
Calculate the mass of copper deposited on the electrode.
Given: I= 2.5 A, t= 30 minutes = 1800 s, z= 2 (valence of copper),
atomic mass of copper = 63.5 g/mol.
First, calculate the charge passed through the electrolyte: Q=I×t=
2.5×1800 = 4500 C
Now, use Faraday’s law to find the number of moles of copper deposited:
n=Q
zF =4500
2×96485 ≈0.0234 moles
Finally, calculate the mass of copper: Mass = n×Atomic mass = 0.0234 ×
63.5≈1.487 g
Therefore, approximately 1.487 grams of copper will be deposited on the
electrode.
Question 22
Question
Consider an electrolysis cell where a copper(I) sulfate solution is electrolyzed
using copper electrodes. The electrolysis cell operates for 1 hour at a constant
current of 2.5 A. Calculate the mass of copper deposited on the cathode during
this time.
21
Given: - The molar mass of copper is 63.55 g/mol. - 1 Faraday (F) is
equivalent to 96,485 C/mol.
Solution
Step 1: Calculate the total charge passed through the cell. Given current I= 2.5
A and time t= 1 hour = 3600 s. Using the formula Q=It, where Qis the
charge passed through the cell.
Charge passed (Q) = 2.5 A ×3600 s = 9000 C
Step 2: Determine the number of moles of copper deposited. Since 1 Fara-
day is equivalent to 96,485 C/mol, we can find the number of moles of copper
deposited using the formula:
Moles of copper = Charge passed (C)
Faraday constant (C/mol)
Moles of copper = 9000 C
96485 C/mol = 0.0934 mol
Step 3: Calculate the mass of copper deposited. Given the molar mass
of copper is 63.55 g/mol, we can find the mass of copper deposited using the
formula:
Mass of copper = Moles of copper ×Molar mass of copper
Mass of copper = 0.0934 mol ×63.55 g/mol = 5.928 g
Therefore, the mass of copper deposited on the cathode during 1 hour of
electrolysis is 5.928 g.
Question 23
Question
A solution containing NaCl is electrolyzed using a current of 3.20 A for 45.0
minutes. If 2.00 g of Na is deposited at the cathode, determine the half-reaction
occurring at the cathode and calculate the number of moles of electrons trans-
ferred during the electrolysis.
Solution
Step 1: Calculate the amount of charge passed through the electrolyte. Given:
Current, I= 3.20 A Time, t= 45.0 minutes
First, we convert the time to seconds:
t= 45.0×60 = 2700 s
22
The amount of charge, Q, passed through the electrolyte is given by:
Q=I×t= 3.20 ×2700 = 8640 C
Step 2: Determine the half-reaction occurring at the cathode. From the
amount of deposited Na, we can determine that the half-reaction at the cathode
is:
Na++e−→Na
Step 3: Calculate the number of moles of electrons transferred during the
electrolysis. To calculate the number of moles of electrons transferred, we need
to find the number of moles of Na deposited at the cathode first.
Given: Mass of Na deposited, m= 2.00 g Molar mass of Na, M= 22.99 g/mol
The number of moles of Na deposited is:
n=m
M=2.00
22.99 = 0.0870 mol
Since 1 mole of Na requires 1 mole of electrons, the number of moles of
electrons transferred is also 0.0870 mol.
Therefore, the number of moles of electrons transferred during the electrol-
ysis is 0.0870 mol.
Question 24
Question
A solution of copper(II) sulfate CuSO4is electrolyzed using a current of 2.0
A for 10 minutes. If the mass of copper deposited at the cathode is 2.45 g,
calculate the molar mass of copper(II) sulfate.
Solution
Step 1: Calculate the total charge passed through the electrolyte. Given that
the current is 2.0 A and the time is 10 minutes, which is equal to 600 seconds,
the total charge passed can be calculated using the formula:
Q=It
Q= 2.0 A ×600 s
Q= 1200 C
Step 2: Determine the number of moles of electrons passed. Since each mole
of copper(II) ions requires 2 moles of electrons for reduction, we can calculate
the number of moles of electrons passed using the formula:
moles of electrons = Q
F
23
where Fis the Faraday’s constant (96500 C mol−1).
moles of electrons = 1200 C
96500 C mol−1
moles of electrons ≈0.0124 mol
Step 3: Calculate the number of moles of copper deposited at the cathode.
From the balanced equation for the electrolysis of copper(II) sulfate (Cu2+ +
2e−→Cu), we see that 1 mole of copper corresponds to 2 moles of electrons.
Therefore, the number of moles of copper deposited can be calculated as follows:
moles of copper = 0.5×moles of electrons
moles of copper = 0.5×0.0124
moles of copper ≈0.0062 mol
Step 4: Calculate the molar mass of copper(II) sulfate. We are given that
the mass of copper deposited is 2.45 g. Therefore, the molar mass of copper(II)
sulfate can be calculated using the formula:
Molar mass of CuSO4=Mass of copper
moles of copper
Molar mass of CuSO4=2.45 g
0.0062 mol
Molar mass of CuSO4≈395 g/mol
Therefore, the molar mass of copper(II) sulfate is approximately 395 g/mol.
Question 25
Question
A solution of copper sulfate (CuSO4) is electrolyzed using platinum electrodes.
If a current of 2.50 A is passed through the solution for 30.0 minutes, what
mass of copper is deposited at the cathode? (Given: 1 F = 96500 C/mol,
M(Cu) = 63.5 g/mol)
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. Given
that the current is I= 2.50 A and the time is t= 30.0 min, first convert the
time to seconds:
t= 30.0×60 = 1800 s.
Then, calculate the total charge passed using:
Q=I×t= 2.50 A ×1800 s.
24
Step 2: Determine the number of moles of electrons passed. Since 1 Faraday
(F) represents 96500 C:
1 F = 96500 C ⇒Q=1
96500 F.
Thus, the number of moles of electrons passed is:
n=Q
96500 mol.
Step 3: Use the stoichiometry to find the moles of copper deposited. Since
each mole of copper requires 2 moles of electrons to be deposited:
2 mol Cu : 2 mol e−⇒1 mol Cu : 1 mol e−.
Therefore, the moles of copper deposited at the cathode is equal to the moles
of electrons passed in Step 2.
Step 4: Calculate the mass of copper deposited. Finally, calculate the mass
of copper deposited using the molar mass of copper:
m=n×M(Cu) = n×63.5 g/mol.
Question 26
Question
Faraday’s laws of electrolysis state that the amount of a substance produced
during electrolysis is proportional to the quantity of electricity passed through
the electrolyte solution.
Consider an electrolysis experiment where a current of 3.2 A is passed
through a copper(II) sulfate solution for 45 minutes. If the mass of copper
deposited on the cathode is found to be 0.72 g, determine the half-reaction that
occurs at the cathode and calculate the Faraday constant (F) based on this
experiment.
Solution
Step 1: Calculate the total charge passed through the electrolyte solution. The
total charge (Q) can be calculated using the formula:
Q=I×t
Given: Current, I= 3.2 A, Time, t= 45 minutes = 2700 seconds
Substitute the given values into the formula:
Q= 3.2 A ×2700 s = 8640 C
25
Step 2: Determine the half-reaction at the cathode. The half-reaction at
the cathode involves the reduction of copper(II) ions to form solid copper. The
balanced half-reaction is:
Cu2+ + 2e−→Cu
So, 2 electrons are required to reduce 1 copper(II) ion.
Step 3: Calculate the number of moles of copper deposited. The number of
moles of copper deposited can be calculated using the formula:
moles = mass
molar mass
Given: Mass of copper deposited, 0.72 g
The molar mass of copper (Cu) is 63.55 g/mol.
Substitute the given values into the formula:
moles = 0.72 g
63.55 g/mol ≈0.01133 mol
Step 4: Calculate the number of electrons transferred. Since 2 electrons are
required to reduce 1 copper(II) ion, the total number of electrons transferred
can be calculated as:
Total electrons = 2 ×moles of copper deposited
Total electrons = 2 ×0.01133 mol = 0.02266 mol
Step 5: Determine the Faraday constant. The Faraday constant (F) repre-
sents the charge of 1 mole of electrons and is equal to the charge of 1 mole of
electrons = 1 Faraday. It can be calculated using the formula:
F=Q
Total electrons
Substitute the calculated values into the formula:
F=8640 C
0.02266 mol ≈381103 C/mol
Therefore, the Faraday constant based on this experiment is approximately
381103 C/mol.
Question 27
Question
An electroplating experiment is set up using a solution of silver nitrate (AgNO3)
with two electrodes connected by a battery. The current passing through the
circuit is 0.5 A for 30 minutes. Calculate the amount of silver deposited on one
of the electrodes during this time.
(Assume the molar mass of silver is 107.87 g/mol, and the Faraday constant
is 96485 C/mol.)
26
Solution
Step 1: Calculate the total charge passed through the circuit. Given that the
current is 0.5 A and the time is 30 minutes, we first convert the time to seconds:
30 minutes = 30 ×60 seconds = 1800 seconds.
The total charge passed through the circuit can be calculated using the
formula Q=I×t, where Iis the current and tis the time: Q= 0.5 A×1800 s =
900 C.
Step 2: Convert the charge to moles of electrons. From Faraday’s laws
of electrolysis, we know that one mole of electrons has a charge of 96485 C.
Therefore, the number of moles of electrons passed through the circuit is given
by: Moles of electrons = 900 C
96485 C/mol = 0.00933 mol.
Step 3: Calculate the amount of silver deposited. In the electroplating pro-
cess of silver, each mole of silver deposited corresponds to one mole of electrons.
Given that the molar mass of silver is 107.87 g/mol, we can calculate the mass
of silver deposited using the formula: Mass of silver = Moles of electrons ×
Molar mass of silver.
Substitute the values to find the mass of silver deposited: Mass of silver =
0.00933 mol ×107.87 g/mol = 1.008 g.
Therefore, during the electrolysis process, 1.008 g of silver will be deposited
on one of the electrodes.
Question 28
Question
Faraday’s laws of electrolysis state that the mass of a substance deposited or lib-
erated at an electrode during electrolysis is directly proportional to the quantity
of electricity passed through the electrolyte.
Consider the electrolysis of molten lead(II) bromide using two inert elec-
trodes. If a current of 2.5 A is passed through the molten lead(II) bromide for
1 hour, calculate:
1. The mass of lead deposited at the cathode.
2. The volume of bromine gas evolved at the anode (at STP).
Given:
Atomic masses: Pb = 207 g/mol; Br = 80 g/mol
1 F (Faraday) of charge = 96500 C
Solution
Let’s start by calculating the number of coulombs of charge passed through the
electrolyte:
Total charge = Current ×Time
27
Total charge = 2.5 A ×3600 s = 9000 C
Step 1: Calculate the amount of substance deposited at the cathode (Pb):
First, find the number of moles of electrons passed:
Moles of electrons = Total charge
96500 C/mol
Moles of electrons = 9000 C
96500 C/mol = 0.09326 mol
Since the molten lead(II) bromide will be reduced at the cathode:
Pb2+ + 2e−→Pb
We can see that 1 mole of Pb requires 2 moles of electrons.
Therefore, the number of moles of Pb deposited at the cathode will be:
Moles of Pb = 0.09326 mol∇ · 2 = 0.04663 mol
Next, calculate the mass of lead deposited at the cathode:
Mass of Pb = Moles of Pb ×Molar mass of Pb
Mass of Pb = 0.04663 mol ×207 g/mol = 9.65 g
Therefore, the mass of lead deposited at the cathode is 9.65 g.
Step 2: Calculate the volume of bromine gas evolved at the anode:
The half-reaction at the anode will be the oxidation of bromide ions:
2Br−→Br2+ 2e−
The moles of electrons passed through the electrolyte will produce half as
many moles of bromine gas:
Moles of Br2=Moles of electrons
2= 0.09326 mol
Since 1 mole of gas occupies 24 L at STP:
Volume of Br2= Moles of Br2×24 L/mol = 0.09326 mol ×24 L/mol = 2.23 L
Therefore, the volume of bromine gas evolved at the anode is 2.23 L at STP.
Question 29
Question
A solution contains two electrolytes, copper sulfate and sodium chloride. If a
current of 2.5 A is passed through the solution for 1 hour, calculate the mass of
copper deposited. Given: atomic masses - Cu = 63.5 g/mol, Na = 23 g/mol, Cl
= 35.5 g/mol; Faraday constant = 96500 C/mol, and 1F = 96500 C. (Assume
100
28
Solution
Step 1: Calculate the total charge passed through the solution. Given current
I= 2.5Aand time t= 1 hour = 3600 seconds, we have
Q=I×t= 2.5A×3600 s= 9000 C
Step 2: Determine the number of moles of electrons passed. Since the charge
on 1 mole of electrons is equal to 1 Faraday, the number of moles of electrons
passed is given by
n=Q
96500 C/mol =9000 C
96500 C/mol ≈0.0933 mol
Step 3: Calculate the number of moles of copper deposited. From the bal-
anced chemical equation for the electrolysis of copper sulfate,
2e−+Cu2+ →Cu
we see that 2 moles of electrons deposited 1 mole of copper. Therefore,
nCu = 0.5×n= 0.5×0.0933 = 0.0467 mol
Step 4: Find the mass of copper deposited. The mass of copper deposited
can be calculated using the formula:
Mass = n×Atomic Mass of Cu = 0.0467 mol ×63.5g/mol = 2.971 g
Therefore, the mass of copper deposited is approximately 2.971 grams.
Question 30
Question
A student conducted an electrolysis experiment using a solution of copper(II)
sulfate, CuSO4, and two copper electrodes. During the experiment, they ob-
served that a current of 2.50 A flowed through the cell for 2.00 hours. The
student also measured the mass of the copper electrode connected to the posi-
tive terminal before and after the experiment and found that it had decreased
by 0.145 g.
Determine the following:
1. The total charge passed through the cell.
2. The number of moles of electrons that flowed through the cell.
3. The mass of copper deposited on the negative electrode during the 2-hour
period.
Given:
Faraday’s constant, F= 96,485 C/mol
Atomic mass of copper, 63.55 g/mol
29
Solution
1. First, let’s find the total charge passed through the cell using the formula:
Total charge = Current ×Time
Total charge = 2.50 A ×2.00 hours ×3600 s/hour
Step 1:
Total charge = 2.50 A ×2.00 hours ×3600 s/hour = 18,000 C
2. Next, to find the number of moles of electrons that flowed through the
cell, we use Faraday’s laws of electrolysis that relate the total charge passed to
the number of moles of electrons:
Total charge = Number of moles of electrons ×F araday′s constant
Number of moles of electrons = Total charge
F araday′s constant
Step 2:
Number of moles of electrons = 18,000 C
96,485 C/mol
Number of moles of electrons ≈0.1866 mol
3. Finally, to find the mass of copper deposited on the negative electrode
during the 2-hour period, we use the stoichiometry of the reaction:
2 moles of electrons + Cu2+ →Cu
Since the molar ratio between moles of electrons and moles of copper is 2:1,
we have:
Moles of copper deposited = 1
2×Number of moles of electrons
Step 3:
Moles of copper deposited = 1
2×0.1866 mol = 0.0933 mol
Now, we can find the mass of copper deposited using the molar mass of
copper:
Mass of copper deposited = Moles of copper deposited ×Molar mass of copper
Mass of copper deposited = 0.0933 mol ×63.55 g/mol
Mass of copper deposited ≈5.93 g
Therefore, the mass of copper deposited on the negative electrode during
the 2-hour period is approximately 5.93 g.
30
Question 31
Question
A solution of copper(II) ions is electrolyzed using a current of 2.00 A for 4.00
hours. If copper is deposited on one of the electrodes, calculate the mass of
copper deposited. Given that the molar mass of copper is 63.55 g/mol and the
Faraday constant is 9.65 ×104C/mol.
Solution
Step 1: Calculate the total charge passed through the solution.
Given: Current, I= 2.00 A
Time, t= 4.00 hours
The total charge, Q, passed through the solution can be calculated using the
formula:
Q=I×t
Q= 2.00 A ×4.00 hours ×3600 s/hour
Q= 2.00 ×4.00 ×3600 C
Q= 28800 C
Step 2: Calculate the number of moles of copper deposited.
The number of moles of electrons involved in the reduction of 1 mole of cop-
per(II) ions is 2. This is because each copper(II) ion gains 2 electrons to form
copper.
The number of moles of copper deposited can be calculated using the formula:
Moles of copper = Q
n×F
where: Q= 28800 C (total charge passed through the solution)
n= 2 (number of moles of electrons involved)
F= 9.65 ×104C/mol (Faraday constant)
Moles of copper = 28800
2×9.65 ×104
Moles of copper = 28800
1.93 ×105
Moles of copper ≈0.1493 mol
Step 3: Calculate the mass of copper deposited.
The mass of copper deposited can be calculated using the formula:
Mass = Moles ×Molar mass
Mass = 0.1493 mol ×63.55 g/mol
Mass = 9.48 g
Therefore, approximately 9.48 g of copper is deposited on one of the elec-
trodes during the electrolysis process.
31
Question 32
Question
Explain Faraday’s laws of electrolysis. An experiment was conducted using a
copper chloride solution with a current of 2.5 A passed through it for 2 hours.
If the mass of copper deposited was 6.78 g, calculate the Faraday constant.
Solution
Faraday’s laws of electrolysis state the following: 1. The mass of a substance
discharged at an electrode during electrolysis is directly proportional to the
quantity of electricity passed through the electrolyte. 2. The masses of different
substances discharged by the same quantity of electricity are proportional to
their chemical equivalent weights.
To calculate the Faraday constant, we will use the formula:
Faraday constant (F) = m×Z
n×F
where: - mis the mass of the substance deposited (in grams), - Zis the valency
of the ions, - nis the number of electrons exchanged in the reaction, - Fis the
Faraday constant (charge on one mole of electrons is 9.65 ×104C/mol).
Given: - Current, I= 2.5 A, - Time, t= 2 hours = 7200 s, - Mass of copper
deposited, m= 6.78 g, - Valency of copper ions, Z= 2.
First, let’s calculate the charge passed through the electrolyte using the
formula Q=It:
Q= 2.5×7200
Q= 18000 C
Next, calculate the number of moles of copper deposited:
Moles of Cu = Mass
Molar mass
The molar mass of copper is 63.55 g/mol, so:
Moles of Cu = 6.78
63.55
Now, we can calculate the Faraday constant:
F=m×Z
n×Q=6.78 ×2
6.78
63.55 ×18000 =13.56
6.78
63.55 ×18000
F=13.56
6.78
63.55 ×18000
F=13.56 ×63.55
6.78 ×18000
F=861.438
122040
F= 0.00706 C/mol
32
Question 33
Question
In an electrolytic cell, a metal plate is connected to the positive terminal of a
battery while a copper plate is connected to the negative terminal. The metal
plate loses mass at a rate of 0.2 grams per minute. If the Faraday constant is
9.65 ×104C/mol, calculate the current passing through the cell.
Solution
Step 1: Determine the number of moles of metal being deposited per minute.
Given that the metal plate loses mass at a rate of 0.2 grams per minute, we can
calculate the number of moles using the molar mass of the metal. Let’s assume
the molar mass of the metal is Mg/mol.
Molar mass of metal = Mg/mol
Number of moles of metal deposited per minute = 0.2 g/min
Mmol/min
Step 2: Determine the charge passing through the cell per minute. The
charge required to deposit one mole of the metal can be calculated using Fara-
day’s constant.
1 mole of metal requires 1 ×Faraday constant = 1 ×9.65 ×104C/mol
Therefore, the charge required to deposit 0.2
Mmoles of metal is:
Charge passing through the cell per minute = 0.2
M×9.65 ×104C/min
Step 3: Calculate the current passing through the cell. The current passing
through the cell can be calculated using the formula:
I=Q
t
where Iis the current, Qis the charge passing through the cell per minute, and
tis the time in minutes.
I=0.2×9.65 ×104
tM A
Question 34
Question
A copper sulfate solution is electrolyzed using inert electrodes. If 0.2 Faraday
of electricity is passed through the solution, what mass of copper is deposited
at the cathode? Assume 100
33
Solution
Faraday’s laws of electrolysis state that the amount of chemical change produced
by a current is proportional to the quantity of electricity passed through the
electrolyte. The amount of substance deposited or reacted can be calculated by
the equation:
Amount of substance = Charge passed
Faraday constant ×Valency
In the case of copper sulfate, the valency of copper is 2 and the Faraday
constant is 96500 C/mol.
Step 1: Calculate the amount of substance deposited at the cathode.
Amount of substance = 0.2 Faraday
96500 C/mol ×2=0.2
193000 mol
Step 2: Calculate the mass of copper deposited using the molar mass of
copper (Cu is 63.5 g/mol).
Mass of copper = Amount of substance ×Molar mass = 0.2
193000 ×63.5 g
Therefore, 0.000328 of copper will be deposited at the cathode.
Question 35
Question
An electric current of 5.0 A is passed through a solution of copper(II) sulfate
for 30 minutes. If the current causes the reduction of copper(II) ions to copper
metal, calculate:
1. The amount of copper deposited.
2. The mass of copper deposited, given that the molar mass of copper is
63.55 g/mol.
Solution
Let’s start by calculating the amount of charge passed through the solution us-
ing Faraday’s laws of electrolysis:
Step 1: Find the amount of charge passed The formula relating charge,
current, and time is given by:
Charge (Q) = Current (I) ×Time (t)
34
We are given that the current is 5.0 A and the time is 30 minutes. Converting
the time to hours:
Time (t) = 30 minutes ×1 hour
60 minutes = 0.5 hours
Now, we can calculate the charge passed through the solution:
Charge (Q) = 5.0 A ×0.5 h = 2.5 C
Step 2: Find the amount of copper deposited The amount of substance
deposited during electrolysis can be determined using the equation:
Amount of substance = Charge
Faraday’s constant ×valency
Given that the valency of copper(II) ions is 2, and Faraday’s constant is 96500 C/mol,
we can calculate the amount of copper deposited:
Amount of copper = 2.5 C
96500 C/mol ×2= 1.30 ×10−5mol
Step 3: Find the mass of copper deposited To find the mass of copper
deposited, we use the formula:
Mass = Amount ×Molar mass
Given that the molar mass of copper is 63.55 g/mol, we can calculate the mass
of copper deposited:
Mass of copper = 1.30 ×10−5mol ×63.55 g/mol = 0.826 ×10−3g=0.826 mg
Therefore, the mass of copper deposited is 0.826 mg.
35
