CHEM 121 - GENERAL CHEMISTRY
I - Galvanic and Electrolytic Cells
Question Bank - Set 4
Liberty University
Question 1
Question
Define a galvanic cell and an electrolytic cell. Explain the main differences
between the two types of cells.
Solution
Galvanic Cell: Step 1: A galvanic cell, also known as a voltaic cell, is an
electrochemical cell that generates electrical energy from spontaneous redox
reactions. Step 2: In a galvanic cell, chemical energy is converted into electrical
energy; the redox reaction releases energy that can be harnessed as electricity.
Step 3: Electrons flow from the anode (where oxidation occurs) to the cathode
(where reduction occurs) through an external circuit. Step 4: A classic example
of a galvanic cell is the Daniell cell, which consists of a zinc electrode (anode),
a copper electrode (cathode), and an electrolyte solution. Step 5: The voltage
produced by a galvanic cell is positive because the reaction is spontaneous.
Electrolytic Cell: Step 6: An electrolytic cell is an electrochemical cell
that uses electrical energy to drive a non-spontaneous redox reaction. Step 7:
In an electrolytic cell, electrical energy is converted into chemical energy; the
non-spontaneous redox reaction is made to occur by external electrical energy.
Step 8: Electrons flow from the external power source to the anode (where
oxidation occurs) and from the cathode (where reduction occurs) back to the
power source. Step 9: A classic example of an electrolytic cell is the electrolysis
of water, which breaks down water into hydrogen and oxygen gas. Step 10:
The voltage required by an electrolytic cell is negative because the reaction is
non-spontaneous and requires an external energy source.
Main differences between Galvanic and Electrolytic Cells: Step 11:
Galvanic cells produce electrical energy from spontaneous reactions, while elec-
trolytic cells require electrical energy to drive non-spontaneous reactions. Step
12: In galvanic cells, electrons flow from the anode to the cathode; in electrolytic
cells, electrons flow from the external source to the anode and from the cathode
back to the external source. Step 13: Galvanic cells have a positive cell potential,
while electrolytic cells have a negative cell potential. Step 14: Galvanic cells are
commonly used in batteries to generate electricity, while electrolytic cells are
used in processes such as electroplating, electrorefining, and electrolysis. Step
15: Overall, galvanic cells convert chemical energy into electrical energy, while
electrolytic cells convert electrical energy into chemical energy.
Question 2
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦
red =−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦
red = +0.34 V
If the initial concentrations are [Zn2+]=1.0 M and [Cu2+]=1.0 M, calculate
the cell potential when the cell is operating.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Determine the cell potential using the standard reduction potentials
provided:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = (+0.34 V) −(−0.76 V) = 1.10 V
Step 3: Apply the Nernst equation to account for non-standard conditions:
The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Cu2+]
Since this is a redox reaction involving the transfer of 2 electrons, n= 2.
Substitute the given concentrations:
Ecell = 1.10 V −0.0592
2log 1.0
1.0
Ecell = 1.10 V
Therefore, the cell potential when the cell is in operation is 1.10 V.
2
Question 3
Question
Consider a galvanic cell with a standard reduction potential of E◦
cell = 0.30 V.
If the cell potential is measured to be 0.45 V under non-standard conditions
where the concentration of Cu2+ ions in the cathode compartment is 0.10 M,
determine the concentration of Cu2+ ions in the anode compartment.
Solution
Step 1: Write the half-reactions for the cell. The overall cell reaction can be
split into two half-reactions: Cathode (Reduction): Cu2+(aq)+2e−→Cu(s)
Anode (Oxidation): Zn(s)→Zn2+(aq)+2e−
Step 2: Calculate the cell potential under non-standard conditions. By using
the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Cu2+]anode
[Cu2+]cathode
Substitute the given values:
0.45 = 0.30 −0.0592
2log [Cu2+]anode
0.10
Step 3: Solve for the concentration of Cu2+ ions in the anode compartment.
Simplify the equation:
0.15 = −0.0296 log [Cu2+]anode
0.10
5.07 = log [Cu2+]anode
0.10
105.07 =[Cu2+]anode
0.10
[Cu2+]anode = 105.07 ×0.10
[Cu2+]anode ≈1.23 ×105M
Therefore, the concentration of Cu2+ ions in the anode compartment is ap-
proximately 1.23 ×105M.
Question 4
Question
Consider a galvanic cell that consists of a copper electrode in a Cu2+ solution
connected by a salt bridge to a silver electrode in a Ag+solution. The standard
reduction potentials are E◦
Cu2+ /Cu = 0.34 V and E◦
Ag+/Ag = 0.80 V. Calculate
the cell potential for this galvanic cell.
3
Solution
Step 1: Write the half-reactions for the redox reactions occurring at each elec-
trode.
Reduction at the copper electrode: Cu2+ + 2e−→Cu E◦= 0.34 V
Reduction at the silver electrode: Ag++e−→Ag E◦= 0.80 V
Step 2: Write the overall cell reaction by summing the two half-reactions.
The cell potential is the difference in standard electrode potentials.
Cu2+ + 2Ag →Cu + 2Ag+
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −0.34 V
E◦
cell = 0.46 V
Therefore, the cell potential for this galvanic cell is 0.46 V.
Question 5
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
Calculate the cell potential at 25
°
C when the concentrations of Zn2+ and Cu2+
are both 0.10 M. Assume both half-reactions are at standard conditions.
Solution
Step 1: Write the overall cell reaction by subtracting the reduction potential of
the Zn half-reaction from the reduction potential of the Cu half-reaction:
Zn2+(aq) + Cu(s)→Cu2+(aq) + Zn(s)
Step 2: Determine the standard cell potential (E◦
cell) using the Nernst equa-
tion:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = (0.34 V) −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the cell potential (Ecell) at 25
°
C using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
4
Step 4: Calculate the cell potential by plugging in the given values:
Ecell = 1.10 V −0.0592
2log 0.10
0.10
Ecell = 1.10 V
Therefore, the cell potential at 25
°
C when the concentrations of Zn2+ and
Cu2+ are both 0.10 M is 1.10 V .
Question 6
Question
Consider a galvanic cell based on the following half-reactions:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
Calculate the standard cell potential at 25
°
C for this galvanic cell.
Solution
Step 1: Write the cell reaction by reversing the anode half-reaction and adding
the cathode half-reaction.
Zn2+(aq) + 2H+(aq)→Zn(s)+H2(g)
Step 2: Write the standard cell potential in terms of standard reduction
potentials (E◦) for the half-reactions involved. The standard cell potential is
given by
E◦
cell =E◦
cathode −E◦
anode
Step 3: Find the standard reduction potentials for the half-reactions from a
standard reduction potential table. The values are:
E◦
Zn2+/Zn =−0.76 V
E◦
2H+/H2= 0.00 V
Step 4: Substitute the values into the formula for the standard cell potential.
E◦
cell = 0.00 V −(−0.76 V)
E◦
cell = 0.76 V
Therefore, the standard cell potential at 25
°
C for the given galvanic cell is
0.76 V.
5
Question 7
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 potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =
−0.76 V, calculate the standard cell potential (E◦
cell) of the galvanic cell.
Solution
Step 1: Write the overall cell reaction by combining the two half-reactions,
ensuring that the number of electrons lost equals the number of electrons gained:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions. The standard cell potential is given
by:
E◦
cell =E◦
cathode −E◦
anode
Substitute the values of E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =−0.76 V into the
equation:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the standard cell potential of the galvanic cell is E◦
cell = 1.10 V.
Question 8
Question
Consider an electrochemical cell consisting of a copper electrode immersed in
1.0 M Cu2+ solution and a silver electrode immersed in 1.0 M Ag+solution.
Write the balanced half-reactions and the overall cell reaction. Is this cell a
galvanic cell or an electrolytic cell? Justify your answer.
Solution
Step 1: Write the balanced half-reactions: At the copper electrode: Cu2+ +
2e−→Cu At the silver electrode: Ag++e−→Ag
Step 2: Write the overall cell reaction:
Cu2+ + 2Ag →Cu + 2Ag+
6
Step 3: Determine if the cell is a galvanic or electrolytic cell: In a galvanic
cell, the redox reaction is spontaneous and produces electrical energy. In this
case, the cell reaction is proceeding from left to right (oxidation at copper elec-
trode, reduction at silver electrode), which means it is a spontaneous process.
Therefore, this cell is a galvanic cell.
Question 9
Question
Consider a galvanic cell constructed with a silver electrode in a 1.0 M AgNO3
solution and a copper electrode in a 1.0 M Cu(NO3)2solution. The standard
reduction potentials are: Ag+(aq) +e−→Ag(s) (E◦= 0.80 V) and Cu2+(aq)+
2e−→Cu(s) (E◦= 0.34 V). Determine the cell potential at 25
°
C.
Solution
Step 1: Write out the overall reaction for the galvanic cell:
Ag(s) + Cu2+(aq)→Cu(s) + Ag+(aq)
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials provided:
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
Step 3: Calculate the non-standard cell potential using the Nernst equation:
E=E◦−0.0592
nlog(Q)
where nis the number of moles of electrons transferred and Qis the reaction
quotient.
Step 4: Calculate the reaction quotient Q:
Q=[products]
[reactants] =[Cu][Ag+]
[Ag][Cu2+]
Since both concentrations are 1.0 M, Q= 1.
Step 5: Substitute the values into the Nernst equation:
E=−0.46 V −0.0592
2log(1)
E=−0.46 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is −0.46 V.
7
Question 10
Question
Consider a Galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Pb2+(aq) + 2e−→Pb(s)E◦=−0.13 V
If the initial concentrations are [Zn2+] = 1.0 M, [Pb2+] = 0.1 M, and the cell
operates until equilibrium is reached, determine the cell potential at equilibrium.
Solution
Step 1: Write the cell reaction using the given half-reactions:
Zn2+(aq) + Pb(s)→Zn(s) + Pb2+(aq)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =−0.13 V −(−0.76 V) = 0.63 V
Step 3: Calculate the reaction quotient (Q) at equilibrium using the initial
concentrations given:
Q=[Zn](Pb2+)
[Zn2+][Pb]
Q=(1.0)(0.1)
(1.0)(0.1) = 1
Step 4: Use the Nernst equation to find the cell potential at equilibrium
(Ecell):
Ecell =E◦
cell −0.0592
nlog(Q)
Since the reaction involves the transfer of 2 electrons, n= 2.
Ecell = 0.63 V −0.0592
2log(1) = 0.63 V
Therefore, the cell potential at equilibrium is 0.63 V.
8
Question 11
Question
Consider the following half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Determine if a spontaneous reaction will occur when a strip of copper metal is
placed in a solution of zinc sulfate. If so, write the overall balanced cell reaction
and calculate the cell potential.
Solution
Step 1: Write the individual half-reactions and find the standard cell potential.
Given the half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
The overall cell reaction will involve the reduction of Cu2+ and the oxidation
of Zn. To determine if a spontaneous reaction will occur, we calculate the overall
standard cell potential, E◦
cell, using the formula:
E◦
cell =E◦
cathode −E◦
anode
where
E◦
cathode = 0.34 V
E◦
anode =−0.76 V
Plugging these values into the formula, we get:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Since the standard cell potential is positive, a spontaneous reaction will
occur.
Step 2: Write the overall balanced cell reaction. The overall cell reaction is
the sum of the two half-reactions.
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 3: Calculate the cell potential. The standard cell potential was already
calculated as 1.10 V in Step 1.
9
Question 12
Question
A voltaic cell is constructed using a silver/silver chloride electrode and a zinc
electrode. The standard reduction potentials are as follows:
AgCl(s) + e−→Ag(s) + Cl−(aq)E◦
cell = 0.22 V
Zn2+(aq)+2e−→Zn(s)E◦
cell =−0.76 V
What is the cell potential when the concentration of Zn2+ is 1.0 M and the
solid AgCl is in contact with a 0.10 M AgNO3solution? Use R= 8.314 J/K mol,
F= 96,485 C/mol, and T= 298 K.
Solution
Step 1: Write the cell reaction and cell potential equation using the given half-
reactions. The cell reaction is the sum of the two half-reactions:
AgCl(s) + Zn2+(aq)+2e−→Ag(s) + Zn(s) + Cl−(aq)
The cell potential equation is:
Ecell =E◦
cell −0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced reaction,
and Qis the reaction quotient.
Step 2: Calculate the cell potential First, calculate Qusing the concentra-
tions provided. Since there are no concentrations for the solid electrodes (Ag
and Zn), their activities are assumed to be 1. Therefore,
Q=[Ag+][Cl−]
[Zn2+]
Q=0.102
1.0= 0.01
Step 3: Plug the values into the formula
Ecell = 0.22V−0.0592
2log 0.01
Ecell = 0.22V−0.0296 ×2×2
Ecell = 0.22V−0.1184 = 0.1016 V
Therefore, the cell potential is 0.1016 V.
10
Question 13
Question
Consider a galvanic cell where the reaction occurring is:
Cd2+(aq) + 2Ag(s)→Cd(s) + 2Ag+(aq)
Determine the cell potential at standard conditions given that E◦
Cd2+/Cd =
−0.40 V and E◦
Ag+/Ag = 0.80 V.
Solution
Step 1: Write the half-reactions and standard electrode potentials. The half-
reactions involved in the cell are:
Reduction half-reaction:
Cd2+(aq) + 2e−→Cd(s)E◦
Cd2+/Cd =−0.40 V
Oxidation half-reaction:
2Ag(s)→2Ag+(aq) + 2e−E◦
Ag+/Ag = 0.80 V
Step 2: Calculate the cell potential. The cell potential E◦
cell can be calculated
using 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.
Given that E◦
Cd2+/Cd =−0.40 V and E◦
Ag+/Ag = 0.80 V, we have:
E◦
cell =E◦
Ag+/Ag −E◦
Cd2+/Cd = 0.80 −(−0.40) = 1.20 V
Therefore, the cell potential at standard conditions is 1.20 V.
Question 14
Question
Consider a galvanic cell with the following half-reactions:
Cu2+ + 2e−→Cu E◦
cell = 0.34 V
Fe3+ +e−→Fe2+ E◦
cell =−0.77 V
Calculate the standard Gibbs free energy change (∆G◦) for the overall cell
reaction and state whether the cell is spontaneous at standard conditions.
11
Solution
Step 1: Calculate the standard cell potential (E◦
cell) for the overall reaction. The
overall cell reaction is the sum of the two half-reactions:
Cu2+ + 2Fe3+ →2Fe2+ + Cu
The standard cell potential is the sum of the standard reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values:
E◦
cell =E◦
Cu −E◦
Fe
E◦
cell = 0.34 V −(−0.77 V) = 1.11 V
Step 2: Calculate the standard Gibbs free energy change (∆G◦) for the
overall reaction. The relationship between standard cell potential and standard
Gibbs free energy change is given by:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Fis the Faraday constant (F= 96485 C/mol).
Since 2 moles of electrons are transferred in the balanced cell reaction, n= 2.
Substitute the values:
∆G◦=−2×96485 C/mol ×1.11 V
∆G◦=−213.59 kJ/mol
Step 3: Determine if the cell is spontaneous at standard conditions. For a
reaction to be spontaneous at standard conditions, ∆G◦must be negative. Since
∆G◦=−213.59 kJ/mol (which is negative), the cell reaction is spontaneous at
standard conditions.
Question 15
Question
Consider a galvanic cell with a standard reduction potential of E◦
cell = 1.23 V.
Assuming the reaction in the cell is spontaneous, calculate the standard Gibbs
free energy change, ∆G◦, for the cell reaction. Also, if the concentration of Mn2+
ions in the cell’s cathode compartment is 0.1 M, determine the cell potential
when the concentration of Mn2+ ions in the cathode compartment is increased
to 1.0 M.
12
Solution
Step 1: Recall the relationship between standard Gibbs free energy change,
standard cell potential, and the Faraday constant:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced redox
reaction, Fis the Faraday constant (96485 C/mol), and E◦
cell is the standard
cell potential.
Step 2: Calculate the number of moles of electrons transferred. The stan-
dard reduction potential (E◦
cell) is related to the number of moles of electrons
transferred (n) by the equation:
E◦
cell =∆G◦
−nF
Given E◦
cell = 1.23 V, and F= 96485 C/mol, we can solve for n.
∆G◦=−nF E◦
cell
n=−∆G◦
F E◦
cell
Step 3: Substitute the values into the equation to solve for n.
n=−∆G◦
F E◦
cell
=−∆G◦
(96485 C/mol)(1.23 V)
Step 4: Calculate the standard Gibbs free energy change (∆G◦). We know
that ∆G◦=−nF E◦
cell and we have found the value for nin the previous step.
∆G◦=−nF E◦
cell =−−∆G◦
(96485 C/mol)(1.23 V)(96485 C/mol)(1.23 V)
1=∆G◦
Therefore, ∆G◦= 1 J.
Step 5: Calculate the cell potential when the concentration of Mn2+ ions
in the cathode compartment is 1.0 M. Assuming the cell reaction is occurring
at equilibrium, we can use the Nernst equation:
E=E◦−0.0592
nlog [Mn2+]cathode
[Mn2+]anode
where [Mn2+]cathode is the concentration of Mn2+ ions in the cathode compart-
ment and [Mn2+]anode is the concentration of Mn2+ ions in the anode compart-
ment.
Step 6: Calculate the cell potential with [Mn2+]cathode = 1.0 M and [Mn2+]anode =
0.1 M. Given that E◦
cell = 1.23 V, n= 2 (as indicated by the balanced redox
reaction), and substituting [Mn2+]cathode = 1.0 M and [Mn2+]anode
13
Question 16
Question
Consider a galvanic cell where the following reaction takes place:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Determine the cell potential at 25
°
C if the standard reduction potentials are
E◦(Zn2+/Zn) = −0.76 V and E◦(Ag+/Ag) = 0.80 V.
Solution
Step 1: Write the half-reactions and find the cell potential. The standard cell
potential (E◦
cell) can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
We have the half-reactions:
Zn2+(aq)+2e−→Zn(s) with E◦
Zn2+/Zn =−0.76 V
2Ag+(aq)+2e−→2Ag(s) with E◦
Ag+/Ag = 0.80 V
The cell potential is:
E◦
cell =E◦
cathode −E◦
anode =E◦
Ag+/Ag −E◦
Zn2+/Zn = 0.80V−(−0.76V)=1.56 V
Therefore, the cell potential at 25
°
C is 1.56 V.
Question 17
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a standard iron (III) ion electrode with an Fe3+/Fe2+ half-cell. The stan-
dard reduction potentials are as follows: E◦(H+/H2) = 0.00 V and E◦(Fe3+/Fe2+)
= -0.77 V. Calculate the cell potential (E◦
cell) for this galvanic cell. Is the reac-
tion spontaneous or non-spontaneous?
Solution
Step 1: Write the overall cell reaction as the sum of the two half-cell reactions.
Overall reaction: 2H++ Fe3+ →H2+ Fe2+
Step 2: Write the two half-cell reactions, making sure the number of electrons
match to cancel out in the overall reaction.
Anode (oxidation): Fe3+ +e−→Fe2+ E◦= -0.77 V
14
Cathode (reduction): 2H++ 2e−→H2E◦= 0.00 V
Step 3: Calculate the overall cell potential using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(−0.77 V)
E◦
cell = 0.77 V
Step 4: Determine the spontaneity of the reaction based on the cell potential.
Since the cell potential is positive (0.77 V), the reaction is spontaneous as the
flow of electrons is favored from the anode to the cathode.
Question 18
Question
A galvanic cell is constructed with a standard hydrogen electrode (Pt|H2(g)|H+(1M))
on one side and an iron electrode (Fe3+(1M)|Fe2+(1M)) on the other side. The
standard reduction potential of the iron electrode is −0.77 V and the standard
reduction potential of the hydrogen electrode is 0 V. Determine the cell poten-
tial, and identify the cathode and anode in this galvanic cell. Also, determine
if the reaction is spontaneous or not.
Solution
Step 1: Write the half-reactions for the two electrodes. The half-reaction of the
hydrogen electrode is the reduction of hydrogen ions to hydrogen gas:
2H++ 2e−→H2(g)
The half-reaction of the iron electrode is the reduction of iron(III) ions to iron(II)
ions:
Fe3+ + e−→Fe2+
Step 2: Determine the standard cell potential by finding the difference be-
tween the standard reduction potentials of the two half-reactions. The standard
cell potential (E◦
cell) can be calculated as:
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 E◦
cathode = 0 V and E◦
anode =−0.77 V, we have:
E◦
cell = 0 V −(−0.77 V) = 0.77 V
Step 3: Identify the cathode and anode in the galvanic cell. The electrode
with the higher standard reduction potential is the cathode, while the electrode
15
with the lower standard reduction potential is the anode. So, in this case, the
hydrogen electrode is the cathode and the iron electrode is the anode.
Step 4: Determine if the reaction is spontaneous or not. The reaction is
spontaneous if E◦
cell is positive. Since E◦
cell = 0.77 V, the reaction is spontaneous.
Question 19
Question
A cell is constructed with a standard hydrogen electrode (SHE) and a zinc
electrode. The standard reduction potential for the hydrogen electrode is 0 V,
and for the zinc electrode is -0.76 V. Calculate the standard cell potential for
this galvanic cell.
Solution
Step 1: Write the half-reactions for the hydrogen electrode and zinc electrode.
The half-reactions are:
H+(aq)+e−→1
2H2(g) (Reduction at SHE)
Zn2+(aq) + 2e−→Zn(s) (Reduction at Zn electrode)
Step 2: Determine the standard cell potential. The standard 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.
In this case, the cathode is the SHE (standard hydrogen electrode) with
E◦
cathode = 0 V, and the anode is the zinc electrode with E◦
anode =−0.76 V.
Therefore, the standard cell potential is:
E◦
cell = 0 V −(−0.76 V) = 0.76 V
So, the standard cell potential for this galvanic cell is 0.76 V.
Question 20
Question
Consider the following cell reaction:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
If the standard reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu =
0.34 V, calculate the standard cell potential (E◦
cell) and state whether this cell
is galvanic or electrolytic.
16
Solution
To calculate the standard cell potential (E◦
cell), we use the formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode (Cu2+/Cu)
and E◦
anode is the standard reduction potential of the anode (Zn2+/Zn).
We are given:
E◦
Zn2+/Zn =−0.76 V, E◦
Cu2+/Cu = 0.34 V
Therefore,
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn = 0.34 V −(−0.76 V) = 1.10 V
Since the standard cell potential (E◦
cell) is positive, the cell is galvanic.
Question 21
Question
Consider a galvanic cell that consists of a zinc electrode in a 1.0 M solution
of Zn2+ ions and a copper electrode in a 1.0 M solution of Cu2+ ions. The
standard reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu = 0.34 V,
respectively. Determine the cell potential at standard conditions and identify
the anode and cathode.
Solution
Step 1: Write the reduction half-reactions and standard cell potential.
The reduction half-reactions are:
Zn2+ + 2e−→Zn (oxidation)
Cu2+ + 2e−→Cu (reduction)
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
Step 2: Calculate the standard cell potential.
E◦
cell = 1.10 V
Therefore, the standard cell potential is 1.10 V.
17
Step 3: Identify the anode and cathode.
Since the cell potential is positive, the reaction is spontaneous, with the copper
electrode acting as the cathode (where reduction occurs) and the zinc electrode
acting as the anode (where oxidation occurs).
Thus, in the given galvanic cell, zinc is the anode and copper is the cathode.
Question 22
Question
Consider a galvanic cell with a zinc electrode in a 1.0 M Zn2+ solution and
a silver electrode in a 1.0 M Ag+solution. The standard reduction potentials
are E◦
Zn2+/Zn =−0.76 V and E◦
Ag+/Ag = 0.80 V. Calculate the cell potential at
standard conditions.
Solution
Step 1: Write the two half-reactions representing the oxidation and reduction
processes. The oxidation half-reaction occurs at the zinc electrode:
Zn(s) →Zn2+(aq)+2e−
The reduction half-reaction occurs at the silver electrode:
Ag+(aq) + e−→Ag(s)
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials. The cell potential is calculated by subtracting the standard
reduction potential of the anode from the standard reduction potential of the
cathode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −(−0.76 V) = 1.56 V
Therefore, the cell potential under standard conditions is 1.56 V.
Question 23
Question
Consider a galvanic cell that consists of a copper electrode in a 1.0 M copper(II)
sulfate solution and a platinum electrode in a 1.0 M hydrogen ion solution.
Suppose the standard reduction potential for the copper half-cell is +0.34 V, and
the standard reduction potential for the hydrogen half-cell is 0.00 V. Determine
the cell potential at 25
°
C for this galvanic cell.
18
Solution
Step 1: Write the half-reactions for each electrode based on the given informa-
tion.
Copper electrode: Cu2+(aq)+2e−→Cu(s)E◦= +0.34 V
Platinum electrode: 2H+(aq)+2e−→H2(g)E◦= 0.00 V
Step 2: Write the overall cell reaction by summing the two half-reactions.
Cu2+(aq)+2H+(aq)→Cu(s)+H2(g)
Step 3: Determine the cell potential by adding the standard reduction po-
tentials of the two half-cells.
Cell potential: E◦
cell =E◦
cathode −E◦
anode = 0.00 V −(+0.34 V) = −0.34 V
Therefore, the cell potential at 25
°
C for this galvanic cell is -0.34 V.
Question 24
Question
Consider a galvanic cell constructed with a standard hydrogen electrode (SHE)
as the anode and a copper electrode as the cathode. If the standard reduction
potential of the copper electrode is +0.34 V, determine the cell potential at
25◦C. Is this cell galvanic or electrolytic?
Solution
Step 1: Write the two half-reactions for the cell: The half-reaction at the anode
involving the standard hydrogen electrode (SHE) is:
2H++ 2e−→H2(g)
The half-reaction at the cathode involving the copper electrode is:
Cu2+ + 2e−→Cu(s)
Step 2: Calculate the standard cell potential (E◦) using the standard reduc-
tion potentials of the half-reactions: Given that the standard reduction potential
of the copper electrode is +0.34 V, and the standard reduction potential of the
standard hydrogen electrode (SHE) is 0 V, the cell potential can be calculated
as:
E◦=E◦
cathode −E◦
anode
E◦= +0.34 V −0 V
E◦= +0.34 V
19
Step 3: Calculate the cell potential at 25◦C using the Nernst equation: The
Nernst equation relates the cell potential (E) to the standard cell potential (E◦),
the reaction quotient (Q), the gas constant (R), the temperature in Kelvin (T),
and the number of electrons transferred in the reaction (n):
E=E◦−RT
nF ln(Q)
At 25◦C, T= 298 K. Since the reaction quotient Q= 1 for standard condi-
tions, the equation simplifies to:
E=E◦
Therefore, the cell potential at 25◦C is +0.34 V.
Step 4: Determine if the cell is galvanic or electrolytic: A cell is galvanic if
the cell potential is positive, indicating a spontaneous redox reaction. Since the
cell potential at 25◦C is +0.34 V (positive), the cell is galvanic.
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Cathode: Ag+(aq)+e−→Ag(s)E◦= 0.80 V
Anode: Zn(s)→Zn2+(aq) + 2e−E◦=−0.76 V
Calculate the cell potential (E◦
cell) and the equilibrium constant (Keq) for the
overall reaction of the galvanic cell.
Solution
Step 1: Calculate E◦
cell by using the formula:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values:
E◦
cell = 0.80 V −(−0.76 V)
E◦
cell = 1.56 V
Step 2: Calculate Keq using the formula:
E◦
cell =RT
nF ln Keq
Solving for Keq:
Keq =e
nF E◦
cell
RT
20
Given that R= 8.314 J/mol K, T= 298 K, F= 96485 C/mol, and n= 2 (from
balancing the half-reactions), we have:
Keq =e2×96485×1.56
8.314×298
Keq =e7.77
Keq ≈2417.89
Question 26
Question
Consider a galvanic cell with the following half-reactions:
Anode: Cd(s) →Cd2+(aq)+2e−
Cathode: Ag+(aq) + e−→Ag(s)
Calculate the cell potential at standard conditions (T= 298 K, P= 1 atm),
given that [Cd2+] = 0.20 M and [Ag+] = 0.15 M.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions.
Cd(s) + 2Ag+(aq)→Cd2+(aq)+2Ag(s)
Step 2: Write the Nernst equation for cell potential:
E=E◦−RT
nF ln(Q)
Where: - Eis the cell potential, - E◦is the standard cell potential, - Ris
the ideal gas constant (8.314 J/(mol·K)), - Tis the temperature in Kelvin (298
K), - nis the number of electrons transferred in the balanced cell reaction, - F
is the Faraday constant (96485 C/mol), - Qis the reaction quotient.
Step 3: Calculate the standard cell potential E◦using the Standard Reduc-
tion Potentials table.
E◦=E◦
cathode −E◦
anode
=E◦
Ag+/Ag −E◦
Cd2+/Cd
= 0.80 V −(−0.40 V)
= 1.20 V
Step 4: Calculate the reaction quotient Qusing the given concentrations.
Q=[Cd2+]
[Ag+]2=0.20
0.152
21
Step 5: Substitute all values into the Nernst equation and solve for the cell
potential E.
E= 1.20 V −(8.314 J/(mol ·K)) ·(298 K)
2·(96485 C/mol) ln 0.20
0.152
E= 1.20 V −0.0268 V ·ln 0.20
0.152
E≈1.16 V
Therefore, the cell potential at standard conditions is approximately 1.16 V.
Question 27
Question
Consider a galvanic cell with the following half-reactions:
Anode: Pb(s)→Pb2+(aq) + 2e−
Cathode: 2Ag+(aq) + 2e−→2Ag(s)
If the standard reduction potentials are E◦(Pb2+/Pb) = −0.13 V and E◦(Ag+/Ag) =
0.80 V, calculate the standard cell potential (E◦
cell) for this galvanic cell.
Solution
Step 1: Determine the overall cell reaction by adding the 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. The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
From the data given, we have:
E◦
anode =−0.13 V and E◦
cathode = 0.80 V
Therefore,
E◦
cell = 0.80 V −(−0.13 V) = 0.93 V
Hence, the standard cell potential for this galvanic cell is 0.93 V.
22
Question 28
Question
Consider the reaction between silver ions and copper metal:
Ag+(aq) + Cu(s)→Ag(s) + Cu2+(aq)
Given that the standard reduction potentials are E◦(Ag+/Ag) = 0.80 V and
E◦(Cu2+/Cu)=0.34 V, determine whether the reaction is spontaneous under
standard conditions. If not, calculate the voltage that would be required to
drive the reaction forward.
Solution
Step 1: Calculate the standard cell potential (E◦
cell) using the standard reduction
potentials.
E◦
cell =E◦(cathode) −E◦(anode)
E◦
cell =E◦(Ag+/Ag)−E◦(Cu2+/Cu)
E◦
cell = 0.80 V −0.34 V
E◦
cell = 0.46 V
Step 2: Determine if the reaction is spontaneous under standard conditions.
Since E◦
cell = 0.46 V, the reaction is spontaneous under standard conditions
because E◦
cell >0.
Therefore, the given reaction proceeds spontaneously to the right.
Step 3: If the reaction is not spontaneous, calculate the voltage required to
drive the reaction forward. Since the reaction is spontaneous under standard
conditions, we do not need to calculate the additional voltage required in this
case.
Question 29
Question
Consider a galvanic cell with the following half-reactions:
Ag+(aq)+e−→Ag(s)E◦= 0.80 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
If the concentration of Ag+is 1.0 M and Cu2+ is 0.1 M, determine the cell
potential for this galvanic cell at 25◦C.
23
Solution
Step 1: Identify the anode and cathode reactions. The half-reaction with the
higher standard reduction potential (E◦) will be reduced (cathode) and the
other will be oxidized (anode).
Here, the reduction potential for the reduction half-reaction of Ag+is 0.80
V, which is higher than 0.34 V for Cu2+. Thus, the reduction half-reaction for
Ag+will occur at the cathode and the oxidation half-reaction for Cu2+ will
occur at the anode.
Step 2: Write the overall cell reaction by adding the anode and cathode
half-reactions. Ensure that the same number of electrons are involved in both
half-reactions so that they can cancel out.
The overall reaction for the galvanic cell is:
2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Step 3: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦−0.0592
nlog(Q)
where: - E◦is the standard cell potential, - nis the number of moles of
electrons transferred in the balanced cell reaction, - Qis the reaction quotient
(ratio of product concentrations to reactant concentrations).
In this case, n= 2 since 2 moles of electrons are transferred in the overall
reaction.
Step 4: Calculate Qusing the given concentrations. Remember, the concen-
trations of solid species do not appear in the expression for Q.
Q=[Ag+]2
[Cu2+]=(1.0)2
0.1= 10
Step 5: Substitute the values of E◦,n, and Qinto the Nernst equation and
solve for Ecell.
Ecell = 0.80 −0.0592
2log(10) = 0.80 −0.0296 ×1=0.7704 V
Therefore, the cell potential for this galvanic cell at 25◦C is 0.7704 V.
Question 30
Question
Consider the following Galvanic and Electrolytic cells:
Galvanic Cell Electrolytic Cell
Zn(s) |Zn2+(aq) || Cu2+(aq) |Cu(s) Cu(s) |Cu2+(aq) || Ag+(aq) |Ag(s)
24
For each cell, determine the following: 1. The half-reaction that occurs at
the cathode and the anode. 2. The overall cell reaction. 3. The direction of
electron flow. 4. The direction of cation flow. 5. Whether the cell functions as
a galvanic cell or an electrolytic cell.
Solution
1. For the Galvanic Cell:
Anode (Zn(s)): Zn(s) →Zn2+(aq) + 2e−
Cathode (Cu2+(aq)): Cu2+(aq) + 2e−→Cu(s)
For the Electrolytic Cell:
Anode (Cu(s)): Cu(s) →Cu2+(aq) + 2e−
Cathode (Ag+(aq)): Ag+(aq) + e−→Ag(s)
2. The overall cell reaction for the Galvanic Cell is the sum of the half-
reactions:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
The overall cell reaction for the Electrolytic Cell is the sum of the half-
reactions:
Cu(s) + Ag+(aq) →Cu2+(aq) + Ag(s)
3. In both cells, electrons flow from the anode to the cathode.
4. In the Galvanic Cell, cations flow from the anode to the cathode.
In the Electrolytic Cell, cations flow from the cathode to the anode.
5. The Galvanic Cell functions as a spontaneous chemical reaction generating
electricity, while the Electrolytic Cell requires an external electrical current to
drive a non-spontaneous chemical reaction.
Question 31
Question
A voltaic cell is constructed with a standard hydrogen electrode and a standard
silver electrode. The standard cell potential is measured to be 0.80 V at 25
°
C.
Calculate the standard free energy change for the cell reaction.
Solution
Step 1: Write the balanced cell reaction for the voltaic cell. The cell reaction
for a voltaic cell with a standard hydrogen electrode (H+(aq),H2(g)) and a
standard silver electrode (Ag+(aq),Ag(s)) is:
2H+(aq) + 2e−→H2(g)E◦= 0 V
25
2Ag+(aq) + 2e−→2Ag(s)E◦= 0.80 V
Overall cell reaction:
2Ag+(aq) + 2H+(aq)→2Ag(s)+H2(g)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials. The standard cell potential (E◦
cell) is the difference between
the standard reduction potentials of the two half-reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −0 V = 0.80 V
Step 3: Calculate the standard free energy change (∆G◦) using the equation:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced cell
reaction, Fis the Faraday constant (96,485 C/mol).
For the given cell reaction, n= 2 moles of electrons transferred. Thus,
∆G◦=−2×96485 C/mol ×0.80 V = −155136 J/mol
Therefore, the standard free energy change for the cell reaction is −155136 J/mol.
Question 32
Question
Consider a galvanic cell consisting of a standard hydrogen electrode (SHE) and
a copper electrode with the following half-reactions:
Cathode: Cu2+ + 2 e−−−→ Cu E◦= 0.34 V
Anode: 2 H++ 2 e−−−→ H2E◦= 0.00 V
Calculate the cell potential at 298 K when the concentration of Cu2+ is 0.10 M,
and the pressure of H2is 1.0 atm.
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction is the sum of the two half-reactions:
Cu2+ + 2 H+−−→ Cu + H2
Step 2: Calculate the cell potential at standard conditions.
The standard cell potential E◦
cell can be calculated using the standard re-
duction potentials:
E◦
cell =E◦
cathode −E◦
anode
26
E◦
cell = 0.34V−0.00V= 0.34 V
Step 3: Consider the non-standard conditions.
Since the concentrations of Cu2+ and the pressure of H2are given, we can
calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the cell reaction and Qis
the reaction quotient.
Step 4: Calculate the reaction quotient Q.
Q=[Cu]
[Cu2+][H+]2
Q=1.0
0.10 ×1.02= 10
Step 5: Substitute values into the Nernst equation and solve for the cell
potential.
Ecell = 0.34 −0.0592
2log(10)
Ecell = 0.34 −0.0296 ×1
Ecell = 0.3104 V
Therefore, the cell potential at 298 K when the concentration of Cu2+ is
0.10 M and the pressure of H2is 1.0 atm is 0.3104 V.
Question 33
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.78 V that
operates under standard conditions. The cell reaction is given by:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Calculate the standard Gibbs free energy change for the cell reaction and
determine if the reaction is spontaneous.
Solution
Step 1: The standard Gibbs free energy change (∆G◦) for the cell reaction can
be calculated using the equation:
∆G◦=−nF E◦
where nis the number of moles of electrons transferred in the balanced redox
equation, and Fis the Faraday constant (96500 C/mol).
From the given cell reaction, we see that 2 moles of electrons are transferred,
so n= 2. Substituting into the equation:
∆G◦=−2×96500 C/mol ×0.78 V
27
∆G◦=−188220 J/mol
Step 2: To determine if the reaction is spontaneous, we can use the relation-
ship between ∆G◦and the equilibrium constant (K):
∆G◦=−RT ln K
where Ris the gas constant (8.314 J/mol ·K) and Tis the temperature in
Kelvin.
At standard conditions, T= 298 K and R= 8.314 J/mol ·K. Thus, we have:
−188220 J/mol = −8.314 J/mol ·K×298 K ln K
Solving for ln K:
ln K=−188220 J/mol
−8.314 J/mol ·K×298 K
ln K= 27.37
Step 3: Exponentiating both sides of the equation to eliminate the natural
logarithm:
K=e27.37
K= 1.17 ×1011
Since K > 1, the reaction is spontaneous.
Question 34
Question
A student sets up a galvanic cell with a standard hydrogen electrode (E◦
H+/H2=
0.00 V) and a copper electrode (E◦
Cu2+/Cu = 0.34 V), connected by a salt bridge.
The initial concentrations are [H+] = 1.0 M and [Cu2+]=0.1 M.
Calculate the cell potential at 25
Step 1: Write the balanced redox reaction and the cell notation.
The balanced redox reaction is:
Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)
The cell notation is:
Pt|H2(1 M)|H+(1 M)||Cu2+(0.1 M)|Cu
Step 2: Calculate the cell potential (E◦
cell) at standard conditions.
28
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
H+/H2= 0.00 V, and E◦
Cu2+/Cu = 0.34 V, we have:
E◦
cell = 0.00 V −0.34 V = −0.34 V
Step 3: Calculate the reaction quotient (Q).
The reaction quotient Qis calculated using the concentrations of the species
involved in the reaction:
Q=[Cu](solid)
[Cu2+][H+]2
Plugging in the values, we get:
Q=1
0.1×12= 10
Step 4: Calculate the cell potential (Ecell) under non-standard conditions.
The Nernst equation relates the cell potential under non-standard conditions
(Ecell) to the standard cell potential (E◦
cell) and the reaction quotient (Q):
Ecell =E◦
cell −0.0592
nlog Q
Since the reaction involves the transfer of 2 moles of electrons, n= 2. Plug-
ging in the values, we get:
Ecell =−0.34 V −0.0592
2log 10
Ecell =−0.34 V −0.0296 log 10
Ecell =−0.34 V −0.0296 ×1
Ecell =−0.34 V −0.0296
Ecell =−0.3696 V
Therefore, the cell potential at 25
Question
Consider a galvanic cell that uses the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential of Cu2+/Cu is 0.34 V, calculate the stan-
dard cell potential (E◦
cell) of this galvanic cell.
29
Question 3
Question
Consider a galvanic cell with a standard reduction potential of E◦
cell = 0.30 V.
If the cell potential is measured to be 0.45 V under non-standard conditions
where the concentration of Cu2+ ions in the cathode compartment is 0.10 M,
determine the concentration of Cu2+ ions in the anode compartment.
Solution
Step 1: Write the half-reactions for the cell. The overall cell reaction can be
split into two half-reactions: Cathode (Reduction): Cu2+(aq)+2e−→Cu(s)
Anode (Oxidation): Zn(s)→Zn2+(aq)+2e−
Step 2: Calculate the cell potential under non-standard conditions. By using
the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Cu2+]anode
[Cu2+]cathode
Substitute the given values:
0.45 = 0.30 −0.0592
2log [Cu2+]anode
0.10
Step 3: Solve for the concentration of Cu2+ ions in the anode compartment.
Simplify the equation:
0.15 = −0.0296 log [Cu2+]anode
0.10
5.07 = log [Cu2+]anode
0.10
105.07 =[Cu2+]anode
0.10
[Cu2+]anode = 105.07 ×0.10
[Cu2+]anode ≈1.23 ×105M
Therefore, the concentration of Cu2+ ions in the anode compartment is ap-
proximately 1.23 ×105M.
Question 4
Question
Consider a galvanic cell that consists of a copper electrode in a Cu2+ solution
connected by a salt bridge to a silver electrode in a Ag+solution. The standard
reduction potentials are E◦
Cu2+ /Cu = 0.34 V and E◦
Ag+/Ag = 0.80 V. Calculate
the cell potential for this galvanic cell.
3
Solution
Step 1: Write the half-reactions for the redox reactions occurring at each elec-
trode.
Reduction at the copper electrode: Cu2+ + 2e−→Cu E◦= 0.34 V
Reduction at the silver electrode: Ag++e−→Ag E◦= 0.80 V
Step 2: Write the overall cell reaction by summing the two half-reactions.
The cell potential is the difference in standard electrode potentials.
Cu2+ + 2Ag →Cu + 2Ag+
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −0.34 V
E◦
cell = 0.46 V
Therefore, the cell potential for this galvanic cell is 0.46 V.
Question 5
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
Calculate the cell potential at 25
°
C when the concentrations of Zn2+ and Cu2+
are both 0.10 M. Assume both half-reactions are at standard conditions.
Solution
Step 1: Write the overall cell reaction by subtracting the reduction potential of
the Zn half-reaction from the reduction potential of the Cu half-reaction:
Zn2+(aq) + Cu(s)→Cu2+(aq) + Zn(s)
Step 2: Determine the standard cell potential (E◦
cell) using the Nernst equa-
tion:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = (0.34 V) −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the cell potential (Ecell) at 25
°
C using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
4
Step 4: Calculate the cell potential by plugging in the given values:
Ecell = 1.10 V −0.0592
2log 0.10
0.10
Ecell = 1.10 V
Therefore, the cell potential at 25
°
C when the concentrations of Zn2+ and
Cu2+ are both 0.10 M is 1.10 V .
Question 6
Question
Consider a galvanic cell based on the following half-reactions:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: 2H+(aq)+2e−→H2(g)
Calculate the standard cell potential at 25
°
C for this galvanic cell.
Solution
Step 1: Write the cell reaction by reversing the anode half-reaction and adding
the cathode half-reaction.
Zn2+(aq) + 2H+(aq)→Zn(s)+H2(g)
Step 2: Write the standard cell potential in terms of standard reduction
potentials (E◦) for the half-reactions involved. The standard cell potential is
given by
E◦
cell =E◦
cathode −E◦
anode
Step 3: Find the standard reduction potentials for the half-reactions from a
standard reduction potential table. The values are:
E◦
Zn2+/Zn =−0.76 V
E◦
2H+/H2= 0.00 V
Step 4: Substitute the values into the formula for the standard cell potential.
E◦
cell = 0.00 V −(−0.76 V)
E◦
cell = 0.76 V
Therefore, the standard cell potential at 25
°
C for the given galvanic cell is
0.76 V.
5
Question 7
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 potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =
−0.76 V, calculate the standard cell potential (E◦
cell) of the galvanic cell.
Solution
Step 1: Write the overall cell reaction by combining the two half-reactions,
ensuring that the number of electrons lost equals the number of electrons gained:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions. The standard cell potential is given
by:
E◦
cell =E◦
cathode −E◦
anode
Substitute the values of E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =−0.76 V into the
equation:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the standard cell potential of the galvanic cell is E◦
cell = 1.10 V.
Question 8
Question
Consider an electrochemical cell consisting of a copper electrode immersed in
1.0 M Cu2+ solution and a silver electrode immersed in 1.0 M Ag+solution.
Write the balanced half-reactions and the overall cell reaction. Is this cell a
galvanic cell or an electrolytic cell? Justify your answer.
Solution
Step 1: Write the balanced half-reactions: At the copper electrode: Cu2+ +
2e−→Cu At the silver electrode: Ag++e−→Ag
Step 2: Write the overall cell reaction:
Cu2+ + 2Ag →Cu + 2Ag+
6
Step 3: Determine if the cell is a galvanic or electrolytic cell: In a galvanic
cell, the redox reaction is spontaneous and produces electrical energy. In this
case, the cell reaction is proceeding from left to right (oxidation at copper elec-
trode, reduction at silver electrode), which means it is a spontaneous process.
Therefore, this cell is a galvanic cell.
Question 9
Question
Consider a galvanic cell constructed with a silver electrode in a 1.0 M AgNO3
solution and a copper electrode in a 1.0 M Cu(NO3)2solution. The standard
reduction potentials are: Ag+(aq) +e−→Ag(s) (E◦= 0.80 V) and Cu2+(aq)+
2e−→Cu(s) (E◦= 0.34 V). Determine the cell potential at 25
°
C.
Solution
Step 1: Write out the overall reaction for the galvanic cell:
Ag(s) + Cu2+(aq)→Cu(s) + Ag+(aq)
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials provided:
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
Step 3: Calculate the non-standard cell potential using the Nernst equation:
E=E◦−0.0592
nlog(Q)
where nis the number of moles of electrons transferred and Qis the reaction
quotient.
Step 4: Calculate the reaction quotient Q:
Q=[products]
[reactants] =[Cu][Ag+]
[Ag][Cu2+]
Since both concentrations are 1.0 M, Q= 1.
Step 5: Substitute the values into the Nernst equation:
E=−0.46 V −0.0592
2log(1)
E=−0.46 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is −0.46 V.
7
Question 10
Question
Consider a Galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Pb2+(aq) + 2e−→Pb(s)E◦=−0.13 V
If the initial concentrations are [Zn2+] = 1.0 M, [Pb2+] = 0.1 M, and the cell
operates until equilibrium is reached, determine the cell potential at equilibrium.
Solution
Step 1: Write the cell reaction using the given half-reactions:
Zn2+(aq) + Pb(s)→Zn(s) + Pb2+(aq)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =−0.13 V −(−0.76 V) = 0.63 V
Step 3: Calculate the reaction quotient (Q) at equilibrium using the initial
concentrations given:
Q=[Zn](Pb2+)
[Zn2+][Pb]
Q=(1.0)(0.1)
(1.0)(0.1) = 1
Step 4: Use the Nernst equation to find the cell potential at equilibrium
(Ecell):
Ecell =E◦
cell −0.0592
nlog(Q)
Since the reaction involves the transfer of 2 electrons, n= 2.
Ecell = 0.63 V −0.0592
2log(1) = 0.63 V
Therefore, the cell potential at equilibrium is 0.63 V.
8
Question 11
Question
Consider the following half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Determine if a spontaneous reaction will occur when a strip of copper metal is
placed in a solution of zinc sulfate. If so, write the overall balanced cell reaction
and calculate the cell potential.
Solution
Step 1: Write the individual half-reactions and find the standard cell potential.
Given the half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
The overall cell reaction will involve the reduction of Cu2+ and the oxidation
of Zn. To determine if a spontaneous reaction will occur, we calculate the overall
standard cell potential, E◦
cell, using the formula:
E◦
cell =E◦
cathode −E◦
anode
where
E◦
cathode = 0.34 V
E◦
anode =−0.76 V
Plugging these values into the formula, we get:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Since the standard cell potential is positive, a spontaneous reaction will
occur.
Step 2: Write the overall balanced cell reaction. The overall cell reaction is
the sum of the two half-reactions.
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 3: Calculate the cell potential. The standard cell potential was already
calculated as 1.10 V in Step 1.
9
Question 12
Question
A voltaic cell is constructed using a silver/silver chloride electrode and a zinc
electrode. The standard reduction potentials are as follows:
AgCl(s) + e−→Ag(s) + Cl−(aq)E◦
cell = 0.22 V
Zn2+(aq)+2e−→Zn(s)E◦
cell =−0.76 V
What is the cell potential when the concentration of Zn2+ is 1.0 M and the
solid AgCl is in contact with a 0.10 M AgNO3solution? Use R= 8.314 J/K mol,
F= 96,485 C/mol, and T= 298 K.
Solution
Step 1: Write the cell reaction and cell potential equation using the given half-
reactions. The cell reaction is the sum of the two half-reactions:
AgCl(s) + Zn2+(aq)+2e−→Ag(s) + Zn(s) + Cl−(aq)
The cell potential equation is:
Ecell =E◦
cell −0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced reaction,
and Qis the reaction quotient.
Step 2: Calculate the cell potential First, calculate Qusing the concentra-
tions provided. Since there are no concentrations for the solid electrodes (Ag
and Zn), their activities are assumed to be 1. Therefore,
Q=[Ag+][Cl−]
[Zn2+]
Q=0.102
1.0= 0.01
Step 3: Plug the values into the formula
Ecell = 0.22V−0.0592
2log 0.01
Ecell = 0.22V−0.0296 ×2×2
Ecell = 0.22V−0.1184 = 0.1016 V
Therefore, the cell potential is 0.1016 V.
10
Question 13
Question
Consider a galvanic cell where the reaction occurring is:
Cd2+(aq) + 2Ag(s)→Cd(s) + 2Ag+(aq)
Determine the cell potential at standard conditions given that E◦
Cd2+/Cd =
−0.40 V and E◦
Ag+/Ag = 0.80 V.
Solution
Step 1: Write the half-reactions and standard electrode potentials. The half-
reactions involved in the cell are:
Reduction half-reaction:
Cd2+(aq) + 2e−→Cd(s)E◦
Cd2+/Cd =−0.40 V
Oxidation half-reaction:
2Ag(s)→2Ag+(aq) + 2e−E◦
Ag+/Ag = 0.80 V
Step 2: Calculate the cell potential. The cell potential E◦
cell can be calculated
using 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.
Given that E◦
Cd2+/Cd =−0.40 V and E◦
Ag+/Ag = 0.80 V, we have:
E◦
cell =E◦
Ag+/Ag −E◦
Cd2+/Cd = 0.80 −(−0.40) = 1.20 V
Therefore, the cell potential at standard conditions is 1.20 V.
Question 14
Question
Consider a galvanic cell with the following half-reactions:
Cu2+ + 2e−→Cu E◦
cell = 0.34 V
Fe3+ +e−→Fe2+ E◦
cell =−0.77 V
Calculate the standard Gibbs free energy change (∆G◦) for the overall cell
reaction and state whether the cell is spontaneous at standard conditions.
11
Solution
Step 1: Calculate the standard cell potential (E◦
cell) for the overall reaction. The
overall cell reaction is the sum of the two half-reactions:
Cu2+ + 2Fe3+ →2Fe2+ + Cu
The standard cell potential is the sum of the standard reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values:
E◦
cell =E◦
Cu −E◦
Fe
E◦
cell = 0.34 V −(−0.77 V) = 1.11 V
Step 2: Calculate the standard Gibbs free energy change (∆G◦) for the
overall reaction. The relationship between standard cell potential and standard
Gibbs free energy change is given by:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Fis the Faraday constant (F= 96485 C/mol).
Since 2 moles of electrons are transferred in the balanced cell reaction, n= 2.
Substitute the values:
∆G◦=−2×96485 C/mol ×1.11 V
∆G◦=−213.59 kJ/mol
Step 3: Determine if the cell is spontaneous at standard conditions. For a
reaction to be spontaneous at standard conditions, ∆G◦must be negative. Since
∆G◦=−213.59 kJ/mol (which is negative), the cell reaction is spontaneous at
standard conditions.
Question 15
Question
Consider a galvanic cell with a standard reduction potential of E◦
cell = 1.23 V.
Assuming the reaction in the cell is spontaneous, calculate the standard Gibbs
free energy change, ∆G◦, for the cell reaction. Also, if the concentration of Mn2+
ions in the cell’s cathode compartment is 0.1 M, determine the cell potential
when the concentration of Mn2+ ions in the cathode compartment is increased
to 1.0 M.
12
Solution
Step 1: Recall the relationship between standard Gibbs free energy change,
standard cell potential, and the Faraday constant:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced redox
reaction, Fis the Faraday constant (96485 C/mol), and E◦
cell is the standard
cell potential.
Step 2: Calculate the number of moles of electrons transferred. The stan-
dard reduction potential (E◦
cell) is related to the number of moles of electrons
transferred (n) by the equation:
E◦
cell =∆G◦
−nF
Given E◦
cell = 1.23 V, and F= 96485 C/mol, we can solve for n.
∆G◦=−nF E◦
cell
n=−∆G◦
F E◦
cell
Step 3: Substitute the values into the equation to solve for n.
n=−∆G◦
F E◦
cell
=−∆G◦
(96485 C/mol)(1.23 V)
Step 4: Calculate the standard Gibbs free energy change (∆G◦). We know
that ∆G◦=−nF E◦
cell and we have found the value for nin the previous step.
∆G◦=−nF E◦
cell =−−∆G◦
(96485 C/mol)(1.23 V)(96485 C/mol)(1.23 V)
1=∆G◦
Therefore, ∆G◦= 1 J.
Step 5: Calculate the cell potential when the concentration of Mn2+ ions
in the cathode compartment is 1.0 M. Assuming the cell reaction is occurring
at equilibrium, we can use the Nernst equation:
E=E◦−0.0592
nlog [Mn2+]cathode
[Mn2+]anode
where [Mn2+]cathode is the concentration of Mn2+ ions in the cathode compart-
ment and [Mn2+]anode is the concentration of Mn2+ ions in the anode compart-
ment.
Step 6: Calculate the cell potential with [Mn2+]cathode = 1.0 M and [Mn2+]anode =
0.1 M. Given that E◦
cell = 1.23 V, n= 2 (as indicated by the balanced redox
reaction), and substituting [Mn2+]cathode = 1.0 M and [Mn2+]anode
13
Question 16
Question
Consider a galvanic cell where the following reaction takes place:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Determine the cell potential at 25
°
C if the standard reduction potentials are
E◦(Zn2+/Zn) = −0.76 V and E◦(Ag+/Ag) = 0.80 V.
Solution
Step 1: Write the half-reactions and find the cell potential. The standard cell
potential (E◦
cell) can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
We have the half-reactions:
Zn2+(aq)+2e−→Zn(s) with E◦
Zn2+/Zn =−0.76 V
2Ag+(aq)+2e−→2Ag(s) with E◦
Ag+/Ag = 0.80 V
The cell potential is:
E◦
cell =E◦
cathode −E◦
anode =E◦
Ag+/Ag −E◦
Zn2+/Zn = 0.80V−(−0.76V)=1.56 V
Therefore, the cell potential at 25
°
C is 1.56 V.
Question 17
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a standard iron (III) ion electrode with an Fe3+/Fe2+ half-cell. The stan-
dard reduction potentials are as follows: E◦(H+/H2) = 0.00 V and E◦(Fe3+/Fe2+)
= -0.77 V. Calculate the cell potential (E◦
cell) for this galvanic cell. Is the reac-
tion spontaneous or non-spontaneous?
Solution
Step 1: Write the overall cell reaction as the sum of the two half-cell reactions.
Overall reaction: 2H++ Fe3+ →H2+ Fe2+
Step 2: Write the two half-cell reactions, making sure the number of electrons
match to cancel out in the overall reaction.
Anode (oxidation): Fe3+ +e−→Fe2+ E◦= -0.77 V
14
Cathode (reduction): 2H++ 2e−→H2E◦= 0.00 V
Step 3: Calculate the overall cell potential using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.00 V −(−0.77 V)
E◦
cell = 0.77 V
Step 4: Determine the spontaneity of the reaction based on the cell potential.
Since the cell potential is positive (0.77 V), the reaction is spontaneous as the
flow of electrons is favored from the anode to the cathode.
Question 18
Question
A galvanic cell is constructed with a standard hydrogen electrode (Pt|H2(g)|H+(1M))
on one side and an iron electrode (Fe3+(1M)|Fe2+(1M)) on the other side. The
standard reduction potential of the iron electrode is −0.77 V and the standard
reduction potential of the hydrogen electrode is 0 V. Determine the cell poten-
tial, and identify the cathode and anode in this galvanic cell. Also, determine
if the reaction is spontaneous or not.
Solution
Step 1: Write the half-reactions for the two electrodes. The half-reaction of the
hydrogen electrode is the reduction of hydrogen ions to hydrogen gas:
2H++ 2e−→H2(g)
The half-reaction of the iron electrode is the reduction of iron(III) ions to iron(II)
ions:
Fe3+ + e−→Fe2+
Step 2: Determine the standard cell potential by finding the difference be-
tween the standard reduction potentials of the two half-reactions. The standard
cell potential (E◦
cell) can be calculated as:
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 E◦
cathode = 0 V and E◦
anode =−0.77 V, we have:
E◦
cell = 0 V −(−0.77 V) = 0.77 V
Step 3: Identify the cathode and anode in the galvanic cell. The electrode
with the higher standard reduction potential is the cathode, while the electrode
15
with the lower standard reduction potential is the anode. So, in this case, the
hydrogen electrode is the cathode and the iron electrode is the anode.
Step 4: Determine if the reaction is spontaneous or not. The reaction is
spontaneous if E◦
cell is positive. Since E◦
cell = 0.77 V, the reaction is spontaneous.
Question 19
Question
A cell is constructed with a standard hydrogen electrode (SHE) and a zinc
electrode. The standard reduction potential for the hydrogen electrode is 0 V,
and for the zinc electrode is -0.76 V. Calculate the standard cell potential for
this galvanic cell.
Solution
Step 1: Write the half-reactions for the hydrogen electrode and zinc electrode.
The half-reactions are:
H+(aq)+e−→1
2H2(g) (Reduction at SHE)
Zn2+(aq) + 2e−→Zn(s) (Reduction at Zn electrode)
Step 2: Determine the standard cell potential. The standard 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.
In this case, the cathode is the SHE (standard hydrogen electrode) with
E◦
cathode = 0 V, and the anode is the zinc electrode with E◦
anode =−0.76 V.
Therefore, the standard cell potential is:
E◦
cell = 0 V −(−0.76 V) = 0.76 V
So, the standard cell potential for this galvanic cell is 0.76 V.
Question 20
Question
Consider the following cell reaction:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
If the standard reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu =
0.34 V, calculate the standard cell potential (E◦
cell) and state whether this cell
is galvanic or electrolytic.
16
Solution
To calculate the standard cell potential (E◦
cell), we use the formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode (Cu2+/Cu)
and E◦
anode is the standard reduction potential of the anode (Zn2+/Zn).
We are given:
E◦
Zn2+/Zn =−0.76 V, E◦
Cu2+/Cu = 0.34 V
Therefore,
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn = 0.34 V −(−0.76 V) = 1.10 V
Since the standard cell potential (E◦
cell) is positive, the cell is galvanic.
Question 21
Question
Consider a galvanic cell that consists of a zinc electrode in a 1.0 M solution
of Zn2+ ions and a copper electrode in a 1.0 M solution of Cu2+ ions. The
standard reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu = 0.34 V,
respectively. Determine the cell potential at standard conditions and identify
the anode and cathode.
Solution
Step 1: Write the reduction half-reactions and standard cell potential.
The reduction half-reactions are:
Zn2+ + 2e−→Zn (oxidation)
Cu2+ + 2e−→Cu (reduction)
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
Step 2: Calculate the standard cell potential.
E◦
cell = 1.10 V
Therefore, the standard cell potential is 1.10 V.
17
Step 3: Identify the anode and cathode.
Since the cell potential is positive, the reaction is spontaneous, with the copper
electrode acting as the cathode (where reduction occurs) and the zinc electrode
acting as the anode (where oxidation occurs).
Thus, in the given galvanic cell, zinc is the anode and copper is the cathode.
Question 22
Question
Consider a galvanic cell with a zinc electrode in a 1.0 M Zn2+ solution and
a silver electrode in a 1.0 M Ag+solution. The standard reduction potentials
are E◦
Zn2+/Zn =−0.76 V and E◦
Ag+/Ag = 0.80 V. Calculate the cell potential at
standard conditions.
Solution
Step 1: Write the two half-reactions representing the oxidation and reduction
processes. The oxidation half-reaction occurs at the zinc electrode:
Zn(s) →Zn2+(aq)+2e−
The reduction half-reaction occurs at the silver electrode:
Ag+(aq) + e−→Ag(s)
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials. The cell potential is calculated by subtracting the standard
reduction potential of the anode from the standard reduction potential of the
cathode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80 V −(−0.76 V) = 1.56 V
Therefore, the cell potential under standard conditions is 1.56 V.
Question 23
Question
Consider a galvanic cell that consists of a copper electrode in a 1.0 M copper(II)
sulfate solution and a platinum electrode in a 1.0 M hydrogen ion solution.
Suppose the standard reduction potential for the copper half-cell is +0.34 V, and
the standard reduction potential for the hydrogen half-cell is 0.00 V. Determine
the cell potential at 25
°
C for this galvanic cell.
18
Solution
Step 1: Write the half-reactions for each electrode based on the given informa-
tion.
Copper electrode: Cu2+(aq)+2e−→Cu(s)E◦= +0.34 V
Platinum electrode: 2H+(aq)+2e−→H2(g)E◦= 0.00 V
Step 2: Write the overall cell reaction by summing the two half-reactions.
Cu2+(aq)+2H+(aq)→Cu(s)+H2(g)
Step 3: Determine the cell potential by adding the standard reduction po-
tentials of the two half-cells.
Cell potential: E◦
cell =E◦
cathode −E◦
anode = 0.00 V −(+0.34 V) = −0.34 V
Therefore, the cell potential at 25
°
C for this galvanic cell is -0.34 V.
Question 24
Question
Consider a galvanic cell constructed with a standard hydrogen electrode (SHE)
as the anode and a copper electrode as the cathode. If the standard reduction
potential of the copper electrode is +0.34 V, determine the cell potential at
25◦C. Is this cell galvanic or electrolytic?
Solution
Step 1: Write the two half-reactions for the cell: The half-reaction at the anode
involving the standard hydrogen electrode (SHE) is:
2H++ 2e−→H2(g)
The half-reaction at the cathode involving the copper electrode is:
Cu2+ + 2e−→Cu(s)
Step 2: Calculate the standard cell potential (E◦) using the standard reduc-
tion potentials of the half-reactions: Given that the standard reduction potential
of the copper electrode is +0.34 V, and the standard reduction potential of the
standard hydrogen electrode (SHE) is 0 V, the cell potential can be calculated
as:
E◦=E◦
cathode −E◦
anode
E◦= +0.34 V −0 V
E◦= +0.34 V
19
Step 3: Calculate the cell potential at 25◦C using the Nernst equation: The
Nernst equation relates the cell potential (E) to the standard cell potential (E◦),
the reaction quotient (Q), the gas constant (R), the temperature in Kelvin (T),
and the number of electrons transferred in the reaction (n):
E=E◦−RT
nF ln(Q)
At 25◦C, T= 298 K. Since the reaction quotient Q= 1 for standard condi-
tions, the equation simplifies to:
E=E◦
Therefore, the cell potential at 25◦C is +0.34 V.
Step 4: Determine if the cell is galvanic or electrolytic: A cell is galvanic if
the cell potential is positive, indicating a spontaneous redox reaction. Since the
cell potential at 25◦C is +0.34 V (positive), the cell is galvanic.
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Cathode: Ag+(aq)+e−→Ag(s)E◦= 0.80 V
Anode: Zn(s)→Zn2+(aq) + 2e−E◦=−0.76 V
Calculate the cell potential (E◦
cell) and the equilibrium constant (Keq) for the
overall reaction of the galvanic cell.
Solution
Step 1: Calculate E◦
cell by using the formula:
E◦
cell =E◦
cathode −E◦
anode
Substitute the given values:
E◦
cell = 0.80 V −(−0.76 V)
E◦
cell = 1.56 V
Step 2: Calculate Keq using the formula:
E◦
cell =RT
nF ln Keq
Solving for Keq:
Keq =e
nF E◦
cell
RT
20
Given that R= 8.314 J/mol K, T= 298 K, F= 96485 C/mol, and n= 2 (from
balancing the half-reactions), we have:
Keq =e2×96485×1.56
8.314×298
Keq =e7.77
Keq ≈2417.89
Question 26
Question
Consider a galvanic cell with the following half-reactions:
Anode: Cd(s) →Cd2+(aq)+2e−
Cathode: Ag+(aq) + e−→Ag(s)
Calculate the cell potential at standard conditions (T= 298 K, P= 1 atm),
given that [Cd2+] = 0.20 M and [Ag+] = 0.15 M.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions.
Cd(s) + 2Ag+(aq)→Cd2+(aq)+2Ag(s)
Step 2: Write the Nernst equation for cell potential:
E=E◦−RT
nF ln(Q)
Where: - Eis the cell potential, - E◦is the standard cell potential, - Ris
the ideal gas constant (8.314 J/(mol·K)), - Tis the temperature in Kelvin (298
K), - nis the number of electrons transferred in the balanced cell reaction, - F
is the Faraday constant (96485 C/mol), - Qis the reaction quotient.
Step 3: Calculate the standard cell potential E◦using the Standard Reduc-
tion Potentials table.
E◦=E◦
cathode −E◦
anode
=E◦
Ag+/Ag −E◦
Cd2+/Cd
= 0.80 V −(−0.40 V)
= 1.20 V
Step 4: Calculate the reaction quotient Qusing the given concentrations.
Q=[Cd2+]
[Ag+]2=0.20
0.152
21
Step 5: Substitute all values into the Nernst equation and solve for the cell
potential E.
E= 1.20 V −(8.314 J/(mol ·K)) ·(298 K)
2·(96485 C/mol) ln 0.20
0.152
E= 1.20 V −0.0268 V ·ln 0.20
0.152
E≈1.16 V
Therefore, the cell potential at standard conditions is approximately 1.16 V.
Question 27
Question
Consider a galvanic cell with the following half-reactions:
Anode: Pb(s)→Pb2+(aq) + 2e−
Cathode: 2Ag+(aq) + 2e−→2Ag(s)
If the standard reduction potentials are E◦(Pb2+/Pb) = −0.13 V and E◦(Ag+/Ag) =
0.80 V, calculate the standard cell potential (E◦
cell) for this galvanic cell.
Solution
Step 1: Determine the overall cell reaction by adding the 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. The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
From the data given, we have:
E◦
anode =−0.13 V and E◦
cathode = 0.80 V
Therefore,
E◦
cell = 0.80 V −(−0.13 V) = 0.93 V
Hence, the standard cell potential for this galvanic cell is 0.93 V.
22
Question 28
Question
Consider the reaction between silver ions and copper metal:
Ag+(aq) + Cu(s)→Ag(s) + Cu2+(aq)
Given that the standard reduction potentials are E◦(Ag+/Ag) = 0.80 V and
E◦(Cu2+/Cu)=0.34 V, determine whether the reaction is spontaneous under
standard conditions. If not, calculate the voltage that would be required to
drive the reaction forward.
Solution
Step 1: Calculate the standard cell potential (E◦
cell) using the standard reduction
potentials.
E◦
cell =E◦(cathode) −E◦(anode)
E◦
cell =E◦(Ag+/Ag)−E◦(Cu2+/Cu)
E◦
cell = 0.80 V −0.34 V
E◦
cell = 0.46 V
Step 2: Determine if the reaction is spontaneous under standard conditions.
Since E◦
cell = 0.46 V, the reaction is spontaneous under standard conditions
because E◦
cell >0.
Therefore, the given reaction proceeds spontaneously to the right.
Step 3: If the reaction is not spontaneous, calculate the voltage required to
drive the reaction forward. Since the reaction is spontaneous under standard
conditions, we do not need to calculate the additional voltage required in this
case.
Question 29
Question
Consider a galvanic cell with the following half-reactions:
Ag+(aq)+e−→Ag(s)E◦= 0.80 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
If the concentration of Ag+is 1.0 M and Cu2+ is 0.1 M, determine the cell
potential for this galvanic cell at 25◦C.
23
Solution
Step 1: Identify the anode and cathode reactions. The half-reaction with the
higher standard reduction potential (E◦) will be reduced (cathode) and the
other will be oxidized (anode).
Here, the reduction potential for the reduction half-reaction of Ag+is 0.80
V, which is higher than 0.34 V for Cu2+. Thus, the reduction half-reaction for
Ag+will occur at the cathode and the oxidation half-reaction for Cu2+ will
occur at the anode.
Step 2: Write the overall cell reaction by adding the anode and cathode
half-reactions. Ensure that the same number of electrons are involved in both
half-reactions so that they can cancel out.
The overall reaction for the galvanic cell is:
2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Step 3: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦−0.0592
nlog(Q)
where: - E◦is the standard cell potential, - nis the number of moles of
electrons transferred in the balanced cell reaction, - Qis the reaction quotient
(ratio of product concentrations to reactant concentrations).
In this case, n= 2 since 2 moles of electrons are transferred in the overall
reaction.
Step 4: Calculate Qusing the given concentrations. Remember, the concen-
trations of solid species do not appear in the expression for Q.
Q=[Ag+]2
[Cu2+]=(1.0)2
0.1= 10
Step 5: Substitute the values of E◦,n, and Qinto the Nernst equation and
solve for Ecell.
Ecell = 0.80 −0.0592
2log(10) = 0.80 −0.0296 ×1=0.7704 V
Therefore, the cell potential for this galvanic cell at 25◦C is 0.7704 V.
Question 30
Question
Consider the following Galvanic and Electrolytic cells:
Galvanic Cell Electrolytic Cell
Zn(s) |Zn2+(aq) || Cu2+(aq) |Cu(s) Cu(s) |Cu2+(aq) || Ag+(aq) |Ag(s)
24
For each cell, determine the following: 1. The half-reaction that occurs at
the cathode and the anode. 2. The overall cell reaction. 3. The direction of
electron flow. 4. The direction of cation flow. 5. Whether the cell functions as
a galvanic cell or an electrolytic cell.
Solution
1. For the Galvanic Cell:
Anode (Zn(s)): Zn(s) →Zn2+(aq) + 2e−
Cathode (Cu2+(aq)): Cu2+(aq) + 2e−→Cu(s)
For the Electrolytic Cell:
Anode (Cu(s)): Cu(s) →Cu2+(aq) + 2e−
Cathode (Ag+(aq)): Ag+(aq) + e−→Ag(s)
2. The overall cell reaction for the Galvanic Cell is the sum of the half-
reactions:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
The overall cell reaction for the Electrolytic Cell is the sum of the half-
reactions:
Cu(s) + Ag+(aq) →Cu2+(aq) + Ag(s)
3. In both cells, electrons flow from the anode to the cathode.
4. In the Galvanic Cell, cations flow from the anode to the cathode.
In the Electrolytic Cell, cations flow from the cathode to the anode.
5. The Galvanic Cell functions as a spontaneous chemical reaction generating
electricity, while the Electrolytic Cell requires an external electrical current to
drive a non-spontaneous chemical reaction.
Question 31
Question
A voltaic cell is constructed with a standard hydrogen electrode and a standard
silver electrode. The standard cell potential is measured to be 0.80 V at 25
°
C.
Calculate the standard free energy change for the cell reaction.
Solution
Step 1: Write the balanced cell reaction for the voltaic cell. The cell reaction
for a voltaic cell with a standard hydrogen electrode (H+(aq),H2(g)) and a
standard silver electrode (Ag+(aq),Ag(s)) is:
2H+(aq) + 2e−→H2(g)E◦= 0 V
25
2Ag+(aq) + 2e−→2Ag(s)E◦= 0.80 V
Overall cell reaction:
2Ag+(aq) + 2H+(aq)→2Ag(s)+H2(g)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials. The standard cell potential (E◦
cell) is the difference between
the standard reduction potentials of the two half-reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −0 V = 0.80 V
Step 3: Calculate the standard free energy change (∆G◦) using the equation:
∆G◦=−nF E◦
cell
where nis the number of moles of electrons transferred in the balanced cell
reaction, Fis the Faraday constant (96,485 C/mol).
For the given cell reaction, n= 2 moles of electrons transferred. Thus,
∆G◦=−2×96485 C/mol ×0.80 V = −155136 J/mol
Therefore, the standard free energy change for the cell reaction is −155136 J/mol.
Question 32
Question
Consider a galvanic cell consisting of a standard hydrogen electrode (SHE) and
a copper electrode with the following half-reactions:
Cathode: Cu2+ + 2 e−−−→ Cu E◦= 0.34 V
Anode: 2 H++ 2 e−−−→ H2E◦= 0.00 V
Calculate the cell potential at 298 K when the concentration of Cu2+ is 0.10 M,
and the pressure of H2is 1.0 atm.
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction is the sum of the two half-reactions:
Cu2+ + 2 H+−−→ Cu + H2
Step 2: Calculate the cell potential at standard conditions.
The standard cell potential E◦
cell can be calculated using the standard re-
duction potentials:
E◦
cell =E◦
cathode −E◦
anode
26
E◦
cell = 0.34V−0.00V= 0.34 V
Step 3: Consider the non-standard conditions.
Since the concentrations of Cu2+ and the pressure of H2are given, we can
calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the cell reaction and Qis
the reaction quotient.
Step 4: Calculate the reaction quotient Q.
Q=[Cu]
[Cu2+][H+]2
Q=1.0
0.10 ×1.02= 10
Step 5: Substitute values into the Nernst equation and solve for the cell
potential.
Ecell = 0.34 −0.0592
2log(10)
Ecell = 0.34 −0.0296 ×1
Ecell = 0.3104 V
Therefore, the cell potential at 298 K when the concentration of Cu2+ is
0.10 M and the pressure of H2is 1.0 atm is 0.3104 V.
Question 33
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.78 V that
operates under standard conditions. The cell reaction is given by:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Calculate the standard Gibbs free energy change for the cell reaction and
determine if the reaction is spontaneous.
Solution
Step 1: The standard Gibbs free energy change (∆G◦) for the cell reaction can
be calculated using the equation:
∆G◦=−nF E◦
where nis the number of moles of electrons transferred in the balanced redox
equation, and Fis the Faraday constant (96500 C/mol).
From the given cell reaction, we see that 2 moles of electrons are transferred,
so n= 2. Substituting into the equation:
∆G◦=−2×96500 C/mol ×0.78 V
27
∆G◦=−188220 J/mol
Step 2: To determine if the reaction is spontaneous, we can use the relation-
ship between ∆G◦and the equilibrium constant (K):
∆G◦=−RT ln K
where Ris the gas constant (8.314 J/mol ·K) and Tis the temperature in
Kelvin.
At standard conditions, T= 298 K and R= 8.314 J/mol ·K. Thus, we have:
−188220 J/mol = −8.314 J/mol ·K×298 K ln K
Solving for ln K:
ln K=−188220 J/mol
−8.314 J/mol ·K×298 K
ln K= 27.37
Step 3: Exponentiating both sides of the equation to eliminate the natural
logarithm:
K=e27.37
K= 1.17 ×1011
Since K > 1, the reaction is spontaneous.
Question 34
Question
A student sets up a galvanic cell with a standard hydrogen electrode (E◦
H+/H2=
0.00 V) and a copper electrode (E◦
Cu2+/Cu = 0.34 V), connected by a salt bridge.
The initial concentrations are [H+] = 1.0 M and [Cu2+]=0.1 M.
Calculate the cell potential at 25
Step 1: Write the balanced redox reaction and the cell notation.
The balanced redox reaction is:
Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)
The cell notation is:
Pt|H2(1 M)|H+(1 M)||Cu2+(0.1 M)|Cu
Step 2: Calculate the cell potential (E◦
cell) at standard conditions.
28
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
H+/H2= 0.00 V, and E◦
Cu2+/Cu = 0.34 V, we have:
E◦
cell = 0.00 V −0.34 V = −0.34 V
Step 3: Calculate the reaction quotient (Q).
The reaction quotient Qis calculated using the concentrations of the species
involved in the reaction:
Q=[Cu](solid)
[Cu2+][H+]2
Plugging in the values, we get:
Q=1
0.1×12= 10
Step 4: Calculate the cell potential (Ecell) under non-standard conditions.
The Nernst equation relates the cell potential under non-standard conditions
(Ecell) to the standard cell potential (E◦
cell) and the reaction quotient (Q):
Ecell =E◦
cell −0.0592
nlog Q
Since the reaction involves the transfer of 2 moles of electrons, n= 2. Plug-
ging in the values, we get:
Ecell =−0.34 V −0.0592
2log 10
Ecell =−0.34 V −0.0296 log 10
Ecell =−0.34 V −0.0296 ×1
Ecell =−0.34 V −0.0296
Ecell =−0.3696 V
Therefore, the cell potential at 25
Question
Consider a galvanic cell that uses the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential of Cu2+/Cu is 0.34 V, calculate the stan-
dard cell potential (E◦
cell) of this galvanic cell.
29
Solution
Step 1: Write the full cell reaction by adding the two half-reactions together.
Ensure that the electrons cancel out.
The full cell reaction will be:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials of Zn2+/Zn and Cu2+/Cu.
The standard cell potential 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.
Given that E◦
cathode (for Cu2+/Cu) is 0.34 V, and E◦
anode (for Zn2+/Zn) is
−0.76 V (since it is the reverse of the reduction potential), we can plug these
values into the formula.
Substitute the values into the formula:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the standard cell potential (E◦
cell) of the galvanic cell is 1.10 V.
30
