CHEM 121 - GENERAL CHEMISTRY
I - Galvanic and Electrolytic Cells
Question Bank - Set 1
Liberty University
Question 1
Question
Consider a galvanic cell with a standard cell potential of 0.90 V. If the cell
potential drops to 0.75 V at a certain point during operation, calculate the
change in Gibbs free energy for this process. Is the reaction still spontaneous?
Justify your answer.
Solution
Step 1: The change in cell potential (∆Ecell) can be calculated using the Nernst
equation:
∆Ecell =E◦
cell −0.0592
nlog [products]
[reactants]
Given that E◦
cell = 0.90 V and ∆Ecell = 0.75 V, we can rearrange the equation
to solve for the ratio of products to reactants:
0.75 = 0.90 −0.0592
nlog [products]
[reactants]
Step 2: Solving for [products]
[reactants]:
0.15 = 0.0592
nlog [products]
[reactants]
2.54 = log [products]
[reactants]
102.54 =[products]
[reactants]
382 = [products]
[reactants]
Step 3: Calculate the change in Gibbs free energy using the equation:
∆G=−nF ∆Ecell
Given that n= 1 (assuming one electron transfer), and F= 96485 C/mol, we
have:
∆G=−1×96485 ×(0.90 −0.75)
∆G=−1×96485 ×0.15
∆G=−14472.75 J/mol
Since ∆Gis negative, the reaction is still spontaneous.
Question 2
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
Determine the cell potential (E◦
cell) at standard conditions.
Solution
Step 1: Identify the oxidation and reduction half-reactions.
Oxidation half-reaction: Zn(s)→Zn2+(aq) + 2e−
Reduction half-reaction: Cu2+(aq) + 2e−→Cu(s)
Step 2: Write down the overall cell reaction.
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 3: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials.
E◦
cell =E◦
reduction, cathode −E◦
oxidation, anode = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the cell potential (E◦
cell) at standard conditions for the given
galvanic cell is 1.10 V.
2
Question 3
Question
Consider a galvanic cell with the following half-reactions:
Cathode: Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Anode: MnO−
4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦= 1.51 V
Calculate the cell potential (E◦
cell) for this galvanic cell at standard conditions.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together,
adjusting coefficients to make the number of electrons equal on both sides.
Overall Cell Reaction: Cu2+(aq)+MnO−
4(aq)+8H+(aq)+5e−→Cu(s)+Mn2+(aq)+4H2O(l)
Step 2: Calculate the cell potential (E◦
cell) using the standard reduction
potentials of the half-reactions.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
MnO−
4/Mn2+
E◦
cell = 0.34 V −1.51 V
E◦
cell =−1.17 V
Therefore, the cell potential for this galvanic cell at standard conditions is
−1.17 V.
Question 4
Question
What are the key differences between galvanic cells and electrolytic cells? Pro-
vide a detailed explanation of how each cell operates and how they are used in
different scenarios.
Solution
To understand the key differences between galvanic cells and electrolytic cells,
let’s first define each type of cell and then compare their operation and appli-
cations.
Galvanic Cells:
Galvanic cells, also known as voltaic cells, are devices that use spontaneous
chemical reactions to generate electrical energy.
3
In a galvanic cell, oxidation occurs at the anode, and reduction occurs at
the cathode. Electrons flow from the anode to the cathode through an
external circuit.
The chemical reactions in a galvanic cell produce an electric potential
difference, commonly referred to as voltage.
Galvanic cells are commonly used in batteries to power various devices,
such as flashlights, remote controls, and cell phones.
Electrolytic Cells:
Electrolytic cells are devices that use an external electric current to drive
a non-spontaneous chemical reaction.
In an electrolytic cell, oxidation occurs at the anode (positive electrode),
where electrons are released, and reduction occurs at the cathode (negative
electrode), where electrons are gained.
The external power source (such as a battery) forces the non-spontaneous
reaction to occur by supplying the necessary energy.
Electrolytic cells are commonly used in processes like electroplating, elec-
trolysis of water to produce hydrogen and oxygen, and in the extraction
of metals from their ores.
Comparison:
Energy Source: Galvanic cells convert chemical energy into electrical en-
ergy, while electrolytic cells use electrical energy to drive non-spontaneous
reactions.
Spontaneity: Galvanic cell reactions are spontaneous, while electrolytic
cell reactions are non-spontaneous and require an external energy source.
Anode/Cathode: In a galvanic cell, the anode is negatively charged
and the cathode is positively charged. In an electrolytic cell, the anode is
positively charged and the cathode is negatively charged.
Applications: Galvanic cells are used in batteries for powering devices,
while electrolytic cells are used in processes like electroplating, electrolysis,
and metal extraction.
In conclusion, galvanic cells generate electrical energy from spontaneous
chemical reactions, while electrolytic cells use electrical energy to drive non-
spontaneous reactions. Both types of cells play crucial roles in various techno-
logical applications.
4
Question 5
Question
In a galvanic cell, the standard cell potential (E◦
cell) is 1.08 V. If the concen-
tration of Mn2+ ions in one half-cell is 0.10 M and the concentration of MnO−
4
ions in the other half-cell is 0.20 M, determine the standard electrode potential
(reduction potential) for the Mn2+/Mn half-reaction.
Solution
Step 1: Write out the cell reaction for the galvanic cell. The cell reaction for
the galvanic cell involving Mn2+ and MnO−
4half-reactions is given by:
Mn2+ + MnO−
4→Mn3+ + MnO2
Step 2: Write out the two half-reactions for the cell reaction. The half-
reaction involving Mn2+ is:
Mn2+ →Mn3+ +e−
The half-reaction involving MnO−
4is:
MnO−
4+ 4H++ 3e−→MnO2+ 2H2O
Step 3: Determine the standard cell potential using the standard electrode
potentials. The standard cell potential (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cell = 1.08V, we can substitute the standard electrode potentials
for the half-reactions to get:
1.08 = E◦
cathode −E◦
anode
Step 4: Calculate the standard electrode potential for the cathode half-
reaction. From experimental data, we know that the standard electrode po-
tential for the MnO−
4/MnO2half-reaction is 1.51V. Therefore, the standard
electrode potential for the Mn2+/Mn half-reaction can be calculated as:
1.08V = 1.51V −E◦
anode
E◦
anode = 1.51V −1.08V = 0.43V
Step 5: Write out the standard electrode potential for the Mn2+/Mn half-
reaction. Therefore, the standard electrode potential (reduction potential) for
the Mn2+/Mn half-reaction is 0.43V .
5
Question 6
Question
Consider a galvanic cell with a standard potential of 0.70 V that is made from
a copper electrode in a 1.0 M CuSO4solution and a silver electrode in a 1.0 M
AgNO3solution. a) Write the balanced cell reaction. b) Calculate the standard
cell potential. c) Determine in which direction the cell reaction will proceed. d)
If the current is allowed to flow until 1.0 mol of electrons have passed through
the external circuit, how many grams of silver will be deposited? (Assume 100
Solution
a) The balanced cell reaction is given by:
Cu2+(aq) + 2e−→Cu(s) at the cathode
2Ag(s)→2Ag+(aq) + 2e−at the anode
Overall cell reaction:
Cu2+(aq) + 2Ag(s)→Cu(s) + 2Ag+(aq)
b) The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cathode = 0.34 V and E◦
anode = 0.80 V, we have:
E◦
cell = 0.34 V −0.80 V = −0.46 V
c) The cell reaction will proceed in the reverse direction as the cell potential
is negative.
d) The total charge passed through the circuit (Q) can be calculated using
the formula:
Q=nF
where nis the number of moles of electrons and Fis the Faraday constant
(96500 C/mol). Given that n= 1.0 mol, we have:
Q= 1.0 mol ×96500 C/mol = 96500 C
The amount of silver deposited can be calculated using the formula:
Amount of substance (Ag) = Q
F
Given that the molar mass of silver (Ag) is 107.87 g/mol, we have:
Amount of substance (Ag) = 96500 C
96500 C/mol = 1.0 mol
Mass (Ag) = 1.0 mol ×107.87 g/mol = 107.87 g
Therefore, 107.87 grams of silver will be deposited.
6
Question 7
Question
Consider a galvanic cell constructed with a standard hydrogen electrode (SHE)
and a copper electrode. The standard reduction potential of the copper electrode
is E◦
Cu = 0.34 V. The cell operates at 25◦C. Given the standard reduction
potential of the SHE is 0 V, calculate the cell potential at standard conditions
and determine whether the cell reaction is spontaneous.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the reduction
half-reaction at the cathode minus the reduction half-reaction at the anode.
Cathode: 2H++ 2e−→H2(g)E◦
H+/H2= 0 V
Anode: Cu2+ + 2e−→Cu(s) E◦
Cu2+/Cu = 0.34 V
The overall reaction is:
Cu2+ + 2H+→Cu(s) + H2(g)
Step 2: Calculate the cell potential, Ecell. Using the Nernst equation and
standard reduction potentials:
Ecell =E◦
cathode −E◦
anode
Ecell = 0V−0.34V=−0.34V
Step 3: Determine spontaneity of the reaction. For the reaction to be spon-
taneous, the cell potential must be positive. Since Ecell =−0.34V < 0, the cell
reaction is not spontaneous.
Question 8
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 1.23 V. If
the concentration of Cu2+ in the cathode compartment is 0.10 M, and the
concentration of Zn2+ in the anode compartment is 1.00 M, determine the cell
potential when the concentration of Zn2+ is decreased to 0.10 M.
Solution
Step 1: Write the balanced cell reaction for the galvanic cell. The cell reaction
for the galvanic cell using a Zn anode and a Cu cathode is:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
7
Step 2: Calculate the standard cell potential for the given concentrations.
The Nernst equation relates the cell potential with the concentrations of the
ions involved:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
where n= 2 is the number of electrons transferred in the balanced equation,
Ecell is the cell potential, and E◦
cell is the standard cell potential.
Substitute the given values into the Nernst equation:
Ecell = 1.23−0.0592
2log 0.10
1.00= 1.23−0.0296×−1=1.23+0.0296 = 1.2596 V
Step 3: Calculate the new cell potential with the decreased concentration of
Zn2+. Using the same Nernst equation and substituting the new concentration
of Zn2+:
Enew = 1.23 −0.0592
2log 0.10
0.10= 1.23 −0.0296 ×0=1.23 V
Therefore, the new cell potential when the concentration of Zn2+ is decreased
to 0.10 M is 1.23 V.
Question 9
Question
An electrochemical cell consists of a zinc metal electrode in a 1.0 M Zn2+ solution
and a copper metal electrode in a 1.0 M Cu2+ solution. The standard reduction
potentials are E◦(Cu2+/Cu) = 0.34 V and E◦(Zn2+/Zn) = −0.76 V. Calculate
the cell potential at 25
°
C when 1.0 mol of Zn2+ ions is reduced at the zinc
electrode.
Solution
Step 1: Write the two half-reactions for the cell reaction and their corresponding
standard reduction potentials. The overall cell reaction is:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
The half-reactions are:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Calculate the cell potential. The cell potential (E◦
cell) can be calcu-
lated using the formula:
E◦
cell =E◦
cathode −E◦
anode
8
Substitute the given E◦values:
E◦
cell = 0.34 V −(−0.76) V = 0.34 V + 0.76 V = 1.10 V
Therefore, the cell potential at 25
°
C when 1.0 mol of Zn2+ ions is reduced
at the zinc electrode is 1.10 V.
Question 10
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a standard zinc electrode connected by a salt bridge. The standard reduc-
tion potential of the hydrogen half-cell is 0.00 V, and the standard reduction
potential of the zinc half-cell is −0.76 V. Determine the cell potential at standard
conditions and predict the spontaneity of the reaction.
Solution
Step 1: Write the overall cell reaction. The overall reaction for the galvanic cell
is the oxidation of zinc and reduction of hydrogen:
Zn (s) + 2H+(aq)→Zn2+(aq)+H2(g)
Step 2: Write the half-reactions and their standard reduction potentials.
Zinc half-reaction: Zn(s)→Zn2+(aq) + 2e−(E◦=−0.76 V)
Hydrogen half-reaction: 2H+(aq) + 2e−→H2(g) (E◦= 0.00 V)
Step 3: Calculate the standard cell potential. The standard cell potential
(E◦
cell) is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(−0.76 V) = 0.76 V
Step 4: Predict the spontaneity of the reaction. Since the standard cell
potential is 0.76 V (which is greater than 0 V), the reaction is spontaneous
under standard conditions.
Question 11
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a copper electrode. The standard reduction potential for the copper elec-
trode is +0.34 V. Calculate the standard cell potential for this galvanic cell.
9
Solution
To calculate the standard cell potential for this galvanic cell, we can use the
equation:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode
is the standard reduction potential of the anode.
Given that the standard reduction potential for the copper electrode is +0.34
V, we have:
Step 1: Identify the cathode and anode reactions. The reduction half-
reaction at the copper electrode is:
Cu2+ + 2e−→Cu
The anode reaction at the SHE is the standard hydrogen electrode:
2H++ 2e−→H2
Step 2: Determine the standard cell potential. The standard reduction
potential for the standard hydrogen electrode is 0 V. So, E◦
cathode = 0 V and
E◦
anode =−0.34 V.
Substitute these values into the equation:
E◦
cell = 0 −(−0.34) = +0.34 V
Therefore, the standard cell potential for this galvanic cell is +0.34 V.
Question 12
Question
Consider a galvanic cell with a standard potential of 1.10 V. If the cell potential
is measured to be 1.00 V under non-standard conditions, calculate the reaction
quotient Q with a given concentration of [Mn+](aq)= 0.10 M and [Mn](s)= 1.0
M. Determine if the reaction is at equilibrium or not.
Solution
Step 1: Write the half-reaction for the reduction of Mn2+ to Mn.
The half-reaction is:
Mn2+(aq)+2e−→Mn(s)
Step 2: Calculate the standard potential for the cell.
The standard potential for the cell is given as 1.10 V.
Step 3: Calculate the reaction quotient Q.
10
The Nernst Equation relates the standard cell potential, the actual cell po-
tential, and the reaction quotient Q:
E=E◦−0.0592
nlog Q
Given that E= 1.00 V and E◦= 1.10 V, substituting these values into the
equation gives:
1.00 = 1.10 −0.0592
2log Q
Solving for log Q:
−0.10 = −0.0298 log Q
log Q= 3.355
Q= 103.355 = 2382.5
So, the reaction quotient Q is 2382.5.
Step 4: Determine if the reaction is at equilibrium.
Since Q>K, the reaction quotient is greater than the equilibrium constant
K. This implies that the reaction is not at equilibrium.
Question 13
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe2+(aq)→Fe3+(aq) + e−E◦=−0.77 V
Calculate the standard cell potential for the galvanic cell formed by connecting
these two half-cells in a complete circuit.
Solution
Step 1: Identify the oxidation and reduction half-reactions. The oxidation half-
reaction is:
Fe2+(aq)→Fe3+(aq) + e−E◦=−0.77 V
The reduction half-reaction is:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Write the overall cell reaction. Adding the two half-reactions to-
gether, we get the overall cell reaction:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
11
Step 3: Calculate the standard cell potential. The standard cell potential
is calculated by adding the standard reduction potentials of the reduction half-
reaction and the oxidaiton half-reaction:
E◦
cell =E◦
reduction +E◦
oxidation
E◦
cell = 0.34 V + (−0.77 V) = −0.43 V
Therefore, the standard cell potential for the galvanic cell formed by con-
necting these two half-cells is −0.43 V.
Question 14
Question
Consider an electrochemical cell with the following half-reactions:
Anode: Pb(s) →Pb2+(aq)+2e−
Cathode: Ag+(aq) + e−→Ag(s)
If the standard reduction potentials are E◦
Pb2+/Pb =−0.13 V and E◦
Ag+/Ag =
0.80 V, determine the cell potential and state whether the cell is galvanic or
electrolytic.
Solution
Step 1: Write the overall cell reaction by summing the two half-reactions:
Pb(s) + 2Ag+(aq)→Pb2+(aq) + 2Ag(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials for each half-reaction:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −(−0.13 V) = 0.93 V
Step 3: Determine whether the cell is galvanic or electrolytic by examining
the sign of E◦
cell. Since E◦
cell is positive (0.93 V), the cell is galvanic because a
spontaneous chemical reaction will occur.
Question 15
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.84 V. The cell
reaction is given by:
12
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
(a) Determine the standard cell potential for the reverse reaction.
(b) Calculate the equilibrium constant, K, for the forward reaction at 25◦C.
Solution
(a) The standard cell potential for the reverse reaction can be determined using
the relationship:
E◦
cell, reverse =−E◦
cell, forward
Therefore,
E◦
cell, reverse =−E◦=−0.84 V
So, the standard cell potential for the reverse reaction is −0.84 V.
(b) The equilibrium constant, K, for the forward reaction can be determined
from the standard cell potential using the Nernst equation:
E=E◦−0.0592
nlog Q
For the standard cell potential, Q=K, so:
E◦= 0.0592 log K
n
Given E◦= 0.84 V, n= 2 (from the coefficients in the balanced equation),
and T= 25◦C, we have:
0.84 = 0.0592 ×log K
2
Solving for log K:
0.84
0.0592 ×2= log K
7.09 = log K
K= 107.09
K≈8.90 ×107
Therefore, the equilibrium constant, K, for the forward reaction at 25◦C is
approximately 8.90 ×107.
13
Question 16
Question
Consider a galvanic cell with a standard potential of E◦= 0.40 V. If the cell
potential is measured to be 0.60 V under non-standard conditions where the
concentrations are [M2+]=0.10 M and [M]=1.0 M, determine the reaction
quotient Qand discuss if the cell reaction is at equilibrium or not. Additionally,
explain whether the cell reaction would be spontaneous in the forward direction
or not.
Solution
Step 1: Write the half-reactions for the galvanic cell. The overall cell reaction
for a galvanic cell based on the given standard potential is:
Oxidation half reaction: M(s)→M2+(aq)+2e−
Reduction half reaction: M2+(aq)+2e−→M(s)
Step 2: Determine the reaction quotient Q. The reaction quotient Qis given
by the expression:
Q=[M2+]
[M]2=0.10
1.02= 0.10
Step 3: Compare the cell potential Eand the standard cell potential E◦.
Given that the cell potential E= 0.60 V and the standard cell potential E◦=
0.40 V, we can compare these values to determine the direction of the reaction. If
E > E◦, the reaction quotient Q<K, and the cell reaction is not at equilibrium.
Step 4: Determine if the cell reaction is spontaneous and in which direction.
Since E > E◦, the reaction is not at equilibrium and will proceed in the direction
to reach equilibrium. In this case, the cell reaction will be spontaneous in the
forward direction.
Therefore, based on the values provided, the reaction quotient Qis 0.10, the
cell reaction is not at equilibrium, and the cell reaction will be spontaneous in
the forward direction.
Question 17
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) →Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Calculate the cell potential at standard conditions (T= 25◦C, P= 1 atm)
for the given galvanic cell. Assume all species are at 1 M concentration.
14
Solution
Step 1: Write the cell reaction by adding the two half-reactions:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Write the Nernst equation for cell potential (Ecell):
Ecell =E◦
cell −0.0592
nlog(Q)
where: - Ecell is the cell potential, - E◦
cell is the standard cell potential, - nis
the number of electrons transferred (from the balanced equation), - Qis the
reaction quotient.
Step 3: Determine the standard cell potential (E◦
cell) using standard reduc-
tion potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
Step 4: Substitute the standard reduction potentials into the equation:
E◦
cell = (0.34 V) −(−0.76 V)
Step 5: Calculate E◦
cell:
E◦
cell = 1.10 V
Step 6: Determine the number of electrons transferred, n= 2.
Step 7: Calculate the reaction quotient Q:
Q=[Zn2+][Cu]
[Zn][Cu2+]
Step 8: Substitute Q,n, and E◦
cell into the Nernst equation and solve for
Ecell to find the cell potential at standard conditions.
Question 18
Question
Consider a galvanic cell that consists of a zinc electrode immersed in a 1.0 M
Zn2+ solution and a cadmium electrode immersed in a 1.0 M Cd2+ solution.
The standard reduction potentials are as follows: Zn2+/Zn with E◦=−0.76 V
and Cd2+/Cd with E◦=−0.40 V. Calculate the cell potential at 298 K.
15
Solution
Step 1: Write the balanced cell reaction and calculate the cell potential. The
balanced cell reaction can be written as:
Zn(s) + Cd2+(aq)→Zn2+(aq) + Cd(s)
The cell potential (E◦
cell) is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode
is the standard reduction potential of the anode.
Substitute the given values to calculate the cell potential:
E◦
cell =E◦
Cd2+/Cd −E◦
Zn2+/Zn
E◦
cell = (−0.40 V) −(−0.76 V)
E◦
cell = 0.36 V
Therefore, the cell potential at 298 K is 0.36 V.
Question 19
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a 1.0 M solution of Cu2+. The standard reduction potential for the
hydrogen electrode is 0.00 V and for the copper electrode is +0.34 V.
a) Write the balanced cell reaction for the galvanic cell.
b) Calculate the cell potential at 25
°
C.
c) If a voltage of 0.50 V is applied to the cell in the reverse direction, will
the copper electrode dissolve or plate out? Justify your answer.
Solution
a) The balanced cell reaction for the galvanic cell is given by the oxidation of
copper at the anode and the reduction of hydrogen at the cathode:
Anode: Cu(s)→Cu2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
Combining the two half-reactions, we get:
Overall Cell Reaction: Cu(s) + 2H+(aq)→Cu2+(aq)+H2(g)
16
b) The cell potential, E◦
cell, can be calculated using the standard reduction
potentials for the half-reactions and the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(+0.34 V) = −0.34 V
c) The cell potential of -0.34 V indicates that the cell reaction is spontaneous
as written above. If a voltage of 0.50 V is applied in the reverse direction, the
cell will be forced to run in reverse. Since the new applied voltage is greater than
the cell potential, the reaction will be non-spontaneous in the reverse direction.
This means that copper will dissolve from the copper electrode instead of plating
out.
Question 20
Question
Consider a galvanic cell that consists of a silver electrode immersed in a 1.0
M solution of AgNO3and a zinc electrode immersed in a 1.0 M solution of
Zn(NO3)2. The standard reduction potentials are Ag++e−→Ag(E◦= 0.80 V)
and Zn2+ + 2e−→Zn(E◦=−0.76 V). Determine the standard cell potential
and identify the anode and cathode.
Solution
Step 1: Write the two half-reactions and determine the standard cell poten-
tial. The spontaneous reaction for the galvanic cell is given by the overall cell
reaction: Zn + Ag+→Zn2+ + Ag.
This reaction can be broken down into two half-reactions: Anode: Zn →
Zn2+ + 2e−(oxidation) Cathode: Ag++ e−→Ag (reduction)
The standard cell potential can be calculated as the sum of the standard
reduction potentials for the two half-reactions: Standard cell potential, E◦
cell =
E◦
cathode −E◦
anode = 0.80 V −(−0.76 V) = 1.56 V.
Therefore, the standard cell potential for the galvanic cell is 1.56 V.
Step 2: Identify the anode and cathode. Since the more negative standard
reduction potential corresponds to the anode, in this case, zinc is the anode and
silver is the cathode.
Therefore, the anode is Zn and the cathode is Ag+.
Question 21
Question
Consider the following galvanic cell:
17
Zn |Zn2+(0.1M)|| Ni2+(0.01 M)|Ni
The standard reduction potentials for the half-reactions are given as follows:
Zn2+(aq)+2e−→Zn(s) with E◦=−0.76 V
Ni2+(aq)+2e−→Ni(s) with E◦=−0.25 V
Calculate the cell potential at 25◦C and determine whether the reaction is
spontaneous or non-spontaneous.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the sum of
the two half-reactions:
Zn(s) + Ni2+(aq)→Zn2+(aq) + Ni(s)
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the standard reduction potentials (E◦) of the
two half-reactions.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.25 V −(−0.76 V)
E◦
cell = 1.01 V
Step 3: Calculate the cell potential at 25◦C using the Nernst Equation. The
Nernst Equation relates the cell potential to the standard cell potential and the
reaction quotient (Q) at non-standard conditions.
Ecell =E◦
cell −0.0592
nlog Q
Step 4: Calculate the reaction quotient (Q). Since Ni2+ is the oxidizing
agent and Zn is the reducing agent, the reaction quotient is given by:
Q=[Zn2+]
[Ni2+]=0.1
0.01 = 10
Step 5: Plug the values into the Nernst equation to find Ecell.
Ecell = 1.01 V −0.0592
2log(10)
Ecell = 1.01 V −(0.0296) ×1
Ecell = 1.01 V −0.0296
Ecell = 0.9804 V
Step 6: Determine if the reaction is spontaneous. Since Ecell is positive, the
reaction is spontaneous as the cell potential is greater than zero.
18
Question 22
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.70 V. Determine
the equilibrium constant, K, for the following cell reaction:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Solution
Step 1: Write the half-reactions for the anode and cathode:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Step 2: Determine the standard cell potential by subtracting the potential
of the anode from that of the cathode:
E◦
cell =E◦
cathode −E◦
anode = 0.34 V
Step 3: Using the formula ∆G◦=−nF E◦, find the standard Gibbs free
energy change:
∆G◦=−nF E◦
cell =−2(96485C/mol)(0.34V) = −65.1 kJ/mol
Step 4: Determine Kusing the equation ∆G◦=−RT ln K:
−65.1 kJ/mol = −(8.314J/(mol ·K))(298K) ln K
ln K= 23.4
K=e23.4= 1.05 ×1010
Therefore, the equilibrium constant for the cell reaction is K= 1.05 ×1010.
Question 23
Question
A student is conducting an experiment involving galvanic and electrolytic cells.
The student is given a cell with a standard cell potential of E◦
cell =−0.56 V.
The student wants to know whether this cell would operate as a galvanic cell
(producing electricity) or as an electrolytic cell (consuming electricity). Deter-
mine whether the cell would function as a galvanic cell or an electrolytic cell,
and justify your answer.
19
Solution
Step 1: Recall that for a galvanic cell, the standard cell potential (E◦
cell) is
positive, indicating spontaneous electron flow from the anode to the cathode.
For an electrolytic cell, the standard cell potential is negative, requiring an
external voltage to drive a non-spontaneous reaction.
Step 2: Given that E◦
cell =−0.56 V, the negative value indicates that the cell
reaction is not spontaneous and would require an external voltage to proceed.
Step 3: Therefore, the cell with a standard cell potential of E◦
cell =−0.56
V would function as an electrolytic cell as it does not naturally generate
electricity and would need an external power source to operate.
Thus, the cell in question would operate as an electrolytic cell.
Question 24
Question
Consider the following half-reactions:
Ag+(aq)+e−→Ag(s) (E◦= 0.80 V)
Ni2+(aq) + 2e−→Ni(s) (E◦=−0.26 V)
Determine the cell potential at 25
°
C for the reaction:
2Ag+(aq) + Ni(s) →2Ag(s) + Ni2+(aq)
Solution
Step 1: Write the half-reactions involved in the cell reaction.
Oxidation half-reaction: Ni(s) →Ni2+(aq) + 2e−
Reduction half-reaction: 2Ag+(aq) + 2e−→2Ag(s)
Step 2: Write the overall cell reaction as the combination of the two half-
reactions.
2Ag+(aq) + Ni(s) →2Ag(s) + Ni2+(aq)
Step 3: Find the standard cell potential by summing the standard reduction
potentials of the half-reactions.
E◦
cell =E◦
reduction(Ag+) + E◦
oxidation(Ni)
E◦
cell = 0.80 V + (−0.26 V)
E◦
cell = 0.54 V
Thus, the cell potential at 25
°
C for the given reaction is 0.54 V.
20
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
Given that the standard reduction potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =
−0.76 V, calculate the standard cell potential E◦
cell. Is this cell galvanic or elec-
trolytic? Justify your answer.
Solution
Step 1: Write the expression for the standard cell potential:
E◦
cell =E◦
cathode −E◦
anode
Step 2: Substitute the standard reduction potentials into the formula:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Since E◦
cell is positive, the cell reaction is spontaneous and the cell is galvanic.
The electrons flow from the anode (Zn) to the cathode (Cu).
Question 26
Question
Consider a galvanic cell constructed with a nickel electrode in contact with
a solution that is 0.10 M in Ni2+ ions and a silver electrode in contact with
a solution that is 1.0 M in Ag+ions. The standard reduction potentials are
E◦(Ni2+/Ni) = −0.25 V and E◦(Ag+/Ag) = 0.80 V. Determine the cell poten-
tial and the direction of the spontaneous reaction.
Solution
Step 1: Write the half-reactions for the cell: In the galvanic cell, the cathode is
the nickel electrode (Ni2+ being reduced) and the anode is the silver electrode
(Ag being oxidized). Cathode half-reaction: Ni2+ + 2e−→Ni
Anode half-reaction: Ag →Ag++e−
Step 2: Determine the overall cell reaction: The standard cell potential,
E◦
cell, is given by: E◦
cell =E◦
cathode −E◦
anode
Substitute the given values into the equation: E◦
cell =E◦
Ni2+/Ni −E◦
Ag+/Ag
E◦
cell =−0.25 V −0.80 V = −1.05 V
21
Step 3: Determine the direction of the spontaneous reaction: Since the cell
potential is negative, the reaction as written is not spontaneous. To make it
spontaneous, reverse the reaction.
Therefore, the spontaneous reaction in the galvanic cell is: Ag + Ni2+ →
Ag++ Ni
Question 27
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential for the reduction of Cu2+(aq) is 0.34 V
and the standard reduction potential for the reduction of Zn2+(aq) is −0.76 V,
calculate the standard cell potential (E◦) of the galvanic cell.
Solution
Step 1: Identify the half-reactions and their standard reduction potentials.
The given half-reactions are: Anode:
Zn(s) −−→ Zn2+(aq) + 2 e−with Eo=−0.76 V
Cathode:
Cu2+(aq) + 2 e−−−→ Cu(s) with Eo= 0.34 V
Step 2: Write the overall cell reaction.
The overall cell reaction is obtained by adding the two half-cell reactions
together so that electrons cancel out:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 3: Calculate the standard cell potential (E◦) using the formula:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values into the formula:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the standard cell potential (E◦) of the galvanic cell is 1.10 V.
22
Question 28
Question
Consider a galvanic cell with the following half-reactions:
Anode: Ag(s)→Ag+(aq) + e−
Cathode: Cu2+(aq)+2e−→Cu(s)
If the standard reduction potentials are E◦
Ag+/Ag = 0.80 V and E◦
Cu2+/Cu =
0.34 V, calculate the standard cell potential (E◦
cell) for the Galvanic Cell.
Solution
Step 1: Write the overall cell reaction.
Ag(s) + Cu2+(aq)→Ag+(aq) + Cu(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Ag+/Ag
E◦
cell = 0.34 V −0.80 V
E◦
cell =−0.46 V
Therefore, the standard cell potential for the Galvanic Cell is −0.46 V.
Question 29
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a silver electrode. The standard reduction potential for the Ag++e−−>
Aghalf −reactionis0.80V.Ifthemeasuredcellpotentialis0.35V, determinethestandardreductionpotentialfortheSHEhalf−
reaction.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction for a galvanic
cell involving a standard hydrogen electrode (SHE) and a silver electrode can
be represented as:
2H+(aq) + 2e−→H2(g) + Ag+(aq)→Ag(s)+H2(g)
23
Step 2: Write the half-reaction at the SHE. The standard reduction potential
for the SHE half-reaction is 0 V by definition. Therefore, the half-reaction at
the SHE is:
2H+(aq) + 2e−→H2(g)
Step 3: Write the half-reaction at the Ag electrode. Given that the standard
reduction potential for the Ag half-reaction is 0.80 V, the half-reaction at the
Ag electrode is:
Ag+(aq)+e−→Ag(s)
Step 4: Use the measured cell potential to find the standard reduction po-
tential for the SHE half-reaction. The measured cell potential is the sum of the
standard reduction potentials for the two half-reactions. Therefore, we have:
Ecell =ESHE −EAg
0.35 V = 0 V −(0.80 V)
ESHE = 0.35 V + 0.80 V
ESHE = 1.15 V
Therefore, the standard reduction potential for the SHE half-reaction is 1.15
V.
Question 30
Question
Consider a galvanic cell with the following half-cell reactions:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
If the standard reduction potential for the Zn2+|Zn half-cell is −0.76 V and
the standard reduction potential for the H+|H2half-cell is 0.00 V, calculate the
standard cell potential (E◦
cell) for this galvanic cell.
Solution
Step 1: Write the overall cell reaction by summing the half-cell reactions. Re-
arrange the half-cell reactions so that electrons cancel out:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
The overall cell reaction is:
Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)
24
Step 2: Calculate the standard cell potential using the standard reduction
potentials. The standard cell potential (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
The standard reduction potential for the cathode reaction is 0.00 V, and for
the anode reaction is −0.76 V. Therefore,
E◦
cell = 0.00 V −(−0.76) V = 0.76 V
Therefore, the standard cell potential for this galvanic cell is 0.76 V.
Question 31
Question
Consider a galvanic cell with a standard voltage of 0.85 V. If the concentration
of Fe2+ ions in the cell is 0.10 M and the concentration of Zn2+ ions is 1.0 M,
determine the standard electrode potential of the Zn2+/Zn half-cell.
Solution
To find the standard electrode potential of the Zn2+/Zn half-cell, we can use
the Nernst equation and the given information about the galvanic cell.
Step 1: Write the overall cell reaction. The galvanic cell is composed of
two half-reactions: Fe2+ + 2e−→Fe and Zn2+ + 2e−→Zn. The overall cell
reaction is the sum of these two half-reactions: Fe2+ + Zn2+ →Fe + Zn2+.
Step 2: Determine the standard cell potential. The standard cell potential,
E◦
cell, is given as 0.85 V.
Step 3: Calculate the cell potential under the given conditions. The Nernst
equation relates the standard cell potential to the actual cell potential and the
concentrations of the species:
Ecell =E◦
cell −0.0592
nlog [Fe2+]
[Zn2+]
Given that Ecell = 0.85 V, [Fe2+] = 0.10 M, and [Zn2+] = 1.0 M, and the
number of electrons involved in the reaction is 2, we can substitute these values
into the Nernst equation to solve for the standard electrode potential of the
Zn2+/Zn half-cell.
Step 4: Calculate the standard electrode potential of the Zn2+/Zn half-cell.
Using the Nernst equation:
0.85 V = E◦
cell −0.0592
2log 0.10
1.0
Solving for E◦
cell:
E◦
cell = 0.85 + 0.0296 log(0.10)
25
E◦
cell = 0.85 + 0.0296(−1.0)
E◦
cell = 0.85 −0.0296
E◦
cell = 0.8204 V
Therefore, the standard electrode potential of the Zn2+/Zn half-cell is 0.8204
V.
Question 32
Question
Consider a galvanic cell with a standard cell potential of 1.23 V consisting of a
standard hydrogen electrode (SHE) and a copper electrode. If the concentration
of Cu2+ ions is 1.0 M in the half-cell containing the copper electrode, calculate
the standard reduction potential of the copper electrode.
Solution
Step 1: Write the cell reaction for the galvanic cell.
The cell reaction is:
Cu2+ + 2e−→Cu
Step 2: Write the cell potential equation.
The standard cell potential, E◦
cell, is given as 1.23 V.
The standard reduction potential of SHE is 0 V.
Using the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Substitute the values:
1.23 = E◦
Cu2+/Cu −0
Step 3: Calculate the standard reduction potential of the copper electrode.
So,
E◦
Cu2+/Cu = 1.23 V
Therefore, the standard reduction potential of the copper electrode is 1.23
V.
Question 33
Question
A galvanic cell has a standard cell potential of E◦= 0.81 V. If the cell operates
until the concentration of Zn2+ ions is decreased to 0.300 M from an initial
concentration of 1.00 M, determine the final potential of the cell. Assume that
the temperature remains constant.
26
Solution
Step 1: Write the balanced cell reaction.
The galvanic cell consists of a zinc electrode in a Zn2+ solution and a copper
electrode in a Cu2+ solution. The cell reaction is:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Identify the half-reactions.
The half-reactions for the cell are: - Oxidation half-reaction: Zn(s)→
Zn2+(aq)+2e−- Reduction half-reaction: Cu2+(aq)+2e−→Cu(s)
Step 3: Calculate the standard cell potential (E◦).
Using the standard reduction potentials for zinc and copper from tables, we
have: E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.76 V) = 1.10 V
Step 4: Use the Nernst equation to calculate the final cell potential.
The Nernst equation relates the non-standard cell potential (E) to the stan-
dard cell potential (E◦), the reaction quotient (Q), the Faraday constant (F),
the gas constant (R), the temperature (T), and the number of electrons trans-
ferred (n) in the cell reaction.
E=E◦−RT ln(Q)
nF
Given that the initial concentration of Zn2+ ions is 1.00 M and the final
concentration is 0.300 M, we can calculate the reaction quotient Qusing these
concentrations. The value of nis 2 for this cell reaction.
Step 5: Calculate the final potential of the cell.
Substitute the values into the Nernst equation:
E= 1.10 V −
(8.314 J/K ·mol)(298 K) ln 0.300
1.00
2(96,485 C/mol) ≈1.07 V
Therefore, the final potential of the cell is approximately 1.07 V.
Question 34
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
If the concentration of Cu2+(aq) is 1.0 M and Fe3+(aq) is 0.1 M, calculate
the cell potential at 25
°
C. Determine the direction of electron flow and identify
the anode and cathode.
27
Solution
Step 1: Write the balanced overall redox reaction. Let’s first write the two
half-reactions in the reduction form:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
To find the overall reaction, we need to add the two half-reactions together:
3Cu2+(aq) + Fe3+(aq)→3Cu(s) + Fe(s)
Step 2: Calculate the standard cell potential (E◦). The standard cell poten-
tial is the difference between the standard reduction potentials of the cathode
and the anode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Fe3+/Fe −E◦
Cu2+/Cu
E◦
cell = (−0.04 V) −(0.34 V)
E◦
cell =−0.38 V
Step 3: Calculate the cell potential at 25
°
C using the Nernst equation.
Ecell =E◦
cell −RT
nF ln Q
K
where Ris the gas constant (8.314 J/(mol
·
K)), Tis the temperature in Kelvin,
nis the number of moles of electrons transferred, Fis Faraday’s constant (96485
C/mol), Qis the reaction quotient, and Kis the equilibrium constant.
Since this is not at equilibrium, Qis not equal to K. We need to calculate
the reaction quotient Q:
Q=[Fe(s)]
[Cu(s)] =1
1/3= 3
Plugging in the values, we get:
Ecell =−0.38 V −(8.314 J/(mol
·
K))(298 K)
3(96485 C/mol) ln(3)
Ecell =−0.38 V −0.004 ln(3)
Ecell =−0.38 V −0.004 ×1.099
Ecell =−0.38 V −0.0044
Ecell ≈ −0.3844 V
Step 4: Determine the direction of electron flow and identify the anode and
cathode. Since the cell potential is negative, the reaction as written is not
spontaneous. The electron will flow from the Fe electrode to the Cu electrode.
Thus, Fe will be the anode and Cu will be the cathode.
28
Question 35
Question
Consider the following cell reaction:
Zn(s) |Zn2+(0.10M)|| Cu2+(2.0M)|Cu(s)
Calculate the cell potential at 25
°
C for the electrochemical cell. Given that
the standard reduction potential for the reduction of Cu2+ to Cu is E◦=−0.34
V and for the reduction of Zn2+ to Zn is E◦=−0.76 V.
Solution
Step 1: Write the cell reaction and find the standard cell potential. The cell
reaction is:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
The standard cell potential, E◦
cell, is the difference in standard reduction
potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = (−0.34 V) −(−0.76 V)
E◦
cell = 0.42 V
Step 2: Calculate the cell potential at 25
°
C using the Nernst equation: The
Nernst equation is:
E=E◦−0.0592
nlog Q
Where: E= cell potential E◦= standard cell potential n= number of electrons
transferred in the balanced cell reaction Q= reaction quotient at non-standard
conditions
Step 3: Determine the reaction quotient, Q.
Q=[Zn2+]
[Cu2+]
Q=0.10
2.0
Q= 0.05
Step 4: Calculate the cell potential, E, at 25
°
C. The balanced cell reaction
involves the transfer of 2 moles of electrons. Substituting the values into the
Nernst equation:
E= 0.42 V −0.0592
2log(0.05)
29
382 = [products]
[reactants]
Step 3: Calculate the change in Gibbs free energy using the equation:
∆G=−nF ∆Ecell
Given that n= 1 (assuming one electron transfer), and F= 96485 C/mol, we
have:
∆G=−1×96485 ×(0.90 −0.75)
∆G=−1×96485 ×0.15
∆G=−14472.75 J/mol
Since ∆Gis negative, the reaction is still spontaneous.
Question 2
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
Determine the cell potential (E◦
cell) at standard conditions.
Solution
Step 1: Identify the oxidation and reduction half-reactions.
Oxidation half-reaction: Zn(s)→Zn2+(aq) + 2e−
Reduction half-reaction: Cu2+(aq) + 2e−→Cu(s)
Step 2: Write down the overall cell reaction.
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 3: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials.
E◦
cell =E◦
reduction, cathode −E◦
oxidation, anode = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the cell potential (E◦
cell) at standard conditions for the given
galvanic cell is 1.10 V.
2
Question 3
Question
Consider a galvanic cell with the following half-reactions:
Cathode: Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Anode: MnO−
4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦= 1.51 V
Calculate the cell potential (E◦
cell) for this galvanic cell at standard conditions.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together,
adjusting coefficients to make the number of electrons equal on both sides.
Overall Cell Reaction: Cu2+(aq)+MnO−
4(aq)+8H+(aq)+5e−→Cu(s)+Mn2+(aq)+4H2O(l)
Step 2: Calculate the cell potential (E◦
cell) using the standard reduction
potentials of the half-reactions.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
MnO−
4/Mn2+
E◦
cell = 0.34 V −1.51 V
E◦
cell =−1.17 V
Therefore, the cell potential for this galvanic cell at standard conditions is
−1.17 V.
Question 4
Question
What are the key differences between galvanic cells and electrolytic cells? Pro-
vide a detailed explanation of how each cell operates and how they are used in
different scenarios.
Solution
To understand the key differences between galvanic cells and electrolytic cells,
let’s first define each type of cell and then compare their operation and appli-
cations.
Galvanic Cells:
Galvanic cells, also known as voltaic cells, are devices that use spontaneous
chemical reactions to generate electrical energy.
3
In a galvanic cell, oxidation occurs at the anode, and reduction occurs at
the cathode. Electrons flow from the anode to the cathode through an
external circuit.
The chemical reactions in a galvanic cell produce an electric potential
difference, commonly referred to as voltage.
Galvanic cells are commonly used in batteries to power various devices,
such as flashlights, remote controls, and cell phones.
Electrolytic Cells:
Electrolytic cells are devices that use an external electric current to drive
a non-spontaneous chemical reaction.
In an electrolytic cell, oxidation occurs at the anode (positive electrode),
where electrons are released, and reduction occurs at the cathode (negative
electrode), where electrons are gained.
The external power source (such as a battery) forces the non-spontaneous
reaction to occur by supplying the necessary energy.
Electrolytic cells are commonly used in processes like electroplating, elec-
trolysis of water to produce hydrogen and oxygen, and in the extraction
of metals from their ores.
Comparison:
Energy Source: Galvanic cells convert chemical energy into electrical en-
ergy, while electrolytic cells use electrical energy to drive non-spontaneous
reactions.
Spontaneity: Galvanic cell reactions are spontaneous, while electrolytic
cell reactions are non-spontaneous and require an external energy source.
Anode/Cathode: In a galvanic cell, the anode is negatively charged
and the cathode is positively charged. In an electrolytic cell, the anode is
positively charged and the cathode is negatively charged.
Applications: Galvanic cells are used in batteries for powering devices,
while electrolytic cells are used in processes like electroplating, electrolysis,
and metal extraction.
In conclusion, galvanic cells generate electrical energy from spontaneous
chemical reactions, while electrolytic cells use electrical energy to drive non-
spontaneous reactions. Both types of cells play crucial roles in various techno-
logical applications.
4
Question 5
Question
In a galvanic cell, the standard cell potential (E◦
cell) is 1.08 V. If the concen-
tration of Mn2+ ions in one half-cell is 0.10 M and the concentration of MnO−
4
ions in the other half-cell is 0.20 M, determine the standard electrode potential
(reduction potential) for the Mn2+/Mn half-reaction.
Solution
Step 1: Write out the cell reaction for the galvanic cell. The cell reaction for
the galvanic cell involving Mn2+ and MnO−
4half-reactions is given by:
Mn2+ + MnO−
4→Mn3+ + MnO2
Step 2: Write out the two half-reactions for the cell reaction. The half-
reaction involving Mn2+ is:
Mn2+ →Mn3+ +e−
The half-reaction involving MnO−
4is:
MnO−
4+ 4H++ 3e−→MnO2+ 2H2O
Step 3: Determine the standard cell potential using the standard electrode
potentials. The standard cell potential (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cell = 1.08V, we can substitute the standard electrode potentials
for the half-reactions to get:
1.08 = E◦
cathode −E◦
anode
Step 4: Calculate the standard electrode potential for the cathode half-
reaction. From experimental data, we know that the standard electrode po-
tential for the MnO−
4/MnO2half-reaction is 1.51V. Therefore, the standard
electrode potential for the Mn2+/Mn half-reaction can be calculated as:
1.08V = 1.51V −E◦
anode
E◦
anode = 1.51V −1.08V = 0.43V
Step 5: Write out the standard electrode potential for the Mn2+/Mn half-
reaction. Therefore, the standard electrode potential (reduction potential) for
the Mn2+/Mn half-reaction is 0.43V .
5
Question 6
Question
Consider a galvanic cell with a standard potential of 0.70 V that is made from
a copper electrode in a 1.0 M CuSO4solution and a silver electrode in a 1.0 M
AgNO3solution. a) Write the balanced cell reaction. b) Calculate the standard
cell potential. c) Determine in which direction the cell reaction will proceed. d)
If the current is allowed to flow until 1.0 mol of electrons have passed through
the external circuit, how many grams of silver will be deposited? (Assume 100
Solution
a) The balanced cell reaction is given by:
Cu2+(aq) + 2e−→Cu(s) at the cathode
2Ag(s)→2Ag+(aq) + 2e−at the anode
Overall cell reaction:
Cu2+(aq) + 2Ag(s)→Cu(s) + 2Ag+(aq)
b) The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cathode = 0.34 V and E◦
anode = 0.80 V, we have:
E◦
cell = 0.34 V −0.80 V = −0.46 V
c) The cell reaction will proceed in the reverse direction as the cell potential
is negative.
d) The total charge passed through the circuit (Q) can be calculated using
the formula:
Q=nF
where nis the number of moles of electrons and Fis the Faraday constant
(96500 C/mol). Given that n= 1.0 mol, we have:
Q= 1.0 mol ×96500 C/mol = 96500 C
The amount of silver deposited can be calculated using the formula:
Amount of substance (Ag) = Q
F
Given that the molar mass of silver (Ag) is 107.87 g/mol, we have:
Amount of substance (Ag) = 96500 C
96500 C/mol = 1.0 mol
Mass (Ag) = 1.0 mol ×107.87 g/mol = 107.87 g
Therefore, 107.87 grams of silver will be deposited.
6
Question 7
Question
Consider a galvanic cell constructed with a standard hydrogen electrode (SHE)
and a copper electrode. The standard reduction potential of the copper electrode
is E◦
Cu = 0.34 V. The cell operates at 25◦C. Given the standard reduction
potential of the SHE is 0 V, calculate the cell potential at standard conditions
and determine whether the cell reaction is spontaneous.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the reduction
half-reaction at the cathode minus the reduction half-reaction at the anode.
Cathode: 2H++ 2e−→H2(g)E◦
H+/H2= 0 V
Anode: Cu2+ + 2e−→Cu(s) E◦
Cu2+/Cu = 0.34 V
The overall reaction is:
Cu2+ + 2H+→Cu(s) + H2(g)
Step 2: Calculate the cell potential, Ecell. Using the Nernst equation and
standard reduction potentials:
Ecell =E◦
cathode −E◦
anode
Ecell = 0V−0.34V=−0.34V
Step 3: Determine spontaneity of the reaction. For the reaction to be spon-
taneous, the cell potential must be positive. Since Ecell =−0.34V < 0, the cell
reaction is not spontaneous.
Question 8
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 1.23 V. If
the concentration of Cu2+ in the cathode compartment is 0.10 M, and the
concentration of Zn2+ in the anode compartment is 1.00 M, determine the cell
potential when the concentration of Zn2+ is decreased to 0.10 M.
Solution
Step 1: Write the balanced cell reaction for the galvanic cell. The cell reaction
for the galvanic cell using a Zn anode and a Cu cathode is:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
7
Step 2: Calculate the standard cell potential for the given concentrations.
The Nernst equation relates the cell potential with the concentrations of the
ions involved:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
where n= 2 is the number of electrons transferred in the balanced equation,
Ecell is the cell potential, and E◦
cell is the standard cell potential.
Substitute the given values into the Nernst equation:
Ecell = 1.23−0.0592
2log 0.10
1.00= 1.23−0.0296×−1=1.23+0.0296 = 1.2596 V
Step 3: Calculate the new cell potential with the decreased concentration of
Zn2+. Using the same Nernst equation and substituting the new concentration
of Zn2+:
Enew = 1.23 −0.0592
2log 0.10
0.10= 1.23 −0.0296 ×0=1.23 V
Therefore, the new cell potential when the concentration of Zn2+ is decreased
to 0.10 M is 1.23 V.
Question 9
Question
An electrochemical cell consists of a zinc metal electrode in a 1.0 M Zn2+ solution
and a copper metal electrode in a 1.0 M Cu2+ solution. The standard reduction
potentials are E◦(Cu2+/Cu) = 0.34 V and E◦(Zn2+/Zn) = −0.76 V. Calculate
the cell potential at 25
°
C when 1.0 mol of Zn2+ ions is reduced at the zinc
electrode.
Solution
Step 1: Write the two half-reactions for the cell reaction and their corresponding
standard reduction potentials. The overall cell reaction is:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
The half-reactions are:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Calculate the cell potential. The cell potential (E◦
cell) can be calcu-
lated using the formula:
E◦
cell =E◦
cathode −E◦
anode
8
Substitute the given E◦values:
E◦
cell = 0.34 V −(−0.76) V = 0.34 V + 0.76 V = 1.10 V
Therefore, the cell potential at 25
°
C when 1.0 mol of Zn2+ ions is reduced
at the zinc electrode is 1.10 V.
Question 10
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a standard zinc electrode connected by a salt bridge. The standard reduc-
tion potential of the hydrogen half-cell is 0.00 V, and the standard reduction
potential of the zinc half-cell is −0.76 V. Determine the cell potential at standard
conditions and predict the spontaneity of the reaction.
Solution
Step 1: Write the overall cell reaction. The overall reaction for the galvanic cell
is the oxidation of zinc and reduction of hydrogen:
Zn (s) + 2H+(aq)→Zn2+(aq)+H2(g)
Step 2: Write the half-reactions and their standard reduction potentials.
Zinc half-reaction: Zn(s)→Zn2+(aq) + 2e−(E◦=−0.76 V)
Hydrogen half-reaction: 2H+(aq) + 2e−→H2(g) (E◦= 0.00 V)
Step 3: Calculate the standard cell potential. The standard cell potential
(E◦
cell) is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(−0.76 V) = 0.76 V
Step 4: Predict the spontaneity of the reaction. Since the standard cell
potential is 0.76 V (which is greater than 0 V), the reaction is spontaneous
under standard conditions.
Question 11
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a copper electrode. The standard reduction potential for the copper elec-
trode is +0.34 V. Calculate the standard cell potential for this galvanic cell.
9
Solution
To calculate the standard cell potential for this galvanic cell, we can use the
equation:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode
is the standard reduction potential of the anode.
Given that the standard reduction potential for the copper electrode is +0.34
V, we have:
Step 1: Identify the cathode and anode reactions. The reduction half-
reaction at the copper electrode is:
Cu2+ + 2e−→Cu
The anode reaction at the SHE is the standard hydrogen electrode:
2H++ 2e−→H2
Step 2: Determine the standard cell potential. The standard reduction
potential for the standard hydrogen electrode is 0 V. So, E◦
cathode = 0 V and
E◦
anode =−0.34 V.
Substitute these values into the equation:
E◦
cell = 0 −(−0.34) = +0.34 V
Therefore, the standard cell potential for this galvanic cell is +0.34 V.
Question 12
Question
Consider a galvanic cell with a standard potential of 1.10 V. If the cell potential
is measured to be 1.00 V under non-standard conditions, calculate the reaction
quotient Q with a given concentration of [Mn+](aq)= 0.10 M and [Mn](s)= 1.0
M. Determine if the reaction is at equilibrium or not.
Solution
Step 1: Write the half-reaction for the reduction of Mn2+ to Mn.
The half-reaction is:
Mn2+(aq)+2e−→Mn(s)
Step 2: Calculate the standard potential for the cell.
The standard potential for the cell is given as 1.10 V.
Step 3: Calculate the reaction quotient Q.
10
The Nernst Equation relates the standard cell potential, the actual cell po-
tential, and the reaction quotient Q:
E=E◦−0.0592
nlog Q
Given that E= 1.00 V and E◦= 1.10 V, substituting these values into the
equation gives:
1.00 = 1.10 −0.0592
2log Q
Solving for log Q:
−0.10 = −0.0298 log Q
log Q= 3.355
Q= 103.355 = 2382.5
So, the reaction quotient Q is 2382.5.
Step 4: Determine if the reaction is at equilibrium.
Since Q>K, the reaction quotient is greater than the equilibrium constant
K. This implies that the reaction is not at equilibrium.
Question 13
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe2+(aq)→Fe3+(aq) + e−E◦=−0.77 V
Calculate the standard cell potential for the galvanic cell formed by connecting
these two half-cells in a complete circuit.
Solution
Step 1: Identify the oxidation and reduction half-reactions. The oxidation half-
reaction is:
Fe2+(aq)→Fe3+(aq) + e−E◦=−0.77 V
The reduction half-reaction is:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Write the overall cell reaction. Adding the two half-reactions to-
gether, we get the overall cell reaction:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
11
Step 3: Calculate the standard cell potential. The standard cell potential
is calculated by adding the standard reduction potentials of the reduction half-
reaction and the oxidaiton half-reaction:
E◦
cell =E◦
reduction +E◦
oxidation
E◦
cell = 0.34 V + (−0.77 V) = −0.43 V
Therefore, the standard cell potential for the galvanic cell formed by con-
necting these two half-cells is −0.43 V.
Question 14
Question
Consider an electrochemical cell with the following half-reactions:
Anode: Pb(s) →Pb2+(aq)+2e−
Cathode: Ag+(aq) + e−→Ag(s)
If the standard reduction potentials are E◦
Pb2+/Pb =−0.13 V and E◦
Ag+/Ag =
0.80 V, determine the cell potential and state whether the cell is galvanic or
electrolytic.
Solution
Step 1: Write the overall cell reaction by summing the two half-reactions:
Pb(s) + 2Ag+(aq)→Pb2+(aq) + 2Ag(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials for each half-reaction:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −(−0.13 V) = 0.93 V
Step 3: Determine whether the cell is galvanic or electrolytic by examining
the sign of E◦
cell. Since E◦
cell is positive (0.93 V), the cell is galvanic because a
spontaneous chemical reaction will occur.
Question 15
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.84 V. The cell
reaction is given by:
12
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
(a) Determine the standard cell potential for the reverse reaction.
(b) Calculate the equilibrium constant, K, for the forward reaction at 25◦C.
Solution
(a) The standard cell potential for the reverse reaction can be determined using
the relationship:
E◦
cell, reverse =−E◦
cell, forward
Therefore,
E◦
cell, reverse =−E◦=−0.84 V
So, the standard cell potential for the reverse reaction is −0.84 V.
(b) The equilibrium constant, K, for the forward reaction can be determined
from the standard cell potential using the Nernst equation:
E=E◦−0.0592
nlog Q
For the standard cell potential, Q=K, so:
E◦= 0.0592 log K
n
Given E◦= 0.84 V, n= 2 (from the coefficients in the balanced equation),
and T= 25◦C, we have:
0.84 = 0.0592 ×log K
2
Solving for log K:
0.84
0.0592 ×2= log K
7.09 = log K
K= 107.09
K≈8.90 ×107
Therefore, the equilibrium constant, K, for the forward reaction at 25◦C is
approximately 8.90 ×107.
13
Question 16
Question
Consider a galvanic cell with a standard potential of E◦= 0.40 V. If the cell
potential is measured to be 0.60 V under non-standard conditions where the
concentrations are [M2+]=0.10 M and [M]=1.0 M, determine the reaction
quotient Qand discuss if the cell reaction is at equilibrium or not. Additionally,
explain whether the cell reaction would be spontaneous in the forward direction
or not.
Solution
Step 1: Write the half-reactions for the galvanic cell. The overall cell reaction
for a galvanic cell based on the given standard potential is:
Oxidation half reaction: M(s)→M2+(aq)+2e−
Reduction half reaction: M2+(aq)+2e−→M(s)
Step 2: Determine the reaction quotient Q. The reaction quotient Qis given
by the expression:
Q=[M2+]
[M]2=0.10
1.02= 0.10
Step 3: Compare the cell potential Eand the standard cell potential E◦.
Given that the cell potential E= 0.60 V and the standard cell potential E◦=
0.40 V, we can compare these values to determine the direction of the reaction. If
E > E◦, the reaction quotient Q<K, and the cell reaction is not at equilibrium.
Step 4: Determine if the cell reaction is spontaneous and in which direction.
Since E > E◦, the reaction is not at equilibrium and will proceed in the direction
to reach equilibrium. In this case, the cell reaction will be spontaneous in the
forward direction.
Therefore, based on the values provided, the reaction quotient Qis 0.10, the
cell reaction is not at equilibrium, and the cell reaction will be spontaneous in
the forward direction.
Question 17
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) →Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Calculate the cell potential at standard conditions (T= 25◦C, P= 1 atm)
for the given galvanic cell. Assume all species are at 1 M concentration.
14
Solution
Step 1: Write the cell reaction by adding the two half-reactions:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Write the Nernst equation for cell potential (Ecell):
Ecell =E◦
cell −0.0592
nlog(Q)
where: - Ecell is the cell potential, - E◦
cell is the standard cell potential, - nis
the number of electrons transferred (from the balanced equation), - Qis the
reaction quotient.
Step 3: Determine the standard cell potential (E◦
cell) using standard reduc-
tion potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
Step 4: Substitute the standard reduction potentials into the equation:
E◦
cell = (0.34 V) −(−0.76 V)
Step 5: Calculate E◦
cell:
E◦
cell = 1.10 V
Step 6: Determine the number of electrons transferred, n= 2.
Step 7: Calculate the reaction quotient Q:
Q=[Zn2+][Cu]
[Zn][Cu2+]
Step 8: Substitute Q,n, and E◦
cell into the Nernst equation and solve for
Ecell to find the cell potential at standard conditions.
Question 18
Question
Consider a galvanic cell that consists of a zinc electrode immersed in a 1.0 M
Zn2+ solution and a cadmium electrode immersed in a 1.0 M Cd2+ solution.
The standard reduction potentials are as follows: Zn2+/Zn with E◦=−0.76 V
and Cd2+/Cd with E◦=−0.40 V. Calculate the cell potential at 298 K.
15
Solution
Step 1: Write the balanced cell reaction and calculate the cell potential. The
balanced cell reaction can be written as:
Zn(s) + Cd2+(aq)→Zn2+(aq) + Cd(s)
The cell potential (E◦
cell) is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode
is the standard reduction potential of the anode.
Substitute the given values to calculate the cell potential:
E◦
cell =E◦
Cd2+/Cd −E◦
Zn2+/Zn
E◦
cell = (−0.40 V) −(−0.76 V)
E◦
cell = 0.36 V
Therefore, the cell potential at 298 K is 0.36 V.
Question 19
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a 1.0 M solution of Cu2+. The standard reduction potential for the
hydrogen electrode is 0.00 V and for the copper electrode is +0.34 V.
a) Write the balanced cell reaction for the galvanic cell.
b) Calculate the cell potential at 25
°
C.
c) If a voltage of 0.50 V is applied to the cell in the reverse direction, will
the copper electrode dissolve or plate out? Justify your answer.
Solution
a) The balanced cell reaction for the galvanic cell is given by the oxidation of
copper at the anode and the reduction of hydrogen at the cathode:
Anode: Cu(s)→Cu2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
Combining the two half-reactions, we get:
Overall Cell Reaction: Cu(s) + 2H+(aq)→Cu2+(aq)+H2(g)
16
b) The cell potential, E◦
cell, can be calculated using the standard reduction
potentials for the half-reactions and the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(+0.34 V) = −0.34 V
c) The cell potential of -0.34 V indicates that the cell reaction is spontaneous
as written above. If a voltage of 0.50 V is applied in the reverse direction, the
cell will be forced to run in reverse. Since the new applied voltage is greater than
the cell potential, the reaction will be non-spontaneous in the reverse direction.
This means that copper will dissolve from the copper electrode instead of plating
out.
Question 20
Question
Consider a galvanic cell that consists of a silver electrode immersed in a 1.0
M solution of AgNO3and a zinc electrode immersed in a 1.0 M solution of
Zn(NO3)2. The standard reduction potentials are Ag++e−→Ag(E◦= 0.80 V)
and Zn2+ + 2e−→Zn(E◦=−0.76 V). Determine the standard cell potential
and identify the anode and cathode.
Solution
Step 1: Write the two half-reactions and determine the standard cell poten-
tial. The spontaneous reaction for the galvanic cell is given by the overall cell
reaction: Zn + Ag+→Zn2+ + Ag.
This reaction can be broken down into two half-reactions: Anode: Zn →
Zn2+ + 2e−(oxidation) Cathode: Ag++ e−→Ag (reduction)
The standard cell potential can be calculated as the sum of the standard
reduction potentials for the two half-reactions: Standard cell potential, E◦
cell =
E◦
cathode −E◦
anode = 0.80 V −(−0.76 V) = 1.56 V.
Therefore, the standard cell potential for the galvanic cell is 1.56 V.
Step 2: Identify the anode and cathode. Since the more negative standard
reduction potential corresponds to the anode, in this case, zinc is the anode and
silver is the cathode.
Therefore, the anode is Zn and the cathode is Ag+.
Question 21
Question
Consider the following galvanic cell:
17
Zn |Zn2+(0.1M)|| Ni2+(0.01 M)|Ni
The standard reduction potentials for the half-reactions are given as follows:
Zn2+(aq)+2e−→Zn(s) with E◦=−0.76 V
Ni2+(aq)+2e−→Ni(s) with E◦=−0.25 V
Calculate the cell potential at 25◦C and determine whether the reaction is
spontaneous or non-spontaneous.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the sum of
the two half-reactions:
Zn(s) + Ni2+(aq)→Zn2+(aq) + Ni(s)
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the standard reduction potentials (E◦) of the
two half-reactions.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.25 V −(−0.76 V)
E◦
cell = 1.01 V
Step 3: Calculate the cell potential at 25◦C using the Nernst Equation. The
Nernst Equation relates the cell potential to the standard cell potential and the
reaction quotient (Q) at non-standard conditions.
Ecell =E◦
cell −0.0592
nlog Q
Step 4: Calculate the reaction quotient (Q). Since Ni2+ is the oxidizing
agent and Zn is the reducing agent, the reaction quotient is given by:
Q=[Zn2+]
[Ni2+]=0.1
0.01 = 10
Step 5: Plug the values into the Nernst equation to find Ecell.
Ecell = 1.01 V −0.0592
2log(10)
Ecell = 1.01 V −(0.0296) ×1
Ecell = 1.01 V −0.0296
Ecell = 0.9804 V
Step 6: Determine if the reaction is spontaneous. Since Ecell is positive, the
reaction is spontaneous as the cell potential is greater than zero.
18
Question 22
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.70 V. Determine
the equilibrium constant, K, for the following cell reaction:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Solution
Step 1: Write the half-reactions for the anode and cathode:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Step 2: Determine the standard cell potential by subtracting the potential
of the anode from that of the cathode:
E◦
cell =E◦
cathode −E◦
anode = 0.34 V
Step 3: Using the formula ∆G◦=−nF E◦, find the standard Gibbs free
energy change:
∆G◦=−nF E◦
cell =−2(96485C/mol)(0.34V) = −65.1 kJ/mol
Step 4: Determine Kusing the equation ∆G◦=−RT ln K:
−65.1 kJ/mol = −(8.314J/(mol ·K))(298K) ln K
ln K= 23.4
K=e23.4= 1.05 ×1010
Therefore, the equilibrium constant for the cell reaction is K= 1.05 ×1010.
Question 23
Question
A student is conducting an experiment involving galvanic and electrolytic cells.
The student is given a cell with a standard cell potential of E◦
cell =−0.56 V.
The student wants to know whether this cell would operate as a galvanic cell
(producing electricity) or as an electrolytic cell (consuming electricity). Deter-
mine whether the cell would function as a galvanic cell or an electrolytic cell,
and justify your answer.
19
Solution
Step 1: Recall that for a galvanic cell, the standard cell potential (E◦
cell) is
positive, indicating spontaneous electron flow from the anode to the cathode.
For an electrolytic cell, the standard cell potential is negative, requiring an
external voltage to drive a non-spontaneous reaction.
Step 2: Given that E◦
cell =−0.56 V, the negative value indicates that the cell
reaction is not spontaneous and would require an external voltage to proceed.
Step 3: Therefore, the cell with a standard cell potential of E◦
cell =−0.56
V would function as an electrolytic cell as it does not naturally generate
electricity and would need an external power source to operate.
Thus, the cell in question would operate as an electrolytic cell.
Question 24
Question
Consider the following half-reactions:
Ag+(aq)+e−→Ag(s) (E◦= 0.80 V)
Ni2+(aq) + 2e−→Ni(s) (E◦=−0.26 V)
Determine the cell potential at 25
°
C for the reaction:
2Ag+(aq) + Ni(s) →2Ag(s) + Ni2+(aq)
Solution
Step 1: Write the half-reactions involved in the cell reaction.
Oxidation half-reaction: Ni(s) →Ni2+(aq) + 2e−
Reduction half-reaction: 2Ag+(aq) + 2e−→2Ag(s)
Step 2: Write the overall cell reaction as the combination of the two half-
reactions.
2Ag+(aq) + Ni(s) →2Ag(s) + Ni2+(aq)
Step 3: Find the standard cell potential by summing the standard reduction
potentials of the half-reactions.
E◦
cell =E◦
reduction(Ag+) + E◦
oxidation(Ni)
E◦
cell = 0.80 V + (−0.26 V)
E◦
cell = 0.54 V
Thus, the cell potential at 25
°
C for the given reaction is 0.54 V.
20
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
Given that the standard reduction potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =
−0.76 V, calculate the standard cell potential E◦
cell. Is this cell galvanic or elec-
trolytic? Justify your answer.
Solution
Step 1: Write the expression for the standard cell potential:
E◦
cell =E◦
cathode −E◦
anode
Step 2: Substitute the standard reduction potentials into the formula:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Since E◦
cell is positive, the cell reaction is spontaneous and the cell is galvanic.
The electrons flow from the anode (Zn) to the cathode (Cu).
Question 26
Question
Consider a galvanic cell constructed with a nickel electrode in contact with
a solution that is 0.10 M in Ni2+ ions and a silver electrode in contact with
a solution that is 1.0 M in Ag+ions. The standard reduction potentials are
E◦(Ni2+/Ni) = −0.25 V and E◦(Ag+/Ag) = 0.80 V. Determine the cell poten-
tial and the direction of the spontaneous reaction.
Solution
Step 1: Write the half-reactions for the cell: In the galvanic cell, the cathode is
the nickel electrode (Ni2+ being reduced) and the anode is the silver electrode
(Ag being oxidized). Cathode half-reaction: Ni2+ + 2e−→Ni
Anode half-reaction: Ag →Ag++e−
Step 2: Determine the overall cell reaction: The standard cell potential,
E◦
cell, is given by: E◦
cell =E◦
cathode −E◦
anode
Substitute the given values into the equation: E◦
cell =E◦
Ni2+/Ni −E◦
Ag+/Ag
E◦
cell =−0.25 V −0.80 V = −1.05 V
21
Step 3: Determine the direction of the spontaneous reaction: Since the cell
potential is negative, the reaction as written is not spontaneous. To make it
spontaneous, reverse the reaction.
Therefore, the spontaneous reaction in the galvanic cell is: Ag + Ni2+ →
Ag++ Ni
Question 27
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential for the reduction of Cu2+(aq) is 0.34 V
and the standard reduction potential for the reduction of Zn2+(aq) is −0.76 V,
calculate the standard cell potential (E◦) of the galvanic cell.
Solution
Step 1: Identify the half-reactions and their standard reduction potentials.
The given half-reactions are: Anode:
Zn(s) −−→ Zn2+(aq) + 2 e−with Eo=−0.76 V
Cathode:
Cu2+(aq) + 2 e−−−→ Cu(s) with Eo= 0.34 V
Step 2: Write the overall cell reaction.
The overall cell reaction is obtained by adding the two half-cell reactions
together so that electrons cancel out:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 3: Calculate the standard cell potential (E◦) using the formula:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values into the formula:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the standard cell potential (E◦) of the galvanic cell is 1.10 V.
22
Question 28
Question
Consider a galvanic cell with the following half-reactions:
Anode: Ag(s)→Ag+(aq) + e−
Cathode: Cu2+(aq)+2e−→Cu(s)
If the standard reduction potentials are E◦
Ag+/Ag = 0.80 V and E◦
Cu2+/Cu =
0.34 V, calculate the standard cell potential (E◦
cell) for the Galvanic Cell.
Solution
Step 1: Write the overall cell reaction.
Ag(s) + Cu2+(aq)→Ag+(aq) + Cu(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Ag+/Ag
E◦
cell = 0.34 V −0.80 V
E◦
cell =−0.46 V
Therefore, the standard cell potential for the Galvanic Cell is −0.46 V.
Question 29
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a silver electrode. The standard reduction potential for the Ag++e−−>
Aghalf −reactionis0.80V.Ifthemeasuredcellpotentialis0.35V, determinethestandardreductionpotentialfortheSHEhalf−
reaction.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction for a galvanic
cell involving a standard hydrogen electrode (SHE) and a silver electrode can
be represented as:
2H+(aq) + 2e−→H2(g) + Ag+(aq)→Ag(s)+H2(g)
23
Step 2: Write the half-reaction at the SHE. The standard reduction potential
for the SHE half-reaction is 0 V by definition. Therefore, the half-reaction at
the SHE is:
2H+(aq) + 2e−→H2(g)
Step 3: Write the half-reaction at the Ag electrode. Given that the standard
reduction potential for the Ag half-reaction is 0.80 V, the half-reaction at the
Ag electrode is:
Ag+(aq)+e−→Ag(s)
Step 4: Use the measured cell potential to find the standard reduction po-
tential for the SHE half-reaction. The measured cell potential is the sum of the
standard reduction potentials for the two half-reactions. Therefore, we have:
Ecell =ESHE −EAg
0.35 V = 0 V −(0.80 V)
ESHE = 0.35 V + 0.80 V
ESHE = 1.15 V
Therefore, the standard reduction potential for the SHE half-reaction is 1.15
V.
Question 30
Question
Consider a galvanic cell with the following half-cell reactions:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
If the standard reduction potential for the Zn2+|Zn half-cell is −0.76 V and
the standard reduction potential for the H+|H2half-cell is 0.00 V, calculate the
standard cell potential (E◦
cell) for this galvanic cell.
Solution
Step 1: Write the overall cell reaction by summing the half-cell reactions. Re-
arrange the half-cell reactions so that electrons cancel out:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
The overall cell reaction is:
Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)
24
Step 2: Calculate the standard cell potential using the standard reduction
potentials. The standard cell potential (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
The standard reduction potential for the cathode reaction is 0.00 V, and for
the anode reaction is −0.76 V. Therefore,
E◦
cell = 0.00 V −(−0.76) V = 0.76 V
Therefore, the standard cell potential for this galvanic cell is 0.76 V.
Question 31
Question
Consider a galvanic cell with a standard voltage of 0.85 V. If the concentration
of Fe2+ ions in the cell is 0.10 M and the concentration of Zn2+ ions is 1.0 M,
determine the standard electrode potential of the Zn2+/Zn half-cell.
Solution
To find the standard electrode potential of the Zn2+/Zn half-cell, we can use
the Nernst equation and the given information about the galvanic cell.
Step 1: Write the overall cell reaction. The galvanic cell is composed of
two half-reactions: Fe2+ + 2e−→Fe and Zn2+ + 2e−→Zn. The overall cell
reaction is the sum of these two half-reactions: Fe2+ + Zn2+ →Fe + Zn2+.
Step 2: Determine the standard cell potential. The standard cell potential,
E◦
cell, is given as 0.85 V.
Step 3: Calculate the cell potential under the given conditions. The Nernst
equation relates the standard cell potential to the actual cell potential and the
concentrations of the species:
Ecell =E◦
cell −0.0592
nlog [Fe2+]
[Zn2+]
Given that Ecell = 0.85 V, [Fe2+] = 0.10 M, and [Zn2+] = 1.0 M, and the
number of electrons involved in the reaction is 2, we can substitute these values
into the Nernst equation to solve for the standard electrode potential of the
Zn2+/Zn half-cell.
Step 4: Calculate the standard electrode potential of the Zn2+/Zn half-cell.
Using the Nernst equation:
0.85 V = E◦
cell −0.0592
2log 0.10
1.0
Solving for E◦
cell:
E◦
cell = 0.85 + 0.0296 log(0.10)
25
E◦
cell = 0.85 + 0.0296(−1.0)
E◦
cell = 0.85 −0.0296
E◦
cell = 0.8204 V
Therefore, the standard electrode potential of the Zn2+/Zn half-cell is 0.8204
V.
Question 32
Question
Consider a galvanic cell with a standard cell potential of 1.23 V consisting of a
standard hydrogen electrode (SHE) and a copper electrode. If the concentration
of Cu2+ ions is 1.0 M in the half-cell containing the copper electrode, calculate
the standard reduction potential of the copper electrode.
Solution
Step 1: Write the cell reaction for the galvanic cell.
The cell reaction is:
Cu2+ + 2e−→Cu
Step 2: Write the cell potential equation.
The standard cell potential, E◦
cell, is given as 1.23 V.
The standard reduction potential of SHE is 0 V.
Using the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Substitute the values:
1.23 = E◦
Cu2+/Cu −0
Step 3: Calculate the standard reduction potential of the copper electrode.
So,
E◦
Cu2+/Cu = 1.23 V
Therefore, the standard reduction potential of the copper electrode is 1.23
V.
Question 33
Question
A galvanic cell has a standard cell potential of E◦= 0.81 V. If the cell operates
until the concentration of Zn2+ ions is decreased to 0.300 M from an initial
concentration of 1.00 M, determine the final potential of the cell. Assume that
the temperature remains constant.
26
Solution
Step 1: Write the balanced cell reaction.
The galvanic cell consists of a zinc electrode in a Zn2+ solution and a copper
electrode in a Cu2+ solution. The cell reaction is:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Identify the half-reactions.
The half-reactions for the cell are: - Oxidation half-reaction: Zn(s)→
Zn2+(aq)+2e−- Reduction half-reaction: Cu2+(aq)+2e−→Cu(s)
Step 3: Calculate the standard cell potential (E◦).
Using the standard reduction potentials for zinc and copper from tables, we
have: E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.76 V) = 1.10 V
Step 4: Use the Nernst equation to calculate the final cell potential.
The Nernst equation relates the non-standard cell potential (E) to the stan-
dard cell potential (E◦), the reaction quotient (Q), the Faraday constant (F),
the gas constant (R), the temperature (T), and the number of electrons trans-
ferred (n) in the cell reaction.
E=E◦−RT ln(Q)
nF
Given that the initial concentration of Zn2+ ions is 1.00 M and the final
concentration is 0.300 M, we can calculate the reaction quotient Qusing these
concentrations. The value of nis 2 for this cell reaction.
Step 5: Calculate the final potential of the cell.
Substitute the values into the Nernst equation:
E= 1.10 V −
(8.314 J/K ·mol)(298 K) ln 0.300
1.00
2(96,485 C/mol) ≈1.07 V
Therefore, the final potential of the cell is approximately 1.07 V.
Question 34
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
If the concentration of Cu2+(aq) is 1.0 M and Fe3+(aq) is 0.1 M, calculate
the cell potential at 25
°
C. Determine the direction of electron flow and identify
the anode and cathode.
27
Solution
Step 1: Write the balanced overall redox reaction. Let’s first write the two
half-reactions in the reduction form:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
To find the overall reaction, we need to add the two half-reactions together:
3Cu2+(aq) + Fe3+(aq)→3Cu(s) + Fe(s)
Step 2: Calculate the standard cell potential (E◦). The standard cell poten-
tial is the difference between the standard reduction potentials of the cathode
and the anode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Fe3+/Fe −E◦
Cu2+/Cu
E◦
cell = (−0.04 V) −(0.34 V)
E◦
cell =−0.38 V
Step 3: Calculate the cell potential at 25
°
C using the Nernst equation.
Ecell =E◦
cell −RT
nF ln Q
K
where Ris the gas constant (8.314 J/(mol
·
K)), Tis the temperature in Kelvin,
nis the number of moles of electrons transferred, Fis Faraday’s constant (96485
C/mol), Qis the reaction quotient, and Kis the equilibrium constant.
Since this is not at equilibrium, Qis not equal to K. We need to calculate
the reaction quotient Q:
Q=[Fe(s)]
[Cu(s)] =1
1/3= 3
Plugging in the values, we get:
Ecell =−0.38 V −(8.314 J/(mol
·
K))(298 K)
3(96485 C/mol) ln(3)
Ecell =−0.38 V −0.004 ln(3)
Ecell =−0.38 V −0.004 ×1.099
Ecell =−0.38 V −0.0044
Ecell ≈ −0.3844 V
Step 4: Determine the direction of electron flow and identify the anode and
cathode. Since the cell potential is negative, the reaction as written is not
spontaneous. The electron will flow from the Fe electrode to the Cu electrode.
Thus, Fe will be the anode and Cu will be the cathode.
28
Question 35
Question
Consider the following cell reaction:
Zn(s) |Zn2+(0.10M)|| Cu2+(2.0M)|Cu(s)
Calculate the cell potential at 25
°
C for the electrochemical cell. Given that
the standard reduction potential for the reduction of Cu2+ to Cu is E◦=−0.34
V and for the reduction of Zn2+ to Zn is E◦=−0.76 V.
Solution
Step 1: Write the cell reaction and find the standard cell potential. The cell
reaction is:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
The standard cell potential, E◦
cell, is the difference in standard reduction
potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = (−0.34 V) −(−0.76 V)
E◦
cell = 0.42 V
Step 2: Calculate the cell potential at 25
°
C using the Nernst equation: The
Nernst equation is:
E=E◦−0.0592
nlog Q
Where: E= cell potential E◦= standard cell potential n= number of electrons
transferred in the balanced cell reaction Q= reaction quotient at non-standard
conditions
Step 3: Determine the reaction quotient, Q.
Q=[Zn2+]
[Cu2+]
Q=0.10
2.0
Q= 0.05
Step 4: Calculate the cell potential, E, at 25
°
C. The balanced cell reaction
involves the transfer of 2 moles of electrons. Substituting the values into the
Nernst equation:
E= 0.42 V −0.0592
2log(0.05)
29
E= 0.42 V −0.0296 log(0.05)
E≈0.361 V
Therefore, the cell potential at 25
°
C for the electrochemical cell is approxi-
mately 0.361 V.
30
