CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Nernst equation
Question Bank - Set 4
Liberty University
Question 1
Question
A voltaic cell consists of a standard hydrogen electrode (SHE) and a zinc elec-
trode immersed in solutions of zinc sulfate solution and hydrochloric acid. The
concentration of Zn2+ is 0.10Mand the concentration of H+ions is 1.0M. Cal-
culate the cell potential at 25◦C.
Given: E◦
cell = 0.76 V
Solution
Step 1: Write down the half-reactions occurring at the electrodes.
At the standard hydrogen electrode:
2 H+(aq) + 2 e−−−→ H2(g)
At the zinc electrode:
Zn −−→ Zn2+ (aq) + 2 e−
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[H+]
where nis the number of moles of electrons transferred in the balanced
equation, and the concentrations [Zn2+] and [H+] are given in the question.
Step 3: Calculate nfrom the balanced equations. For the reaction at the
zinc electrode, 2 moles of electrons are transferred.
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.76 −0.0592
2log 0.10
1.0
Step 5: Calculate the cell potential at 25◦C:
Ecell = 0.76 −0.0592
2log (0.10) = 0.76 −0.0296 ×1=0.7304 V
Therefore, the cell potential at 25◦C is 0.7304 V.
Question 2
Question
Calculate the standard cell potential (E◦
cell) for the following reaction at 25
°
C:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Given: E◦
cell(Zn2+/Zn) = −0.76 V, E◦
cell(Ag+/Ag) = 0.80 V
Solution
Step 1: Write the two half-cell reactions for the reaction:
Zn(s)→Zn2+(aq)+2e−
2Ag+(aq)+2e−→2Ag(s)
Step 2: Calculate the overall cell potential (Ecell) using the Nernst equation:
Ecell =E◦
cell +0.0592
nlog [Zn2+]
[Ag+]2
Step 3: Given that the standard cell potential (E◦
cell) is the sum of the two
half-cell potentials for the reduction reactions, we have:
E◦
cell =E◦
cell(Zn2+/Zn) + E◦
cell(Ag+/Ag)
E◦
cell =−0.76 V + 0.80 V = 0.04 V
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.04 + 0.0592
2log [Zn2+]
[Ag+]2
Ecell = 0.04 + 0.0296 log [Zn2+]
[Ag+]2
2
Step 5: Since the reaction is at equilibrium, Ecell = 0, and the concentrations
of products over reactants is equal to the equilibrium constant:
[Zn2+]
[Ag+]2=Keq = 100= 1
Step 6: Substitute this equilibrium constant into the equation:
0.04 + 0.0296 log 1 = 0
0.04 + 0 = 0
Therefore, the standard cell potential is 0 V .
Question 3
Question
A concentration cell is set up with two Ag/AgCl electrodes. One half-cell con-
tains a 0.01 M solution of AgCl and the other half-cell contains a 0.1 M solution
of AgCl. Calculate the cell potential at 25
°
C. (Given: E◦
cell = 0.46 V)
Solution
Step 1: Write the half-reactions for the cell:
Anode: AgCl(s)→Ag(s) + Cl−(aq)
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Write the Nernst equation for the cell:
Ecell =E◦
cell −0.0592
nlog [reduced form]
[oxidized form]
Step 3: Calculate the reduction potential difference (E0
cell) using the concen-
trations given:
E0
cell = 0.46 V
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.46 V −0.0592
1log 0.1
0.01
Step 5: Solve for Ecell:
Ecell = 0.46 V −0.0592 log(10) = 0.46 V −0.0592 = 0.4008 V
Therefore, the cell potential at 25
°
C is 0.4008 V.
3
Question 4
Question
Calculate the cell potential at 25
°
C for the following reaction:
Cr2O2−
7(aq) + 14H+(aq) + 6Cl−(aq)→2Cr3+(aq) + 6Cl2(g) + 7H2O(l)
given that the standard reduction potential for Cr2O2−
7(aq) + 14H+(aq) +
6e−→2Cr3+(aq) + 7H2O(l) is +1.33 V.
Solution
Step 1: Write the half-reactions for the redox reaction:
Cr2O2−
7(aq) + 14H+(aq) + 6e−→2Cr3+(aq) + 7H2O(l)
6Cl−(aq)→6Cl2(g) + 12e−
Step 2: Calculate the cell potential at standard conditions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0V−(+1.33V) = −1.33V
Step 3: Calculate the Q value based on the concentrations given in the
reaction.
Step 4: Calculate the cell potential at non-standard conditions using the
Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the balanced redox
equation.
Step 5: Substitute the values into the Nernst equation and calculate the cell
potential at 25
°
C.
Question 5
Question
A voltaic cell is constructed with a silver electrode and a zinc electrode. The half-
reaction for the reduction of silver is Ag++e−→Ag with an electrode potential
of +0.80 V, while the half-reaction for the reduction of zinc is Zn2+ + 2e−→Zn
with an electrode potential of -0.76 V. Calculate the cell potential when the
concentration of Ag+is 0.50 M and the concentration of Zn2+ is 1.00 M.
4
Solution
Step 1: Write the cell reaction: The overall reaction for the voltaic cell is the
sum of the reduction half-reactions for silver and zinc:
Zn2+ + 2Ag →Zn + 2Ag+
Step 2: Calculate the standard cell potential, E
°
: The standard cell potential,
E
°
, is given by the difference in standard reduction potentials for the two half-
reactions:
Ecell =Ecathode −Eanode
Ecell = 0.80 V −(−0.76 V) = 1.56 V
Step 3: Write the Nernst equation: The Nernst equation relates the cell
potential under non-standard conditions to the standard cell potential:
Ecell =Ecell −0.0592
nlog [Ag+]2
[Zn2+]
where n is the number of moles of electrons transferred in the cell reaction.
Step 4: Calculate the cell potential under the given conditions: Since 2 moles
of electrons are transferred in the cell reaction, n = 2.
Ecell = 1.56 V −0.0592
2log (0.50 M)2
(1.00 M)
Ecell = 1.56 V −0.0592
2log(0.25)
Ecell = 1.56 V −0.0592
2(−0.602)
Ecell = 1.56 V + 0.018 = 1.578 V
Therefore, the cell potential when the concentration of Ag+is 0.50 M and
the concentration of Zn2+ is 1.00 M is 1.578 V.
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) as the anode and
a nickel metal electrode as the cathode. Given that the cell potential is 0.44 V
at standard conditions and the concentration of Ni2+ ions in the cathode com-
partment is 0.1 M, calculate the cell potential of the cell when the concentration
of Ni2+ ions is 0.01 M. Assume the temperature is constant.
5
Solution
Step 1: Write the half-reactions for the cell. The overall cell reaction is:
Ni2+(aq)+2e−→Ni(s)
The standard reduction potential of the nickel half-reaction is −0.25 V(given).
Step 2: Calculate the standard cell potential, E◦
cell. Using the Nernst equa-
tion:
Ecell =E◦
cell −0.0592
nlog [Ni2+]cathode
[Ni2+]◦
cathode
Given that E◦
cell = 0.44 V,n= 2, [Ni2+]◦
cathode = 1 M, [Ni2+]cathode = 0.1M,
and R= 0.0592 Vat room temperature, we can substitute these values to find
Ecell.
Step 3: Calculate the cell potential, Ecell, at the new concentration. Substi-
tute [Ni2+]cathode = 0.01 Minto the Nernst equation to calculate the new cell
potential.
Ecell = 0.44 V−0.0592
2log 0.01
1
Ecell = 0.44V−0.0296 log(0.01)
Ecell = 0.44V−0.0296 ×(−2)
Ecell = 0.44V+ 0.0592
Ecell = 0.4992 V
Therefore, the cell potential of the galvanic cell when [Ni2+]cathode is 0.01 M
is 0.4992 V.
Question 7
Question
A lead-acid battery has an electrolyte solution consisting of PbSO4and H2SO4
with concentrations of Pb2+= 0.050 M and [H2SO4]=1.0 M. Calculate the
cell potential at 25◦C for the reaction:
Pb(s) + PbSO4(s) + 2 H2SO4(aq) −−→ 2 PbSO4(s) + 2 H2O(l)
Given: - Standard reduction potential of PbSO4(s)+2 e−−−→ Pb(s)+SO42−(aq)
is −0.36 V. - Standard reduction potential of H3O+(aq) + 2 e−−−→ 1
2H2(g) +
H2O(l) at 25◦C is −0.036 V. - Faraday’s constant F= 96485 C/mol.
6
Solution
Step 1: Write the overall cell reaction.
Pb(s) + PbSO4(s) + 2 H2SO4(aq) −−→ 2 PbSO4(s) + 2 H2O(l)
Step 2: Write the half-reactions. Cathode: PbSO4(s) + 2 e−−−→ Pb(s) +
SO42−(aq) Anode: 2 H+(aq) + 2 e−−−→ H2(g)
Step 3: Calculate the standard cell potential (E◦
cell) using the Nernst equa-
tion:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =−0.36 V −(−0.036 V) = −0.324 V
Step 4: Calculate the cell potential at non-standard conditions using the
Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
where Qis the reaction quotient and Kis the equilibrium constant, and nis
the number of moles of electrons transferred in the reaction.
Step 5: Calculate the reaction quotient (Q):
Q=1
Pb2+2H+2=1
(0.050)2(1.0)2= 400
Step 6: Calculate the cell potential (Ecell) at 25◦C:
Ecell =−0.324 V −0.0592
2log(400)
Ecell =−0.324 V −0.0296 log(400)
Ecell =−0.324 V −0.0296 ×2.602 = −0.402 V
Therefore, the cell potential at 25◦C for the given reaction is −0.402 V.
Question 8
Question
Use the Nernst equation to calculate the cell potential (Ecell) for the following
reaction at 25
°
C:
2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Given that [Cu2+] = 0.020 M, [Ag+]=0.100 M, and the standard reduction
potentials are E◦
Ag+/Ag = 0.80 Vand E◦
Cu2+ /Cu = 0.34 V.
7
Solution
Step 1: Write the half-reactions and identify the oxidation and reduction half-
reactions. The half-reactions for this cell are: Oxidation: Cu(s)→Cu2+(aq) +
2e−Reduction: 2Ag+(aq)+2e−→2Ag(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions: 2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Step 3: Determine the cell potential at standard condition. The cell potential
at standard condition is calculated using the formula: E◦
cell =E◦
reduction, cathode−
E◦
oxidation, anode E◦
cell =E◦
Ag+/Ag −E◦
Cu2+ /Cu E◦
cell = 0.80 V−0.34 V E◦
cell =
0.46 V
Step 4: Apply the Nernst equation to calculate the cell potential under
non-standard condition. The Nernst equation is given by: Ecell =E◦
cell −
0.0592
nlog [Cu2+ ]2
[Ag+]2where nis the number of electrons transferred in the cell re-
action.
For the given reaction, n= 2: Ecell = 0.46 V−0.0592
2log 0.0202
0.1002Ecell =
0.46 V−0.0296 log 0.0004
0.01 Ecell = 0.46 V−0.0296 log 0.04 Ecell = 0.46 V−0.0296×
(−1.3979) Ecell = 0.46 V+ 0.0414 Ecell ≈0.5014 V
Therefore, the cell potential for the given reaction under the given conditions
is approximately 0.5014 V.
Question 9
Question
A galvanic cell is set up with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Ag+(aq) + e−−−→ Ag(s)
If the standard reduction potential for the Ag+(aq) + e−→Ag(s) half-
reaction is +0.80 V and the standard reduction potential for the Zn2+(aq) +
2e−→Zn(s) half-reaction is −0.76 V, calculate the cell potential when the
concentration of Zn2+ is 0.10 M and the concentration of Ag+is 2.0 M. (Assume
both solutions are at 25
°
C.)
Solution
Step 1: Calculate the cell potential using the Nernst equation:
The Nernst equation for the cell potential (Ecell) is given by:
Ecell =E◦
cell −0.0592
nlog [oxidized form]
[reduced form]
where: - E◦
cell is the standard cell potential, - nis the number of moles
of electrons transferred in the balanced cell reaction, - [oxidized form] is the
8
concentration of the oxidized form, - [reduced form] is the concentration of the
reduced form.
Step 2: Identify the number of moles of electrons transferred in the cell
reaction: From the given half-reactions, we can see that 2 moles of electrons are
transferred.
Step 3: Calculate the cell potential at standard conditions: The standard
cell potential (E◦
cell) can be 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
Step 4: Calculate the cell potential under given conditions:
Ecell = +1.56 V −0.0592
2log 0.10
2.0
Ecell = +1.56 V −0.0296 log 0.05
Ecell = +1.56 V −0.0296 ×(−1.301) = +1.6 V
Therefore, the cell potential when the concentration of Zn2+ is 0.10 M and
the concentration of Ag+is 2.0 M is +1.6 V.
Question 10
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a silver-silver chloride electrode (AgCl/Ag). The concentration of [Ag+] in the
AgCl/Ag electrode is 0.10 M. If the cell potential is 0.69 V at 25◦C, calculate
the [H+] concentration in the SHE when the cell is at equilibrium.
Given: E◦
cell = 0.80 V.
Solution
Step 1: Write the half-reactions for the cell.
The overall reaction for the cell is the combination of the reduction half-reaction
at the AgCl electrode and the oxidation half-reaction at the standard hydrogen
electrode.
Reduction half-reaction: AgCl(s) + e−→Ag(s) + Cl−(aq)
Oxidation half-reaction: 2H+(aq)+2e−→H2(g)
Step 2: Write the cell equation.
The cell equation can be obtained by adding the two half-reactions together.
The cell potential is given as 0.69 V.
0.69 = E◦
cell −0.0592
2log [H+]2
1
9
Step 3: Solve for [H+].
Plug in the given values (E◦
cell = 0.80 V) and solve for [H+].
0.69 = 0.80−0.0296 log[H+]⇒log[H+] = 0.80 −0.69
0.0296 ⇒log[H+] = 3.72 ⇒[H+]=5.10×10−4M
Therefore, the [H+] concentration in the standard hydrogen electrode when
the cell is at equilibrium is 5.10 ×10−4M.
Question 11
Question
A galvanic cell with a copper electrode and a silver electrode is set up with
copper(II) sulfate and silver nitrate solutions. The initial concentrations are
Cu2+= 0.10 M,Ag+= 0.0010 M. The reduction potentials are E◦(Cu2+/Cu) =
0.34 Vand E◦(Ag+/Ag) = 0.80 V. Determine the cell potential when the con-
centration of Cu2+is decreased to 0.0050 Mand the concentration of Ag+
is increased to 0.10 M.
Solution
Step 1: Write the half-reactions and Nernst equation.
The half-reactions for the two electrodes are:
At Cu electrode: Cu2+ + 2e−→Cu
At Ag electrode: Ag++e−→Ag
The Nernst equation relates the cell potential Eto the standard cell potential
E◦, the reaction quotient Q, and the temperature T:
E=E◦−0.0592
nlog Q
Where nis the number of moles of electrons transferred and Qis the reaction
quotient, given by:
Q=[products]coefficient of products
[reactants]coefficient of reactants
Step 2: Calculate Qfor the initial concentration conditions.
For the initial concentrations, Q=[Cu2+]1[Ag+]1
[Cu]1[Ag]2. Substitute the given concen-
trations into Q:Q=(0.10)(0.0010)
1(1) = 0.00010.
Step 3: Calculate the cell potential Einitial.
Given E◦(Cu2+/Cu) = 0.34 Vand E◦(Ag+/Ag) = 0.80 V, the overall cell reac-
tion is: Cu2+ + 2Ag →Cu + 2Ag+.
10
The standard cell potential E◦
cell can be calculated using the standard reduc-
tion potentials: E◦
cell =E◦(Cu2+/Cu) −E◦(Ag+/Ag) = 0.34 −0.80 = −0.46 V.
Now, plug the values into the Nernst equation: Einitial =−0.46−0.0592
2log(0.00010) ≈
−0.46 V.
Step 4: Calculate Qfor the final concentration conditions.
For the final concentrations, Q=[Cu2+]1[Ag+]2
[Cu]1[Ag]2. Substitute the given concentra-
tions into Q:Q=(0.0050)(0.10)2
1(1)2= 0.00050.
Step 5: Calculate the cell potential Efinal.
Using the same Nernst equation and overall cell reaction, we have: Efinal =
−0.46 −0.0592
2log(0.00050) ≈ −0.46 −0.048 = −0.508 V.
Therefore, the cell potential when [Cu2+]=0.0050
Question 12
Question
Calculate the cell potential at 25
°
C for a concentration cell with two half-cells,
each containing a zinc electrode. One half-cell has a zinc ion concentration of
0.10 M, while the other half-cell has a zinc ion concentration of 0.0010 M. The
standard reduction potential for the zinc electrode is -0.76 V.
Solution
Step 1: Write the balanced half-reaction for the cell. The half-reaction for the
reduction of zinc is:
Zn2+(aq)+2e−→Zn(s)
Step 2: Determine the cell reaction. In this concentration cell, each half-cell
contains the same substance (zinc). Therefore, the cell reaction consists of both
half-reactions:
Zn2+
0.10M(aq)+2e−→Zn0.10M(s)
Zn2+
0.0010M(aq)+2e−→Zn0.0010M(s)
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is:
E=Eo−0.0592
nlog [Zn2+]2
[Zn2+]1
where: - E= cell potential - Eo= standard cell potential - n= number of
electrons transferred (2 in this case) - [Zn2+]2= zinc ion concentration in the
second half-cell - [Zn2+]1= zinc ion concentration in the first half-cell
Plugging in the values:
E=−0.76 V −0.0592
2log 0.0010 M
0.10 M
E=−0.76 V −0.0296 log(0.01)
11
E=−0.76 V −0.0296 ×(−2)
E=−0.76 V + 0.0592
E=−0.7008 V
Therefore, the cell potential at 25
°
C for this concentration cell is -0.7008 V.
Question 13
Question
Given the Nernst equation: E=E◦−0.0592
nlog [Cu2+ ]
[Cu], calculate the cell
potential (E) for the following reaction at 25
°
C:
Cu2+(aq)+2e−→Cu(s)
Given E◦= 0.34 V, [Cu2+] = 0.10 M, [Cu] = 1.0 M. (Assume n= 2)
Solution
Step 1: Identify the given values: Given values: E◦= 0.34 V
[Cu2+] = 0.10 M
[Cu]=1.0 M
n= 2
We need to find E.
Step 2: Substitute the values into the Nernst equation:
E=E◦−0.0592
nlog [Cu2+]
[Cu]
E= 0.34 −0.0592
2log 0.10
1.0
Step 3: Calculate the natural logarithm:
E= 0.34 −0.0296 log (0.10)
E= 0.34 −0.0296 ×(−1)
E= 0.34 + 0.0296
E= 0.3696 V
Therefore, the cell potential (E) for the given reaction at 25
°
C is 0.3696 V.
12
Question 14
Question
A galvanic cell consists of a silver-silver chloride electrode and a platinum elec-
trode in a solution where the concentration of [Cl−] is 0.10 M. The half-cell
reaction at the Ag-AgCl electrode is given by:
AgCl(s) + e−→Ag(s) + Cl−
Calculate the cell potential at 25
°
C when the concentration of [Ag+] is 0.0010
M. (Given: E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the overall cell reaction involving Ag-AgCl and Pt electrodes:
AgCl(s) + e−→Ag(s) + Cl−+e−→Pt(s)
Step 2: Calculate the cell potential using the Nernst equation:
The Nernst equation is given by:
E=E◦−0.0592
nlog [Ag+]
[AgCl][Cl−]
where: - Eis the cell potential - E◦is the standard cell potential - nis
the number of moles of electrons transferred in the cell reaction - [Ag+] is the
concentration of silver ions - [AgCl] is the concentration of silver chloride - [Cl−]
is the concentration of chloride ions
Step 3: Calculate the number of moles of electrons transferred in the cell
reaction (n):
From the balanced cell reaction, n= 1 (1 mole of electrons transferred in
the reaction).
Step 4: Plug in the given values and calculate the cell potential:
E= 0.80V−0.0592
1log 0.0010
1×0.10
E= 0.80V−0.0592 ×log(0.0010/0.10)
E= 0.80V−0.0592 ×log(0.01)
E= 0.80V−0.0592 ×(−2)
E= 0.80V+ 0.1184
E= 0.9184 V
Therefore, the cell potential at 25
°
C when the concentration of [Ag+] is
0.0010 M is 0.9184 V.
13
Question 15
Question
A zinc-silver cell has the following half-reactions:
Anode: Zn(s) →Zn2+(aq) + 2e−
Cathode: Ag+(aq)+e−→Ag(s)
At 25
°
C, the zinc ion concentration is 0.1 M, while the silver ion concentration
is 1.0 ×10−3M. Calculate the cell potential of this zinc-silver cell. (Given:
E◦
Zn2+/Zn =−0.76 V and E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the overall cell reaction and calculate the standard cell potential
(E◦
cell).
Zn(s) + 2Ag+(aq)→Zn2+(aq)+2Ag(s)
E◦
cell =E◦
cathode −E◦
anode E◦
cell = 0.80 V −(−0.76 V) E◦
cell = 1.56 V
Step 2: Calculate the cell potential (Ecell) using the Nernst equation: Ecell =
E◦
cell −0.0592
nlog(Q)n= 2 (number of electrons transferred) Q=[Zn2+ ]
[Ag+]2
Q=0.1
(1.0×10−3)2Q= 1000 Ecell = 1.56 V −0.0592
2log(1000)Ecell = 1.56 V −
0.0592
2×3Ecell = 1.56 V −0.089 Ecell = 1.471 V
Therefore, the cell potential of the zinc-silver cell is 1.471 V.
Question 16
Question
A concentration cell is set up with two half-cells. The first half-cell has a copper
electrode in a 0.10 M Cu2+ solution, and the second half-cell has a copper
electrode in a 0.0010 M Cu2+ solution. Given that the standard reduction
potential of Cu2+/Cu is 0.34 V, calculate the cell potential at 25
°
C using the
Nernst equation.
Solution
Step 1: Write the balanced half-reaction for the reduction of Cu2+ to copper:
Cu2+ + 2e−→Cu
Step 2: Write the Nernst equation:
Ecell =E◦
cell −RT
nF ln(Q)
14
Step 3: Calculate the standard cell potential E◦
cell using the standard reduc-
tion potential of Cu2+/Cu:
E◦
cell =E◦
cathode −E◦
anode = 0 −0.34 = −0.34 V
Step 4: Calculate the reaction quotient Q:
Q=[Cu2+]lower conc
[Cu2+]higher conc
=0.0010
0.10 = 0.010
Step 5: Calculate the cell potential Ecell:
Ecell =−0.34 −(8.314 J/mol ·K)(298 K)
2(96485 C/mol)ln(0.010)
Ecell =−0.34 −2470.072
192970 ln(0.010)
Ecell =−0.34 −(0.0128) ln(0.010)
Step 6: Calculate Ecell:
Ecell ≈ −0.34 −(0.0128)(−4.605)
Ecell ≈ −0.34 + 0.0590
Ecell ≈ −0.2810 V
Therefore, the cell potential at 25
°
C using the Nernst equation is approxi-
mately -0.2810 V.
Question 17
Question
Calculate the cell potential at 25
°
C for a cell in which the following half-reactions
occur:
Fe3+ + 3e−→Fe E◦=−0.036 V
Ag++ e−→Ag E◦= 0.80 V
Given that [Fe3+]=0.15 M and [Ag+]=0.30 M.
Solution
Step 1: Write the balanced cell reaction using the half-reactions provided:
Fe3+ + 3Ag →Fe + 3Ag+
Step 2: Calculate the standard cell potential, E◦
cell, using the standard re-
duction potentials:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −(−0.036 V) = 0.836 V
15
Step 3: Calculate the reaction quotient, Q, using the concentrations pro-
vided:
Q=[Fe][Ag+]3
[Fe3+]=(0.15)(0.30)3
0.30 = 0.027 M3
Step 4: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of moles of electrons transferred in the balanced cell
reaction. In this case, n= 3 because 3 electrons are transferred in the reaction.
Step 5: Substitute the values into the equation to solve for Ecell:
Ecell = 0.836 V −0.0592
3log(0.027) ≈0.832 V
Therefore, the cell potential at 25
°
C for the given cell is approximately 0.832
V.
Question 18
Question
A fuel cell operates at 25
°
C with a hydrogen gas concentration of 1.5×10−2M
on one side and 5.0×10−4M on the other side. Calculate the cell potential
at these conditions using the Nernst equation. Given that the standard cell
potential is 0.44 V.
Solution
Step 1: Write the Nernst equation which relates the cell potential to the con-
centrations of the reactants and products:
E=E◦−0.0592
nlog [H2]0.5
low
[H2]0.5
high !
Step 2: Identify the relevant concentrations and values: - E◦= 0.44 V -
T= 25C= 298 K - n= 2 (since the reaction involves the oxidation of hydrogen
gas) - [H2]high = 1.5×10−2M - [H2]low = 5.0×10−4M
Step 3: Substitute the values into the Nernst equation and solve for the cell
potential:
E= 0.44 −0.0592
2log (5.0×10−4)0.5
(1.5×10−2)0.5
Step 4: Perform the calculations:
E= 0.44 −0.0592
2log 0.02236
0.12247
16
E= 0.44 −0.0296 log(0.1826)
E= 0.44 −0.0296 ×(−0.7395)
E= 0.44 + 0.0219
E= 0.4619 V
Therefore, the cell potential at these conditions is 0.4619 V.
Question 19
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a silver-
silver chloride electrode. The standard reduction potential for the AgCl/Ag
half-cell is 0.222 V. Calculate the potential of the cell when the concentrations
of Ag+and Cl−are both 0.045 M. (Given: E◦
H+/H2= 0 V at 25◦C)
Solution
Step 1: Write the half-reaction equations for the electrodes involved. The half-
reaction for the standard hydrogen electrode is the reduction of hydrogen ions:
2H++ 2e−→H2E◦′ = 0 V
The half-reaction for the silver-silver chloride electrode is the reduction of silver
ions:
Ag++ e−→Ag E◦′ = 0.222 V
Step 2: Write the overall cell reaction. Since the cell reaction is the sum of
the half-reactions, the overall cell reaction is:
2AgCl + 2H+→2Ag + 2HCl
Step 3: Calculate the cell potential using the Nernst equation:
E=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, and Qis the reaction quotient for the
cell.
Step 4: Calculate the reaction quotient Qfor the cell under the given con-
ditions by plugging in the concentrations of Ag+and Cl−:
Q=[Ag+]2
[AgCl]2=(0.045)2
(1)2= 0.002025
17
Step 5: Calculate the cell potential by plugging in the values for E◦,n, and
Qinto the Nernst equation:
E= 0.222V−0.0592
2log 0.002025 = 0.222V+ 0.0296 = 0.2516 V
Therefore, the potential of the cell when the concentrations of Ag+and Cl−
are both 0.045 M is 0.2516 V.
Question 20
Question
A voltaic cell is set up with a silver electrode in a 1.00 M AgNO3solution and
a zinc electrode in a 1.00 M Zn(N O3)2solution. Given that the temperature
is 25
°
C and that the standard reduction potential for the Ag+/Ag half-cell is
+0.80 V and for the Zn2+/Zn half-cell is -0.76 V, calculate the cell potential
at this temperature.
Solution
Step 1: Write the half-reactions for the cell: Ag++e−→Ag Zn2+ + 2e−→Zn
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials: E◦
cell =E◦
cathode −E◦
anode E◦
cell =E◦
Ag+/Ag −E◦
Zn2+ /Zn
E◦
cell = (+0.80 V) −(−0.76 V) E◦
cell = 1.56 V
Step 3: Calculate the actual cell potential using the Nernst equation: Ecell =
E◦
cell −0.0592 V
nlog Q
Kwhere n= total number of moles of electrons trans-
ferred in the balanced redox reaction, Q= reaction quotient, K= equilibrium
constant, 0.0592 V is the value of RT
Fat 25
°
C.
Step 4: Calculate Qfor the given cell: Q=[Ag+]
[Zn2+]Q=1.00
1.00 = 1.00
Step 5: Substitute the values into the Nernst equation: Ecell = 1.56 V −
0.0592 V
2log(1.00) Ecell = 1.56 V
Therefore, the cell potential at 25
°
C is 1.56 V .
Question 21
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell has a silver electrode in a 0.10 M AgNO3 solution, while the other half-
cell has a silver chloride electrode in a 0.0010 M AgNO3 solution. Calculate the
cell potential at 25
°
C. Given E◦
Ag+/AgCl = 0.222 V and E◦
Ag+/Ag = 0.80 V.
18
Solution
Step 1: Write the half-reactions for each half-cell:
Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−
Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)
Step 2: Write the Nernst equation for each half-cell: For the cathode:
Ecathode =E◦
Ag+/Ag −0.0592
1log [Ag+]
[
For the anode:
Eanode =E◦
Ag+/AgCl −0.0592
1log [Ag+]
[AgCl−]
Step 3: Calculate the cell potential by subtracting the anode potential from
the cathode potential:
Ecell =Ecathode −Eanode
Substitute the given values:
Ecathode = 0.80 V −0.0592
1log 0.10
0.0010
Eanode = 0.222 V −0.0592
1log 0.10
0.0010
Ecell =Ec−Ea
Calculate the cell potential.
Question 22
Question
Calculate the cell potential (Ecell) for a galvanic cell where the following reac-
tions occur:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Given that the concentrations of Zn2+ and Cu2+ ions are 0.1 M each, at 25
°
C.
(Given: E◦
Zn2+/Zn =−0.76 V, E◦
Cu2+/Cu = 0.34 V)
19
Solution
Step 1: Write the cell reaction. The overall cell reaction can be constructed by
adding the anode and cathode half-reactions:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
Step 2: Write the cell potential expression using the Nernst equation. The
Nernst equation relates the cell potential (Ecell) to the standard cell potential
(E◦
cell), the reaction quotient (Q), the gas constant (R), the temperature (T),
and the number of electrons transferred (n). The general form of the Nernst
equation for this cell is:
Ecell =E◦
cell −0.0592
nlog(Q)
Step 3: Calculate the cell potential for the given cell. Given: E◦
Zn2+/Zn =
−0.76 V, E◦
Cu2+/Cu = 0.34 V, [Zn2+]=0.1 M, and [Cu2+] = 0.1 M.
The reaction quotient Qcan be calculated using the concentrations of the
ions involved:
Q=[Zn2+][Cu]
[Zn][Cu2+]=(0.1)(0.1)
(1)(1) = 0.01
Substitute the given values and the calculated Qinto the Nernst equation:
Ecell = (0.34 −(−0.76)) −0.0592
2log(0.01)
Ecell = 1.10 −0.0296 × −2
Ecell = 1.10 + 0.0592
Ecell = 1.16 V
Therefore, the cell potential for the given galvanic cell is 1.16 V.
Question 23
Question
A concentration cell is set up at 25
°
C in which both half-cells contain Cr3+/Cr
couple. One half-cell contains a chromium electrode in 0.010 M Cr3+ ion, and
the other contains a chromium electrode in 0.10 M Cr3+ ion. Calculate the cell
potential for this concentration cell. Given that the standard cell potential for
the Cr3+/Cr couple is +0.74 V.
20
Solution
Step 1: Write the half-cell reactions.
The half-cell reactions for the Cr3+/Cr couple are:
Cr3+ + 3e−→Cr (Reduction)
Step 2: Calculate the standard cell potential E◦
cell.
Given: E◦
cell = +0.74 V
Since both half-cells are the same, the standard cell potential is the difference
in concentration: ∆E◦
cell =E◦
cell, cathode −E◦
cell, anode
∆E◦
cell = 0.0592 V log 0.10
0.010
∆E◦
cell = 0.0592 V log(10)
∆E◦
cell = 0.0592 V ×1
∆E◦
cell = 0.0592 V
Step 3: Calculate the cell potential Ecell.
The Nernst equation is given by: Ecell =E◦
cell −0.0592
nlog Q
where nis the number of electrons in the balanced half-cell reaction and Qis
the reaction quotient.
For the given cell, n= 3 (since 3 electrons are involved in the half-reaction)
Q=[Cr3+]cathode
[Cr3+]anode
=0.10
0.010 = 10
Ecell = 0.74 −0.0592
3log(10)
Ecell = 0.74 −0.01973
Ecell = 0.7203 V
Therefore, the cell potential for this concentration cell is 0.7203 V.
Question 24
Question
A galvanic cell consists of a standard hydrogen electrode and a copper electrode.
The standard reduction potential for the copper electrode is Eo
Cu2+/Cu = 0.34 V.
At 25
°
C, the concentration of Cu2+ ions is 0.10 M and the pressure of hydrogen
gas (H2) is 1.0 atm.
Calculate the cell potential for this galvanic cell at these conditions.
Solution
The cell potential (Ecell) can be calculated using the Nernst equation:
Ecell =Eo
cell −0.0592
nlog(Q)
21
Where: Ecell = cell potential Eo
cell = standard cell potential n= number of
electrons transferred in the cell reaction Q= reaction quotient
First, let’s write the balanced cell reaction for this galvanic cell:
Cu2+ + 2e−→Cu
Since 2 electrons are transferred in this reaction, n= 2.
Next, let’s calculate the reaction quotient (Q):
Q=[Cu]
[H+]
2
At standard conditions, [Cu2+]=0.10 M and PH2= 1.0 atm. Since the standard
hydrogen electrode is the reference electrode, [H+] = 1 M. So, Q= (0.10)/(1)2=
0.10.
Now, we can calculate Ecell:
Ecell = 0.34V−0.0592
2log(0.10)
Ecell = 0.34V−0.0296 log(0.10)
Ecell = 0.34V−0.0296 × −1
Ecell = 0.34V+ 0.0296
Ecell = 0.3696 V
Therefore, the cell potential for this galvanic cell at these conditions is 0.3696
V.
Question 25
Question
A voltaic cell consists of a silver-silver chloride electrode (Ag|AgCl) and a plat-
inum electrode in 1.0 M KCl. The standard reduction potential of the silver-
silver chloride electrode is 0.222 V. If the concentration of Cl−in the solution
is 0.10 M, what is the cell potential at 25
°
C?
Solution
Step 1: Write the half-reaction for the silver-silver chloride electrode:
AgCl(s) + e−→Ag(s) + Cl−
Step 2: Write the Nernst equation for the cell potential (E):
E=E◦−0.0592
nlog Q
22
Step 3: Calculate the reaction quotient (Q) for the cell at 25
°
C:
Q=[Ag][Cl−]
[AgCl] =1
[AgCl] =1
Ksp
Step 4: Calculate the cell potential using the Nernst equation:
E= 0.222 V −0.0592
1log 1
Ksp
Step 5: Determine the value of the solubility product constant (Ksp) for
silver chloride (AgCl):
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−] = x(0.1) = x×0.1 = x×10−1
Step 6: Substitute the expression for Ksp into the Nernst equation:
E= 0.222 V −0.0592
1logx×10−1
Step 7: Solve for the value of Eat 25
°
C.
Question 26
Question
A chemist is studying a reaction that involves the transfer of two electrons. The
chemist sets up an electrochemical cell with a standard hydrogen electrode as
the reference electrode. At 25
°
C, the chemist measures that the concentration
of Mn3+ is 0.10 M and the concentration of Mn2+ is 0.0010 M. What is the cell
potential for this reaction? (Standard reduction potential for Mn3+/Mn2+ is
+1.57 V at 25
°
C)
Solution
Step 1: Write the balanced half-reaction for the reduction of Mn3+ to Mn2+:
Mn3+ + 2e−→Mn2+
Step 2: Write the expression for the cell potential, Ecell, using the Nernst
equation:
Ecell =E0
cell −0.0592
nlog [Mn2+]
[Mn3+]2
Step 3: Substitute the given values into the Nernst equation:
Ecell = 1.57 V −0.0592
2log 0.0010
(0.10)2
23
Step 4: Calculate the cell potential:
Ecell = 1.57 V −0.0592
2log(0.0010/0.0100)
Ecell = 1.57 V −0.0592
2log(0.01)
Ecell = 1.57 V −0.0592
2×(−2)
Ecell = 1.57 V + 0.0592 V
Ecell = 1.6292 V
Therefore, the cell potential for this reaction is 1.6292 V.
Question 27
Question
A Daniell cell consists of a copper electrode immersed in a 1.0 M Cu2+ solu-
tion and a zinc electrode immersed in a 1.0 M Zn2+ solution. If the standard
electrode potential of the copper electrode is E◦= 0.34 V and the standard
electrode potential of the zinc electrode is E◦=−0.76 V, calculate the cell
potential (Ecell) at 25
°
C.
Solution
Step 1: Write the overall cell reaction for the Daniell cell:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials of the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+ /Cu −E◦
Zn2+ /Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the equilibrium constant, K, for the cell reaction using
the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
K= 10(nE◦
cell/0.0592)
24
Step 4: Calculate the cell potential, Ecell, at 25
°
C using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
Given that [Cu2+]=1.0 M and [Zn2+] = 1.0 M, we have:
Ecell = 1.10 V −0.0592
2log(1.0/1.0)
Ecell = 1.10 V
Question 28
Question
At 25
°
C, a concentration cell is set up using two half-cells where one half-cell
contains a 0.2 M Fe2+ solution and the other half-cell contains a 0.02 M Fe2+
solution. Calculate the cell potential for this concentration cell.
Solution
Step 1: Write the half-reactions for the oxidation and reduction half-cells:
Oxidation (Anode): Fe →Fe2+ + 2e−
Reduction (Cathode): Fe2+ + 2e−→Fe
Step 2: Write the Nernst equation for the cell potential (Ecell):
Ecell =E0
cell −RT
nF ln Q
K
where E0
cell is the standard cell potential, Ris the gas constant (8.314 J/(mol·K)),
Tis the temperature in Kelvin, nis the number of moles of electrons transferred
in the balanced redox reaction, Fis the Faraday constant (96485 C/mol), Qis
the reaction quotient, and Kis the equilibrium constant.
Step 3: Calculate the standard cell potential (E0
cell) using standard reduction
potentials:
E0
cell =E0
cathode −E0
anode
where E0
cathode = +0.44 V and E0
anode =−0.44 V.
Step 4: Calculate the reaction quotient (Q) for the concentration cell:
Q=[Fe2+]anode
[Fe2+]cathode
=0.2
0.02 = 10
Step 5: Substitute the known values into the Nernst equation to solve for
the cell potential (Ecell):
Ecell = 0.44 V −(8.314 J/(mol*K)) ×(298 K)
2×(96485 C/mol) ln(10)
Step 6: Calculate Ecell to find the cell potential for this concentration cell.
25
Question 29
Question
A voltaic cell is set up with a standard hydrogen electrode (SHE) as the anode
and a copper electrode as the cathode. The standard reduction potential for
the copper electrode is +0.34 V. The cell operates under standard conditions at
25
°
C. Calculate the cell potential if the concentration of Cu2+ in the cathode
compartment is 0.10 M. (Hint: Use the Nernst equation)
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the reduction
of copper ions by hydrogen gas: Cu2+ (aq) + 2e−→Cu (s)
Step 2: Write the half-reaction at the cathode. Cu2+ (aq) + 2e−→Cu (s)
Step 3: Determine the standard cell potential. The standard cell potential,
E◦, can be calculated using the standard reduction potentials: E◦= E◦
cathode -
E◦
anode = +0.34V−0V= +0.34V
Step 4: Calculate the Q value for the cell reaction. Q =
[Cu2+]
1
= 0.10 M
Step 5: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by: E = E◦-
0.0592
nlogQ
K
where E is the cell potential, E◦is the standard cell potential, n is the number
of electrons transferred, Q is the reaction quotient, and K is the equilibrium
constant.
For this cell, n = 2 (from the balanced cell reaction), E = +0.34 V -
0.0592
2log0.10
1
= +0.34 V - 0.0296 V ×1 = +0.34 V - 0.0296 V = +0.3104 V
Therefore, the cell potential is +0.3104 V when the concentration of Cu2+
in the cathode compartment is 0.10 M.
Question 30
Question
Calculate the cell potential for the following reaction at 25
°
C given the concen-
trations of the ions in the half-cells are as follows:
Zn2+(0.10M)∥Cu2+(1.0×10−3M)
26
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard reduction potential for Cu2+/Cu is E◦= 0.34 V and for
Zn2+/Zn is E◦=−0.76 V.
Solution
Step 1: Write the half-reactions for the reduction that occur at each electrode.
The reduction half-reactions are:
Cu2+ + 2e−→Cu E◦= 0.34 V
Zn2+ + 2e−→Zn E◦=−0.76 V
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials for the half-reactions.
The standard cell potential can be calculated using the equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the reaction quotient (Q) for the cell reaction using the
concentrations of the ions given.
The reaction quotient is given by:
Q=[Zn2+]
[Cu2+]=0.10
1.0×10−3= 100
Step 4: Calculate the cell potential (Ecell) using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
Where nis the number of moles of electrons transferred in the cell reaction. In
this case, two moles of electrons are transferred.
Ecell = 1.10 V −0.0592
2log(100)
Ecell = 1.10 V −0.0296 ×2×2
Ecell = 1.10 V −0.236 V
Ecell = 0.864 V
Therefore, the cell potential for the given reaction at 25
°
C is 0.864 V.
27
Question 31
Question
Calculate the cell potential for the following reaction at 25
°
C given that [F e3+] =
0.010M, [F e2+] = 0.050M, and E◦
cell = 0.771 V.
2F e3+(aq)+2e−→2F e2+(aq)
Solution
Step 1: Write the half-cell reactions.
Oxidation: F e2+ →F e3+ +e−E◦=−0.771 V
Reduction: F e3+ +e−→F e2+ E◦= 0.771 V
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [F e3+]2
[F e2+]2
Step 3: Substitute the given values into the Nernst equation.
Ecell = 0.771 −0.0592
2log 0.0102
0.0502
Step 4: Calculate the cell potential.
Ecell = 0.771 −(0.0296) log(0.04)
Ecell = 0.771 −(0.0296)(−1.3979)
Ecell = 0.771 + 0.0414
Ecell = 0.812 V
Question 32
Question
A voltaic cell is constructed with a copper strip dipping into a 1.0 M Cu2+
solution and a silver strip dipping into a 0.10 M Ag+solution. The measured
cell potential at standard conditions is 0.46 V. Calculate the cell potential at
25
°
C when the concentration of Cu2+ is reduced to 0.10 M and the concentration
of Ag+is increased to 1.0 M. Assume the temperature coefficient for the cell
reaction is −2.20 ×10−4V/
°
C.
28
Solution
Step 1: Write the balanced cell reaction and the Nernst equation.
The balanced cell reaction is:
Cu2+(aq) + 2e−→Cu(s)
The Nernst equation is:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential at the non-standard conditions, - E◦is
the standard cell potential (given as 0.46 V), - nis the number of moles of
electrons transferred in the balanced cell reaction (2 in this case), - Qis the
reaction quotient which is the ratio of the concentrations of the products over
the concentrations of the reactants raised to their stoichiometric coefficients.
Step 2: Calculate the cell potential at the new concentrations using the
Nernst equation.
Given concentrations:
[Cu2+] = 0.10 M
[Ag+]=1.0 M
Calculate Q:
Q=[Cu](s)
[Cu2+]=1
0.102= 10
Now substitute the given values into the Nernst equation:
E= 0.46 V −0.0592
2log(10) = 0.46 V −0.0592 ×1=0.46 V −0.0592 = 0.40 V
Therefore, the cell potential at 25
°
C when the concentration of Cu2+ is
reduced to 0.10 M and the concentration of Ag+is increased to 1.0 M is 0.40
V.
Question 33
Question
A voltaic cell consists of a copper electrode in a 1.0 M solution of copper(II)
sulfate and a zinc electrode in a 1.0 M solution of zinc sulfate. Given that the
standard reduction potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =−0.76 V,
calculate the cell potential when the copper electrode is at 298 K and the zinc
electrode is at 400 K.
29
Solution
Step 1: Write the cell reaction and overall cell potential at 298 K. The cell
reaction is as follows:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard cell potential E◦
cell at 298 K can be calculated using the Nernst
equation, which takes the form:
Ecell =E◦
cell −0.0592
nlog Q
where Ecell = cell potential at any temperature, E◦
cell = standard cell potential
at 298 K, n= number of moles of electrons transferred (in this case, 2), Q=
reaction quotient.
The standard cell potential E◦
cell is calculated using the standard reduction
potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn = 0.34V−(−0.76V)=1.10V
Step 2: Determine the reaction quotient. The reaction quotient Qcan be
determined using the concentrations of the species involved in the reaction.
Since both solutions are 1.0 M, the reaction quotient is 1.
Step 3: Calculate the cell potential at 400 K. Substitute the values into the
Nernst equation:
Ecell = 1.10V−0.0592
2log(1) = 1.10V−0=1.10V
Therefore, the cell potential at 400 K is 1.10 V.
Question 34
Question
A concentration cell is set up using two silver electrodes with a Ag+concentra-
tion of 1.0 M in one half-cell and a Ag+concentration of 0.001 M in the other
half-cell. If the standard reduction potential of Ag+reduction to Ag is 0.80 V,
calculate the cell potential at 25
°
C using the Nernst equation.
Solution
Step 1: Write the half-reactions and the balanced overall cell reaction. The
half-reaction for the reduction of Ag+to Ag is: Ag++ e−→Ag The overall
cell reaction is: 2Ag+→2Ag
Step 2: Calculate the cell potential using the standard reduction potential
and the Nernst equation. The standard cell potential, E◦
cell, can be calculated
30
using the standard reduction potentials of the half-reactions: E◦
cell = E◦
cathode -
E◦
anode E◦
cell = 0 V - (-0.80 V) E◦
cell = 0.80 V
Step 3: Apply the Nernst equation to calculate the cell potential at 25
°
C.
The Nernst equation is given by: E = E◦-0.0592
n×log Q
K
Where: E = cell potential E◦= standard cell potential n = number of
electrons transferred in the balanced half-reaction Q = reaction quotient K =
equilibrium constant
For the given cell with two silver half-cells, n = 2.
Step 4: Calculate the reaction quotient (Q) for the cell. Q = [Ag+]0.001/[Ag+]1
Q = 0.001/1.0 Q = 0.001
Step 5: Substitute the values into the Nernst equation and solve for the cell
potential. E = 0.80 V - 0.0592
2×log(0.001) E = 0.80 V - 0.0296 ×(-3) E = 0.80
V + 0.0888 V
E= 0.89 V
Question 35
Question
A voltaic cell consists of a standard hydrogen electrode as the anode and a
copper electrode as the cathode. The initial concentrations of H+ions and
Cu2+ ions are 0.10 M and 0.20 M, respectively. The cell potential is measured
to be 0.78 V at 25
°
C. Calculate the equilibrium constant, K, for the following
redox reaction:
Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)
Given: E◦
cell = 0.34 V, F= 96,500 C/mol, and R= 8.314 J/mol ·K.
Solution
Step 1: Write the half-reactions for the anode and cathode.
At the anode:
2H+(aq)+2e−→H2(g)E◦
anode = 0.00 V
At the cathode:
Cu2+(aq)+2e−→Cu(s)E◦
cathode = 0.34 V
Step 2: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Cu2+]
[H+]2
0.78 V = 0.34 V −0.0592
2log 0.20
0.102
31
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.76 −0.0592
2log 0.10
1.0
Step 5: Calculate the cell potential at 25◦C:
Ecell = 0.76 −0.0592
2log (0.10) = 0.76 −0.0296 ×1=0.7304 V
Therefore, the cell potential at 25◦C is 0.7304 V.
Question 2
Question
Calculate the standard cell potential (E◦
cell) for the following reaction at 25
°
C:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Given: E◦
cell(Zn2+/Zn) = −0.76 V, E◦
cell(Ag+/Ag) = 0.80 V
Solution
Step 1: Write the two half-cell reactions for the reaction:
Zn(s)→Zn2+(aq)+2e−
2Ag+(aq)+2e−→2Ag(s)
Step 2: Calculate the overall cell potential (Ecell) using the Nernst equation:
Ecell =E◦
cell +0.0592
nlog [Zn2+]
[Ag+]2
Step 3: Given that the standard cell potential (E◦
cell) is the sum of the two
half-cell potentials for the reduction reactions, we have:
E◦
cell =E◦
cell(Zn2+/Zn) + E◦
cell(Ag+/Ag)
E◦
cell =−0.76 V + 0.80 V = 0.04 V
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.04 + 0.0592
2log [Zn2+]
[Ag+]2
Ecell = 0.04 + 0.0296 log [Zn2+]
[Ag+]2
2
Step 5: Since the reaction is at equilibrium, Ecell = 0, and the concentrations
of products over reactants is equal to the equilibrium constant:
[Zn2+]
[Ag+]2=Keq = 100= 1
Step 6: Substitute this equilibrium constant into the equation:
0.04 + 0.0296 log 1 = 0
0.04 + 0 = 0
Therefore, the standard cell potential is 0 V .
Question 3
Question
A concentration cell is set up with two Ag/AgCl electrodes. One half-cell con-
tains a 0.01 M solution of AgCl and the other half-cell contains a 0.1 M solution
of AgCl. Calculate the cell potential at 25
°
C. (Given: E◦
cell = 0.46 V)
Solution
Step 1: Write the half-reactions for the cell:
Anode: AgCl(s)→Ag(s) + Cl−(aq)
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Write the Nernst equation for the cell:
Ecell =E◦
cell −0.0592
nlog [reduced form]
[oxidized form]
Step 3: Calculate the reduction potential difference (E0
cell) using the concen-
trations given:
E0
cell = 0.46 V
Step 4: Substitute the given values into the Nernst equation:
Ecell = 0.46 V −0.0592
1log 0.1
0.01
Step 5: Solve for Ecell:
Ecell = 0.46 V −0.0592 log(10) = 0.46 V −0.0592 = 0.4008 V
Therefore, the cell potential at 25
°
C is 0.4008 V.
3
Question 4
Question
Calculate the cell potential at 25
°
C for the following reaction:
Cr2O2−
7(aq) + 14H+(aq) + 6Cl−(aq)→2Cr3+(aq) + 6Cl2(g) + 7H2O(l)
given that the standard reduction potential for Cr2O2−
7(aq) + 14H+(aq) +
6e−→2Cr3+(aq) + 7H2O(l) is +1.33 V.
Solution
Step 1: Write the half-reactions for the redox reaction:
Cr2O2−
7(aq) + 14H+(aq) + 6e−→2Cr3+(aq) + 7H2O(l)
6Cl−(aq)→6Cl2(g) + 12e−
Step 2: Calculate the cell potential at standard conditions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0V−(+1.33V) = −1.33V
Step 3: Calculate the Q value based on the concentrations given in the
reaction.
Step 4: Calculate the cell potential at non-standard conditions using the
Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the balanced redox
equation.
Step 5: Substitute the values into the Nernst equation and calculate the cell
potential at 25
°
C.
Question 5
Question
A voltaic cell is constructed with a silver electrode and a zinc electrode. The half-
reaction for the reduction of silver is Ag++e−→Ag with an electrode potential
of +0.80 V, while the half-reaction for the reduction of zinc is Zn2+ + 2e−→Zn
with an electrode potential of -0.76 V. Calculate the cell potential when the
concentration of Ag+is 0.50 M and the concentration of Zn2+ is 1.00 M.
4
Solution
Step 1: Write the cell reaction: The overall reaction for the voltaic cell is the
sum of the reduction half-reactions for silver and zinc:
Zn2+ + 2Ag →Zn + 2Ag+
Step 2: Calculate the standard cell potential, E
°
: The standard cell potential,
E
°
, is given by the difference in standard reduction potentials for the two half-
reactions:
Ecell =Ecathode −Eanode
Ecell = 0.80 V −(−0.76 V) = 1.56 V
Step 3: Write the Nernst equation: The Nernst equation relates the cell
potential under non-standard conditions to the standard cell potential:
Ecell =Ecell −0.0592
nlog [Ag+]2
[Zn2+]
where n is the number of moles of electrons transferred in the cell reaction.
Step 4: Calculate the cell potential under the given conditions: Since 2 moles
of electrons are transferred in the cell reaction, n = 2.
Ecell = 1.56 V −0.0592
2log (0.50 M)2
(1.00 M)
Ecell = 1.56 V −0.0592
2log(0.25)
Ecell = 1.56 V −0.0592
2(−0.602)
Ecell = 1.56 V + 0.018 = 1.578 V
Therefore, the cell potential when the concentration of Ag+is 0.50 M and
the concentration of Zn2+ is 1.00 M is 1.578 V.
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) as the anode and
a nickel metal electrode as the cathode. Given that the cell potential is 0.44 V
at standard conditions and the concentration of Ni2+ ions in the cathode com-
partment is 0.1 M, calculate the cell potential of the cell when the concentration
of Ni2+ ions is 0.01 M. Assume the temperature is constant.
5
Solution
Step 1: Write the half-reactions for the cell. The overall cell reaction is:
Ni2+(aq)+2e−→Ni(s)
The standard reduction potential of the nickel half-reaction is −0.25 V(given).
Step 2: Calculate the standard cell potential, E◦
cell. Using the Nernst equa-
tion:
Ecell =E◦
cell −0.0592
nlog [Ni2+]cathode
[Ni2+]◦
cathode
Given that E◦
cell = 0.44 V,n= 2, [Ni2+]◦
cathode = 1 M, [Ni2+]cathode = 0.1M,
and R= 0.0592 Vat room temperature, we can substitute these values to find
Ecell.
Step 3: Calculate the cell potential, Ecell, at the new concentration. Substi-
tute [Ni2+]cathode = 0.01 Minto the Nernst equation to calculate the new cell
potential.
Ecell = 0.44 V−0.0592
2log 0.01
1
Ecell = 0.44V−0.0296 log(0.01)
Ecell = 0.44V−0.0296 ×(−2)
Ecell = 0.44V+ 0.0592
Ecell = 0.4992 V
Therefore, the cell potential of the galvanic cell when [Ni2+]cathode is 0.01 M
is 0.4992 V.
Question 7
Question
A lead-acid battery has an electrolyte solution consisting of PbSO4and H2SO4
with concentrations of Pb2+= 0.050 M and [H2SO4]=1.0 M. Calculate the
cell potential at 25◦C for the reaction:
Pb(s) + PbSO4(s) + 2 H2SO4(aq) −−→ 2 PbSO4(s) + 2 H2O(l)
Given: - Standard reduction potential of PbSO4(s)+2 e−−−→ Pb(s)+SO42−(aq)
is −0.36 V. - Standard reduction potential of H3O+(aq) + 2 e−−−→ 1
2H2(g) +
H2O(l) at 25◦C is −0.036 V. - Faraday’s constant F= 96485 C/mol.
6
Solution
Step 1: Write the overall cell reaction.
Pb(s) + PbSO4(s) + 2 H2SO4(aq) −−→ 2 PbSO4(s) + 2 H2O(l)
Step 2: Write the half-reactions. Cathode: PbSO4(s) + 2 e−−−→ Pb(s) +
SO42−(aq) Anode: 2 H+(aq) + 2 e−−−→ H2(g)
Step 3: Calculate the standard cell potential (E◦
cell) using the Nernst equa-
tion:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =−0.36 V −(−0.036 V) = −0.324 V
Step 4: Calculate the cell potential at non-standard conditions using the
Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
where Qis the reaction quotient and Kis the equilibrium constant, and nis
the number of moles of electrons transferred in the reaction.
Step 5: Calculate the reaction quotient (Q):
Q=1
Pb2+2H+2=1
(0.050)2(1.0)2= 400
Step 6: Calculate the cell potential (Ecell) at 25◦C:
Ecell =−0.324 V −0.0592
2log(400)
Ecell =−0.324 V −0.0296 log(400)
Ecell =−0.324 V −0.0296 ×2.602 = −0.402 V
Therefore, the cell potential at 25◦C for the given reaction is −0.402 V.
Question 8
Question
Use the Nernst equation to calculate the cell potential (Ecell) for the following
reaction at 25
°
C:
2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Given that [Cu2+] = 0.020 M, [Ag+]=0.100 M, and the standard reduction
potentials are E◦
Ag+/Ag = 0.80 Vand E◦
Cu2+ /Cu = 0.34 V.
7
Solution
Step 1: Write the half-reactions and identify the oxidation and reduction half-
reactions. The half-reactions for this cell are: Oxidation: Cu(s)→Cu2+(aq) +
2e−Reduction: 2Ag+(aq)+2e−→2Ag(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions: 2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)
Step 3: Determine the cell potential at standard condition. The cell potential
at standard condition is calculated using the formula: E◦
cell =E◦
reduction, cathode−
E◦
oxidation, anode E◦
cell =E◦
Ag+/Ag −E◦
Cu2+ /Cu E◦
cell = 0.80 V−0.34 V E◦
cell =
0.46 V
Step 4: Apply the Nernst equation to calculate the cell potential under
non-standard condition. The Nernst equation is given by: Ecell =E◦
cell −
0.0592
nlog [Cu2+ ]2
[Ag+]2where nis the number of electrons transferred in the cell re-
action.
For the given reaction, n= 2: Ecell = 0.46 V−0.0592
2log 0.0202
0.1002Ecell =
0.46 V−0.0296 log 0.0004
0.01 Ecell = 0.46 V−0.0296 log 0.04 Ecell = 0.46 V−0.0296×
(−1.3979) Ecell = 0.46 V+ 0.0414 Ecell ≈0.5014 V
Therefore, the cell potential for the given reaction under the given conditions
is approximately 0.5014 V.
Question 9
Question
A galvanic cell is set up with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Ag+(aq) + e−−−→ Ag(s)
If the standard reduction potential for the Ag+(aq) + e−→Ag(s) half-
reaction is +0.80 V and the standard reduction potential for the Zn2+(aq) +
2e−→Zn(s) half-reaction is −0.76 V, calculate the cell potential when the
concentration of Zn2+ is 0.10 M and the concentration of Ag+is 2.0 M. (Assume
both solutions are at 25
°
C.)
Solution
Step 1: Calculate the cell potential using the Nernst equation:
The Nernst equation for the cell potential (Ecell) is given by:
Ecell =E◦
cell −0.0592
nlog [oxidized form]
[reduced form]
where: - E◦
cell is the standard cell potential, - nis the number of moles
of electrons transferred in the balanced cell reaction, - [oxidized form] is the
8
concentration of the oxidized form, - [reduced form] is the concentration of the
reduced form.
Step 2: Identify the number of moles of electrons transferred in the cell
reaction: From the given half-reactions, we can see that 2 moles of electrons are
transferred.
Step 3: Calculate the cell potential at standard conditions: The standard
cell potential (E◦
cell) can be 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
Step 4: Calculate the cell potential under given conditions:
Ecell = +1.56 V −0.0592
2log 0.10
2.0
Ecell = +1.56 V −0.0296 log 0.05
Ecell = +1.56 V −0.0296 ×(−1.301) = +1.6 V
Therefore, the cell potential when the concentration of Zn2+ is 0.10 M and
the concentration of Ag+is 2.0 M is +1.6 V.
Question 10
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a silver-silver chloride electrode (AgCl/Ag). The concentration of [Ag+] in the
AgCl/Ag electrode is 0.10 M. If the cell potential is 0.69 V at 25◦C, calculate
the [H+] concentration in the SHE when the cell is at equilibrium.
Given: E◦
cell = 0.80 V.
Solution
Step 1: Write the half-reactions for the cell.
The overall reaction for the cell is the combination of the reduction half-reaction
at the AgCl electrode and the oxidation half-reaction at the standard hydrogen
electrode.
Reduction half-reaction: AgCl(s) + e−→Ag(s) + Cl−(aq)
Oxidation half-reaction: 2H+(aq)+2e−→H2(g)
Step 2: Write the cell equation.
The cell equation can be obtained by adding the two half-reactions together.
The cell potential is given as 0.69 V.
0.69 = E◦
cell −0.0592
2log [H+]2
1
9
Step 3: Solve for [H+].
Plug in the given values (E◦
cell = 0.80 V) and solve for [H+].
0.69 = 0.80−0.0296 log[H+]⇒log[H+] = 0.80 −0.69
0.0296 ⇒log[H+] = 3.72 ⇒[H+]=5.10×10−4M
Therefore, the [H+] concentration in the standard hydrogen electrode when
the cell is at equilibrium is 5.10 ×10−4M.
Question 11
Question
A galvanic cell with a copper electrode and a silver electrode is set up with
copper(II) sulfate and silver nitrate solutions. The initial concentrations are
Cu2+= 0.10 M,Ag+= 0.0010 M. The reduction potentials are E◦(Cu2+/Cu) =
0.34 Vand E◦(Ag+/Ag) = 0.80 V. Determine the cell potential when the con-
centration of Cu2+is decreased to 0.0050 Mand the concentration of Ag+
is increased to 0.10 M.
Solution
Step 1: Write the half-reactions and Nernst equation.
The half-reactions for the two electrodes are:
At Cu electrode: Cu2+ + 2e−→Cu
At Ag electrode: Ag++e−→Ag
The Nernst equation relates the cell potential Eto the standard cell potential
E◦, the reaction quotient Q, and the temperature T:
E=E◦−0.0592
nlog Q
Where nis the number of moles of electrons transferred and Qis the reaction
quotient, given by:
Q=[products]coefficient of products
[reactants]coefficient of reactants
Step 2: Calculate Qfor the initial concentration conditions.
For the initial concentrations, Q=[Cu2+]1[Ag+]1
[Cu]1[Ag]2. Substitute the given concen-
trations into Q:Q=(0.10)(0.0010)
1(1) = 0.00010.
Step 3: Calculate the cell potential Einitial.
Given E◦(Cu2+/Cu) = 0.34 Vand E◦(Ag+/Ag) = 0.80 V, the overall cell reac-
tion is: Cu2+ + 2Ag →Cu + 2Ag+.
10
The standard cell potential E◦
cell can be calculated using the standard reduc-
tion potentials: E◦
cell =E◦(Cu2+/Cu) −E◦(Ag+/Ag) = 0.34 −0.80 = −0.46 V.
Now, plug the values into the Nernst equation: Einitial =−0.46−0.0592
2log(0.00010) ≈
−0.46 V.
Step 4: Calculate Qfor the final concentration conditions.
For the final concentrations, Q=[Cu2+]1[Ag+]2
[Cu]1[Ag]2. Substitute the given concentra-
tions into Q:Q=(0.0050)(0.10)2
1(1)2= 0.00050.
Step 5: Calculate the cell potential Efinal.
Using the same Nernst equation and overall cell reaction, we have: Efinal =
−0.46 −0.0592
2log(0.00050) ≈ −0.46 −0.048 = −0.508 V.
Therefore, the cell potential when [Cu2+]=0.0050
Question 12
Question
Calculate the cell potential at 25
°
C for a concentration cell with two half-cells,
each containing a zinc electrode. One half-cell has a zinc ion concentration of
0.10 M, while the other half-cell has a zinc ion concentration of 0.0010 M. The
standard reduction potential for the zinc electrode is -0.76 V.
Solution
Step 1: Write the balanced half-reaction for the cell. The half-reaction for the
reduction of zinc is:
Zn2+(aq)+2e−→Zn(s)
Step 2: Determine the cell reaction. In this concentration cell, each half-cell
contains the same substance (zinc). Therefore, the cell reaction consists of both
half-reactions:
Zn2+
0.10M(aq)+2e−→Zn0.10M(s)
Zn2+
0.0010M(aq)+2e−→Zn0.0010M(s)
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is:
E=Eo−0.0592
nlog [Zn2+]2
[Zn2+]1
where: - E= cell potential - Eo= standard cell potential - n= number of
electrons transferred (2 in this case) - [Zn2+]2= zinc ion concentration in the
second half-cell - [Zn2+]1= zinc ion concentration in the first half-cell
Plugging in the values:
E=−0.76 V −0.0592
2log 0.0010 M
0.10 M
E=−0.76 V −0.0296 log(0.01)
11
E=−0.76 V −0.0296 ×(−2)
E=−0.76 V + 0.0592
E=−0.7008 V
Therefore, the cell potential at 25
°
C for this concentration cell is -0.7008 V.
Question 13
Question
Given the Nernst equation: E=E◦−0.0592
nlog [Cu2+ ]
[Cu], calculate the cell
potential (E) for the following reaction at 25
°
C:
Cu2+(aq)+2e−→Cu(s)
Given E◦= 0.34 V, [Cu2+] = 0.10 M, [Cu] = 1.0 M. (Assume n= 2)
Solution
Step 1: Identify the given values: Given values: E◦= 0.34 V
[Cu2+] = 0.10 M
[Cu]=1.0 M
n= 2
We need to find E.
Step 2: Substitute the values into the Nernst equation:
E=E◦−0.0592
nlog [Cu2+]
[Cu]
E= 0.34 −0.0592
2log 0.10
1.0
Step 3: Calculate the natural logarithm:
E= 0.34 −0.0296 log (0.10)
E= 0.34 −0.0296 ×(−1)
E= 0.34 + 0.0296
E= 0.3696 V
Therefore, the cell potential (E) for the given reaction at 25
°
C is 0.3696 V.
12
Question 14
Question
A galvanic cell consists of a silver-silver chloride electrode and a platinum elec-
trode in a solution where the concentration of [Cl−] is 0.10 M. The half-cell
reaction at the Ag-AgCl electrode is given by:
AgCl(s) + e−→Ag(s) + Cl−
Calculate the cell potential at 25
°
C when the concentration of [Ag+] is 0.0010
M. (Given: E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the overall cell reaction involving Ag-AgCl and Pt electrodes:
AgCl(s) + e−→Ag(s) + Cl−+e−→Pt(s)
Step 2: Calculate the cell potential using the Nernst equation:
The Nernst equation is given by:
E=E◦−0.0592
nlog [Ag+]
[AgCl][Cl−]
where: - Eis the cell potential - E◦is the standard cell potential - nis
the number of moles of electrons transferred in the cell reaction - [Ag+] is the
concentration of silver ions - [AgCl] is the concentration of silver chloride - [Cl−]
is the concentration of chloride ions
Step 3: Calculate the number of moles of electrons transferred in the cell
reaction (n):
From the balanced cell reaction, n= 1 (1 mole of electrons transferred in
the reaction).
Step 4: Plug in the given values and calculate the cell potential:
E= 0.80V−0.0592
1log 0.0010
1×0.10
E= 0.80V−0.0592 ×log(0.0010/0.10)
E= 0.80V−0.0592 ×log(0.01)
E= 0.80V−0.0592 ×(−2)
E= 0.80V+ 0.1184
E= 0.9184 V
Therefore, the cell potential at 25
°
C when the concentration of [Ag+] is
0.0010 M is 0.9184 V.
13
Question 15
Question
A zinc-silver cell has the following half-reactions:
Anode: Zn(s) →Zn2+(aq) + 2e−
Cathode: Ag+(aq)+e−→Ag(s)
At 25
°
C, the zinc ion concentration is 0.1 M, while the silver ion concentration
is 1.0 ×10−3M. Calculate the cell potential of this zinc-silver cell. (Given:
E◦
Zn2+/Zn =−0.76 V and E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the overall cell reaction and calculate the standard cell potential
(E◦
cell).
Zn(s) + 2Ag+(aq)→Zn2+(aq)+2Ag(s)
E◦
cell =E◦
cathode −E◦
anode E◦
cell = 0.80 V −(−0.76 V) E◦
cell = 1.56 V
Step 2: Calculate the cell potential (Ecell) using the Nernst equation: Ecell =
E◦
cell −0.0592
nlog(Q)n= 2 (number of electrons transferred) Q=[Zn2+ ]
[Ag+]2
Q=0.1
(1.0×10−3)2Q= 1000 Ecell = 1.56 V −0.0592
2log(1000)Ecell = 1.56 V −
0.0592
2×3Ecell = 1.56 V −0.089 Ecell = 1.471 V
Therefore, the cell potential of the zinc-silver cell is 1.471 V.
Question 16
Question
A concentration cell is set up with two half-cells. The first half-cell has a copper
electrode in a 0.10 M Cu2+ solution, and the second half-cell has a copper
electrode in a 0.0010 M Cu2+ solution. Given that the standard reduction
potential of Cu2+/Cu is 0.34 V, calculate the cell potential at 25
°
C using the
Nernst equation.
Solution
Step 1: Write the balanced half-reaction for the reduction of Cu2+ to copper:
Cu2+ + 2e−→Cu
Step 2: Write the Nernst equation:
Ecell =E◦
cell −RT
nF ln(Q)
14
Step 3: Calculate the standard cell potential E◦
cell using the standard reduc-
tion potential of Cu2+/Cu:
E◦
cell =E◦
cathode −E◦
anode = 0 −0.34 = −0.34 V
Step 4: Calculate the reaction quotient Q:
Q=[Cu2+]lower conc
[Cu2+]higher conc
=0.0010
0.10 = 0.010
Step 5: Calculate the cell potential Ecell:
Ecell =−0.34 −(8.314 J/mol ·K)(298 K)
2(96485 C/mol)ln(0.010)
Ecell =−0.34 −2470.072
192970 ln(0.010)
Ecell =−0.34 −(0.0128) ln(0.010)
Step 6: Calculate Ecell:
Ecell ≈ −0.34 −(0.0128)(−4.605)
Ecell ≈ −0.34 + 0.0590
Ecell ≈ −0.2810 V
Therefore, the cell potential at 25
°
C using the Nernst equation is approxi-
mately -0.2810 V.
Question 17
Question
Calculate the cell potential at 25
°
C for a cell in which the following half-reactions
occur:
Fe3+ + 3e−→Fe E◦=−0.036 V
Ag++ e−→Ag E◦= 0.80 V
Given that [Fe3+]=0.15 M and [Ag+]=0.30 M.
Solution
Step 1: Write the balanced cell reaction using the half-reactions provided:
Fe3+ + 3Ag →Fe + 3Ag+
Step 2: Calculate the standard cell potential, E◦
cell, using the standard re-
duction potentials:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −(−0.036 V) = 0.836 V
15
Step 3: Calculate the reaction quotient, Q, using the concentrations pro-
vided:
Q=[Fe][Ag+]3
[Fe3+]=(0.15)(0.30)3
0.30 = 0.027 M3
Step 4: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of moles of electrons transferred in the balanced cell
reaction. In this case, n= 3 because 3 electrons are transferred in the reaction.
Step 5: Substitute the values into the equation to solve for Ecell:
Ecell = 0.836 V −0.0592
3log(0.027) ≈0.832 V
Therefore, the cell potential at 25
°
C for the given cell is approximately 0.832
V.
Question 18
Question
A fuel cell operates at 25
°
C with a hydrogen gas concentration of 1.5×10−2M
on one side and 5.0×10−4M on the other side. Calculate the cell potential
at these conditions using the Nernst equation. Given that the standard cell
potential is 0.44 V.
Solution
Step 1: Write the Nernst equation which relates the cell potential to the con-
centrations of the reactants and products:
E=E◦−0.0592
nlog [H2]0.5
low
[H2]0.5
high !
Step 2: Identify the relevant concentrations and values: - E◦= 0.44 V -
T= 25C= 298 K - n= 2 (since the reaction involves the oxidation of hydrogen
gas) - [H2]high = 1.5×10−2M - [H2]low = 5.0×10−4M
Step 3: Substitute the values into the Nernst equation and solve for the cell
potential:
E= 0.44 −0.0592
2log (5.0×10−4)0.5
(1.5×10−2)0.5
Step 4: Perform the calculations:
E= 0.44 −0.0592
2log 0.02236
0.12247
16
E= 0.44 −0.0296 log(0.1826)
E= 0.44 −0.0296 ×(−0.7395)
E= 0.44 + 0.0219
E= 0.4619 V
Therefore, the cell potential at these conditions is 0.4619 V.
Question 19
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a silver-
silver chloride electrode. The standard reduction potential for the AgCl/Ag
half-cell is 0.222 V. Calculate the potential of the cell when the concentrations
of Ag+and Cl−are both 0.045 M. (Given: E◦
H+/H2= 0 V at 25◦C)
Solution
Step 1: Write the half-reaction equations for the electrodes involved. The half-
reaction for the standard hydrogen electrode is the reduction of hydrogen ions:
2H++ 2e−→H2E◦′ = 0 V
The half-reaction for the silver-silver chloride electrode is the reduction of silver
ions:
Ag++ e−→Ag E◦′ = 0.222 V
Step 2: Write the overall cell reaction. Since the cell reaction is the sum of
the half-reactions, the overall cell reaction is:
2AgCl + 2H+→2Ag + 2HCl
Step 3: Calculate the cell potential using the Nernst equation:
E=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, and Qis the reaction quotient for the
cell.
Step 4: Calculate the reaction quotient Qfor the cell under the given con-
ditions by plugging in the concentrations of Ag+and Cl−:
Q=[Ag+]2
[AgCl]2=(0.045)2
(1)2= 0.002025
17
Step 5: Calculate the cell potential by plugging in the values for E◦,n, and
Qinto the Nernst equation:
E= 0.222V−0.0592
2log 0.002025 = 0.222V+ 0.0296 = 0.2516 V
Therefore, the potential of the cell when the concentrations of Ag+and Cl−
are both 0.045 M is 0.2516 V.
Question 20
Question
A voltaic cell is set up with a silver electrode in a 1.00 M AgNO3solution and
a zinc electrode in a 1.00 M Zn(N O3)2solution. Given that the temperature
is 25
°
C and that the standard reduction potential for the Ag+/Ag half-cell is
+0.80 V and for the Zn2+/Zn half-cell is -0.76 V, calculate the cell potential
at this temperature.
Solution
Step 1: Write the half-reactions for the cell: Ag++e−→Ag Zn2+ + 2e−→Zn
Step 2: Determine the standard cell potential (E◦
cell) using the standard
reduction potentials: E◦
cell =E◦
cathode −E◦
anode E◦
cell =E◦
Ag+/Ag −E◦
Zn2+ /Zn
E◦
cell = (+0.80 V) −(−0.76 V) E◦
cell = 1.56 V
Step 3: Calculate the actual cell potential using the Nernst equation: Ecell =
E◦
cell −0.0592 V
nlog Q
Kwhere n= total number of moles of electrons trans-
ferred in the balanced redox reaction, Q= reaction quotient, K= equilibrium
constant, 0.0592 V is the value of RT
Fat 25
°
C.
Step 4: Calculate Qfor the given cell: Q=[Ag+]
[Zn2+]Q=1.00
1.00 = 1.00
Step 5: Substitute the values into the Nernst equation: Ecell = 1.56 V −
0.0592 V
2log(1.00) Ecell = 1.56 V
Therefore, the cell potential at 25
°
C is 1.56 V .
Question 21
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell has a silver electrode in a 0.10 M AgNO3 solution, while the other half-
cell has a silver chloride electrode in a 0.0010 M AgNO3 solution. Calculate the
cell potential at 25
°
C. Given E◦
Ag+/AgCl = 0.222 V and E◦
Ag+/Ag = 0.80 V.
18
Solution
Step 1: Write the half-reactions for each half-cell:
Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−Cathode: AgCl(s) + e−→Ag(s) + Cl−
Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)Anode: Ag+(aq) + e−→Ag(s)
Step 2: Write the Nernst equation for each half-cell: For the cathode:
Ecathode =E◦
Ag+/Ag −0.0592
1log [Ag+]
[
For the anode:
Eanode =E◦
Ag+/AgCl −0.0592
1log [Ag+]
[AgCl−]
Step 3: Calculate the cell potential by subtracting the anode potential from
the cathode potential:
Ecell =Ecathode −Eanode
Substitute the given values:
Ecathode = 0.80 V −0.0592
1log 0.10
0.0010
Eanode = 0.222 V −0.0592
1log 0.10
0.0010
Ecell =Ec−Ea
Calculate the cell potential.
Question 22
Question
Calculate the cell potential (Ecell) for a galvanic cell where the following reac-
tions occur:
Anode: Zn(s)→Zn2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Given that the concentrations of Zn2+ and Cu2+ ions are 0.1 M each, at 25
°
C.
(Given: E◦
Zn2+/Zn =−0.76 V, E◦
Cu2+/Cu = 0.34 V)
19
Solution
Step 1: Write the cell reaction. The overall cell reaction can be constructed by
adding the anode and cathode half-reactions:
Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)
Step 2: Write the cell potential expression using the Nernst equation. The
Nernst equation relates the cell potential (Ecell) to the standard cell potential
(E◦
cell), the reaction quotient (Q), the gas constant (R), the temperature (T),
and the number of electrons transferred (n). The general form of the Nernst
equation for this cell is:
Ecell =E◦
cell −0.0592
nlog(Q)
Step 3: Calculate the cell potential for the given cell. Given: E◦
Zn2+/Zn =
−0.76 V, E◦
Cu2+/Cu = 0.34 V, [Zn2+]=0.1 M, and [Cu2+] = 0.1 M.
The reaction quotient Qcan be calculated using the concentrations of the
ions involved:
Q=[Zn2+][Cu]
[Zn][Cu2+]=(0.1)(0.1)
(1)(1) = 0.01
Substitute the given values and the calculated Qinto the Nernst equation:
Ecell = (0.34 −(−0.76)) −0.0592
2log(0.01)
Ecell = 1.10 −0.0296 × −2
Ecell = 1.10 + 0.0592
Ecell = 1.16 V
Therefore, the cell potential for the given galvanic cell is 1.16 V.
Question 23
Question
A concentration cell is set up at 25
°
C in which both half-cells contain Cr3+/Cr
couple. One half-cell contains a chromium electrode in 0.010 M Cr3+ ion, and
the other contains a chromium electrode in 0.10 M Cr3+ ion. Calculate the cell
potential for this concentration cell. Given that the standard cell potential for
the Cr3+/Cr couple is +0.74 V.
20
Solution
Step 1: Write the half-cell reactions.
The half-cell reactions for the Cr3+/Cr couple are:
Cr3+ + 3e−→Cr (Reduction)
Step 2: Calculate the standard cell potential E◦
cell.
Given: E◦
cell = +0.74 V
Since both half-cells are the same, the standard cell potential is the difference
in concentration: ∆E◦
cell =E◦
cell, cathode −E◦
cell, anode
∆E◦
cell = 0.0592 V log 0.10
0.010
∆E◦
cell = 0.0592 V log(10)
∆E◦
cell = 0.0592 V ×1
∆E◦
cell = 0.0592 V
Step 3: Calculate the cell potential Ecell.
The Nernst equation is given by: Ecell =E◦
cell −0.0592
nlog Q
where nis the number of electrons in the balanced half-cell reaction and Qis
the reaction quotient.
For the given cell, n= 3 (since 3 electrons are involved in the half-reaction)
Q=[Cr3+]cathode
[Cr3+]anode
=0.10
0.010 = 10
Ecell = 0.74 −0.0592
3log(10)
Ecell = 0.74 −0.01973
Ecell = 0.7203 V
Therefore, the cell potential for this concentration cell is 0.7203 V.
Question 24
Question
A galvanic cell consists of a standard hydrogen electrode and a copper electrode.
The standard reduction potential for the copper electrode is Eo
Cu2+/Cu = 0.34 V.
At 25
°
C, the concentration of Cu2+ ions is 0.10 M and the pressure of hydrogen
gas (H2) is 1.0 atm.
Calculate the cell potential for this galvanic cell at these conditions.
Solution
The cell potential (Ecell) can be calculated using the Nernst equation:
Ecell =Eo
cell −0.0592
nlog(Q)
21
Where: Ecell = cell potential Eo
cell = standard cell potential n= number of
electrons transferred in the cell reaction Q= reaction quotient
First, let’s write the balanced cell reaction for this galvanic cell:
Cu2+ + 2e−→Cu
Since 2 electrons are transferred in this reaction, n= 2.
Next, let’s calculate the reaction quotient (Q):
Q=[Cu]
[H+]
2
At standard conditions, [Cu2+]=0.10 M and PH2= 1.0 atm. Since the standard
hydrogen electrode is the reference electrode, [H+] = 1 M. So, Q= (0.10)/(1)2=
0.10.
Now, we can calculate Ecell:
Ecell = 0.34V−0.0592
2log(0.10)
Ecell = 0.34V−0.0296 log(0.10)
Ecell = 0.34V−0.0296 × −1
Ecell = 0.34V+ 0.0296
Ecell = 0.3696 V
Therefore, the cell potential for this galvanic cell at these conditions is 0.3696
V.
Question 25
Question
A voltaic cell consists of a silver-silver chloride electrode (Ag|AgCl) and a plat-
inum electrode in 1.0 M KCl. The standard reduction potential of the silver-
silver chloride electrode is 0.222 V. If the concentration of Cl−in the solution
is 0.10 M, what is the cell potential at 25
°
C?
Solution
Step 1: Write the half-reaction for the silver-silver chloride electrode:
AgCl(s) + e−→Ag(s) + Cl−
Step 2: Write the Nernst equation for the cell potential (E):
E=E◦−0.0592
nlog Q
22
Step 3: Calculate the reaction quotient (Q) for the cell at 25
°
C:
Q=[Ag][Cl−]
[AgCl] =1
[AgCl] =1
Ksp
Step 4: Calculate the cell potential using the Nernst equation:
E= 0.222 V −0.0592
1log 1
Ksp
Step 5: Determine the value of the solubility product constant (Ksp) for
silver chloride (AgCl):
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−] = x(0.1) = x×0.1 = x×10−1
Step 6: Substitute the expression for Ksp into the Nernst equation:
E= 0.222 V −0.0592
1logx×10−1
Step 7: Solve for the value of Eat 25
°
C.
Question 26
Question
A chemist is studying a reaction that involves the transfer of two electrons. The
chemist sets up an electrochemical cell with a standard hydrogen electrode as
the reference electrode. At 25
°
C, the chemist measures that the concentration
of Mn3+ is 0.10 M and the concentration of Mn2+ is 0.0010 M. What is the cell
potential for this reaction? (Standard reduction potential for Mn3+/Mn2+ is
+1.57 V at 25
°
C)
Solution
Step 1: Write the balanced half-reaction for the reduction of Mn3+ to Mn2+:
Mn3+ + 2e−→Mn2+
Step 2: Write the expression for the cell potential, Ecell, using the Nernst
equation:
Ecell =E0
cell −0.0592
nlog [Mn2+]
[Mn3+]2
Step 3: Substitute the given values into the Nernst equation:
Ecell = 1.57 V −0.0592
2log 0.0010
(0.10)2
23
Step 4: Calculate the cell potential:
Ecell = 1.57 V −0.0592
2log(0.0010/0.0100)
Ecell = 1.57 V −0.0592
2log(0.01)
Ecell = 1.57 V −0.0592
2×(−2)
Ecell = 1.57 V + 0.0592 V
Ecell = 1.6292 V
Therefore, the cell potential for this reaction is 1.6292 V.
Question 27
Question
A Daniell cell consists of a copper electrode immersed in a 1.0 M Cu2+ solu-
tion and a zinc electrode immersed in a 1.0 M Zn2+ solution. If the standard
electrode potential of the copper electrode is E◦= 0.34 V and the standard
electrode potential of the zinc electrode is E◦=−0.76 V, calculate the cell
potential (Ecell) at 25
°
C.
Solution
Step 1: Write the overall cell reaction for the Daniell cell:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials of the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+ /Cu −E◦
Zn2+ /Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the equilibrium constant, K, for the cell reaction using
the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
K= 10(nE◦
cell/0.0592)
24
Step 4: Calculate the cell potential, Ecell, at 25
°
C using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
Given that [Cu2+]=1.0 M and [Zn2+] = 1.0 M, we have:
Ecell = 1.10 V −0.0592
2log(1.0/1.0)
Ecell = 1.10 V
Question 28
Question
At 25
°
C, a concentration cell is set up using two half-cells where one half-cell
contains a 0.2 M Fe2+ solution and the other half-cell contains a 0.02 M Fe2+
solution. Calculate the cell potential for this concentration cell.
Solution
Step 1: Write the half-reactions for the oxidation and reduction half-cells:
Oxidation (Anode): Fe →Fe2+ + 2e−
Reduction (Cathode): Fe2+ + 2e−→Fe
Step 2: Write the Nernst equation for the cell potential (Ecell):
Ecell =E0
cell −RT
nF ln Q
K
where E0
cell is the standard cell potential, Ris the gas constant (8.314 J/(mol·K)),
Tis the temperature in Kelvin, nis the number of moles of electrons transferred
in the balanced redox reaction, Fis the Faraday constant (96485 C/mol), Qis
the reaction quotient, and Kis the equilibrium constant.
Step 3: Calculate the standard cell potential (E0
cell) using standard reduction
potentials:
E0
cell =E0
cathode −E0
anode
where E0
cathode = +0.44 V and E0
anode =−0.44 V.
Step 4: Calculate the reaction quotient (Q) for the concentration cell:
Q=[Fe2+]anode
[Fe2+]cathode
=0.2
0.02 = 10
Step 5: Substitute the known values into the Nernst equation to solve for
the cell potential (Ecell):
Ecell = 0.44 V −(8.314 J/(mol*K)) ×(298 K)
2×(96485 C/mol) ln(10)
Step 6: Calculate Ecell to find the cell potential for this concentration cell.
25
Question 29
Question
A voltaic cell is set up with a standard hydrogen electrode (SHE) as the anode
and a copper electrode as the cathode. The standard reduction potential for
the copper electrode is +0.34 V. The cell operates under standard conditions at
25
°
C. Calculate the cell potential if the concentration of Cu2+ in the cathode
compartment is 0.10 M. (Hint: Use the Nernst equation)
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is the reduction
of copper ions by hydrogen gas: Cu2+ (aq) + 2e−→Cu (s)
Step 2: Write the half-reaction at the cathode. Cu2+ (aq) + 2e−→Cu (s)
Step 3: Determine the standard cell potential. The standard cell potential,
E◦, can be calculated using the standard reduction potentials: E◦= E◦
cathode -
E◦
anode = +0.34V−0V= +0.34V
Step 4: Calculate the Q value for the cell reaction. Q =
[Cu2+]
1
= 0.10 M
Step 5: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by: E = E◦-
0.0592
nlogQ
K
where E is the cell potential, E◦is the standard cell potential, n is the number
of electrons transferred, Q is the reaction quotient, and K is the equilibrium
constant.
For this cell, n = 2 (from the balanced cell reaction), E = +0.34 V -
0.0592
2log0.10
1
= +0.34 V - 0.0296 V ×1 = +0.34 V - 0.0296 V = +0.3104 V
Therefore, the cell potential is +0.3104 V when the concentration of Cu2+
in the cathode compartment is 0.10 M.
Question 30
Question
Calculate the cell potential for the following reaction at 25
°
C given the concen-
trations of the ions in the half-cells are as follows:
Zn2+(0.10M)∥Cu2+(1.0×10−3M)
26
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard reduction potential for Cu2+/Cu is E◦= 0.34 V and for
Zn2+/Zn is E◦=−0.76 V.
Solution
Step 1: Write the half-reactions for the reduction that occur at each electrode.
The reduction half-reactions are:
Cu2+ + 2e−→Cu E◦= 0.34 V
Zn2+ + 2e−→Zn E◦=−0.76 V
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials for the half-reactions.
The standard cell potential can be calculated using the equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Step 3: Calculate the reaction quotient (Q) for the cell reaction using the
concentrations of the ions given.
The reaction quotient is given by:
Q=[Zn2+]
[Cu2+]=0.10
1.0×10−3= 100
Step 4: Calculate the cell potential (Ecell) using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
Where nis the number of moles of electrons transferred in the cell reaction. In
this case, two moles of electrons are transferred.
Ecell = 1.10 V −0.0592
2log(100)
Ecell = 1.10 V −0.0296 ×2×2
Ecell = 1.10 V −0.236 V
Ecell = 0.864 V
Therefore, the cell potential for the given reaction at 25
°
C is 0.864 V.
27
Question 31
Question
Calculate the cell potential for the following reaction at 25
°
C given that [F e3+] =
0.010M, [F e2+] = 0.050M, and E◦
cell = 0.771 V.
2F e3+(aq)+2e−→2F e2+(aq)
Solution
Step 1: Write the half-cell reactions.
Oxidation: F e2+ →F e3+ +e−E◦=−0.771 V
Reduction: F e3+ +e−→F e2+ E◦= 0.771 V
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [F e3+]2
[F e2+]2
Step 3: Substitute the given values into the Nernst equation.
Ecell = 0.771 −0.0592
2log 0.0102
0.0502
Step 4: Calculate the cell potential.
Ecell = 0.771 −(0.0296) log(0.04)
Ecell = 0.771 −(0.0296)(−1.3979)
Ecell = 0.771 + 0.0414
Ecell = 0.812 V
Question 32
Question
A voltaic cell is constructed with a copper strip dipping into a 1.0 M Cu2+
solution and a silver strip dipping into a 0.10 M Ag+solution. The measured
cell potential at standard conditions is 0.46 V. Calculate the cell potential at
25
°
C when the concentration of Cu2+ is reduced to 0.10 M and the concentration
of Ag+is increased to 1.0 M. Assume the temperature coefficient for the cell
reaction is −2.20 ×10−4V/
°
C.
28
Solution
Step 1: Write the balanced cell reaction and the Nernst equation.
The balanced cell reaction is:
Cu2+(aq) + 2e−→Cu(s)
The Nernst equation is:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential at the non-standard conditions, - E◦is
the standard cell potential (given as 0.46 V), - nis the number of moles of
electrons transferred in the balanced cell reaction (2 in this case), - Qis the
reaction quotient which is the ratio of the concentrations of the products over
the concentrations of the reactants raised to their stoichiometric coefficients.
Step 2: Calculate the cell potential at the new concentrations using the
Nernst equation.
Given concentrations:
[Cu2+] = 0.10 M
[Ag+]=1.0 M
Calculate Q:
Q=[Cu](s)
[Cu2+]=1
0.102= 10
Now substitute the given values into the Nernst equation:
E= 0.46 V −0.0592
2log(10) = 0.46 V −0.0592 ×1=0.46 V −0.0592 = 0.40 V
Therefore, the cell potential at 25
°
C when the concentration of Cu2+ is
reduced to 0.10 M and the concentration of Ag+is increased to 1.0 M is 0.40
V.
Question 33
Question
A voltaic cell consists of a copper electrode in a 1.0 M solution of copper(II)
sulfate and a zinc electrode in a 1.0 M solution of zinc sulfate. Given that the
standard reduction potentials are E◦
Cu2+/Cu = 0.34 V and E◦
Zn2+/Zn =−0.76 V,
calculate the cell potential when the copper electrode is at 298 K and the zinc
electrode is at 400 K.
29
Solution
Step 1: Write the cell reaction and overall cell potential at 298 K. The cell
reaction is as follows:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard cell potential E◦
cell at 298 K can be calculated using the Nernst
equation, which takes the form:
Ecell =E◦
cell −0.0592
nlog Q
where Ecell = cell potential at any temperature, E◦
cell = standard cell potential
at 298 K, n= number of moles of electrons transferred (in this case, 2), Q=
reaction quotient.
The standard cell potential E◦
cell is calculated using the standard reduction
potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn = 0.34V−(−0.76V)=1.10V
Step 2: Determine the reaction quotient. The reaction quotient Qcan be
determined using the concentrations of the species involved in the reaction.
Since both solutions are 1.0 M, the reaction quotient is 1.
Step 3: Calculate the cell potential at 400 K. Substitute the values into the
Nernst equation:
Ecell = 1.10V−0.0592
2log(1) = 1.10V−0=1.10V
Therefore, the cell potential at 400 K is 1.10 V.
Question 34
Question
A concentration cell is set up using two silver electrodes with a Ag+concentra-
tion of 1.0 M in one half-cell and a Ag+concentration of 0.001 M in the other
half-cell. If the standard reduction potential of Ag+reduction to Ag is 0.80 V,
calculate the cell potential at 25
°
C using the Nernst equation.
Solution
Step 1: Write the half-reactions and the balanced overall cell reaction. The
half-reaction for the reduction of Ag+to Ag is: Ag++ e−→Ag The overall
cell reaction is: 2Ag+→2Ag
Step 2: Calculate the cell potential using the standard reduction potential
and the Nernst equation. The standard cell potential, E◦
cell, can be calculated
30
using the standard reduction potentials of the half-reactions: E◦
cell = E◦
cathode -
E◦
anode E◦
cell = 0 V - (-0.80 V) E◦
cell = 0.80 V
Step 3: Apply the Nernst equation to calculate the cell potential at 25
°
C.
The Nernst equation is given by: E = E◦-0.0592
n×log Q
K
Where: E = cell potential E◦= standard cell potential n = number of
electrons transferred in the balanced half-reaction Q = reaction quotient K =
equilibrium constant
For the given cell with two silver half-cells, n = 2.
Step 4: Calculate the reaction quotient (Q) for the cell. Q = [Ag+]0.001/[Ag+]1
Q = 0.001/1.0 Q = 0.001
Step 5: Substitute the values into the Nernst equation and solve for the cell
potential. E = 0.80 V - 0.0592
2×log(0.001) E = 0.80 V - 0.0296 ×(-3) E = 0.80
V + 0.0888 V
E= 0.89 V
Question 35
Question
A voltaic cell consists of a standard hydrogen electrode as the anode and a
copper electrode as the cathode. The initial concentrations of H+ions and
Cu2+ ions are 0.10 M and 0.20 M, respectively. The cell potential is measured
to be 0.78 V at 25
°
C. Calculate the equilibrium constant, K, for the following
redox reaction:
Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)
Given: E◦
cell = 0.34 V, F= 96,500 C/mol, and R= 8.314 J/mol ·K.
Solution
Step 1: Write the half-reactions for the anode and cathode.
At the anode:
2H+(aq)+2e−→H2(g)E◦
anode = 0.00 V
At the cathode:
Cu2+(aq)+2e−→Cu(s)E◦
cathode = 0.34 V
Step 2: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Cu2+]
[H+]2
0.78 V = 0.34 V −0.0592
2log 0.20
0.102
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Step 3: Solve for Kusing the cell potential and the equilibrium constant
expression:
Ecell =RT
2Fln K
0.78 V = (8.314 J/mol ·K)(298 K)
2(96,500 C/mol) ln K
K= exp 2(0.78 V)(96500 C/mol)
8.314 J/mol ·K·298
K= 1.34 ×107
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