CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Nernst equation
Question Bank - Set 5
Liberty University
Question 1
Question
A voltaic cell consists of a standard hydrogen electrode (volts) and a copper
electrode immersed in a Cu(NO3)2solution with a concentration of 0.20 M.
If the standard reduction potential of Cu2+/Cu is 0.34 V, calculate the cell
potential at 25◦C when the concentration of hydrogen ions is 0.10 M. (Assume
T = 298 K)
Solution
Step 1: Write the half-reactions for the redox process occurring in the cell. The
oxidation half-reaction at the anode is the standard hydrogen electrode:
H2(g)→2H+(aq) + 2e−
The reduction half-reaction at the cathode is the reduction of copper ions:
Cu2+(aq) + 2e−→Cu(s)
Step 2: Write the cell reaction by combining the half-reactions and determine
the cell potential at standard conditions.
By combining the two half-reactions, we obtain the cell reaction:
Cu2+(aq) + 2H+(aq)→Cu(s) + H2(g)
The standard cell potential can be calculated by finding the difference be-
tween the standard reduction potentials of the cathode and anode:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cathode = 0.34 V and E◦
anode = 0 V, we find:
E◦
cell = 0.34 V
Step 3: Apply the Nernst equation to determine the cell potential under
non-standard conditions.
The Nernst equation is given by:
E=E◦−0.0592
nlog Q
K
Where: - Eis the cell potential under non-standard conditions. - E◦is the
standard cell potential. - nis the number of moles of electrons transferred in the
balanced chemical equation. - Qis the reaction quotient. - Kis the equilibrium
constant. For this reaction, it is equal to the solubility product constant Ksp.
Step 4: Calculate the reaction quotient, Q.
The reaction quotient Qis calculated using the concentrations of the species
involved in the cell reaction:
Q=[Cu](H+)2
H2
Given: - [Cu] = 0.20 M (initial concentration of copper ions), - (H+)=0.10
M, - (H2) = PH2/KH2= 1/10−7atm/1 atm (since hydrogen gas has a pressure
of 1 atm at standard conditions), we can calculate Qas:
Q=(0.20)(0.10)2
1/10−7= 2 ×105
Step 5: Calculate the cell potential, E.
Since the reaction involves 2 moles of electrons (n= 2), we can substitute
the given values into the Nernst equation:
E= 0.34 −0.0592
2log 2×105
E= 0.34 −0.0296 log 2×105
E= 0.34 −0.0296 ×5.30
E= 0.34 −0.1568
E= 0.1832 V
Therefore, the cell potential at 25◦C when the concentration of hydrogen
ions is 0.10 M is 0.1832 V.
2
Question 2
Question
A galvanic cell consists of a standard hydrogen electrode (H2at 1 atm and
[H+] = 1M) and a copper electrode in a Cu2+ solution with [Cu2+] = 0.01M.
Calculate the cell potential at 25
°
C. Given: E◦
cell(Cu2+/Cu)=0.34V
Solution
Step 1: Write the half-reactions for the cell
The half-reactions for the cell are:
Reduction at copper electrode: Cu2+ + 2e−→Cu(s)
Oxidation at standard hydrogen electrode: 2H++ 2e−→H2(g)
Step 2: Calculate the cell potential using the Nernst equation
The cell potential at a given temperature can be calculated using the Nernst
equation:
Ecell =E◦
cell −0.0592V
nlog Q
where: Ecell = cell potential E◦
cell = standard cell potential n= number of
electrons transferred in the cell reaction Q= reaction quotient
Since both half-reactions involve two electrons, n= 2.
The reaction quotient Qfor this cell is given by:
Q=[Cu]1
[H+]2
Substitute the given values and calculate Q:
Q=0.01
12= 0.01
Now, substitute E◦
cell = 0.34V,n= 2, and Q= 0.01 into the Nernst equation
to find the cell potential Ecell:
Ecell = 0.34V−0.0592V
2log(0.01)
Ecell = 0.34V−0.0296V×(−2) log(0.01)
Ecell = 0.34V+ 0.0592V×2×2
Ecell = 0.34V+ 0.1184V
Ecell = 0.4584V
Therefore, the cell potential at 25
°
C is 0.4584 V.
3
Question 3
Question
A voltaic cell consists of a silver-silver chloride electrode (Ag/AgCl) immersed
in a 0.10 M solution of silver nitrate, connected by a salt bridge to a standard
hydrogen electrode (Pt/H2(g), 1.0 atm, pH 0). Given that the standard re-
duction potential for the Ag+(aq)+e−→Ag(s) half-cell reaction is 0.80 V,
determine the cell potential when the concentration of Cl−in the AgCl half-cell
is 0.050 M. (Assume the temperature is 25
°
C and that the pH remains constant
throughout.)
Solution
Step 1: Write the half-cell reactions.
The half-cell reactions are as follows:
AgCl(s) →Ag+(aq) + Cl−(aq) (E◦= ? V)
2H+
2(aq) + 2e−→H2(g) (E◦= 0.00 V)
Step 2: Calculate the standard cell potential.
The standard cell potential is given by the Nernst equation:
E=E◦−0.0592
nlog Q
where
Q=[products]
[reactants]
n= number of electrons transferred = 1
Since the concentrations and the reaction quotient (Q) are not given, we
cannot calculate the cell potential at this point.
Step 3: Derive the expression for Q.
At equilibrium, the cell potential is zero, so E= 0:
0 = E◦−0.0592
1log Q
Step 4: Solve for Q.
Q= products = [Ag+]
[Cl−]= [Ag+] = 0.10 M
Step 5: Determine the cell potential at the given conditions.
Substitute Q= 0.10 M into the Nernst equation and calculate the cell potential:
E=E◦−0.0592
1log 0.10
4
E= 0.80 −0.0592 log 0.10
E= 0.80 −0.0592(−1)
E= 0.86 V
Therefore, the cell potential under these conditions is 0.86 V.
Question 4
Question
Calculate the cell potential for the following reaction at 298 K:
Zn(s) + Cu2+(0.10 M) →Zn2+(0.50 M) + Cu(s)
Given:
E◦
Zn2+/Zn =−0.76 V
E◦
Cu2+/Cu = 0.34 V
Solution
Step 1: Write out the half-reactions involved in the cell reaction.
Anode (oxidation) : Zn(s) →Zn2+(aq)+2e−
E◦
Zn2+/Zn =−0.76 V
Cathode (reduction) : Cu2+(aq)+2e−→Cu(s)
E◦
Cu2+/Cu = 0.34 V
Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Cu2+][Zn]
[Zn2+][Cu]
where - Eis the cell potential, - E◦is the standard cell potential, - nis the
number of moles of electrons transferred (in this case, n= 2), - [Cu2+] and [Zn]
are the concentrations of Cu2+ and Zn, respectively, - [Zn2+] and [Cu] are the
concentrations of Zn2+ and Cu, respectively.
Step 3: Substitute the given values into the Nernst equation and solve for
E.
E= (0.34 V) −0.0592
2log 0.10 ×0.50
1
E= 0.34 V −0.0296 log(0.05)
E= 0.34 V −0.0296 ×(−1.30)
E≈0.38 V
5
Therefore, the cell potential for the given reaction at 298 K is approximately
0.38 V.
Question 5
Question
A concentration cell consists of two half-cells, one with a silver electrode in a
0.10 M Ag+solution and the other with a silver electrode in a 0.0010 M Ag+
solution. Calculate the cell potential at 25
°
C. (Given: E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the half-reactions for the cell. The two half-reactions are:
Ag++e−→Ag (cathode)
Ag →Ag++e−(anode)
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog [Ag+cathode]
[Ag+anode]
Given: E◦
Ag+/Ag = 0.80 V, [Ag+cathode] = 0.10 M, [Ag+anode] = 0.0010
M, and n= 1 (1 electron transfer)
Substitute the given values into the Nernst equation:
E= 0.80 −0.0592
1log 0.10
0.0010
E= 0.80 −0.0592 ×2
E= 0.80 −0.1184
E= 0.6816 V
Therefore, the cell potential at 25
°
C for the concentration cell is 0.6816 V.
6
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode (E◦= 0 V) and a
nickel electrode. The nickel electrode reaction is given by:
Ni2+(aq)+2e−→Ni(s)
If the concentration of Ni2+ is 0.10 M and the cell potential is measured to be
0.35 V at 25◦C, calculate E◦for the nickel electrode reaction.
Solution
Step 1: Write the cell reaction for the galvanic cell. The cell reaction for the
galvanic cell involving a standard hydrogen electrode and a nickel electrode is:
2H+(aq) + 2e−→H2(g)
Ni2+(aq) + 2e−→Ni(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions:
2H+(aq) + Ni2+(aq)→H2(g) + Ni(s)
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−RT
nF ln Q
where - Eis the measured cell potential (0.35 V), - E◦is the standard cell
potential (unknown), - Ris the gas constant (8.314 J/mol ·K), - Tis the tem-
perature in Kelvin (298 K), - nis the number of moles of electrons transferred
(2 moles in this case), - Fis the Faraday constant (96485 C/mol), and - Qis
the reaction quotient.
Step 4: Calculate the reaction quotient Qand standard cell potential E◦.
Given that the concentration of Ni2+ is 0.10 M and at standard conditions
[H+] = 1.0 M, we can calculate Q:
Q=[H2]
[Ni2+]=1
0.10 = 10
Now we can rearrange the Nernst equation to solve for E◦:
E◦=E+RT
nF ln Q
E◦= 0.35 + (8.314 ×298)
(2 ×96485) ln 10
7
E◦= 0.35 + (0.0245) ln 10
E◦= 0.35 + 0.0567
E◦= 0.4067 V
Therefore, the standard cell potential E◦for the nickel electrode reaction is
0.4067 V.
Question 7
Question
Calculate the cell potential for a galvanic cell with the following half-reactions:
Zn2+ + 2e−→Zn(s) E◦=−0.76 V
Cu2+ + 2e−→Cu(s) E◦= +0.34 V
Given that the initial concentration of Zn2+ is 1.0 M and the initial concen-
tration of Cu2+ is 0.0010 M at 25◦C. (Given: R= 8.314 J/(mol ·K), F=
96,485 C/mol)
Solution
Step 1: Write the overall cell reaction.
Zn(s) + Cu2+ →Zn2+ + Cu(s)
Step 2: Determine the standard cell potential, E◦
cell, using the standard
reduction potentials provided.
E◦
cell =E◦
cathode −E◦
anode = (+0.34) −(−0.76) = +1.10 V
Step 3: Calculate the reaction quotient, Q, using the initial concentrations
of the reactants.
Q=[Zn2+]
[Cu2+]=1.0
0.0010 = 1000
Step 4: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
2log Q
Ecell = 1.10 −0.0592
2log 1000 = 1.10 −0.0907 = 1.01 V
Therefore, the cell potential for the galvanic cell is 1.01 V.
8
Question 8
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the reduction half-
reaction is Fe3+ + e−−−→ Fe2+ with a standard reduction potential of +0.77
V, and the oxidation half-reaction is Pb(s) −−→ Pb2+ + 2 e−with a standard
reduction potential of -0.13 V. The concentration of Fe3+ in the cell is 0.10 M,
and the concentration of Pb2+ is 0.50 M.
Solution
Step 1: Write the overall balanced cell reaction based on the two half-reactions
given:
3 Fe2+ + 2 Pb(s) −−→ 3 Fe3+ + 2 Pb2+
Step 2: Write the Nernst equation for the cell potential:
E=E◦−0.0592
nlog [Fe3+]3[Pb2+]2
[Fe2+]3[Pb(s)2]
Step 3: Calculate the standard cell potential (E◦) using the standard reduc-
tion potentials given:
E◦=E◦
cathode −E◦
anode = 0.77 V −(−0.13 V) = 0.90 V
Step 4: Calculate the cell potential at 25
°
C by plugging in the values into
the Nernst equation:
E= 0.90 V −0.0592
5log (0.10)3(0.50)2
(1)3
E= 0.90 V −0.0119 log(0.025)
E= 0.90 V −0.0119(−1.60)
E= 0.90 V + 0.0190
E= 0.9190 V
Therefore, the cell potential at 25
°
C for this galvanic cell is 0.9190 V.
Question 9
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn2+(0.010 M) + Cu(s)→Zn(s) + Cu2+(0.030 M)
Given: E◦
cell = 1.10 V R= 8.314 J/(mol ·K) F= 96485 C/mol
9
Solution
Step 1: Write the half-reactions. The half-reactions for this cell are:
Zn2+ + 2e−→Zn(s)
E◦
Zn2+/Zn =−0.76 V
Cu2+ + 2e−→Cu(s)
E◦
Cu2+/Cu = 0.34 V
Step 2: Write the overall cell reaction. The overall cell reaction is the com-
bination of the two half-reactions:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
Step 3: Calculate Ecell using the Nernst equation. The Nernst equation is
given by:
Ecell =E◦
cell −RT
nF ln(Q)
where: Ecell is the cell potential, E◦
cell is the standard cell potential, Ris
the gas constant, Tis the temperature in Kelvin, nis the number of electrons
transferred in the cell reaction, Fis Faraday’s constant, and Qis the reaction
quotient.
In this case, since the standard cell potential is given, we can use the sim-
plified Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
Step 4: Calculate the reaction quotient Q. Given that the concentrations
are: [Zn2+] = 0.010 M [Cu2+] = 0.030 M
The reaction quotient, Q, is calculated as:
Q=[product]coefficient
[reactant]coefficient
Q=[Zn](s)×[Cu2+]
[Zn2+]×[Cu] =1×0.030
0.010 ×1= 3
Step 5: Calculate the cell potential, Ecell. Substitute the given values into
the simplified Nernst equation:
Ecell = 1.10 −0.0592
2log(3)
Ecell = 1.10 −0.0296 log(3)
Ecell = 1.10 −0.0296 ×0.4771
Ecell = 1.10 −0.0142
Ecell = 1.0858 V
Therefore, the cell potential for the given reaction at 25
°
C is 1.0858 V.
10
Question 10
Question
A redox reaction involves the oxidation of ferrous ions (Fe2+) to ferric ions
(Fe3+). Given the following half-reactions:
Fe2+ →Fe3+ + e−E◦= 0.77 V
Determine the equilibrium constant, K, at 25
°
C for the reaction:
Fe2+ + e−→Fe3+
Given that the standard reduction potential for the reaction is 0.77 V.
Solution
Step 1: To find the equilibrium constant, K, we can use the Nernst equation:
E=E◦−0.0592
nlog Q
where Eis the cell potential under non-standard conditions, E◦is the standard
cell potential, nis the number of moles of electrons transferred in the balanced
redox reaction, and Qis the reaction quotient.
Step 2: First, we need to find the value of nin the balanced redox reaction.
From the half-reaction given, we can see that 1 mole of electron is transferred.
Step 3: The reaction quotient, Q, is given by the following expression:
Q=[Fe3+]
[Fe2+]
Step 4: Since we are trying to find the equilibrium constant, K, which is
related to the reaction quotient by the equation:
K=enE◦
0.0592
Step 5: Now, substituting the given values into the equation, we have:
K=e(1)(0.77)
0.0592
Step 6: Calculating the value of K:
K=e13.03 ≈4.09 ×105
Therefore, the equilibrium constant Kfor the reaction at 25
°
C is approxi-
mately 4.09 ×105.
11
Question 11
Question
A galvanic cell consists of a Ag/AgCl half-cell with [AgCl] = 1.0×10−2M and
a standard hydrogen electrode (SHE). If the cell potential is measured to be
0.30 V at 25
°
C, calculate the concentration of [Ag+] in the cell with the Nernst
equation.
Solution
Step 1: Write the reduction half-reaction for the Ag/AgCl electrode.
AgCl(s) + e- −→ Ag(s) + Cl-
Step 2: Write the overall cell reaction.
AgCl(s) + e- −→ Ag(s) + Cl-
Step 3: Write the Nernst equation.
E=E◦−0.0592
nlog [products]
[reactants]
Step 4: Identify E,E◦,n, [products], and [reactants]. Given: E= 0.30 V,
E◦= 0.80 V, n= 1 (since it involves the transfer of 1 electron), and [products] =
[Ag+] and [reactants] = [Cl-].
Step 5: Substitute the values into the Nernst equation.
0.30 = 0.80 −0.0592
1log [Ag+]
1.0×10−2
Step 6: Solve for [Ag+].
0.0592
1log [Ag+]
1.0×10−2= 0.80 −0.30
log [Ag+]
1.0×10−2=0.50
0.0592
log [Ag+]
1.0×10−2≈8.45
Step 7: Solve for [Ag+].
[Ag+]
1.0×10−2≈108.45
[Ag+] ≈3.50 ×108M
Therefore, the concentration of [Ag+] in the cell is 3.50 ×108M.
12
Question 12
Question
A voltaic cell consists of a standard hydrogen electrode (Pt — H2(g) — H+(aq))
and a copper electrode (Cu — Cu2+(aq)). The standard reduction potential of
the copper electrode is +0.34 V. If the concentration of Cu2+ is 0.10 M and the
pH of the solution is 3.0, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction. The cell reaction for the voltaic cell can
be represented as:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the half-reaction for the standard hydrogen electrode. The
half-reaction for the standard hydrogen electrode is:
2H+(aq)+2e−→H2(g)
Step 3: Calculate the standard cell potential (E◦). Using the given stan-
dard reduction potential of the copper electrode (+0.34 V) and the standard
reduction potential of the standard hydrogen electrode (0 V), we can calculate
the standard cell potential:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34V −0V
E◦
cell = 0.34V
Step 4: Account for non-standard conditions using the Nernst equation. The
Nernst equation is given by:
E=E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the cell reaction, and
Q is the reaction quotient. In this case, the cell reaction involves 2 moles of
electrons.
Step 5: Calculate the reaction quotient (Q). The reaction quotient (Q) for
the cell reaction is given by:
Q=[Cu]
[Cu2+]
Q=1
0.10 = 10
Step 6: Calculate the cell potential under non-standard conditions. Plugging
in the values into the Nernst equation:
E= 0.34V −0.0592
2log(10)
13
E= 0.34V −0.0296 log(10)
E= 0.34V −0.0296 ×1
E= 0.3104V
Therefore, the cell potential under the given conditions is 0.3104 V.
Question 13
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the reduction half-
reaction is Fe3+(aq) + 3e−→Fe(s) with a standard potential of 0.77 V, and
the oxidation half-reaction is I−(aq)→I2(s) + 2e−with a standard potential of
0.54 V. The concentration of Fe3+ is 0.10 M and the concentration of I−is 0.50
M.
Solution
Step 1: Write the overall cell reaction.
Fe3+(aq) + 2I−(aq)→Fe(s)+I2(s)
Step 2: Determine the standard cell potential.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.77 V −0.54 V
E◦
cell = 0.23 V
Step 3: Calculate the reaction quotient, Q.
Q=[Fe]1[I2]
[Fe3+]1[I−]2
Q=(1)(1)
(0.10)(0.50)2
Q=1
0.025
Q= 40
Step 4: Use the Nernst equation to find the cell potential.
Ecell =E◦
cell −0.0592
nlog(Q)
Where n is the number of moles of electrons transferred, which is 3 in this case.
Ecell = 0.23 V −0.0592
3log(40)
14
Ecell = 0.23 V −(0.01973)(3.69)
Ecell = 0.23 V −0.0727
Ecell = 0.157 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is 0.157 V.
Question 14
Question
A galvanic cell consists of a silver electrode in a 1.0 M AgN O3solution and a
lead electrode in a 0.1 M P b(NO3)2solution. Given that the standard reduction
potential for the Ag+/Ag couple is Eo= 0.80 V and the standard reduction
potential for the P b2+/P b couple is Eo=−0.13 V, calculate the cell potential
at 25
°
C.
Solution
Step 1: Write the half-reactions for each electrode process.
The half-reaction for the silver electrode is: Ag++e−→Ag
The half-reaction for the lead electrode is: P b2+ + 2e−→P b
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation relates the standard cell potential (Eo), the cell potential (E), the re-
action quotient (Q), the gas constant (R), the temperature (T), and the number
of electrons transferred (n) through the equation:
E=Eo−RT
nF ln Q
Given: Eo
Ag+/Ag = 0.80 V, Eo
P b2+/P b =−0.13 V, [Ag+]=1.0 M, [P b2+] =
0.1 M
Balancing the half reactions: Ag++e−→Ag P b2+ + 2e−→P b
The cell reaction is: 2Ag++P b2+ →2Ag +P b
Therefore, the cell potential can be calculated as:
E=Eo
cell −RT
2Fln [Ag]2
[P b2+]
Step 3: Substitute the given values into the Nernst equation and calculate
the cell potential.
E= (0.80 V−(−0.13 V)) −(8.314 J/mol ·K)(298 K)
(2)(96485 C/mol)ln (1.0M)2
0.1M
E= 0.93 V−8.314 ×298
2×96485 ln 1.0
0.1= 0.93 −2474.952
192970 ln 10 V
15
E0.93 −0.0318 V0.8982 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is approxi-
mately 0.8982 V.
Question 15
Question
A concentration cell is set up using two half-cells with silver electrodes. One half-
cell has 1 M Ag+ions, while the other half-cell has 0.1 M Ag+ions. Calculate
the cell potential at 25◦C for this concentration cell. Given that the standard
reduction potential for Ag+ions is E◦= 0.80 V and that the gas constant
R= 8.314 J/(mol·K), and at 25◦C, the Faraday constant F= 96,485 C/mol.
Solution
Step 1: Write the half-reactions for the two half-cells: The half-reaction for the
reduction of Ag+ions to Ag is:
Ag++e−→Ag
Step 2: Write the Nernst equation for the cell potential (Ecell): The Nernst
equation is given by:
Ecell =E◦−RT
nF ln(Q)
where: - Ecell = cell potential - E◦= standard reduction potential - R= gas
constant - T= temperature in Kelvin - n= number of electrons transferred -
F= Faraday constant - Q= reaction quotient
Step 3: Determine the number of electrons transferred (n): Since the half-
reaction involves the transfer of 1 electron, n= 1.
Step 4: Calculate the cell potential using the Nernst equation: For the 1 M
Ag+half-cell:
Q1 M =[products]
[reactants] =1
1= 1
For the 0.1 M Ag+half-cell:
Q0.1 M =[products]
[reactants] =0.1
1= 0.1
Plugging the values into the Nernst equation:
Ecell = 0.80 V −(8.314 J/(mol ·K))(298 K)
(1)(96485 C/mol) ln 1
0.1
Ecell = 0.80 V −(8.314)(298)
96485 ln(10)
16
Ecell = 0.80 V −0.02569 ln(10)
Ecell = 0.80 V −0.02569(2.303)
Ecell = 0.80 V −0.05912
Ecell = 0.7409 V
Therefore, the cell potential for the concentration cell with 1 M Ag+and 0.1
M Ag+half-cells is 0.7409 V.
Question 16
Question
For the half-reaction Fe3+ + e−⇌Fe2+, calculate the cell potential at 25◦C
when [Fe3+] = 0.20 M, [Fe2+] = 0.50 M, and E◦= 0.77 V. Given R = 8.314
J/(mol·K), F = 96485 C/mol, and T = 298 K.
Solution
Step 1: Write the Nernst equation for the cell potential under non-standard
conditions:
E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 2: Determine the number of electrons transferred in the half-reaction.
The number of electrons transferred (n) in the given half-reaction Fe3+ + e−⇌
Fe2+ is 1.
Step 3: Plug in the given values into the Nernst equation:
E= 0.77 −0.0592
1log 0.50
0.20
Step 4: Calculate the cell potential E.
E= 0.77 −0.0592 log(2.50)
E= 0.77 −0.0592(0.3979)
E0.7471 V
Therefore, the cell potential under the given conditions is approximately
0.7471 V.
17
Question 17
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the following half-
reactions occur:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.036 V
Given that the concentration of Cu2+ is 0.10 M and the concentration of Fe3+
is 0.25 M. (Hint: Nernst equation)
Solution
Step 1: Write the overall cell reaction and determine the cell potential using
standard reduction potentials. The overall cell reaction is:
Cu2+(aq) + Fe(s)→Cu(s) + Fe3+(aq)
To determine the cell potential, we subtract the standard reduction potential
of the Fe half-reaction from the standard reduction potential of the Cu half-
reaction:
E◦
cell =E◦
Cu2+/Cu −E◦
Fe3+ /Fe
E◦
cell = 0.34 V −(−0.036 V) = 0.376 V
Step 2: Calculate the reaction quotient, Q, and the Nernst equation. The
reaction quotient is given by:
Q=[Cu2+]
[Fe3+]
Q=0.10
0.25 = 0.40
The Nernst equation is:
Ecell =E◦
cell −0.0592
nlog(Q)
Where nis the number of electrons transferred in the balanced equation.
Step 3: Calculate Ecell using the Nernst equation. Since 2 electrons are
transferred in the balanced equation, n= 2.
Ecell = 0.376 V −0.0592
2log(0.40)
Ecell = 0.376 V −0.0296 log(0.40)
Ecell = 0.376 V −0.0296 ×(−0.3979)
Ecell = 0.376 V + 0.0112
Ecell = 0.387 V
Therefore, the cell potential at 25
°
C for the galvanic cell is 0.387 V.
18
Question 18
Question
A concentration cell is set up with two half-cells as follows:
Half-cell 1: Zn2+(0.10 M) ∥Zn2+(xM)
The measured cell potential is found to be 0.06 V. Calculate the concentration
of Zn2+ in half-cell 2.
Given: Standard reduction potential of Zn2+/Zn = -0.76 V Gas constant
R = 8.314 J/(mol·K) Temperature T = 298 K Faraday’s constant F = 96485
C/mol
Solution
Step 1: Write the Nernst equation for the concentration cell setup:
E=E◦−0.0592
nlog [Zn2+]2
[Zn2+]1
where: - E= measured cell potential = 0.06 V - E◦= standard cell potential
= -0.76 V - n= number of electrons involved in the reaction, which is 2 for
this cell - [Zn2+]1= concentration of Zn2+ in half-cell 1 = 0.10 M - [Zn2+]2=
concentration of Zn2+ in half-cell 2 (to be calculated)
Step 2: Substitute the given values into the Nernst equation:
0.06 = −0.76 −0.0592
2log [Zn2+]2
0.10
Step 3: Solve for [Zn2+]2:
0.06 = −0.76 −0.0296 log [Zn2+]2
0.10
0.82 = −0.0296 log [Zn2+]2
0.10
log [Zn2+]2
0.10 =0.82
−0.0296 =−27.70
[Zn2+]2
0.10 = 10−27.70
[Zn2+]2= 0.10 ×10−27.70
[Zn2+]2= 10−27.70 M
Therefore, the concentration of Zn2+ in half-cell 2 is 10−27.70 M.
19
Question 19
Question
Calculate the cell potential for a galvanic cell with a standard cell potential of
1.15 V at 25
°
C, when [Cu2+] = 0.20 M, [Zn2+] = 1.0 ×10−4M, and [Zn2+/Zn]
= 0.10.
Solution
Step 1: Write the half-reactions for the galvanic cell. The given cell is: Zn(s)
— Zn2+(aq, 1.0 ×10−4M) —— Cu2+(aq, 0.20 M) — Cu(s)
The two half-reactions are: Zn(s) →Zn2+(aq) + 2e−Cu2+(aq) + 2e−→
Cu(s)
Step 2: Write the Nernst equation. The Nernst equation is given by: E =
E◦-0.0592
nlog [Cu2+ ]
[Zn2+ ]
where: E = cell potential at non-standard conditions E◦= standard cell
potential n = number of electrons involved in the cell reaction [Cu2+] = con-
centration of copper ions [Zn2+] = concentration of zinc ions
Step 3: Calculate the cell potential using the Nernst equation. Substitute
the given values into the equation: E = 1.15V - 0.0592
2log 0.20
1.0×10−4
E = 1.15V - 0.0592
2log(2000) E = 1.15V - 0.0592
2×3.301
E = 1.15V - 0.0978 E 1.05 V
Therefore, the cell potential for the given galvanic cell is approximately 1.05
V at 25
°
C.
Question 20
Question
Calculate the cell potential for a concentration cell at 25
°
C that consists of two
half-cells, each with a silver electrode. One half-cell contains a 0.20 M solution
of AgNO3and the other half-cell contains a 0.025 M solution of AgNO3. The
standard reduction potential for the Ag++ e−→Ag half-reaction is 0.80 V.
Solution
Step 1: Write the balanced redox reaction for the concentration cell. The half-
reactions for the silver electrodes are: 1. Anode (oxidation): Ag(s) →Ag++
e−2. Cathode (reduction): Ag++ e−→Ag(s)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
E=E◦−RT
nF ln Q
20
Where: - Eis the cell potential - E◦is the standard cell potential (given:
0.80 V) - Ris the ideal gas constant (8.314 J/(mol*K)) - Tis the temperature in
Kelvin (25
°
C = 298 K) - nis the number of electrons transferred in the balanced
redox reaction (1 for this reaction) - Fis the Faraday constant (96485 C/mol)
-Qis the reaction quotient
Step 3: Calculate the reaction quotient Qfor this cell.
Q=[Ag+]anode
[Ag+]cathode
Q=0.20
0.025
Step 4: Substitute the given values into the Nernst equation and solve for
E.
E= 0.80 −(8.314 ∗298)
1∗96485 ln 0.20
0.025
Step 5: Calculate the value of E.
E= 0.80 −2474.072
96485 ln(8)
E≈0.72 V
Therefore, the cell potential for the concentration cell is approximately 0.72
V.
Question 21
Question
Calculate the cell potential at 25
°
C for 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
Given that the concentrations of Zn2+ and Cu2+ in the cell are 0.10 M and 1.00
M, respectively.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is obtained by
adding the two half-reactions:
Zn(s) + Cu2+(aq)→Cu(s) + Zn2+(aq)
Step 2: Determine 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
21
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 3: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation relates the cell potential under non-
standard conditions to the standard cell potential:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]2
Step 4: Substitute the given values into the Nernst equation.
Ecell = 1.10 V −0.0592
2log 1.00
0.102
Ecell = 1.10 V −0.0298 ×2×log(100) = 1.10 V + 0.0596 = 1.16 V
Thus, the cell potential at 25
°
C for the given galvanic cell is 1.16 V.
Question 22
Question
Calculate the cell potential at 25
°
C for a galvanic cell consisting of a standard
hydrogen electrode (SHE) and a silver electrode, given the following concen-
trations: [Ag+] = 1.0×10−3M and [H+]=0.10 M. The standard reduction
potential for the silver electrode is E0
Ag+/Ag = 0.80 V, and the reduction half-
reaction is Ag++e−→Ag. (Assume one electron is transferred in the cell
reaction).
Solution
Step 1: Write down the balanced redox reaction for the cell:
2H++ 2Ag →2Ag++ H2
Step 2: Determine the standard cell potential using the Standard Gibb’s
Free Energy and the Nernst equation.
The Standard Gibb’s Free Energy (∆G0) is related to the standard cell
potential by the equation:
∆G0=−nF ·E0
cell
where nis the number of moles of electrons transferred (2 in this case) and F
is the Faraday constant (96485 C/mol).
Given the standard reduction potentials: E0
Ag+/Ag = 0.80 V E0
H+/H2= 0
The standard cell potential is:
E0
cell =E0
cathode −E0
anode =E0
Ag+/Ag −E0
H+/H2= 0.80 −0 = 0.80 V
22
Step 3: Calculate the reaction quotient (Q) using the concentrations of the
species involved in the cell reaction:
Q=[Ag+]2
[H+]2
Step 4: Use the Nernst equation to calculate the cell potential at 25
°
C:
Ecell =E0
cell −0.0592
nlog(Q)
Substitute the values we’ve calculated:
Ecell = 0.80 −0.0592
2log (1.0×10−3)2
(0.10)2
Step 5: Calculate the cell potential:
Ecell = 0.80 −0.0296 log (1.0×10−3)2
(0.10)2
Ecell = 0.80 −0.0296 log(0.010)
Ecell = 0.80 −0.0296 × −2
Ecell = 0.80 −0.0592
Ecell = 0.74 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is 0.74 V.
Question 23
Question
Calculate the cell potential for a galvanic cell in which the following reaction
occurs:
Fe2+(0.10M) + Ag+(0.20M)→Fe3+(0.05M) + Ag(s)
Given that the standard cell potential E◦
cell = 0.77 V at 25
°
C, determine the
cell potential when Q= 4.0.
Solution
Step 1: Write the balanced redox reaction for the cell. The given reaction is:
Fe2+ + Ag+→Fe3+ + Ag
Step 2: Write the half-reactions for the cell. The half-reactions are: Anode:
Fe2+ →Fe3+ +e−(oxidation) Cathode: Ag++e−→Ag (reduction)
23
Step 3: Determine the standard cell potential using the standard reduction
potentials. Using the standard reduction potentials, we find E◦
cell =E◦
cathode −
E◦
anode = 0.80 V −(−0.44) V = 1.24 V
Step 4: Calculate the reaction quotient Q. The reaction quotient Q=
[Fe3+][Ag]
[Fe2+][Ag+]=(0.05)(1)
(0.10)(0.20) = 2.5
Step 5: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation is Ecell =E◦
cell −0.0592
nlog(Q). Since
the reaction involves the transfer of 1 electron, the number of moles of electrons
(n) is 1.
Substitute the given values:
Ecell = 1.24 V −0.0592
1log(4.0) = 1.24 V −0.0862 = 1.16 V
Therefore, the cell potential under non-standard conditions is 1.16 V.
Question 24
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a 1.0 M solution of Cu2+. The initial concentration of Cu2+ is 0.10
M, and the cell operates at 25◦C. Calculate the cell potential after the concen-
tration of Cu2+ has decreased to 0.050 M. The standard reduction potential for
Cu2+ + 2e−→Cu is 0.34 V.
Solution
Step 1: Write the cell reaction. The cell reaction for the galvanic cell composed
of a standard hydrogen electrode (SHE) and a copper electrode is:
Cu2+(0.10 M) + 2e−→Cu(s)
Step 2: Determine the initial cell potential 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 (T),
and the number of electrons transferred in the cell reaction (n). The Nernst
equation is given by:
E=E◦−0.0592
nlog(Q)
Given: E◦= 0.34 V T= 25◦C= 298 K [Cu2+]initial = 0.10 M
The reaction quotient Qcan be written as:
Q=[Cu]
1= 0.10 M
24
Plug in the values into the Nernst equation:
Einitial = 0.34 −0.0592
2log(0.10)
Einitial = 0.34 −0.0296 log(0.10)
Einitial = 0.34 −0.0296(−1)
Einitial = 0.34 + 0.0296 = 0.3696 V
Question 25
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to a
nickel electrode. The concentrations of Ni2+ and H+ions in the cell are 0.10 M
and 1.0M, respectively. Given that the standard reduction potential for Ni2+
is −0.25 V, calculate the cell potential at 25◦.
Solution
To find the cell potential, we will use the Nernst equation:
E=E◦−0.0592
nlog(Q)
where: E= cell potential E◦= standard cell potential n= number of electrons
transferred Q= reaction quotient
Step 1: Write the balanced cell reaction and the expression for Q.
The balanced cell reaction is:
Ni2+ + 2e−→Ni(s)
The expression for Qis:
Q=[Ni2+]
[H+]2
Step 2: Calculate the value of Q.
Q=0.10
(1.0)2= 0.10
Step 3: Determine the number of electrons transferred (n). From the bal-
anced cell reaction, we see that 2 electrons are transferred during the reaction.
Step 4: Calculate the cell potential (E). Using the Nernst equation:
E=−0.25 V−0.0592
2log(0.10)
25
E=−0.25 V−0.0592
2× −1.00
E=−0.25 V−(−0.0296)
E=−0.25 V+ 0.0296
E=−0.2204 V
Therefore, the cell potential at 25◦is −0.2204 V.
Question 26
Question
Calculate the cell potential at 25
°
C for a cell in which the reaction is:
2Cr3+(aq) + 2Mn2+(aq)→2Cr2+(aq) + 2Mn3+(aq)
Given the following half-reactions and their standard reduction potentials:
Cr3+ + 3e−→Cr2+ E◦=−0.74 V
Mn3+ +e−→Mn2+ E◦=−1.18 V
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together.
Then, determine the standard cell potential E◦.
2Cr3+(aq) + 2Mn2+(aq)→2Cr2+(aq) + 2Mn3+(aq)
The standard cell potential E◦can be calculated using the standard reduc-
tion potentials:
E◦
cell =E◦
cathode −E◦
anode
Given E◦
Cr2+/Cr3+ =−0.74 V and E◦
Mn2+/Mn3+ =−1.18 V, we can calculate
E◦
cell.
E◦
cell = (−0.74 V) −(−1.18 V) = 0.44 V
Step 2: Calculate the reaction quotient Q to use in the Nernst equation.
Q=[Cr2+]2[Mn3+]2
[Cr3+]2[Mn2+]2
Step 3: Use the Nernst equation to calculate the cell potential at 25
°
C.
E=E◦−0.0592
nlog(Q)
In this case, n= 4 because of the balanced coefficients in the overall reaction.
Since the reaction is at 25
°
C, T= 25 + 273 = 298 K.
E= 0.44 −0.0592
4log(Q) (at 25
°
C)
26
Question 27
Question
A voltaic cell consists of a standard hydrogen electrode (SHE) in one half-cell
and a silver-silver chloride electrode in the other half-cell. The standard reduc-
tion potential for the silver-silver chloride electrode is 0.222 V. If the concen-
tration of AgCl in the cell is 1.0 M, calculate the cell potential at 25
°
C. (Given:
E◦
Step 1: Write the overall cell reaction The overall cell reaction is the sum of the
two half-cell reactions. The half-cell reactions are:
Anode: H2(g)→2H+(aq) + 2e−E◦= 0.000 V
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)E◦= 0.222 V
The overall cell reaction is:
2H2(g) + 2AgCl(s)→2Ag(s) + 2H+(aq) + 2Cl−(aq)
Step 2: Calculate the cell potential The cell potential can be calculated using
the Nernst equation:
E=E◦−0.0592
nlog [Ag+][H+]
[AgCl]
where - nis the number of moles of electrons transferred in the balanced
equation (in this case, 2), - E◦is the standard cell potential, - [Ag+] and [H+]
are the concentrations of silver ions and hydrogen ions, respectively, in the cell,
- [AgCl] is the concentration of silver chloride.
Since the reaction is at equilibrium, Q=[Ag+][H+]
[AgCl] .
Given the concentrations of AgCl and E◦
cell, we can solve for Eusing the
Nernst equation. Substituting the values:
E= (0.222 V) −0.0592
2log [Ag+][H+]
[AgCl]
E= (0.222 V) −0.0296 log [Ag+][H+]
1.0
Step 3: Calculate the cell potential Since the cell is at equilibrium, Q=K
and the reaction quotient Qis equal to the equilibrium constant Keq = 10−10
for the dissociation of water. We can use the equilibrium constant expression
for water to relate the concentrations of Ag+and H+.
Keq = [H+][OH−] = 10−14 M2
Given that [H+] = xM and [Ag+] = xM, the equation becomes:
x2= 10−14
27
x= 10−7
Therefore, [H+] = [Ag+] = 10−7M. Substituting this back into the Nernst
equation:
E= (0.222 V) −0.0296 log 10−7×10−7
1.0
E= 0.222 V −0.0296 log10−14
E= 0.
Question 28
Question
A half-cell reaction for a particular redox reaction involving silver is given by:
Ag+(aq)+e−→Ag(s)
If the standard reduction potential for this half-cell is +0.80 V, calculate the
cell potential at 25C when the concentration of Ag+is 0.10 M. Given that the
universal gas constant is 8.31 J/(mol·K) and the Faraday constant is 96,485
C/mol.
Solution
Step 1: Write the half-cell reaction and the Nernst equation.
Half-cell reaction: Ag+(aq)+e−→Ag(s)
Nernst equation: E=E◦−0.0591
nlog(Q) where:
–Eis the cell potential
–E◦is the standard cell potential
–nis the number of moles of electrons transferred in the balanced
redox reaction
–Qis the reaction quotient
Step 2: Calculate the standard cell potential (E◦) using the given standard
reduction potential.
Standard reduction potential for Ag half-cell: E◦
Ag+/Ag = +0.80 V
Therefore, E◦=E◦
Ag+/Ag = +0.80 V
Step 3: Calculate the reaction quotient (Q) using the concentration of Ag+.
Given concentration of Ag+([Ag+]): 0.10 M
28
Since the half-cell reaction involves the reduction of Ag+,Q= [Ag+].
Therefore, Q= 0.10 M
Step 4: Calculate the cell potential (E) at 25C using the Nernst equation.
Number of moles of electrons transferred (n) in the half-cell reaction: 1
Gas constant (R) = 8.31 J/(mol·K)
Temperature (T) = 25C = (25 + 273) K = 298 K
Plugging in the values into the Nernst equation:
E=E◦−0.0591
nlog(Q)
E= 0.80 −0.0591
1log(0.10)
E= 0.80 −0.0591 ×log(0.10)
E≈0.80 −0.0591 ×(−1.00)
E≈0.80 + 0.0591
E≈0.8591 V
Therefore, the cell potential at 25C when the concentration of Ag+is 0.10
M is approximately 0.8591 V.
Question 29
Question
A fuel cell uses the reaction 2H2(g) + O2(g)→2H2O(l) to generate electricity.
At 25
°
C, the concentration of H2O(l) is 1.0 M, O2(g) is 0.20 M, and H2(g) is
0.30 M.
Calculate the cell potential at standard conditions for the fuel cell reaction.
Given: E◦
cell = 1.23 V, R= 8.314 J/(mol·K), and F= 96485 C/mol.
Solution
Step 1: Write the half reactions and determine the cell potential under standard
conditions.
The oxidation half reaction is:
2H2(g)→4H+(aq)+4e−
The reduction half reaction is:
O2(g)+4H+(aq)+4e−→2H2O(l)
29
The cell potential under standard conditions can be calculated as:
E◦
cell =E◦
reduction −E◦
oxidation
Given E◦
cell = 1.23 V, we can calculate E◦
reduction and E◦
oxidation.
Step 2: Calculate the cell potential under standard conditions.
Now, we substitute the standard reduction potentials for the reactions to
find E◦
cell. Given that E◦
cell =E◦
reduction −E◦
oxidation, we have:
1.23 = E◦
reduction −E◦
oxidation
The standard reduction potential for the reduction half reaction is E◦
reduction =
1.23 V. The standard reduction potential for the oxidation half reaction is
E◦
oxidation = 0 V (by definition).
Therefore, 1.23 = 1.23 −0 V,
E◦
cell = 1.23 V
Question 30
Question
Calculate the cell potential at 25
°
C for a galvanic cell with a silver electrode
immersed in a 1.0 M AgNO3solution connected to a hydrogen electrode at pH
2. The standard reduction potential for the Ag+/ Ag electrode is +0.80 V, and
the standard reduction potential for the H+/ H2electrode is 0.00 V. Assume
that the activity coefficient for the silver ion is 0.85.
Solution
Step 1: Write the overall cell reaction. The cell reaction consists of the reduc-
tion half-reaction at the Ag electrode and the oxidation half-reaction at the H2
electrode. The overall reaction is:
Ag++ e−→Ag (E◦= +0.80 V)
2H++ 2e−→H2(E◦= 0.00 V)
Since the Ag half-reaction has a higher standard reduction potential, it will
be the reduction half-reaction in this cell.
Step 2: Write the Nernst equation for the cell potential. The Nernst equa-
tion relates the standard cell potential to the cell potential under non-standard
conditions:
E=E◦−0.0592
nlog Q
K
30
where Eis the cell potential under non-standard conditions, E◦is the stan-
dard cell potential, nis the number of electrons transferred in the balanced cell
reaction, Qis the reaction quotient, and Kis the equilibrium constant.
Step 3: Calculate the cell potential. First, calculate the reaction quotient,
Q, for the cell reaction:
Q=[Ag][H2]2
[Ag+][H+]2=1
1.0×(0.01)2= 100
Next, substitute the given values into the Nernst equation and solve for E:
E= 0.80 V −0.0592
1log 100
1= 0.80 V −0.0592 ×2=0.6816 V
Therefore, the cell potential at 25
°
C for the galvanic cell described is 0.6816
V.
Question 31
Question
A voltaic cell consists of a silver electrode immersed in a 0.20 M AgNO3solution
and a zinc electrode immersed in a 0.10 M Zn(NO3)2solution. The measured
cell potential is 1.56 V at 25◦C. Calculate the solubility of silver ion in water at
this temperature.
Solution
Step 1: Write the half-reactions for the cell reaction. The cell reaction can be
represented as:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Step 2: Write the Nernst equation for the cell reaction. The Nernst equation
for the cell reaction is:
E=E◦−0.0592
nlog [Zn2+]
[Ag+]2
Step 3: Substitute the given values into the Nernst equation. Given: E=
1.56 V
E◦= 1.56 V (since the cell is at standard conditions)
[Zn2+]=0.10 M
[Ag+]=0.20 M
n= 2
Substitute these values into the Nernst equation:
1.56 = 1.56 −0.0592
2log 0.10
0.202
31
Step 4: Solve for the solubility of silver ion. Simplify the equation:
0.0592
2log 0.10
0.202= 0
log 0.10
0.202= 0
log 0.10
0.04 = 0
log 2.5
1= 0
log 2.5 = 0
Therefore, log 2.5 = 0 at 25◦C means that the solubility of silver ion in water
at this temperature is 2.5 M.
Question 32
Question
A concentration cell is set up with two half-cells containing silver electrodes. One
half-cell has a silver electrode in a 0.10 M solution of AgNO3, while the other
half-cell has a silver electrode in a 0.0010 M solution of AgNO3. Calculate the
cell potential at 25
°
C for this concentration cell. (Given: E◦
cell(Ag+/Ag) = 0.80
V and R= 8.314 J/mol ·K)
Solution
Step 1: Write the half-reactions for the cathode and anode.
Cathode: Ag++e−→Ag E◦= +0.80 V
Anode: Ag →Ag++e−E◦=−0.80 V
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Reductant]cathode
[Oxidant]anode
where
Ecell = cell potential (unknown)
E◦
cell = standard cell potential = 0.80 V
0.0592 = constant
n←(1 at 25
°
C)
[Reductant]cathode = 0.10 M ←(higher concentration)
[Oxidant]anode = 0.0010 M ←(lower concentration)
32
Step 3: Calculate the cell potential.
∆Ecell = 0.80 −0.0592
1log 0.10
0.0010
∆Ecell = 0.80 −0.0592 log(100) = 0.80 −0.0592 ×2≈0.68 V
Thus, the cell potential for this concentration cell at 25
°
C is approximately
0.68 V.
Question 33
Question
A galvanic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and a copper electrode in a 0.10 M Cu2+ solution as the cathode. At
25
°
C, the standard reduction potential for the Cu2+/Cu half-reaction is +0.34
V. Calculate the cell potential when the concentration of Cu2+ at the cathode
is 0.050 M. (Assume all other concentrations are 1.0 M.)
Solution
Step 1: Write the half-reaction for each electrode. The half-reaction for the
standard hydrogen electrode (anode) is:
2H++ 2e−→H2(g)
The half-reaction for the copper electrode (cathode) is:
Cu2+ + 2e−→Cu(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions:
2H+(aq) + Cu2+(aq) →H2(g) + Cu(s)
Step 3: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Given that the standard reduction potential for the Cu2+/Cu half-reaction is
+0.34 V, and the standard reduction potential for the standard hydrogen elec-
trode is 0 V, we have:
E◦
cell = +0.34 V −0 V = +0.34 V
33
Step 4: Calculate the cell potential under non-standard conditions. The
Nernst equation relates the cell potential at non-standard conditions to the
standard cell potential:
Ecell =E◦
cell −0.0592
nlog Q
where - Ecell is the cell potential under non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of electrons transferred in the balanced
cell reaction, - Qis the reaction quotient.
In this case, n= 2 since 2 electrons are transferred in the overall cell reaction.
The reaction quotient Qis given by:
Q=[Cu]
[H+]2
At equilibrium, Q=K= 1 based on the balanced equation. Substituting the
known values into the Nernst equation:
Ecell = +0.34 V −0.0592
2log0.0502/12
Ecell = +0.34 V −0.0296 log(0.050)
Ecell = +0.34 V −0.0296 × −1.3010
Ecell = +0.34 V + 0.0387
Ecell = +0.3787 V
Therefore, the cell potential when the concentration of Cu2+ at the cathode
is 0.050 M is +0.3787 V.
Question 34
Question
A concentration cell is set up with two half-cells as follows: the first half-cell
contains a silver electrode dipping into a 0.10 M solution of AgNO3, and the
second half-cell contains a silver electrode dipping into a 0.0010 M solution
of AgNO3. Calculate the voltage of the concentration cell at 25
°
C. (Given:
E◦
Ag+/Ag = 0.80 V at 25
°
C)
Solution
Step 1: Write the half-reactions occurring in each half-cell. In this case, both
half-cells involve the reduction of Ag+ion to Ag(s):
Ag++e−→Ag
34
Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Ag+]
[Ag+]standard
where - Eis the cell potential, - E◦is the standard cell potential, - nis the
number of moles of electrons transferred in the balanced half-reaction, and -
[Ag+] and [Ag+]standard are the concentrations of Ag+ions in the two half-cells.
Step 3: Calculate the standard cell potential (E◦) using the given standard
reduction potential of Ag+:
E◦=E◦
Ag+/Ag = 0.80 V
Step 4: Calculate the cell potential (E) for the concentration cell using the
Nernst equation: For the first half-cell: [Ag+]=0.10 M For the second half-cell:
[Ag+]=0.0010 M
E1= 0.80 −0.0592
1log 0.10
1.0= 0.80 −0.0592 log 0.10
E2= 0.80 −0.0592
1log 0.0010
1.0= 0.80 −0.0592 log 0.0010
The cell potential is given by the difference between both half-cells:
Ecell =E1−E2= (0.80 −0.0592 log 0.10) −(0.80 −0.0592 log 0.0010)
Question 35
Question
The standard reduction potential for the half-reaction F e3+(aq)+e−→F e2+(aq)
is given as +0.77 V. Calculate the equilibrium constant (K) for the cell reaction
F e3+(aq) + F e(s)→F e2+(aq) + F e2+(aq) at 25
°
C using the Nernst equation.
Solution
Step 1: Write the cell reaction and the overall cell equation. The cell reaction
is:
F e3+(aq) + F e(s)→F e2+(aq) + F e(s)
The overall cell equation is:
F e3+(aq) + F e(s)→F e2+(aq) + F e2+(aq)
Step 2: Identify the half-reactions. The half-reactions involved in the cell
reaction are: 1. Oxidation half-reaction: F e(s)→F e2+(aq)+2e−2. Reduction
half-reaction: F e3+(aq) + e−→F e2+(aq)
35
Step 3: Calculate the standard cell potential. The standard cell potential
(E◦
cell) can be calculated as the difference between the standard reduction po-
tentials of the reduction and oxidation half-reactions:
E◦
cell =E◦
red −E◦
ox
E◦
cell = (+0.77V)−0
E◦
cell = +0.77V
Step 4: Calculate the equilibrium constant (K) using the Nernst equation.
The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog Q
where: - Ecell is the cell potential at non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of moles of electrons transferred, - Q
is the reaction quotient.
For this cell reaction, n= 1 because one electron is transferred. And, at
equilibrium, Q=K.
Substitute the values into the Nernst equation:
0.77V= 0.77V−0.0592
1log K
0 = −0.0592 log K
log K= 0
K= 100
K= 1
Therefore, the equilibrium constant (K) for the cell reaction is 1 at 25
°
C.
36
Question 2
Question
A galvanic cell consists of a standard hydrogen electrode (H2at 1 atm and
[H+] = 1M) and a copper electrode in a Cu2+ solution with [Cu2+] = 0.01M.
Calculate the cell potential at 25
°
C. Given: E◦
cell(Cu2+/Cu)=0.34V
Solution
Step 1: Write the half-reactions for the cell
The half-reactions for the cell are:
Reduction at copper electrode: Cu2+ + 2e−→Cu(s)
Oxidation at standard hydrogen electrode: 2H++ 2e−→H2(g)
Step 2: Calculate the cell potential using the Nernst equation
The cell potential at a given temperature can be calculated using the Nernst
equation:
Ecell =E◦
cell −0.0592V
nlog Q
where: Ecell = cell potential E◦
cell = standard cell potential n= number of
electrons transferred in the cell reaction Q= reaction quotient
Since both half-reactions involve two electrons, n= 2.
The reaction quotient Qfor this cell is given by:
Q=[Cu]1
[H+]2
Substitute the given values and calculate Q:
Q=0.01
12= 0.01
Now, substitute E◦
cell = 0.34V,n= 2, and Q= 0.01 into the Nernst equation
to find the cell potential Ecell:
Ecell = 0.34V−0.0592V
2log(0.01)
Ecell = 0.34V−0.0296V×(−2) log(0.01)
Ecell = 0.34V+ 0.0592V×2×2
Ecell = 0.34V+ 0.1184V
Ecell = 0.4584V
Therefore, the cell potential at 25
°
C is 0.4584 V.
3
Question 3
Question
A voltaic cell consists of a silver-silver chloride electrode (Ag/AgCl) immersed
in a 0.10 M solution of silver nitrate, connected by a salt bridge to a standard
hydrogen electrode (Pt/H2(g), 1.0 atm, pH 0). Given that the standard re-
duction potential for the Ag+(aq)+e−→Ag(s) half-cell reaction is 0.80 V,
determine the cell potential when the concentration of Cl−in the AgCl half-cell
is 0.050 M. (Assume the temperature is 25
°
C and that the pH remains constant
throughout.)
Solution
Step 1: Write the half-cell reactions.
The half-cell reactions are as follows:
AgCl(s) →Ag+(aq) + Cl−(aq) (E◦= ? V)
2H+
2(aq) + 2e−→H2(g) (E◦= 0.00 V)
Step 2: Calculate the standard cell potential.
The standard cell potential is given by the Nernst equation:
E=E◦−0.0592
nlog Q
where
Q=[products]
[reactants]
n= number of electrons transferred = 1
Since the concentrations and the reaction quotient (Q) are not given, we
cannot calculate the cell potential at this point.
Step 3: Derive the expression for Q.
At equilibrium, the cell potential is zero, so E= 0:
0 = E◦−0.0592
1log Q
Step 4: Solve for Q.
Q= products = [Ag+]
[Cl−]= [Ag+] = 0.10 M
Step 5: Determine the cell potential at the given conditions.
Substitute Q= 0.10 M into the Nernst equation and calculate the cell potential:
E=E◦−0.0592
1log 0.10
4
E= 0.80 −0.0592 log 0.10
E= 0.80 −0.0592(−1)
E= 0.86 V
Therefore, the cell potential under these conditions is 0.86 V.
Question 4
Question
Calculate the cell potential for the following reaction at 298 K:
Zn(s) + Cu2+(0.10 M) →Zn2+(0.50 M) + Cu(s)
Given:
E◦
Zn2+/Zn =−0.76 V
E◦
Cu2+/Cu = 0.34 V
Solution
Step 1: Write out the half-reactions involved in the cell reaction.
Anode (oxidation) : Zn(s) →Zn2+(aq)+2e−
E◦
Zn2+/Zn =−0.76 V
Cathode (reduction) : Cu2+(aq)+2e−→Cu(s)
E◦
Cu2+/Cu = 0.34 V
Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Cu2+][Zn]
[Zn2+][Cu]
where - Eis the cell potential, - E◦is the standard cell potential, - nis the
number of moles of electrons transferred (in this case, n= 2), - [Cu2+] and [Zn]
are the concentrations of Cu2+ and Zn, respectively, - [Zn2+] and [Cu] are the
concentrations of Zn2+ and Cu, respectively.
Step 3: Substitute the given values into the Nernst equation and solve for
E.
E= (0.34 V) −0.0592
2log 0.10 ×0.50
1
E= 0.34 V −0.0296 log(0.05)
E= 0.34 V −0.0296 ×(−1.30)
E≈0.38 V
5
Therefore, the cell potential for the given reaction at 298 K is approximately
0.38 V.
Question 5
Question
A concentration cell consists of two half-cells, one with a silver electrode in a
0.10 M Ag+solution and the other with a silver electrode in a 0.0010 M Ag+
solution. Calculate the cell potential at 25
°
C. (Given: E◦
Ag+/Ag = 0.80 V)
Solution
Step 1: Write the half-reactions for the cell. The two half-reactions are:
Ag++e−→Ag (cathode)
Ag →Ag++e−(anode)
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog [Ag+cathode]
[Ag+anode]
Given: E◦
Ag+/Ag = 0.80 V, [Ag+cathode] = 0.10 M, [Ag+anode] = 0.0010
M, and n= 1 (1 electron transfer)
Substitute the given values into the Nernst equation:
E= 0.80 −0.0592
1log 0.10
0.0010
E= 0.80 −0.0592 ×2
E= 0.80 −0.1184
E= 0.6816 V
Therefore, the cell potential at 25
°
C for the concentration cell is 0.6816 V.
6
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode (E◦= 0 V) and a
nickel electrode. The nickel electrode reaction is given by:
Ni2+(aq)+2e−→Ni(s)
If the concentration of Ni2+ is 0.10 M and the cell potential is measured to be
0.35 V at 25◦C, calculate E◦for the nickel electrode reaction.
Solution
Step 1: Write the cell reaction for the galvanic cell. The cell reaction for the
galvanic cell involving a standard hydrogen electrode and a nickel electrode is:
2H+(aq) + 2e−→H2(g)
Ni2+(aq) + 2e−→Ni(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions:
2H+(aq) + Ni2+(aq)→H2(g) + Ni(s)
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−RT
nF ln Q
where - Eis the measured cell potential (0.35 V), - E◦is the standard cell
potential (unknown), - Ris the gas constant (8.314 J/mol ·K), - Tis the tem-
perature in Kelvin (298 K), - nis the number of moles of electrons transferred
(2 moles in this case), - Fis the Faraday constant (96485 C/mol), and - Qis
the reaction quotient.
Step 4: Calculate the reaction quotient Qand standard cell potential E◦.
Given that the concentration of Ni2+ is 0.10 M and at standard conditions
[H+] = 1.0 M, we can calculate Q:
Q=[H2]
[Ni2+]=1
0.10 = 10
Now we can rearrange the Nernst equation to solve for E◦:
E◦=E+RT
nF ln Q
E◦= 0.35 + (8.314 ×298)
(2 ×96485) ln 10
7
E◦= 0.35 + (0.0245) ln 10
E◦= 0.35 + 0.0567
E◦= 0.4067 V
Therefore, the standard cell potential E◦for the nickel electrode reaction is
0.4067 V.
Question 7
Question
Calculate the cell potential for a galvanic cell with the following half-reactions:
Zn2+ + 2e−→Zn(s) E◦=−0.76 V
Cu2+ + 2e−→Cu(s) E◦= +0.34 V
Given that the initial concentration of Zn2+ is 1.0 M and the initial concen-
tration of Cu2+ is 0.0010 M at 25◦C. (Given: R= 8.314 J/(mol ·K), F=
96,485 C/mol)
Solution
Step 1: Write the overall cell reaction.
Zn(s) + Cu2+ →Zn2+ + Cu(s)
Step 2: Determine the standard cell potential, E◦
cell, using the standard
reduction potentials provided.
E◦
cell =E◦
cathode −E◦
anode = (+0.34) −(−0.76) = +1.10 V
Step 3: Calculate the reaction quotient, Q, using the initial concentrations
of the reactants.
Q=[Zn2+]
[Cu2+]=1.0
0.0010 = 1000
Step 4: Calculate the cell potential, Ecell, using the Nernst equation:
Ecell =E◦
cell −0.0592
2log Q
Ecell = 1.10 −0.0592
2log 1000 = 1.10 −0.0907 = 1.01 V
Therefore, the cell potential for the galvanic cell is 1.01 V.
8
Question 8
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the reduction half-
reaction is Fe3+ + e−−−→ Fe2+ with a standard reduction potential of +0.77
V, and the oxidation half-reaction is Pb(s) −−→ Pb2+ + 2 e−with a standard
reduction potential of -0.13 V. The concentration of Fe3+ in the cell is 0.10 M,
and the concentration of Pb2+ is 0.50 M.
Solution
Step 1: Write the overall balanced cell reaction based on the two half-reactions
given:
3 Fe2+ + 2 Pb(s) −−→ 3 Fe3+ + 2 Pb2+
Step 2: Write the Nernst equation for the cell potential:
E=E◦−0.0592
nlog [Fe3+]3[Pb2+]2
[Fe2+]3[Pb(s)2]
Step 3: Calculate the standard cell potential (E◦) using the standard reduc-
tion potentials given:
E◦=E◦
cathode −E◦
anode = 0.77 V −(−0.13 V) = 0.90 V
Step 4: Calculate the cell potential at 25
°
C by plugging in the values into
the Nernst equation:
E= 0.90 V −0.0592
5log (0.10)3(0.50)2
(1)3
E= 0.90 V −0.0119 log(0.025)
E= 0.90 V −0.0119(−1.60)
E= 0.90 V + 0.0190
E= 0.9190 V
Therefore, the cell potential at 25
°
C for this galvanic cell is 0.9190 V.
Question 9
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn2+(0.010 M) + Cu(s)→Zn(s) + Cu2+(0.030 M)
Given: E◦
cell = 1.10 V R= 8.314 J/(mol ·K) F= 96485 C/mol
9
Solution
Step 1: Write the half-reactions. The half-reactions for this cell are:
Zn2+ + 2e−→Zn(s)
E◦
Zn2+/Zn =−0.76 V
Cu2+ + 2e−→Cu(s)
E◦
Cu2+/Cu = 0.34 V
Step 2: Write the overall cell reaction. The overall cell reaction is the com-
bination of the two half-reactions:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
Step 3: Calculate Ecell using the Nernst equation. The Nernst equation is
given by:
Ecell =E◦
cell −RT
nF ln(Q)
where: Ecell is the cell potential, E◦
cell is the standard cell potential, Ris
the gas constant, Tis the temperature in Kelvin, nis the number of electrons
transferred in the cell reaction, Fis Faraday’s constant, and Qis the reaction
quotient.
In this case, since the standard cell potential is given, we can use the sim-
plified Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
Step 4: Calculate the reaction quotient Q. Given that the concentrations
are: [Zn2+] = 0.010 M [Cu2+] = 0.030 M
The reaction quotient, Q, is calculated as:
Q=[product]coefficient
[reactant]coefficient
Q=[Zn](s)×[Cu2+]
[Zn2+]×[Cu] =1×0.030
0.010 ×1= 3
Step 5: Calculate the cell potential, Ecell. Substitute the given values into
the simplified Nernst equation:
Ecell = 1.10 −0.0592
2log(3)
Ecell = 1.10 −0.0296 log(3)
Ecell = 1.10 −0.0296 ×0.4771
Ecell = 1.10 −0.0142
Ecell = 1.0858 V
Therefore, the cell potential for the given reaction at 25
°
C is 1.0858 V.
10
Question 10
Question
A redox reaction involves the oxidation of ferrous ions (Fe2+) to ferric ions
(Fe3+). Given the following half-reactions:
Fe2+ →Fe3+ + e−E◦= 0.77 V
Determine the equilibrium constant, K, at 25
°
C for the reaction:
Fe2+ + e−→Fe3+
Given that the standard reduction potential for the reaction is 0.77 V.
Solution
Step 1: To find the equilibrium constant, K, we can use the Nernst equation:
E=E◦−0.0592
nlog Q
where Eis the cell potential under non-standard conditions, E◦is the standard
cell potential, nis the number of moles of electrons transferred in the balanced
redox reaction, and Qis the reaction quotient.
Step 2: First, we need to find the value of nin the balanced redox reaction.
From the half-reaction given, we can see that 1 mole of electron is transferred.
Step 3: The reaction quotient, Q, is given by the following expression:
Q=[Fe3+]
[Fe2+]
Step 4: Since we are trying to find the equilibrium constant, K, which is
related to the reaction quotient by the equation:
K=enE◦
0.0592
Step 5: Now, substituting the given values into the equation, we have:
K=e(1)(0.77)
0.0592
Step 6: Calculating the value of K:
K=e13.03 ≈4.09 ×105
Therefore, the equilibrium constant Kfor the reaction at 25
°
C is approxi-
mately 4.09 ×105.
11
Question 11
Question
A galvanic cell consists of a Ag/AgCl half-cell with [AgCl] = 1.0×10−2M and
a standard hydrogen electrode (SHE). If the cell potential is measured to be
0.30 V at 25
°
C, calculate the concentration of [Ag+] in the cell with the Nernst
equation.
Solution
Step 1: Write the reduction half-reaction for the Ag/AgCl electrode.
AgCl(s) + e- −→ Ag(s) + Cl-
Step 2: Write the overall cell reaction.
AgCl(s) + e- −→ Ag(s) + Cl-
Step 3: Write the Nernst equation.
E=E◦−0.0592
nlog [products]
[reactants]
Step 4: Identify E,E◦,n, [products], and [reactants]. Given: E= 0.30 V,
E◦= 0.80 V, n= 1 (since it involves the transfer of 1 electron), and [products] =
[Ag+] and [reactants] = [Cl-].
Step 5: Substitute the values into the Nernst equation.
0.30 = 0.80 −0.0592
1log [Ag+]
1.0×10−2
Step 6: Solve for [Ag+].
0.0592
1log [Ag+]
1.0×10−2= 0.80 −0.30
log [Ag+]
1.0×10−2=0.50
0.0592
log [Ag+]
1.0×10−2≈8.45
Step 7: Solve for [Ag+].
[Ag+]
1.0×10−2≈108.45
[Ag+] ≈3.50 ×108M
Therefore, the concentration of [Ag+] in the cell is 3.50 ×108M.
12
Question 12
Question
A voltaic cell consists of a standard hydrogen electrode (Pt — H2(g) — H+(aq))
and a copper electrode (Cu — Cu2+(aq)). The standard reduction potential of
the copper electrode is +0.34 V. If the concentration of Cu2+ is 0.10 M and the
pH of the solution is 3.0, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction. The cell reaction for the voltaic cell can
be represented as:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the half-reaction for the standard hydrogen electrode. The
half-reaction for the standard hydrogen electrode is:
2H+(aq)+2e−→H2(g)
Step 3: Calculate the standard cell potential (E◦). Using the given stan-
dard reduction potential of the copper electrode (+0.34 V) and the standard
reduction potential of the standard hydrogen electrode (0 V), we can calculate
the standard cell potential:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34V −0V
E◦
cell = 0.34V
Step 4: Account for non-standard conditions using the Nernst equation. The
Nernst equation is given by:
E=E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the cell reaction, and
Q is the reaction quotient. In this case, the cell reaction involves 2 moles of
electrons.
Step 5: Calculate the reaction quotient (Q). The reaction quotient (Q) for
the cell reaction is given by:
Q=[Cu]
[Cu2+]
Q=1
0.10 = 10
Step 6: Calculate the cell potential under non-standard conditions. Plugging
in the values into the Nernst equation:
E= 0.34V −0.0592
2log(10)
13
E= 0.34V −0.0296 log(10)
E= 0.34V −0.0296 ×1
E= 0.3104V
Therefore, the cell potential under the given conditions is 0.3104 V.
Question 13
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the reduction half-
reaction is Fe3+(aq) + 3e−→Fe(s) with a standard potential of 0.77 V, and
the oxidation half-reaction is I−(aq)→I2(s) + 2e−with a standard potential of
0.54 V. The concentration of Fe3+ is 0.10 M and the concentration of I−is 0.50
M.
Solution
Step 1: Write the overall cell reaction.
Fe3+(aq) + 2I−(aq)→Fe(s)+I2(s)
Step 2: Determine the standard cell potential.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.77 V −0.54 V
E◦
cell = 0.23 V
Step 3: Calculate the reaction quotient, Q.
Q=[Fe]1[I2]
[Fe3+]1[I−]2
Q=(1)(1)
(0.10)(0.50)2
Q=1
0.025
Q= 40
Step 4: Use the Nernst equation to find the cell potential.
Ecell =E◦
cell −0.0592
nlog(Q)
Where n is the number of moles of electrons transferred, which is 3 in this case.
Ecell = 0.23 V −0.0592
3log(40)
14
Ecell = 0.23 V −(0.01973)(3.69)
Ecell = 0.23 V −0.0727
Ecell = 0.157 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is 0.157 V.
Question 14
Question
A galvanic cell consists of a silver electrode in a 1.0 M AgN O3solution and a
lead electrode in a 0.1 M P b(NO3)2solution. Given that the standard reduction
potential for the Ag+/Ag couple is Eo= 0.80 V and the standard reduction
potential for the P b2+/P b couple is Eo=−0.13 V, calculate the cell potential
at 25
°
C.
Solution
Step 1: Write the half-reactions for each electrode process.
The half-reaction for the silver electrode is: Ag++e−→Ag
The half-reaction for the lead electrode is: P b2+ + 2e−→P b
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation relates the standard cell potential (Eo), the cell potential (E), the re-
action quotient (Q), the gas constant (R), the temperature (T), and the number
of electrons transferred (n) through the equation:
E=Eo−RT
nF ln Q
Given: Eo
Ag+/Ag = 0.80 V, Eo
P b2+/P b =−0.13 V, [Ag+]=1.0 M, [P b2+] =
0.1 M
Balancing the half reactions: Ag++e−→Ag P b2+ + 2e−→P b
The cell reaction is: 2Ag++P b2+ →2Ag +P b
Therefore, the cell potential can be calculated as:
E=Eo
cell −RT
2Fln [Ag]2
[P b2+]
Step 3: Substitute the given values into the Nernst equation and calculate
the cell potential.
E= (0.80 V−(−0.13 V)) −(8.314 J/mol ·K)(298 K)
(2)(96485 C/mol)ln (1.0M)2
0.1M
E= 0.93 V−8.314 ×298
2×96485 ln 1.0
0.1= 0.93 −2474.952
192970 ln 10 V
15
E0.93 −0.0318 V0.8982 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is approxi-
mately 0.8982 V.
Question 15
Question
A concentration cell is set up using two half-cells with silver electrodes. One half-
cell has 1 M Ag+ions, while the other half-cell has 0.1 M Ag+ions. Calculate
the cell potential at 25◦C for this concentration cell. Given that the standard
reduction potential for Ag+ions is E◦= 0.80 V and that the gas constant
R= 8.314 J/(mol·K), and at 25◦C, the Faraday constant F= 96,485 C/mol.
Solution
Step 1: Write the half-reactions for the two half-cells: The half-reaction for the
reduction of Ag+ions to Ag is:
Ag++e−→Ag
Step 2: Write the Nernst equation for the cell potential (Ecell): The Nernst
equation is given by:
Ecell =E◦−RT
nF ln(Q)
where: - Ecell = cell potential - E◦= standard reduction potential - R= gas
constant - T= temperature in Kelvin - n= number of electrons transferred -
F= Faraday constant - Q= reaction quotient
Step 3: Determine the number of electrons transferred (n): Since the half-
reaction involves the transfer of 1 electron, n= 1.
Step 4: Calculate the cell potential using the Nernst equation: For the 1 M
Ag+half-cell:
Q1 M =[products]
[reactants] =1
1= 1
For the 0.1 M Ag+half-cell:
Q0.1 M =[products]
[reactants] =0.1
1= 0.1
Plugging the values into the Nernst equation:
Ecell = 0.80 V −(8.314 J/(mol ·K))(298 K)
(1)(96485 C/mol) ln 1
0.1
Ecell = 0.80 V −(8.314)(298)
96485 ln(10)
16
Ecell = 0.80 V −0.02569 ln(10)
Ecell = 0.80 V −0.02569(2.303)
Ecell = 0.80 V −0.05912
Ecell = 0.7409 V
Therefore, the cell potential for the concentration cell with 1 M Ag+and 0.1
M Ag+half-cells is 0.7409 V.
Question 16
Question
For the half-reaction Fe3+ + e−⇌Fe2+, calculate the cell potential at 25◦C
when [Fe3+] = 0.20 M, [Fe2+] = 0.50 M, and E◦= 0.77 V. Given R = 8.314
J/(mol·K), F = 96485 C/mol, and T = 298 K.
Solution
Step 1: Write the Nernst equation for the cell potential under non-standard
conditions:
E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 2: Determine the number of electrons transferred in the half-reaction.
The number of electrons transferred (n) in the given half-reaction Fe3+ + e−⇌
Fe2+ is 1.
Step 3: Plug in the given values into the Nernst equation:
E= 0.77 −0.0592
1log 0.50
0.20
Step 4: Calculate the cell potential E.
E= 0.77 −0.0592 log(2.50)
E= 0.77 −0.0592(0.3979)
E0.7471 V
Therefore, the cell potential under the given conditions is approximately
0.7471 V.
17
Question 17
Question
Calculate the cell potential at 25
°
C for a galvanic cell where the following half-
reactions occur:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.036 V
Given that the concentration of Cu2+ is 0.10 M and the concentration of Fe3+
is 0.25 M. (Hint: Nernst equation)
Solution
Step 1: Write the overall cell reaction and determine the cell potential using
standard reduction potentials. The overall cell reaction is:
Cu2+(aq) + Fe(s)→Cu(s) + Fe3+(aq)
To determine the cell potential, we subtract the standard reduction potential
of the Fe half-reaction from the standard reduction potential of the Cu half-
reaction:
E◦
cell =E◦
Cu2+/Cu −E◦
Fe3+ /Fe
E◦
cell = 0.34 V −(−0.036 V) = 0.376 V
Step 2: Calculate the reaction quotient, Q, and the Nernst equation. The
reaction quotient is given by:
Q=[Cu2+]
[Fe3+]
Q=0.10
0.25 = 0.40
The Nernst equation is:
Ecell =E◦
cell −0.0592
nlog(Q)
Where nis the number of electrons transferred in the balanced equation.
Step 3: Calculate Ecell using the Nernst equation. Since 2 electrons are
transferred in the balanced equation, n= 2.
Ecell = 0.376 V −0.0592
2log(0.40)
Ecell = 0.376 V −0.0296 log(0.40)
Ecell = 0.376 V −0.0296 ×(−0.3979)
Ecell = 0.376 V + 0.0112
Ecell = 0.387 V
Therefore, the cell potential at 25
°
C for the galvanic cell is 0.387 V.
18
Question 18
Question
A concentration cell is set up with two half-cells as follows:
Half-cell 1: Zn2+(0.10 M) ∥Zn2+(xM)
The measured cell potential is found to be 0.06 V. Calculate the concentration
of Zn2+ in half-cell 2.
Given: Standard reduction potential of Zn2+/Zn = -0.76 V Gas constant
R = 8.314 J/(mol·K) Temperature T = 298 K Faraday’s constant F = 96485
C/mol
Solution
Step 1: Write the Nernst equation for the concentration cell setup:
E=E◦−0.0592
nlog [Zn2+]2
[Zn2+]1
where: - E= measured cell potential = 0.06 V - E◦= standard cell potential
= -0.76 V - n= number of electrons involved in the reaction, which is 2 for
this cell - [Zn2+]1= concentration of Zn2+ in half-cell 1 = 0.10 M - [Zn2+]2=
concentration of Zn2+ in half-cell 2 (to be calculated)
Step 2: Substitute the given values into the Nernst equation:
0.06 = −0.76 −0.0592
2log [Zn2+]2
0.10
Step 3: Solve for [Zn2+]2:
0.06 = −0.76 −0.0296 log [Zn2+]2
0.10
0.82 = −0.0296 log [Zn2+]2
0.10
log [Zn2+]2
0.10 =0.82
−0.0296 =−27.70
[Zn2+]2
0.10 = 10−27.70
[Zn2+]2= 0.10 ×10−27.70
[Zn2+]2= 10−27.70 M
Therefore, the concentration of Zn2+ in half-cell 2 is 10−27.70 M.
19
Question 19
Question
Calculate the cell potential for a galvanic cell with a standard cell potential of
1.15 V at 25
°
C, when [Cu2+] = 0.20 M, [Zn2+] = 1.0 ×10−4M, and [Zn2+/Zn]
= 0.10.
Solution
Step 1: Write the half-reactions for the galvanic cell. The given cell is: Zn(s)
— Zn2+(aq, 1.0 ×10−4M) —— Cu2+(aq, 0.20 M) — Cu(s)
The two half-reactions are: Zn(s) →Zn2+(aq) + 2e−Cu2+(aq) + 2e−→
Cu(s)
Step 2: Write the Nernst equation. The Nernst equation is given by: E =
E◦-0.0592
nlog [Cu2+ ]
[Zn2+ ]
where: E = cell potential at non-standard conditions E◦= standard cell
potential n = number of electrons involved in the cell reaction [Cu2+] = con-
centration of copper ions [Zn2+] = concentration of zinc ions
Step 3: Calculate the cell potential using the Nernst equation. Substitute
the given values into the equation: E = 1.15V - 0.0592
2log 0.20
1.0×10−4
E = 1.15V - 0.0592
2log(2000) E = 1.15V - 0.0592
2×3.301
E = 1.15V - 0.0978 E 1.05 V
Therefore, the cell potential for the given galvanic cell is approximately 1.05
V at 25
°
C.
Question 20
Question
Calculate the cell potential for a concentration cell at 25
°
C that consists of two
half-cells, each with a silver electrode. One half-cell contains a 0.20 M solution
of AgNO3and the other half-cell contains a 0.025 M solution of AgNO3. The
standard reduction potential for the Ag++ e−→Ag half-reaction is 0.80 V.
Solution
Step 1: Write the balanced redox reaction for the concentration cell. The half-
reactions for the silver electrodes are: 1. Anode (oxidation): Ag(s) →Ag++
e−2. Cathode (reduction): Ag++ e−→Ag(s)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
E=E◦−RT
nF ln Q
20
Where: - Eis the cell potential - E◦is the standard cell potential (given:
0.80 V) - Ris the ideal gas constant (8.314 J/(mol*K)) - Tis the temperature in
Kelvin (25
°
C = 298 K) - nis the number of electrons transferred in the balanced
redox reaction (1 for this reaction) - Fis the Faraday constant (96485 C/mol)
-Qis the reaction quotient
Step 3: Calculate the reaction quotient Qfor this cell.
Q=[Ag+]anode
[Ag+]cathode
Q=0.20
0.025
Step 4: Substitute the given values into the Nernst equation and solve for
E.
E= 0.80 −(8.314 ∗298)
1∗96485 ln 0.20
0.025
Step 5: Calculate the value of E.
E= 0.80 −2474.072
96485 ln(8)
E≈0.72 V
Therefore, the cell potential for the concentration cell is approximately 0.72
V.
Question 21
Question
Calculate the cell potential at 25
°
C for 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
Given that the concentrations of Zn2+ and Cu2+ in the cell are 0.10 M and 1.00
M, respectively.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is obtained by
adding the two half-reactions:
Zn(s) + Cu2+(aq)→Cu(s) + Zn2+(aq)
Step 2: Determine 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
21
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 3: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation relates the cell potential under non-
standard conditions to the standard cell potential:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]2
Step 4: Substitute the given values into the Nernst equation.
Ecell = 1.10 V −0.0592
2log 1.00
0.102
Ecell = 1.10 V −0.0298 ×2×log(100) = 1.10 V + 0.0596 = 1.16 V
Thus, the cell potential at 25
°
C for the given galvanic cell is 1.16 V.
Question 22
Question
Calculate the cell potential at 25
°
C for a galvanic cell consisting of a standard
hydrogen electrode (SHE) and a silver electrode, given the following concen-
trations: [Ag+] = 1.0×10−3M and [H+]=0.10 M. The standard reduction
potential for the silver electrode is E0
Ag+/Ag = 0.80 V, and the reduction half-
reaction is Ag++e−→Ag. (Assume one electron is transferred in the cell
reaction).
Solution
Step 1: Write down the balanced redox reaction for the cell:
2H++ 2Ag →2Ag++ H2
Step 2: Determine the standard cell potential using the Standard Gibb’s
Free Energy and the Nernst equation.
The Standard Gibb’s Free Energy (∆G0) is related to the standard cell
potential by the equation:
∆G0=−nF ·E0
cell
where nis the number of moles of electrons transferred (2 in this case) and F
is the Faraday constant (96485 C/mol).
Given the standard reduction potentials: E0
Ag+/Ag = 0.80 V E0
H+/H2= 0
The standard cell potential is:
E0
cell =E0
cathode −E0
anode =E0
Ag+/Ag −E0
H+/H2= 0.80 −0 = 0.80 V
22
Step 3: Calculate the reaction quotient (Q) using the concentrations of the
species involved in the cell reaction:
Q=[Ag+]2
[H+]2
Step 4: Use the Nernst equation to calculate the cell potential at 25
°
C:
Ecell =E0
cell −0.0592
nlog(Q)
Substitute the values we’ve calculated:
Ecell = 0.80 −0.0592
2log (1.0×10−3)2
(0.10)2
Step 5: Calculate the cell potential:
Ecell = 0.80 −0.0296 log (1.0×10−3)2
(0.10)2
Ecell = 0.80 −0.0296 log(0.010)
Ecell = 0.80 −0.0296 × −2
Ecell = 0.80 −0.0592
Ecell = 0.74 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is 0.74 V.
Question 23
Question
Calculate the cell potential for a galvanic cell in which the following reaction
occurs:
Fe2+(0.10M) + Ag+(0.20M)→Fe3+(0.05M) + Ag(s)
Given that the standard cell potential E◦
cell = 0.77 V at 25
°
C, determine the
cell potential when Q= 4.0.
Solution
Step 1: Write the balanced redox reaction for the cell. The given reaction is:
Fe2+ + Ag+→Fe3+ + Ag
Step 2: Write the half-reactions for the cell. The half-reactions are: Anode:
Fe2+ →Fe3+ +e−(oxidation) Cathode: Ag++e−→Ag (reduction)
23
Step 3: Determine the standard cell potential using the standard reduction
potentials. Using the standard reduction potentials, we find E◦
cell =E◦
cathode −
E◦
anode = 0.80 V −(−0.44) V = 1.24 V
Step 4: Calculate the reaction quotient Q. The reaction quotient Q=
[Fe3+][Ag]
[Fe2+][Ag+]=(0.05)(1)
(0.10)(0.20) = 2.5
Step 5: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation is Ecell =E◦
cell −0.0592
nlog(Q). Since
the reaction involves the transfer of 1 electron, the number of moles of electrons
(n) is 1.
Substitute the given values:
Ecell = 1.24 V −0.0592
1log(4.0) = 1.24 V −0.0862 = 1.16 V
Therefore, the cell potential under non-standard conditions is 1.16 V.
Question 24
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a 1.0 M solution of Cu2+. The initial concentration of Cu2+ is 0.10
M, and the cell operates at 25◦C. Calculate the cell potential after the concen-
tration of Cu2+ has decreased to 0.050 M. The standard reduction potential for
Cu2+ + 2e−→Cu is 0.34 V.
Solution
Step 1: Write the cell reaction. The cell reaction for the galvanic cell composed
of a standard hydrogen electrode (SHE) and a copper electrode is:
Cu2+(0.10 M) + 2e−→Cu(s)
Step 2: Determine the initial cell potential 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 (T),
and the number of electrons transferred in the cell reaction (n). The Nernst
equation is given by:
E=E◦−0.0592
nlog(Q)
Given: E◦= 0.34 V T= 25◦C= 298 K [Cu2+]initial = 0.10 M
The reaction quotient Qcan be written as:
Q=[Cu]
1= 0.10 M
24
Plug in the values into the Nernst equation:
Einitial = 0.34 −0.0592
2log(0.10)
Einitial = 0.34 −0.0296 log(0.10)
Einitial = 0.34 −0.0296(−1)
Einitial = 0.34 + 0.0296 = 0.3696 V
Question 25
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to a
nickel electrode. The concentrations of Ni2+ and H+ions in the cell are 0.10 M
and 1.0M, respectively. Given that the standard reduction potential for Ni2+
is −0.25 V, calculate the cell potential at 25◦.
Solution
To find the cell potential, we will use the Nernst equation:
E=E◦−0.0592
nlog(Q)
where: E= cell potential E◦= standard cell potential n= number of electrons
transferred Q= reaction quotient
Step 1: Write the balanced cell reaction and the expression for Q.
The balanced cell reaction is:
Ni2+ + 2e−→Ni(s)
The expression for Qis:
Q=[Ni2+]
[H+]2
Step 2: Calculate the value of Q.
Q=0.10
(1.0)2= 0.10
Step 3: Determine the number of electrons transferred (n). From the bal-
anced cell reaction, we see that 2 electrons are transferred during the reaction.
Step 4: Calculate the cell potential (E). Using the Nernst equation:
E=−0.25 V−0.0592
2log(0.10)
25
E=−0.25 V−0.0592
2× −1.00
E=−0.25 V−(−0.0296)
E=−0.25 V+ 0.0296
E=−0.2204 V
Therefore, the cell potential at 25◦is −0.2204 V.
Question 26
Question
Calculate the cell potential at 25
°
C for a cell in which the reaction is:
2Cr3+(aq) + 2Mn2+(aq)→2Cr2+(aq) + 2Mn3+(aq)
Given the following half-reactions and their standard reduction potentials:
Cr3+ + 3e−→Cr2+ E◦=−0.74 V
Mn3+ +e−→Mn2+ E◦=−1.18 V
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together.
Then, determine the standard cell potential E◦.
2Cr3+(aq) + 2Mn2+(aq)→2Cr2+(aq) + 2Mn3+(aq)
The standard cell potential E◦can be calculated using the standard reduc-
tion potentials:
E◦
cell =E◦
cathode −E◦
anode
Given E◦
Cr2+/Cr3+ =−0.74 V and E◦
Mn2+/Mn3+ =−1.18 V, we can calculate
E◦
cell.
E◦
cell = (−0.74 V) −(−1.18 V) = 0.44 V
Step 2: Calculate the reaction quotient Q to use in the Nernst equation.
Q=[Cr2+]2[Mn3+]2
[Cr3+]2[Mn2+]2
Step 3: Use the Nernst equation to calculate the cell potential at 25
°
C.
E=E◦−0.0592
nlog(Q)
In this case, n= 4 because of the balanced coefficients in the overall reaction.
Since the reaction is at 25
°
C, T= 25 + 273 = 298 K.
E= 0.44 −0.0592
4log(Q) (at 25
°
C)
26
Question 27
Question
A voltaic cell consists of a standard hydrogen electrode (SHE) in one half-cell
and a silver-silver chloride electrode in the other half-cell. The standard reduc-
tion potential for the silver-silver chloride electrode is 0.222 V. If the concen-
tration of AgCl in the cell is 1.0 M, calculate the cell potential at 25
°
C. (Given:
E◦
Step 1: Write the overall cell reaction The overall cell reaction is the sum of the
two half-cell reactions. The half-cell reactions are:
Anode: H2(g)→2H+(aq) + 2e−E◦= 0.000 V
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)E◦= 0.222 V
The overall cell reaction is:
2H2(g) + 2AgCl(s)→2Ag(s) + 2H+(aq) + 2Cl−(aq)
Step 2: Calculate the cell potential The cell potential can be calculated using
the Nernst equation:
E=E◦−0.0592
nlog [Ag+][H+]
[AgCl]
where - nis the number of moles of electrons transferred in the balanced
equation (in this case, 2), - E◦is the standard cell potential, - [Ag+] and [H+]
are the concentrations of silver ions and hydrogen ions, respectively, in the cell,
- [AgCl] is the concentration of silver chloride.
Since the reaction is at equilibrium, Q=[Ag+][H+]
[AgCl] .
Given the concentrations of AgCl and E◦
cell, we can solve for Eusing the
Nernst equation. Substituting the values:
E= (0.222 V) −0.0592
2log [Ag+][H+]
[AgCl]
E= (0.222 V) −0.0296 log [Ag+][H+]
1.0
Step 3: Calculate the cell potential Since the cell is at equilibrium, Q=K
and the reaction quotient Qis equal to the equilibrium constant Keq = 10−10
for the dissociation of water. We can use the equilibrium constant expression
for water to relate the concentrations of Ag+and H+.
Keq = [H+][OH−] = 10−14 M2
Given that [H+] = xM and [Ag+] = xM, the equation becomes:
x2= 10−14
27
x= 10−7
Therefore, [H+] = [Ag+] = 10−7M. Substituting this back into the Nernst
equation:
E= (0.222 V) −0.0296 log 10−7×10−7
1.0
E= 0.222 V −0.0296 log10−14
E= 0.
Question 28
Question
A half-cell reaction for a particular redox reaction involving silver is given by:
Ag+(aq)+e−→Ag(s)
If the standard reduction potential for this half-cell is +0.80 V, calculate the
cell potential at 25C when the concentration of Ag+is 0.10 M. Given that the
universal gas constant is 8.31 J/(mol·K) and the Faraday constant is 96,485
C/mol.
Solution
Step 1: Write the half-cell reaction and the Nernst equation.
Half-cell reaction: Ag+(aq)+e−→Ag(s)
Nernst equation: E=E◦−0.0591
nlog(Q) where:
–Eis the cell potential
–E◦is the standard cell potential
–nis the number of moles of electrons transferred in the balanced
redox reaction
–Qis the reaction quotient
Step 2: Calculate the standard cell potential (E◦) using the given standard
reduction potential.
Standard reduction potential for Ag half-cell: E◦
Ag+/Ag = +0.80 V
Therefore, E◦=E◦
Ag+/Ag = +0.80 V
Step 3: Calculate the reaction quotient (Q) using the concentration of Ag+.
Given concentration of Ag+([Ag+]): 0.10 M
28
Since the half-cell reaction involves the reduction of Ag+,Q= [Ag+].
Therefore, Q= 0.10 M
Step 4: Calculate the cell potential (E) at 25C using the Nernst equation.
Number of moles of electrons transferred (n) in the half-cell reaction: 1
Gas constant (R) = 8.31 J/(mol·K)
Temperature (T) = 25C = (25 + 273) K = 298 K
Plugging in the values into the Nernst equation:
E=E◦−0.0591
nlog(Q)
E= 0.80 −0.0591
1log(0.10)
E= 0.80 −0.0591 ×log(0.10)
E≈0.80 −0.0591 ×(−1.00)
E≈0.80 + 0.0591
E≈0.8591 V
Therefore, the cell potential at 25C when the concentration of Ag+is 0.10
M is approximately 0.8591 V.
Question 29
Question
A fuel cell uses the reaction 2H2(g) + O2(g)→2H2O(l) to generate electricity.
At 25
°
C, the concentration of H2O(l) is 1.0 M, O2(g) is 0.20 M, and H2(g) is
0.30 M.
Calculate the cell potential at standard conditions for the fuel cell reaction.
Given: E◦
cell = 1.23 V, R= 8.314 J/(mol·K), and F= 96485 C/mol.
Solution
Step 1: Write the half reactions and determine the cell potential under standard
conditions.
The oxidation half reaction is:
2H2(g)→4H+(aq)+4e−
The reduction half reaction is:
O2(g)+4H+(aq)+4e−→2H2O(l)
29
The cell potential under standard conditions can be calculated as:
E◦
cell =E◦
reduction −E◦
oxidation
Given E◦
cell = 1.23 V, we can calculate E◦
reduction and E◦
oxidation.
Step 2: Calculate the cell potential under standard conditions.
Now, we substitute the standard reduction potentials for the reactions to
find E◦
cell. Given that E◦
cell =E◦
reduction −E◦
oxidation, we have:
1.23 = E◦
reduction −E◦
oxidation
The standard reduction potential for the reduction half reaction is E◦
reduction =
1.23 V. The standard reduction potential for the oxidation half reaction is
E◦
oxidation = 0 V (by definition).
Therefore, 1.23 = 1.23 −0 V,
E◦
cell = 1.23 V
Question 30
Question
Calculate the cell potential at 25
°
C for a galvanic cell with a silver electrode
immersed in a 1.0 M AgNO3solution connected to a hydrogen electrode at pH
2. The standard reduction potential for the Ag+/ Ag electrode is +0.80 V, and
the standard reduction potential for the H+/ H2electrode is 0.00 V. Assume
that the activity coefficient for the silver ion is 0.85.
Solution
Step 1: Write the overall cell reaction. The cell reaction consists of the reduc-
tion half-reaction at the Ag electrode and the oxidation half-reaction at the H2
electrode. The overall reaction is:
Ag++ e−→Ag (E◦= +0.80 V)
2H++ 2e−→H2(E◦= 0.00 V)
Since the Ag half-reaction has a higher standard reduction potential, it will
be the reduction half-reaction in this cell.
Step 2: Write the Nernst equation for the cell potential. The Nernst equa-
tion relates the standard cell potential to the cell potential under non-standard
conditions:
E=E◦−0.0592
nlog Q
K
30
where Eis the cell potential under non-standard conditions, E◦is the stan-
dard cell potential, nis the number of electrons transferred in the balanced cell
reaction, Qis the reaction quotient, and Kis the equilibrium constant.
Step 3: Calculate the cell potential. First, calculate the reaction quotient,
Q, for the cell reaction:
Q=[Ag][H2]2
[Ag+][H+]2=1
1.0×(0.01)2= 100
Next, substitute the given values into the Nernst equation and solve for E:
E= 0.80 V −0.0592
1log 100
1= 0.80 V −0.0592 ×2=0.6816 V
Therefore, the cell potential at 25
°
C for the galvanic cell described is 0.6816
V.
Question 31
Question
A voltaic cell consists of a silver electrode immersed in a 0.20 M AgNO3solution
and a zinc electrode immersed in a 0.10 M Zn(NO3)2solution. The measured
cell potential is 1.56 V at 25◦C. Calculate the solubility of silver ion in water at
this temperature.
Solution
Step 1: Write the half-reactions for the cell reaction. The cell reaction can be
represented as:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Step 2: Write the Nernst equation for the cell reaction. The Nernst equation
for the cell reaction is:
E=E◦−0.0592
nlog [Zn2+]
[Ag+]2
Step 3: Substitute the given values into the Nernst equation. Given: E=
1.56 V
E◦= 1.56 V (since the cell is at standard conditions)
[Zn2+]=0.10 M
[Ag+]=0.20 M
n= 2
Substitute these values into the Nernst equation:
1.56 = 1.56 −0.0592
2log 0.10
0.202
31
Step 4: Solve for the solubility of silver ion. Simplify the equation:
0.0592
2log 0.10
0.202= 0
log 0.10
0.202= 0
log 0.10
0.04 = 0
log 2.5
1= 0
log 2.5 = 0
Therefore, log 2.5 = 0 at 25◦C means that the solubility of silver ion in water
at this temperature is 2.5 M.
Question 32
Question
A concentration cell is set up with two half-cells containing silver electrodes. One
half-cell has a silver electrode in a 0.10 M solution of AgNO3, while the other
half-cell has a silver electrode in a 0.0010 M solution of AgNO3. Calculate the
cell potential at 25
°
C for this concentration cell. (Given: E◦
cell(Ag+/Ag) = 0.80
V and R= 8.314 J/mol ·K)
Solution
Step 1: Write the half-reactions for the cathode and anode.
Cathode: Ag++e−→Ag E◦= +0.80 V
Anode: Ag →Ag++e−E◦=−0.80 V
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Reductant]cathode
[Oxidant]anode
where
Ecell = cell potential (unknown)
E◦
cell = standard cell potential = 0.80 V
0.0592 = constant
n←(1 at 25
°
C)
[Reductant]cathode = 0.10 M ←(higher concentration)
[Oxidant]anode = 0.0010 M ←(lower concentration)
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Step 3: Calculate the cell potential.
∆Ecell = 0.80 −0.0592
1log 0.10
0.0010
∆Ecell = 0.80 −0.0592 log(100) = 0.80 −0.0592 ×2≈0.68 V
Thus, the cell potential for this concentration cell at 25
°
C is approximately
0.68 V.
Question 33
Question
A galvanic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and a copper electrode in a 0.10 M Cu2+ solution as the cathode. At
25
°
C, the standard reduction potential for the Cu2+/Cu half-reaction is +0.34
V. Calculate the cell potential when the concentration of Cu2+ at the cathode
is 0.050 M. (Assume all other concentrations are 1.0 M.)
Solution
Step 1: Write the half-reaction for each electrode. The half-reaction for the
standard hydrogen electrode (anode) is:
2H++ 2e−→H2(g)
The half-reaction for the copper electrode (cathode) is:
Cu2+ + 2e−→Cu(s)
Step 2: Write the overall cell reaction. The overall cell reaction is the sum
of the two half-reactions:
2H+(aq) + Cu2+(aq) →H2(g) + Cu(s)
Step 3: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Given that the standard reduction potential for the Cu2+/Cu half-reaction is
+0.34 V, and the standard reduction potential for the standard hydrogen elec-
trode is 0 V, we have:
E◦
cell = +0.34 V −0 V = +0.34 V
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Step 4: Calculate the cell potential under non-standard conditions. The
Nernst equation relates the cell potential at non-standard conditions to the
standard cell potential:
Ecell =E◦
cell −0.0592
nlog Q
where - Ecell is the cell potential under non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of electrons transferred in the balanced
cell reaction, - Qis the reaction quotient.
In this case, n= 2 since 2 electrons are transferred in the overall cell reaction.
The reaction quotient Qis given by:
Q=[Cu]
[H+]2
At equilibrium, Q=K= 1 based on the balanced equation. Substituting the
known values into the Nernst equation:
Ecell = +0.34 V −0.0592
2log0.0502/12
Ecell = +0.34 V −0.0296 log(0.050)
Ecell = +0.34 V −0.0296 × −1.3010
Ecell = +0.34 V + 0.0387
Ecell = +0.3787 V
Therefore, the cell potential when the concentration of Cu2+ at the cathode
is 0.050 M is +0.3787 V.
Question 34
Question
A concentration cell is set up with two half-cells as follows: the first half-cell
contains a silver electrode dipping into a 0.10 M solution of AgNO3, and the
second half-cell contains a silver electrode dipping into a 0.0010 M solution
of AgNO3. Calculate the voltage of the concentration cell at 25
°
C. (Given:
E◦
Ag+/Ag = 0.80 V at 25
°
C)
Solution
Step 1: Write the half-reactions occurring in each half-cell. In this case, both
half-cells involve the reduction of Ag+ion to Ag(s):
Ag++e−→Ag
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Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Ag+]
[Ag+]standard
where - Eis the cell potential, - E◦is the standard cell potential, - nis the
number of moles of electrons transferred in the balanced half-reaction, and -
[Ag+] and [Ag+]standard are the concentrations of Ag+ions in the two half-cells.
Step 3: Calculate the standard cell potential (E◦) using the given standard
reduction potential of Ag+:
E◦=E◦
Ag+/Ag = 0.80 V
Step 4: Calculate the cell potential (E) for the concentration cell using the
Nernst equation: For the first half-cell: [Ag+]=0.10 M For the second half-cell:
[Ag+]=0.0010 M
E1= 0.80 −0.0592
1log 0.10
1.0= 0.80 −0.0592 log 0.10
E2= 0.80 −0.0592
1log 0.0010
1.0= 0.80 −0.0592 log 0.0010
The cell potential is given by the difference between both half-cells:
Ecell =E1−E2= (0.80 −0.0592 log 0.10) −(0.80 −0.0592 log 0.0010)
Question 35
Question
The standard reduction potential for the half-reaction F e3+(aq)+e−→F e2+(aq)
is given as +0.77 V. Calculate the equilibrium constant (K) for the cell reaction
F e3+(aq) + F e(s)→F e2+(aq) + F e2+(aq) at 25
°
C using the Nernst equation.
Solution
Step 1: Write the cell reaction and the overall cell equation. The cell reaction
is:
F e3+(aq) + F e(s)→F e2+(aq) + F e(s)
The overall cell equation is:
F e3+(aq) + F e(s)→F e2+(aq) + F e2+(aq)
Step 2: Identify the half-reactions. The half-reactions involved in the cell
reaction are: 1. Oxidation half-reaction: F e(s)→F e2+(aq)+2e−2. Reduction
half-reaction: F e3+(aq) + e−→F e2+(aq)
35
Step 3: Calculate the standard cell potential. The standard cell potential
(E◦
cell) can be calculated as the difference between the standard reduction po-
tentials of the reduction and oxidation half-reactions:
E◦
cell =E◦
red −E◦
ox
E◦
cell = (+0.77V)−0
E◦
cell = +0.77V
Step 4: Calculate the equilibrium constant (K) using the Nernst equation.
The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog Q
where: - Ecell is the cell potential at non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of moles of electrons transferred, - Q
is the reaction quotient.
For this cell reaction, n= 1 because one electron is transferred. And, at
equilibrium, Q=K.
Substitute the values into the Nernst equation:
0.77V= 0.77V−0.0592
1log K
0 = −0.0592 log K
log K= 0
K= 100
K= 1
Therefore, the equilibrium constant (K) for the cell reaction is 1 at 25
°
C.
36
