CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Nernst equation
Question Bank - Set 3
Liberty University
Question 1
Question
Calculate the cell potential for a galvanic cell where the following half-reactions
occur:
Zn2+(aq) + 2e−→Zn(s) E◦=−0.76 V
Cu2+(aq) + 2e−→Cu(s) E◦= 0.34 V
Given that the concentration of Zn2+ is 0.010 M and the concentration of Cu2+
is 0.100 M. (Assume the temperature is 25
°
C)
Solution
Step 1: Write the overall cell reaction and determine the cell potential at stan-
dard conditions. The overall cell reaction is obtained by adding the two half-
reactions together:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard cell potential (E◦) can be found by subtracting the reduction
potential of the anode from the reduction potential of the cathode:
E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.76 V) = 1.10 V
Step 2: Calculate the cell potential under non-standard conditions using
the Nernst equation. The Nernst equation relates the cell potential under non-
standard conditions to the standard cell potential:
Ecell =E◦
cell −0.0592
nlog Q
Where nis the number of electrons transferred in the cell reaction and Qis the
reaction quotient.
For the given cell reaction, n= 2 since two electrons are transferred. The
reaction quotient Qis calculated by dividing the concentrations of the products
raised to their stoichiometric coefficients by the concentrations of the reactants
raised to their stoichiometric coefficients:
Q=[Zn2+]
[Cu2+]=0.010
0.100 = 0.100
Now, plug in the values into the Nernst equation:
Ecell = 1.10 V −0.0592
2log(0.100)
Ecell = 1.10 V −0.0296 log(0.100) = 1.10 V −(−0.0296 ×1) = 1.1296 V
Therefore, the cell potential under the given conditions is 1.13 V.
Question 2
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode. The standard reduction potential for the copper electrode is E◦=
0.34 V. The concentration of Cu2+ ions is 0.10 M in the half-cell containing
the copper electrode. Given that the hydrogen electrode operates at standard
conditions ([H+] = 1.0 M, PH2= 1.0 atm), calculate the cell potential at 25◦C.
Solution
Step 1: Write the half-reactions for the cell The overall reaction for the galvanic
cell can be represented as:
2H+(aq) + 2e−→H2(g)
Cu2+(aq) + 2e−→Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard reduc-
tion potentials The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cathode = 0 V for the SHE, and E◦
anode = 0.34 V for the copper
electrode:
E◦
cell = 0 −0.34 = −0.34 V
Step 3: Calculate the reaction quotient (Q) Given that [Cu2+] = 0.10 M in
the copper half-cell, and [H+]=1.0 M in the hydrogen half-cell:
Q=[Cu2+]
1= 0.10
2
Step 4: Calculate the cell potential (Ecell) at non-standard conditions using
the Nernst equation The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the balanced redox reaction.
Step 5: Calculate the number of moles of electrons transferred From the
balanced half-reactions, we see that 2 moles of electrons are transferred.
Step 6: Substitute the values into the Nernst equation
Ecell =−0.34 −0.0592
2log(0.10) = −0.34 −0.0296 ×(≈ −1) = −0.0344 V
Therefore, the cell potential at 25◦C for the given galvanic cell is approxi-
mately −0.0344 V.
Question 3
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a solution containing Cu2+ ions. The standard reduction potential
for the half-reaction Cu2+(aq) + 2e−→Cu(s) is E◦= 0.34 V. If the concentra-
tion of Cu2+ ions in the solution is 0.010 M, calculate the cell potential at 25
°
C
using the Nernst equation.
Solution
Step 1: Write the half-reaction for the cell and the Nernst equation.
The half-reaction for the cell is:
2H+(aq)+2e−→H2(g)
The Nernst equation relates the cell potential (Ecell) to the standard cell po-
tential (E◦), the reaction quotient (Q), the number of electrons transferred (n),
the Faraday constant (F), and the temperature (T):
Ecell =E◦−RT
nF ln(Q)
Step 2: Calculate the reaction quotient (Q).
The reaction quotient Qis calculated using the concentrations of the species
involved in the cell reaction. For our cell:
Q=[H2]
[H+]2= 1
Step 3: Calculate the cell potential (Ecell).
Given: E◦= 0.34 V
3
T= 298 K
n= 2 (since 2 electrons are transferred in the cell reaction)
F= 96485 C/mol
Using the Nernst equation:
Ecell = 0.34 −(8.314 J/mol·K)(298 K)
(2)(96485 C/mol) ln(1)
Ecell = 0.34 −(8.314)(298)
2(96485) ×0
Ecell = 0.34 V
Therefore, the cell potential of the galvanic cell at 25
°
C is 0.34 V.
Question 4
Question
Calculate the cell potential for the following reaction at 25
°
C:
2F e3+(aq)+2e−→2F e2+(aq)
Given that the standard reduction potentials are E◦
Fe3+ /Fe2+ = 0.77 V and the
concentration of Fe3+ is 0.1 M while the concentration of Fe2+ is 0.01 M.
Solution
Step 1: Write the Nernst equation for the cell potential:
E=E◦−RT
nF ln(Q)
where - Eis the cell potential, - E◦is the standard cell potential, - Ris the
ideal gas constant (8.314 J/(mol
·
K)), - Tis the temperature in Kelvin (25
°
C =
298 K), - nis the number of moles of electrons transferred (2 in this case), - F
is the Faraday constant (96485 C/mol), - Qis the reaction quotient.
Step 2: Calculate the cell potential using the Nernst equation: Plugging in
the values we have:
E= 0.77 −(8.314)(298)
2(96485) ln (0.01)2
0.12
E= 0.77 −2465.972
192970 ln(0.001)
E= 0.77 −0.0127 ln(0.001)
E= 0.77 −0.0127(−6.907)
E= 0.77 + 0.0874
E= 0.8574 V
Therefore, the cell potential for the given reaction at 25
°
C is 0.8574 V.
4
Question 5
Question
A galvanic cell consists of a silver electrode in 1.0 M Ag+ions and a copper
electrode in 1.0 M Cu2+ ions. The cell has an emf of 1.02 V at 25
°
C. Calculate
the standard cell potential, E◦
cell, for this galvanic cell. Given: E◦
Ag+/Ag = 0.80
V and E◦
Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions and their standard electrode potentials. The
standard cell potential can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
where Q is the reaction quotient and n is the number of moles of electrons
transferred in the balanced cell reaction.
The two half-reactions involved are:
Ag++ e−→Ag E◦
Ag+/Ag = 0.80 V
Cu2+ + 2e−→Cu E◦
Cu2+/Cu = 0.34 V
Step 2: Identify the balanced cell reaction and calculate the standard cell
potential. The balanced cell reaction is the sum of the two half-reactions with
multiplication factors to equalize the electrons:
2Ag++ Cu →2Ag + Cu2+
The standard cell potential is the difference between the standard reduction
potentials of the two half-reactions:
E◦
cell = E◦
Ag+/Ag - E◦
Cu2+/Cu
E◦
cell = 0.80 V −0.34 V = 0.46 V
Therefore, the standard cell potential for this galvanic cell is 0.46 V.
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode and a zinc electrode.
The concentration of Zn2+ ions is 0.10 M, and the pressure of hydrogen gas is
0.80 atm. Calculate the cell potential at 25
°
C using the Nernst equation. Given
that the standard reduction potential for the reduction of Zn2+ to Zn is -0.76
V.
5
Solution
Step 1: Write the overall cell reaction and the Nernst equation. The overall cell
reaction is given by:
Zn2+ + 2e−→Zn(s)
The Nernst equation is given by:
E=Eo−0.0592
nlog Q
where: - Eis the cell potential, - Eois the standard cell potential, - nis the
number of moles of electrons exchanged in the balanced cell reaction, and - Q
is the reaction quotient.
Step 2: Calculate the reaction quotient (Q). The reaction quotient is given
by:
Q=[Zn2+]
P(H2)2
Substitute the given values:
Q=0.10
0.802
Step 3: Calculate the cell potential (E). Given: - Eo=−0.76 V, - n= 2
(from the balanced cell reaction), - Q=0.10
0.802.
Substitute the values in the Nernst equation:
E=−0.76 −0.0592
2log 0.10
0.802
Step 4: Calculate the cell potential.
E=−0.76 −0.0296 log 0.10
0.802
Finally, solve for the cell potential.
Question 7
Question
A concentration cell is constructed using two silver/silver chloride half-cells.
One half-cell contains a solution with [Cl−]=0.10 M, and the other contains
a solution with [Cl−] = 1.0 M. If the standard reduction potential for the
Ag/AgCl electrode is E◦= 0.22 V, calculate the cell potential at 25◦C.
6
Solution
Step 1: Write the half-reactions and the Nernst equation for each half-cell.
Anode: AgCl(s) + e−→Ag(s) + Cl−Eanode =E◦
Cathode: AgCl(s) + e−→Ag(s) + Cl−Ecathode =E◦
The Nernst equation is:
E=E◦−0.0592
nlog [products]
[reactants]
where nis the number of electrons transferred.
Step 2: Calculate the cell potential for each half-cell using the Nernst equa-
tion.
Eanode = 0.22 V−0.0592
1log 0.10
1.0= 0.22 V−0.0592 log(0.10) = 0.22 V−0.0592×(−1) = 0.2792 V
Ecathode = 0.22 V−0.0592
1log 1.0
0.10= 0.22 V−0.0592 log(10) = 0.22 V−0.0592×1=0.1608 V
Step 3: Calculate the overall cell potential. The overall cell potential is the
difference between the cathode potential and the anode potential:
Ecell =Ecathode −Eanode = 0.1608 V −0.2792 V = −0.1184 V
Therefore, the cell potential at 25◦C for the concentration cell with differing
chloride ion concentrations is −0.1184 V.
Question 8
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a copper electrode. The concentration of Cu2+ ions is 0.010 M in the copper
half-cell. The cell potential at 25
°
C is measured to be 0.47 V. Calculate the
standard reduction potential of the copper half-cell.
Given: E◦
SHE = 0.00 V, n= 2
Solution
Step 1: Write the overall cell reaction for the galvanic cell. The cell reaction is:
Cu2+ + 2e−→Cu(s)
Step 2: Write the Nernst equation. The Nernst equation is:
E=E◦−0.0592
nlog Q
7
where: - Eis the cell potential at 25
°
C (0.47 V), - E◦is the standard cell
potential, - nis the number of moles of electrons transferred in the balanced
cell reaction (2 in this case), - Qis the reaction quotient.
Step 3: Calculate the reaction quotient, Q. Since the only ions present in
the copper half-cell are Cu2+ and Cu(s), we have:
Q=[Cu]
[Cu2+]=1
0.010 = 100
Step 4: Substitute the given values into the Nernst equation and solve for
E◦.
0.47 = E◦−0.0592
2log 100
0.47 = E◦−0.0592 ×2×2
0.47 = E◦−0.2368
E◦= 0.47 + 0.2368 = 0.7068
Therefore, the standard reduction potential of the copper half-cell is 0.7068
V.
Question 9
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a silver-
silver chloride electrode. The standard reduction potential for the silver-silver
chloride electrode is E◦= 0.222 V. If the concentration of chloride ions is 0.1
M and the pH of the solution is 3, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction involving the silver-silver chloride elec-
trode and standard hydrogen electrode (SHE). The overall cell reaction is:
AgCl(s) + e−→Ag(s) + Cl−
2H++ 2e−→H2(g)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials. The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦(Ag++ e−→Ag) −E◦(H++ e−→H2)
E◦
cell = 0.222 V −0.000 V
E◦
cell = 0.222 V
8
Step 3: Calculate the cell potential (Ecell) using the Nernst equation. The
Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where Qis the reaction quotient and nis the number of electrons transferred
in the cell reaction.
Step 4: Calculate the reaction quotient (Q) for the cell reaction. Since the
half-reactions are already balanced, Qsimplifies to:
Q=[Ag+][H+]2
[Cl−]2
Plugging in the given values:
Q=[0][10−3]2
[10−1]2
Q= 0
Step 5: Substitute the values of E◦
cell,Q, and ninto the Nernst equation and
solve for Ecell. Since Qis zero, the term 0.0592/n log(Q) becomes zero. Thus:
Ecell =E◦
cell −0=0.222 V
Therefore, the cell potential at 25
°
C is 0.222 V.
Question 10
Question
A concentration cell is set up using two Ag/AgCl electrodes at 25
°
C. One half-
cell contains a 0.1 M solution of NaCl and the other half-cell contains a 0.01 M
solution of NaCl. Calculate the cell potential of this concentration cell.
Given: Standard electrode potential of Ag/AgCl electrode (E
°
) = 0.222 V
Temperature (T) = 25
°
C
Solution
Step 1: Write the half-reactions for the cell: The half-cell reactions for the
concentration cell are:
Anode: AgCl(s)→Ag(s) + Cl−(aq)
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
E=E−0.0592
nlog [Ox]
[Red]
9
where: E= cell potential E= standard cell potential n= number of moles of
electrons transferred in the balanced equation [Ox] = concentration of oxidized
form [Red] = concentration of reduced form
Step 3: Calculate the cell potential for the anode half-cell: For the anode
half-cell with a 0.1 M solution of NaCl: n= 1 (since one electron is transferred
in the half-reaction) [Ox] = [Ag+]=0.1 M [Red] = [Ag+]=0.01 M Substitute
the values into the Nernst equation:
Eanode = 0.222 −0.0592
1log 0.1
0.01
Eanode = 0.222−0.0592 log(10) = 0.222−0.0592×1=0.222−0.0592 = 0.1628 V
Step 4: Calculate the cell potential by considering the difference in electrode
potentials: The total cell potential is the difference between the cathode and
anode potentials:
Ecell =Ecathode −Eanode
Since the cathode and anode are identical in this case, Ecathode = 0.222 V
Therefore, the cell potential is:
Ecell = 0.222 −0.1628 = 0.0592 V
Therefore, the cell potential of this concentration cell is 0.0592 V.
Question 11
Question
A concentration cell is set up using two half-cells. The first half-cell contains
a copper electrode dipped in a 1.0 M Cu2+ solution at 25
°
C, while the second
half-cell contains a copper electrode dipped in a 0.1 M Cu2+ solution at the
same temperature. Calculate the cell potential of this concentration cell.
Given: E◦
cell = 0.34 V, R= 8.31 J/mol ·K, and T= 298 K.
Solution
Step 1: Write the overall cell reaction and equation for cell potential. The overall
cell reaction for the concentration cell can be represented as: Cu2+(1.0 M,cathode) →
Cu2+(0.1 M,anode).
The cell potential can be calculated using the Nernst equation: Ecell =
E◦
cell −0.0592
nlog [Cu2+]anode
[Cu2+]cathode
Given that n= 2 for this cell.
Step 2: Calculate the cell potential. We can now plug in the values into the
Nernst equation: Ecell = 0.34 V −0.0592
2log 0.1
1.0
Ecell = 0.34 V −0.0296 log(0.1)
10
Ecell = 0.34 V −0.0296 ×(−1)
Ecell = 0.34 V + 0.0296
Ecell = 0.3696 V
Therefore, the cell potential of the concentration cell is 0.3696 V.
Question 12
Question
Calculate the cell potential for the reaction occurring in a galvanic cell with the
following half-reactions at 25
°
C:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
Given that the initial concentrations are Zn2+ = 1.0 M, Fe3+ = 0.1 M, and
[Fe2+]=1.0×10−3M. Assume the temperature is 25
°
C.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together.
The overall cell reaction is:
Zn2+(aq) + Fe3+(aq)→Zn(s) + Fe2+(aq)
Step 2: Find the standard cell potential, E◦
cell, using the standard reduction
potentials.
E◦
cell =E◦
cathode −E◦
anode
=E◦
Fe3+ /Fe2+ −E◦
Zn2+/Zn
= (−0.04 V) −(−0.76 V)
= 0.72 V
Step 3: Calculate the reaction quotient, Q, using the initial concentrations
given.
Q=[Fe2+]final
[Zn2+]final[Fe3+]final
Since [Zn2+] is used up in the cell, we can assume that [Zn2+]final = 0. This
simplifies the expression to:
Q=[Fe2+]final
[Fe3+]final
Converting the mole ratio of Fe to Fe2+ from the balanced equation, Fe3+ +
3e−→Fe, we find that the moles of Fe that are produced are equal to the moles
of Fe2+ produced. Therefore,
[Fe2+]final = [Fe3+]initial −[Fe3+]final
11
Substitute the given values into the equation and calculate the final concentra-
tion of Fe2+.
Step 4: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced equation.
Since this cell involves the transfer of 3 moles of electrons, n= 3. Substitute
the values into the Nernst equation, along with the calculated value of Q, to
find the cell potential at 25
°
C.
Question 13
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn2+(0.10M) + Cu(s)→Zn(s) + Cu2+(1.0M)
Given: E◦
cell = 1.10 V, E◦
Zn2+/Zn =−0.76 V, and E◦
Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions for the oxidation and reduction processes:
Zn(s)→Zn2+(aq)+2e−E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Write the overall cell reaction equation:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
Step 3: Calculate E◦
cell using the standard cell potential values:
E◦
cell =E◦
reduction −E◦
oxidation
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 4: The Nernst equation relates the cell potential to the reaction quotient
Qand the standard cell potential:
E=E◦
cell −0.0592
nlog(Q)
where nis the number of moles of electrons transferred in the balanced redox
reaction.
Step 5: Calculate the reaction quotient Qusing the initial concentrations:
Q=[Zn]
[Cu2+]=0.10
1.0= 0.10
12
Step 6: Substitute the given values and calculated Qinto the Nernst equation
to find the cell potential:
E= 1.10 −0.0592
2log(0.10)
E= 1.10 −0.0296 ×(−1)
E= 1.10 + 0.0296 = 1.13 V
Therefore, the cell potential for the reaction at 25
°
C is 1.13 V.
Question 14
Question
A voltaic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and a silver ion electrode as the cathode. The concentration of Ag+(aq)
is 1.0×10−4M in the cathode compartment. If the measured cell potential is
0.54 V at 25◦C, calculate the concentration of H+(aq) in the anode compart-
ment.
Solution
Step 1: Write the cell reaction that occurs in the voltaic cell. The cell reaction
is:
2H+(aq) + 2e−→H2(g)
Step 2: Determine the standard cell potential (E◦) for the cell reaction.
Given that the standard electrode potential for SHE is 0.00 V and for Ag+(aq)|Ag(s)
is 0.80 V, the standard cell potential is:
E◦=E◦
cathode −E◦
anode = 0.80 V −0.00 V = 0.80 V
Step 3: Use the Nernst equation to relate cell potential to concentrations of
ions. The Nernst equation is given by:
E=E◦−0.0592
nlog [H+(aq)]anode
[H+(aq)]cathode
Where Eis the cell potential, E◦is the standard cell potential, nis the number
of electrons transferred, and [H+(aq)]anode and [H+(aq)]cathode are the concen-
trations of H+(aq) in the anode and cathode compartments, respectively.
Step 4: Substitute the known values into the Nernst equation. Given that
E= 0.54 V, E◦= 0.80 V, n= 2 (since 2 electrons are transferred in the cell
13
reaction), and [H+(aq)]cathode = [H+(aq)]SHE = 1.0×10−7M (standard pH
scale for SHE), we can solve for [H+(aq)]anode.
0.54 = 0.80 −0.0592
2log [H+(aq)]anode
1.0×10−7
0.54 = 0.80 −0.0296 log [H+(aq)]anode
1.0×10−7
0.0296 log [H+(aq)]anode
1.0×10−7= 0.26
log [H+(aq)]anode
1.0×10−7=0.26
0.0296 ≈8.78
[H+(aq)]anode
1.0×10−7= 108.78
[H+(aq)]anode ≈108.78 ×1.0×10−7
[H+(aq)]anode ≈2.51 ×101M
Therefore,
Question 15
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) and
a copper electrode. The concentration of Cu2+ ions in the copper half-cell is
0.10 M. The cell potential is measured to be 0.46 V at 25
°
C. Calculate the
standard reduction potential for the copper half-cell.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction for the electro-
chemical cell is:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the Nernst equation. The Nernst equation relates the cell
potential (Ecell) to the standard cell potential (E◦), the gas constant (R), the
temperature (T), the number of electrons transferred (n), and the concentrations
of the species involved. The Nernst equation is given by:
Ecell =E◦−RT
nF ln (Q)
where Ecell = 0.46 V, R= 8.314 J/mol ·K, T= 298 K, n= 2 (number of
electrons transferred), F= 96,485 C/mol, and Qis the reaction quotient.
Step 3: Calculate the reaction quotient. The reaction quotient Qcan be cal-
culated using the concentrations of products and reactants. Since the standard
14
hydrogen electrode has a hydrogen ion concentration of 1.0 M and a pressure of
1 atm, the reaction quotient is:
Q=[Cu](H+)2
[Cu2+]=1×(1.0)2
0.10 = 10
Step 4: Substitute values into the Nernst equation. Now, substitute the
given values and the reaction quotient into the Nernst equation:
0.46 = E◦−(8.314 ·298)
2·96485 ln(10)
Step 5: Solve for the standard reduction potential. Solve the equation for
E◦to find the standard reduction potential for the copper half-cell:
E◦= 0.46 + (8.314 ·298)
2·96485 ln(10)
E◦= 0.46 + 2471.672
192970 ln(10)
E◦= 0.46 + 0.03173 ln(10)
E◦= 0.46 + 0.03173 ×2.303
E◦≈0.535 V
Therefore, the standard reduction potential for the copper half-cell is ap-
proximately 0.535 V.
Question 16
Question
A redox reaction with an equilibrium constant K= 1.0×10−5occurs in a
galvanic cell at 25◦C. If the concentration of Mn2+ is 0.010 M and the concen-
tration of Mn3+ is 0.040 M, calculate the cell potential at standard conditions
and the cell potential at the given concentrations using the Nernst equation.
Solution
Step 1: Write the half-reactions and calculate the standard cell potential: The
half reactions are:
Mn3+ +e−→Mn2+ E◦=−1.18 V
We start by calculating the standard cell potential, E◦
cell:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 −(−1.18) = 1.18 V
15
Step 2: Calculate the cell potential at non-standard conditions using the
Nernst equation: The Nernst equation is given by:
Ecell =E◦
cell −0.0591
nlog [Mn2+]
[Mn3+]
where nis the number of electrons transferred in the balanced half-reaction. In
this case, n = 1.
Substitute the given values into the Nernst equation:
Ecell = 1.18 −0.0591
1log 0.010
0.040
Ecell = 1.18 −0.0591 log(0.25) = 1.18 −0.0591 ×(−0.602) = 1.215 V
Therefore, the cell potential at standard conditions is 1.18 V and at the given
concentrations, it is 1.215 V.
Question 17
Question
A redox reaction occurs in a cell according to the equation: Cu2+ +P b −→
Cu +P b2+. Given that the standard reduction potentials are E◦
Cu2+/Cu = 0.34
V and E◦
Pb2+/Pb =−0.13 V, calculate the cell potential when the concentrations
of Cu2+ and Pb2+ ions are 1.0 M and 0.1 M, respectively.
Solution
Step 1: Write the half-reactions and determine the cell potential in standard
conditions.
Cathode: Cu2+ + 2e−−→ Cu E◦= 0.34 V
Anode: Pb −→ Pb2+ + 2e−E◦=−0.13 V
The cell potential in standard conditions can be calculated as:
E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.13 V) = 0.47 V
Step 2: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation is given by:
Ecell =E◦
cell −0.0592 V
nlog [Pb2+]
[Cu2+]
where nis the number of electrons involved in the reaction. In this case, n= 2.
16
Substitute the given concentrations into the Nernst equation:
Ecell = 0.47 V −0.0592 V
2log 0.1
1.0
Ecell = 0.47 V −0.0296 V log(0.1)
Ecell = 0.47 V −0.0296 V ×(−1)
Ecell = 0.47 V + 0.0296 V = 0.4996 V
Therefore, the cell potential when the concentrations of Cu2+ and Pb2+ ions
are 1.0 M and 0.1 M, respectively, is 0.4996 V.
Question 18
Question
A galvanic cell is constructed with a standard hydrogen electrode (SHE) on the
left side and a Ag/AgCl electrode on the right side. The standard reduction
potential for the Ag/AgCl electrode is 0.222 V. The cell is operated at 298 K.
Calculate the cell potential when the [Cl-] concentration is 0.01 M and the [H+]
concentration is 1.0 M.
Solution
Step 1: Write the cell reaction representing the given information. The cell
reaction for the galvanic cell can be represented as:
2H+(aq)+2e−→H2(g)
AgCl(s) + e−→Ag(s) + Cl−(aq)
Step 2: Calculate the cell potential at standard conditions. The standard
cell potential (E◦
cell) can be calculated using the formula:
E◦
cell =E◦
right −E◦
left = 0.222 V −0 V = 0.222 V
Step 3: Calculate the Nernst equation. The Nernst equation relates the cell
potential at non-standard conditions to the standard cell potential:
Ecell =E◦
cell −0.0592
nlog Q
Pn
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Qis the reaction quotient.
Step 4: Calculate the reaction quotient, Q. The reaction quotient for the
given cell can be calculated using the concentrations provided:
Q=[H+]2[Cl−]
PH2·PAg
=(1.0)2(0.01)
1·1= 0.01
17
Step 5: Substitute values into the Nernst equation. Substitute the given
values and calculated reaction quotient into the Nernst equation to find the cell
potential under non-standard conditions:
Ecell = 0.222 −0.0592
2log 0.01
Step 6: Calculate the cell potential. Now, solve for the cell potential:
Ecell = 0.222 −0.0592
2× −2=0.222 + 0.0592 = 0.2812 V
Therefore, the cell potential when the [Cl-] concentration is 0.01 M and the
[H+] concentration is 1.0 M is 0.2812 V.
Question 19
Question
Calculate the cell potential at 25
°
C for a cell in which the following reaction
occurs:
Fe3+(aq)+e−⇌Fe2+(aq)
Given that the standard reduction potential for the Fe3+/Fe2+ couple is
+0.77V, the concentration of Fe3+ is 0.10M, and the concentration of Fe2+ is
1.0M. Assume that the temperature is 25
°
C and that the Faraday constant, F,
is 96485 C/mol.
Solution
Step 1: Write the half-reaction equation for the Fe3+/Fe2+ couple:
Fe3+(aq)+e−⇌Fe2+(aq)
Step 2: Write the expression for the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 3: Calculate the cell potential using the given values and the Nernst
equation: Given:
E◦= +0.77 V
[Fe3+] = 0.10 M
[Fe2+]=1.0M
n= 1 (number of electrons transferred in the reaction)
T= 25C= 298K
18
F= 96485 C/mol
Plugging in the values:
E= 0.77V−0.0592
1log 1.0
0.10
E= 0.77V−0.0592 log(10)
E= 0.77V−0.0592 ×1
E= 0.77V−0.0592
E= 0.7108V
Therefore, the cell potential at 25
°
C for the given reaction is 0.7108V.
Question 20
Question
A concentration cell is constructed with two silver-silver chloride electrodes.
Electrode 1 has a [Ag+] concentration of 0.10 M and electrode 2 has a [Ag+]
concentration of 0.0010 M. If the standard reduction potential for the reaction
Ag+(0.10 M) + e−→Ag(s) is 0.80 V, calculate the cell potential for this
concentration cell at 25◦C.
Solution
Step 1: Write the half-reaction involved in the concentration cell: The half-
reaction for the silver-silver chloride electrode is:
AgCl(s) + e−→Ag(s) + Cl−
Step 2: Write the Nernst equation for the cell potential (Ecell): The Nernst
equation is given by:
Ecell =E◦
cell −0.0592
nlog [Ag+]cathode
[Ag+]anode
Step 3: Calculate the number of electrons transferred (n): Since the reaction
involves the transfer of 1 electron, n= 1.
Step 4: Calculate the standard cell potential (E◦
cell): Given that the standard
reduction potential for the reaction is 0.80 V, E◦
cell = 0.80 V.
Step 5: Substitute the given values into the Nernst equation:
Ecell = 0.80V−0.0592
1log 0.0010
0.10
Ecell = 0.80V−0.59 log 0.010
19
Step 6: Solve for Ecell:
Ecell = 0.80V−0.59(−2)
Ecell = 0.80V+ 1.18V
Ecell = 1.98 V
Therefore, the cell potential for this concentration cell is 1.98 V at 25◦C.
Question 21
Question
Calculate the cell potential for a galvanic cell at 25
°
C in which the concentration
of Zn2+ is 0.10 M and the concentration of Cu2+ is 1.0 ×10−3M. The standard
reduction potentials are Zn2+/Zn = -0.76 V and Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions involved in the cell reaction. The overall cell
reaction is: Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
The half-reactions are: Zn2+(aq) + 2e−→Zn(s) (oxidation) Cu2+ + 2e−
→Cu(s) (reduction)
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potentials. E◦
cell = E◦
reduction, cathode - E◦
oxidation, anode
E◦
cell = E◦
reduction, cathode - E◦
oxidation, anode E◦
cell = (0.34 V) - (-0.76 V) E◦
cell
= 1.10 V
Step 3: Calculate the Nernst equation to find the cell potential under non-
standard conditions. E = E◦-0.0592
n×log10 [Cu2+]
[Zn2+]
Where: E = cell potential under non-standard conditions E◦= cell potential
at standard conditions (1.10 V) n = number of electrons transferred in the
balanced chemical equation (2) [Cu2+] = concentrationof Cu2+ ions (1.0 ×
10−3M) [Zn2+] = concentrationof Zn2+ ions (0.10 M)
E = 1.10 V - 0.0592
2×log10 1.0×10−3
0.10 E = 1.10 V - 0.0296 ×log1010 E = 1.10
V - 0.0296 ×1 E = 1.10 V - 0.0296 E = 1.07 V
Therefore, the cell potential under the given conditions is 1.07 V.
Question 22
Question
A concentration cell is set up with two Ag—AgCl electrodes. One half-cell
contains a 10−2M concentration of Cl−ions, while the other half-cell contains
a 10−4M concentration of Cl−ions. Calculate the cell potential at 25
°
C using
the Nernst equation. Given: E◦
cell = 0.22 V.
20
Solution
Step 1: Write the two half-reactions that occur in the concentration cell.
Anode: Ag →Ag++e−
Cathode: Ag++e−→Ag
Step 2: Write the overall cell reaction.
Overall: 2Ag →2Ag+
Step 3: Calculate the cell potential at standard conditions using the standard
reduction potentials for each half-reaction. Given: E◦
cell = 0.22 V
Step 4: Substitute the known information into the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Ag+]2
cathode
[Ag+]2
anode
Step 5: Convert the concentrations to molarities.
[Ag+]cathode = 10−4M
[Ag+]anode = 10−2M
Step 6: Substitute the molarities into the Nernst equation and solve for the
cell potential.
Ecell = 0.22 −0.0592
2log (10−4)2
(10−2)2
Step 7: Calculate the cell potential.
Ecell = 0.22 −0.0592
2log 10−8
Ecell = 0.22 −0.0592 ×4≈0.22 −0.2368 ≈ −0.0168 V
Therefore, the cell potential at 25
°
C using the Nernst equation is approxi-
mately -0.0168 V.
Question 23
Question
A galvanic cell is constructed with an Al/Al3+ half-cell and a Br2/Br−half-cell.
The initial concentrations are [Al3+]=1.0 M, [Br−] = 0.10 M, and [Br2]=0.20
M. Given that the standard reduction potentials for the Al3+/Al and Br2/Br−
half-reactions are E◦
cell,Al =−1.68 V and E◦
cell,Br2= 1.09 V, calculate the cell
potential at t= 0.
21
Solution
Step 1: The Nernst equation is given by:
E=E◦−0.0592
nlog Q
where Eis the cell potential, E◦is the standard cell potential, nis the number
of moles of electrons transferred, and Qis the reaction quotient.
Step 2: The balanced overall cell reaction is:
Al + Br2→Al3+ + 2Br−
Step 3: From the standard reduction potentials given, the standard cell
potential E◦
cell can be calculated as:
E◦
cell =E◦
cell, Br2−E◦
cell, Al
Step 4: Calculating the standard cell potential gives:
E◦
cell = 1.09 V −(−1.68 V) = 2.77 V
Step 5: The initial reaction quotient Qcan be determined using the initial
concentrations:
Q=[Al3+][Br−]2
[Al][Br2]=(1.0)(0.10)2
1(0.20) = 0.50
Step 6: Substituting values into the Nernst equation gives:
E= 2.77 −0.0592
1log 0.50
Step 7: Calculating the cell potential at t= 0:
E= 2.77 −0.0592(−0.3010) = 2.79 V
Therefore, the cell potential at t= 0 is 2.79 V.
Question 24
Question
Calculate the potential of a standard hydrogen electrode at 25
°
C when the
concentration of hydrogen ions in the solution is 1.0×10−2M. The standard
reduction potential of the hydrogen electrode is 0.00 V.
22
Solution
Step 1: Write the Nernst equation. The Nernst equation relates the standard
electrode potential to the actual cell potential under non-standard conditions.
The Nernst equation is given by:
E=E◦−0.0592
nlog(Q)
where: - Eis the cell potential under non-standard conditions, - E◦is the
standard electrode potential, - nis the number of electrons involved in the
reaction, - Qis the reaction quotient.
Step 2: Identify the values given in the question. - E◦= 0.00 V (standard
reduction potential of the hydrogen electrode) - T= 25
°
C (298 K) - [H+] =
1.0×10−2M
Step 3: Determine the number of electrons involved in the reaction. Since
the standard hydrogen electrode involves the reduction of hydrogen ions to
hydrogen gas, the reaction is:
2H++ 2e−→H2
Therefore, n= 2 (2 electrons are involved in the reaction).
Step 4: Calculate the reaction quotient Q.
Q=[products]
[reactants] =1
1.0×10−2= 100
Step 5: Substitute the values into the Nernst equation and solve for E.
E= 0.00−0.0592
2log(100) = 0.00 −0.0592 ×2×2=0.00 −0.2368 = −0.2368 V
Therefore, the potential of a standard hydrogen electrode at 25
°
C with 1.0×
10−2M hydrogen ions is −0.2368 V.
Question 25
Question
A galvanic cell consists of a silver electrode in a 0.10 M solution of AgNO3and
a copper electrode in a 0.20 M solution of CuSO4. The standard reduction
potential for Cu2+(aq)+2e−→Cu(s) is 0.34 V, while the standard reduction
potential for Ag+(aq) + e−→Ag(s) is 0.80 V. Calculate the cell potential at
25
°
C. (Given: R= 8.31 J/(mol ·K), F= 96,485 C/mol)
Solution
Step 1: Write the overall cell reaction and find the cell potential at standard
conditions. The overall reaction for the cell is: Cu2+(aq)+2Ag(s)→Cu(s) +
2Ag+(aq)
23
The standard cell potential, Ecell, is given by: Ecell =Ecathode −Eanode
Given that Ecathode is 0.80 V for Ag+(aq) + e−→Ag(s) and Eanode is 0.34
V for Cu2+(aq)+2e−→Cu(s), we have: Ecell = 0.80V−0.34V= 0.46V
Step 2: Calculate the reaction quotient, Q. The reaction quotient, Q, is given
by: Q=[Ag+]2
[Cu2+ ]
Given that [Ag+]=0.10Mand [Cu2+]=0.20M, we have: Q=(0.10)2
0.20 =
0.05
Step 3: Use the Nernst equation to calculate the cell potential at 25
°
C. The
Nernst equation is given by: Ecell =Ecell −0.0592V
nlog(Q)
Since the reaction involves the transfer of 2 moles of electrons, n= 2. Sub-
stituting the values of Ecell,n, and Qinto the equation: Ecell = 0.46V−
0.0592V
2log(0.05) Ecell = 0.46V−0.0592V
2×(−1.3010) Ecell = 0.46V+ 0.0389V
Ecell = 0.4989V
Therefore, the cell potential at 25
°
C for the given galvanic cell setup is 0.4989
V.
Question 26
Question
A half-cell consists of a copper electrode in a 0.05 M CuSO4solution. The
standard reduction potential for the reaction is E◦= 0.34 V. At what pH will
the potential of this half-cell be 0.18 V?
Solution
Step 1: Write the half-reaction and the balanced redox reaction.
The half-reaction for the reduction of copper is: Cu2+ + 2e−→Cu
The balanced redox reaction is: Cu2+ + 2e−→Cu
Step 2: Write the Nernst equation.
The Nernst equation relates the cell potential to the standard cell potential and
the concentrations of the reactants and products:
E=E◦−0.0592
nlog(Q)
where Eis the cell potential, E◦is the standard cell potential, nis the number
of electrons transferred in the balanced equation, and Q=[products]
[reactants].
Step 3: Identify the values given and calculate Q.
Given: E◦= 0.34 V, E= 0.18 V, [Cu2+]=0.05 M
Since the concentration of Cu is not given, we can assume it to be 1 M. Thus,
Q=1
0.052= 400
24
Step 4: Substitute the values into the Nernst equation and solve for the pH.
Substitute the given values into the Nernst equation:
0.18 = 0.34 −0.0592
2log(400)
0.18 = 0.34 −0.0296 log(400)
Now, solve for log(400):
log(400) = 0.34 −0.18
0.0296
log(400) = 0.16
0.0296
log(400)5.41
Step 5: Find the pH.
To find the pH, we need to use the relationship between [H+] and pH: pH =
−log[H+]
Since the concentration of H+can be calculated from the given Q, we have:
[H+] = pQ=√40020
pH = −log(20)1.70
Therefore, the pH at which the potential of the half-cell is 0.18 V is approx-
imately 1.70.
Question 27
Question
Calculate the cell potential at 25
°
C for a galvanic cell with a standard hydrogen
electrode (SHE) on the left and a copper(II) sulfate solution on the right, where
the concentration of Cu2+ is 0.10 M. Given: E◦
Cu2+/Cu = 0.34 V and E◦
H+/H2=
0 V.
Solution
Step 1: Write the overall balanced cell reaction. The overall cell reaction can
be written as:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the half-reactions for the anode and cathode. The half-
reactions at the anode and cathode are as follows: Anode (oxidation):
Cu(s)→Cu2+(aq)+2e−
25
Cathode (reduction):
2H+(aq)+2e−→H2(g)
Step 3: Write the cell potential at standard conditions. The cell potential
at standard conditions (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 V −0.34 V = −0.34 V
Step 4: Determine the reaction quotient, Q. The reaction quotient, Q, is
given by:
Q=[Cu2+]
[H+]
2
Given that [Cu2+]=0.10 M and [H+]=1.0 M (from the SHE), we have:
Q=0.10
1.0
2
= 0.01
Step 5: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the balanced cell reaction.
Since 2 electrons are transferred, n= 2.
Ecell =−0.34 V −0.0592
2log(0.01)
Ecell =−0.34 V −0.0296 log(0.01)
Ecell =−0.34 V −0.0296 ×(−2)
Ecell =−0.34 V + 0.0592
Ecell =−0.2808 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is -0.2808 V.
Question 28
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn(s)|Zn2+(0.10M)||Cu2+(1.0M)|Cu(s)
Given: E◦
cell = 1.10 V R= 8.314 J/(mol ·K) T= 298 K
26
Solution
Step 1: Write the half-reactions for the redox reaction and determine the stan-
dard cell potential. The half-reactions for the redox reaction are:
Zn2+(aq)+2e−→Zn(s) E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s) E◦= 0.34 V
The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 2: Write the Nernst equation. The Nernst equation is given by:
Ecell =E◦
cell −RT
nF ln Q
K
Where: Ecell = cell potential at non-standard conditions E◦
cell = standard cell
potential R= gas constant (8.314 J/(mol
·
K)) T= temperature in Kelvin (298
K) n= number of moles of electrons exchanged in the balanced redox reaction
F= Faraday’s constant (96485 C/mol) Q= reaction quotient (ratio of products
to reactant concentrations) K= equilibrium constant
Step 3: Calculate the number of moles of electrons exchanged (n). From the
balanced redox reaction, 2 moles of electrons are exchanged by both Zn and Cu:
n= 2
Step 4: Calculate the reaction quotient (Q). The reaction quotient is calcu-
lated using the concentrations of the ions involved in the redox reaction:
Q=[Cu2+]
[Zn2+]
Q=1.0
0.10 = 10
Step 5: Substitute the given values into the Nernst equation and solve for
Ecell.
Ecell = 1.10 V −(8.314 J/mol ·K)(298 K)
(2)(96485 C/mol) ln 10
K
Ecell = 1.10 V −0.0257 ln 10
KV
Therefore, the cell potential at 25
°
C for the given redox reaction is Ecell =
1.10 V −0.0257 ln 10
KV.
27
Question 29
Question
A half-cell consists of a silver metal electrode in contact with a 0.05 M silver
nitrate solution. If the standard electrode potential for the Ag/Ag+half-cell is
E◦= 0.80 V, calculate the potential of this half-cell at 25◦C when [Ag+] = 0.01
M. (Given: R= 8.314 J/(mol ·K), F= 96485 C/mol)
Solution
Step 1: Write the Nernst equation: The Nernst equation relates the measured
cell potential (E) to the standard cell potential (E◦), the reaction quotient (Q),
the gas constant (R), the temperature (T), and the Faraday constant (F). The
Nernst equation for the given half-cell can be written as:
E=E◦−RT
nF ln(Q)
Step 2: Calculate the reaction quotient (Q): In this case, the reaction quo-
tient (Q) is the ratio of product concentrations to reactant concentrations, raised
to the power of their stoichiometric coefficients. The given half-cell reaction is:
Ag(s)→Ag+(aq) + e−
Thus, Q=[Ag+]
1.
Step 3: Plug in the given values to solve for E: Given: E◦= 0.80 V, T=
25◦C = 298 K, [Ag+] = 0.01 M. Plugging these values into the Nernst equation,
we get:
E= 0.80 V −(8.314 J/(mol ·K)) ×(298 K)
1×(96485 C/mol) ln(0.01)
Step 4: Calculate E: Calculating the Natural Logarithm first:
ln(0.01) = ln1×10−2=−2 ln(10) ≈ −4.605
Now plug this value back into the Nernst equation:
E= 0.80 V −(8.314 J/(mol ·K)) ×(298 K)
1×(96485 C/mol) × −4.605
E≈0.80 V −(−0.056)
E≈0.856 V
Therefore, the potential of the half-cell at 25◦C when [Ag+]=0.01 M is
approximately 0.856 V.
28
Question 30
Question
A voltaic cell consists of a half-cell with a copper electrode in a 1.0 M Cu2+
solution and a half-cell with a silver electrode in a 0.010 M Ag+solution. Cal-
culate the cell potential at 25
°
C for this cell. Given that the standard reduction
potentials are: Cu2+ + 2e−→Cu(s) with E◦= +0.34 V and Ag++ e−→
Ag(s) with E◦= +0.80 V.
Solution
Step 1: Write the two half-reactions with their respective standard reduction
potentials. The two half-reactions are: Cu2+ + 2e−→Cu(s) E◦
red = +0.34
V Ag++ e−→Ag(s) E◦
red = +0.80 V
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog [Cu2+]
[Ag+]
Step 3: Determine the number of electrons transferred (n) in the overall
cell reaction. From the two half-reactions provided, we can see that 2 moles
of electrons are transferred for the reduction of Cu2+ and 1 mole of electron is
transferred for the reduction of Ag+. So, the least common multiple of 2 and 1
is 2. Hence, n = 2.
Step 4: Substitute the given values into the Nernst equation.
E= 0.80V−0.0592
2log 1.0
0.010
E= 0.80V−0.0296 log(100)
E= 0.80V−0.0296 ×2
E= 0.80V−0.0592
E= 0.74 V
Therefore, the cell potential at 25
°
C for this cell is 0.74 V.
Question 31
Question
A cell is constructed with a standard hydrogen electrode (SHE) on one side
and a copper electrode on the other side. The standard reduction potential
of hydrogen electrode is 0 V and the standard reduction potential of copper
electrode is +0.34 V. Calculate the cell potential at 25◦C.
29
Solution
Step 1: Write the half-reactions for the electrodes involved. The half-reaction
for the standard hydrogen electrode is:
2H++ 2e−→H2E◦= 0 V
The half-reaction for the copper electrode is:
Cu2+ + 2e−→Cu E◦= +0.34 V
Step 2: Determine the overall cell reaction. To find the overall cell reaction,
we add the two half-reactions together. Since the reduction potential of copper
is higher, the copper half-reaction will be the reduction:
Cu2+ + 2H++ 2e−→Cu + H2
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog(Q)
Where: - Eis the cell potential - E◦is the standard cell potential - nis the
number of electrons transferred in the cell reaction - Qis the reaction quotient,
which is the ratio of product concentrations to reactant concentrations
For this cell, E◦= +0.34 V and n= 2.
Step 4: Calculate the reaction quotient (Q). Since concentrations are not
given, Qcannot be calculated.
Therefore, the cell potential of this cell cannot be calculated without knowing
the concentrations of the reactants and products.
Question 32
Question
A voltaic cell consists of a platinum electrode dipping in a solution of 0.01
M Ag+ions and a silver electrode dipping in a solution of 1.0 M Ag+ions.
Calculate the cell potential at 25
°
C given that the standard reduction potential
for the silver/silver ion half-cell is +0.80 V and that the Nernst constant (E◦)
is 0.0592 V.
Solution
Step 1: Write the half-reactions for the cell.
The anode half-reaction is the oxidation of Ag to Ag+:
Ag(s)→Ag+(aq) + e−
30
The cathode half-reaction is the reduction of Ag+to Ag:
Ag+(aq) + e−→Ag(s)
Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
1log [products]
[reactants]
The standard reduction potential (E◦) for the silver/silver ion half-cell is
given as +0.80 V.
Step 3: Calculate the concentrations of Ag+ions at both electrodes.
For the anode, [Ag+] = 0.01MF orthecathode, [Ag+]=1.0M
Step 4: Substitute the values into the Nernst equation and solve for the cell
potential.
E= 0.80V−0.0592
1log 1.0
0.01
E= 0.80V−0.0592 ×log(100)
E= 0.80V−0.0592 ×2
E= 0.80V−0.1184
E= 0.6816 V
Therefore, the cell potential at 25
°
C is 0.6816 V.
Question 33
Question
A cell consists of a standard hydrogen electrode (SHE) and a nickel electrode
in a 1.0 M Ni2+ solution. Given that the standard reduction potential of the
Ni2+/Ni electrode is −0.25 V, calculate the cell potential at 25
°
C when the
concentration of Ni2+ is 0.10 M. (F= 96485 C/mol)
Solution
Step 1: Write the half-reaction for the standard hydrogen electrode (SHE) and
the nickel electrode. The half-reaction for the standard hydrogen electrode is:
2H++ 2e−→H2(g)E◦= 0 V
The half-reaction for the nickel electrode is:
Ni2+ + 2e−→Ni(s) E◦=−0.25 V
31
Step 2: Write the overall cell reaction and find E◦
cell. The overall cell reaction
is the sum of the two half-reactions:
2H++ Ni2+ →H2(g) + Ni(s)
Using the standard reduction potentials, E◦
cell can be calculated as:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 V −(−0.25 V) = 0.25 V
Step 3: Use the Nernst equation to find the cell potential at non-standard
conditions. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
Fn
Where: - E= cell potential at non-standard conditions - E◦= standard cell
potential - n= number of electrons transferred in the balanced cell reaction -
Q= reaction quotient - F= Faraday’s constant
Step 4: Calculate the reaction quotient Q. Since the concentrations of all
species are 1 M in the standard state, the reaction quotient Qcan be calculated
as:
Q=[H2][Ni(s)]
[H+]2[Ni2+]= 1
Step 5: Substitute the values into the Nernst equation. Substitute the values
into the Nernst equation:
E= 0.25 V −0.0592
2log(1)
E= 0.25 V
Therefore, the cell potential at 25
°
C when the concentration of Ni2+ is 0.10
M is 0.25 V.
Question 34
Question
A redox reaction involving the reduction of copper(II) ions to copper metal is
represented by the equation:
Cu2+ + 2e−→Cu
If the standard reduction potential (E◦) for this reaction is +0.34 V, calculate
the cell potential when the concentration of Cu2+ is 0.10 M and the concentra-
tion of Cu is 1.0 M. Assume the temperature is 298 K.
32
Solution
Step 1: Write the Nernst equation for the cell potential E:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential, - E◦is the standard reduction potential, - nis
the number of moles of electrons transferred, - Qis the reaction quotient.
Step 2: Determine the number of moles of electrons transferred (n) from the
balanced redox reaction. In this case, since 2 moles of electrons are involved in
the reduction of 1 mole of Cu2+ to Cu,n= 2.
Step 3: Calculate the reaction quotient Qusing the given concentrations of
Cu2+ and Cu:
Q=[Cu]1
[Cu2+]2
Q=1.0
(0.10)2
Q= 100
Step 4: Plug the values of E◦,n, and Qinto the Nernst equation to solve
for E:
E= 0.34 −0.0592
2log(100)
E= 0.34 −0.0296 ×2
E= 0.34 −0.0592
E= 0.2808 V
Therefore, the cell potential when the concentration of Cu2+ is 0.10 M and
the concentration of Cu is 1.0 M is 0.2808 V.
Question 35
Question
A voltaic cell is constructed with two half-cells. One half-cell consists of a
standard hydrogen electrode (SHE) with H+(aq) at 1.00Mconcentration and
hydrogen gas at 1.00atm. The other half-cell consists of Cu2+(aq) at 0.010M
and Cu(s). Calculate the cell potential at 25◦Cgiven that E◦
cell = 0.34V.
Assume R= 8.314J/(mol ·K) and F= 96485C/mol.
33
Solution
Step 1: Write the half-reactions for the given cell.
Anode (oxidation): Cu(s)→Cu2+(aq)+2e−
Cathode (reduction): 2H+(aq)+2e−→H2(g)
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592V
nlog Q
where nis the total number of electrons transferred and Qis the reaction quo-
tient for the cell.
Step 3: Determine the reaction quotient Qfor the cell.
Q=[H+]2
[Cu2+]
Plugging in the given concentrations:
Q=(1.00)2
0.010 = 100
Step 4: Calculate the cell potential using the Nernst equation.
Ecell = 0.34V−0.0592V
2log 100
Ecell = 0.34V−(0.0296V)×2
Ecell = 0.2808V
Therefore, the cell potential at 25◦Cis 0.2808V.
34
Where nis the number of electrons transferred in the cell reaction and Qis the
reaction quotient.
For the given cell reaction, n= 2 since two electrons are transferred. The
reaction quotient Qis calculated by dividing the concentrations of the products
raised to their stoichiometric coefficients by the concentrations of the reactants
raised to their stoichiometric coefficients:
Q=[Zn2+]
[Cu2+]=0.010
0.100 = 0.100
Now, plug in the values into the Nernst equation:
Ecell = 1.10 V −0.0592
2log(0.100)
Ecell = 1.10 V −0.0296 log(0.100) = 1.10 V −(−0.0296 ×1) = 1.1296 V
Therefore, the cell potential under the given conditions is 1.13 V.
Question 2
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode. The standard reduction potential for the copper electrode is E◦=
0.34 V. The concentration of Cu2+ ions is 0.10 M in the half-cell containing
the copper electrode. Given that the hydrogen electrode operates at standard
conditions ([H+] = 1.0 M, PH2= 1.0 atm), calculate the cell potential at 25◦C.
Solution
Step 1: Write the half-reactions for the cell The overall reaction for the galvanic
cell can be represented as:
2H+(aq) + 2e−→H2(g)
Cu2+(aq) + 2e−→Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard reduc-
tion potentials The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cathode = 0 V for the SHE, and E◦
anode = 0.34 V for the copper
electrode:
E◦
cell = 0 −0.34 = −0.34 V
Step 3: Calculate the reaction quotient (Q) Given that [Cu2+] = 0.10 M in
the copper half-cell, and [H+]=1.0 M in the hydrogen half-cell:
Q=[Cu2+]
1= 0.10
2
Step 4: Calculate the cell potential (Ecell) at non-standard conditions using
the Nernst equation The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the balanced redox reaction.
Step 5: Calculate the number of moles of electrons transferred From the
balanced half-reactions, we see that 2 moles of electrons are transferred.
Step 6: Substitute the values into the Nernst equation
Ecell =−0.34 −0.0592
2log(0.10) = −0.34 −0.0296 ×(≈ −1) = −0.0344 V
Therefore, the cell potential at 25◦C for the given galvanic cell is approxi-
mately −0.0344 V.
Question 3
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper
electrode in a solution containing Cu2+ ions. The standard reduction potential
for the half-reaction Cu2+(aq) + 2e−→Cu(s) is E◦= 0.34 V. If the concentra-
tion of Cu2+ ions in the solution is 0.010 M, calculate the cell potential at 25
°
C
using the Nernst equation.
Solution
Step 1: Write the half-reaction for the cell and the Nernst equation.
The half-reaction for the cell is:
2H+(aq)+2e−→H2(g)
The Nernst equation relates the cell potential (Ecell) to the standard cell po-
tential (E◦), the reaction quotient (Q), the number of electrons transferred (n),
the Faraday constant (F), and the temperature (T):
Ecell =E◦−RT
nF ln(Q)
Step 2: Calculate the reaction quotient (Q).
The reaction quotient Qis calculated using the concentrations of the species
involved in the cell reaction. For our cell:
Q=[H2]
[H+]2= 1
Step 3: Calculate the cell potential (Ecell).
Given: E◦= 0.34 V
3
T= 298 K
n= 2 (since 2 electrons are transferred in the cell reaction)
F= 96485 C/mol
Using the Nernst equation:
Ecell = 0.34 −(8.314 J/mol·K)(298 K)
(2)(96485 C/mol) ln(1)
Ecell = 0.34 −(8.314)(298)
2(96485) ×0
Ecell = 0.34 V
Therefore, the cell potential of the galvanic cell at 25
°
C is 0.34 V.
Question 4
Question
Calculate the cell potential for the following reaction at 25
°
C:
2F e3+(aq)+2e−→2F e2+(aq)
Given that the standard reduction potentials are E◦
Fe3+ /Fe2+ = 0.77 V and the
concentration of Fe3+ is 0.1 M while the concentration of Fe2+ is 0.01 M.
Solution
Step 1: Write the Nernst equation for the cell potential:
E=E◦−RT
nF ln(Q)
where - Eis the cell potential, - E◦is the standard cell potential, - Ris the
ideal gas constant (8.314 J/(mol
·
K)), - Tis the temperature in Kelvin (25
°
C =
298 K), - nis the number of moles of electrons transferred (2 in this case), - F
is the Faraday constant (96485 C/mol), - Qis the reaction quotient.
Step 2: Calculate the cell potential using the Nernst equation: Plugging in
the values we have:
E= 0.77 −(8.314)(298)
2(96485) ln (0.01)2
0.12
E= 0.77 −2465.972
192970 ln(0.001)
E= 0.77 −0.0127 ln(0.001)
E= 0.77 −0.0127(−6.907)
E= 0.77 + 0.0874
E= 0.8574 V
Therefore, the cell potential for the given reaction at 25
°
C is 0.8574 V.
4
Question 5
Question
A galvanic cell consists of a silver electrode in 1.0 M Ag+ions and a copper
electrode in 1.0 M Cu2+ ions. The cell has an emf of 1.02 V at 25
°
C. Calculate
the standard cell potential, E◦
cell, for this galvanic cell. Given: E◦
Ag+/Ag = 0.80
V and E◦
Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions and their standard electrode potentials. The
standard cell potential can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
where Q is the reaction quotient and n is the number of moles of electrons
transferred in the balanced cell reaction.
The two half-reactions involved are:
Ag++ e−→Ag E◦
Ag+/Ag = 0.80 V
Cu2+ + 2e−→Cu E◦
Cu2+/Cu = 0.34 V
Step 2: Identify the balanced cell reaction and calculate the standard cell
potential. The balanced cell reaction is the sum of the two half-reactions with
multiplication factors to equalize the electrons:
2Ag++ Cu →2Ag + Cu2+
The standard cell potential is the difference between the standard reduction
potentials of the two half-reactions:
E◦
cell = E◦
Ag+/Ag - E◦
Cu2+/Cu
E◦
cell = 0.80 V −0.34 V = 0.46 V
Therefore, the standard cell potential for this galvanic cell is 0.46 V.
Question 6
Question
A galvanic cell consists of a standard hydrogen electrode and a zinc electrode.
The concentration of Zn2+ ions is 0.10 M, and the pressure of hydrogen gas is
0.80 atm. Calculate the cell potential at 25
°
C using the Nernst equation. Given
that the standard reduction potential for the reduction of Zn2+ to Zn is -0.76
V.
5
Solution
Step 1: Write the overall cell reaction and the Nernst equation. The overall cell
reaction is given by:
Zn2+ + 2e−→Zn(s)
The Nernst equation is given by:
E=Eo−0.0592
nlog Q
where: - Eis the cell potential, - Eois the standard cell potential, - nis the
number of moles of electrons exchanged in the balanced cell reaction, and - Q
is the reaction quotient.
Step 2: Calculate the reaction quotient (Q). The reaction quotient is given
by:
Q=[Zn2+]
P(H2)2
Substitute the given values:
Q=0.10
0.802
Step 3: Calculate the cell potential (E). Given: - Eo=−0.76 V, - n= 2
(from the balanced cell reaction), - Q=0.10
0.802.
Substitute the values in the Nernst equation:
E=−0.76 −0.0592
2log 0.10
0.802
Step 4: Calculate the cell potential.
E=−0.76 −0.0296 log 0.10
0.802
Finally, solve for the cell potential.
Question 7
Question
A concentration cell is constructed using two silver/silver chloride half-cells.
One half-cell contains a solution with [Cl−]=0.10 M, and the other contains
a solution with [Cl−] = 1.0 M. If the standard reduction potential for the
Ag/AgCl electrode is E◦= 0.22 V, calculate the cell potential at 25◦C.
6
Solution
Step 1: Write the half-reactions and the Nernst equation for each half-cell.
Anode: AgCl(s) + e−→Ag(s) + Cl−Eanode =E◦
Cathode: AgCl(s) + e−→Ag(s) + Cl−Ecathode =E◦
The Nernst equation is:
E=E◦−0.0592
nlog [products]
[reactants]
where nis the number of electrons transferred.
Step 2: Calculate the cell potential for each half-cell using the Nernst equa-
tion.
Eanode = 0.22 V−0.0592
1log 0.10
1.0= 0.22 V−0.0592 log(0.10) = 0.22 V−0.0592×(−1) = 0.2792 V
Ecathode = 0.22 V−0.0592
1log 1.0
0.10= 0.22 V−0.0592 log(10) = 0.22 V−0.0592×1=0.1608 V
Step 3: Calculate the overall cell potential. The overall cell potential is the
difference between the cathode potential and the anode potential:
Ecell =Ecathode −Eanode = 0.1608 V −0.2792 V = −0.1184 V
Therefore, the cell potential at 25◦C for the concentration cell with differing
chloride ion concentrations is −0.1184 V.
Question 8
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a copper electrode. The concentration of Cu2+ ions is 0.010 M in the copper
half-cell. The cell potential at 25
°
C is measured to be 0.47 V. Calculate the
standard reduction potential of the copper half-cell.
Given: E◦
SHE = 0.00 V, n= 2
Solution
Step 1: Write the overall cell reaction for the galvanic cell. The cell reaction is:
Cu2+ + 2e−→Cu(s)
Step 2: Write the Nernst equation. The Nernst equation is:
E=E◦−0.0592
nlog Q
7
where: - Eis the cell potential at 25
°
C (0.47 V), - E◦is the standard cell
potential, - nis the number of moles of electrons transferred in the balanced
cell reaction (2 in this case), - Qis the reaction quotient.
Step 3: Calculate the reaction quotient, Q. Since the only ions present in
the copper half-cell are Cu2+ and Cu(s), we have:
Q=[Cu]
[Cu2+]=1
0.010 = 100
Step 4: Substitute the given values into the Nernst equation and solve for
E◦.
0.47 = E◦−0.0592
2log 100
0.47 = E◦−0.0592 ×2×2
0.47 = E◦−0.2368
E◦= 0.47 + 0.2368 = 0.7068
Therefore, the standard reduction potential of the copper half-cell is 0.7068
V.
Question 9
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) and a silver-
silver chloride electrode. The standard reduction potential for the silver-silver
chloride electrode is E◦= 0.222 V. If the concentration of chloride ions is 0.1
M and the pH of the solution is 3, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction involving the silver-silver chloride elec-
trode and standard hydrogen electrode (SHE). The overall cell reaction is:
AgCl(s) + e−→Ag(s) + Cl−
2H++ 2e−→H2(g)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials. The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦(Ag++ e−→Ag) −E◦(H++ e−→H2)
E◦
cell = 0.222 V −0.000 V
E◦
cell = 0.222 V
8
Step 3: Calculate the cell potential (Ecell) using the Nernst equation. The
Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where Qis the reaction quotient and nis the number of electrons transferred
in the cell reaction.
Step 4: Calculate the reaction quotient (Q) for the cell reaction. Since the
half-reactions are already balanced, Qsimplifies to:
Q=[Ag+][H+]2
[Cl−]2
Plugging in the given values:
Q=[0][10−3]2
[10−1]2
Q= 0
Step 5: Substitute the values of E◦
cell,Q, and ninto the Nernst equation and
solve for Ecell. Since Qis zero, the term 0.0592/n log(Q) becomes zero. Thus:
Ecell =E◦
cell −0=0.222 V
Therefore, the cell potential at 25
°
C is 0.222 V.
Question 10
Question
A concentration cell is set up using two Ag/AgCl electrodes at 25
°
C. One half-
cell contains a 0.1 M solution of NaCl and the other half-cell contains a 0.01 M
solution of NaCl. Calculate the cell potential of this concentration cell.
Given: Standard electrode potential of Ag/AgCl electrode (E
°
) = 0.222 V
Temperature (T) = 25
°
C
Solution
Step 1: Write the half-reactions for the cell: The half-cell reactions for the
concentration cell are:
Anode: AgCl(s)→Ag(s) + Cl−(aq)
Cathode: AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
E=E−0.0592
nlog [Ox]
[Red]
9
where: E= cell potential E= standard cell potential n= number of moles of
electrons transferred in the balanced equation [Ox] = concentration of oxidized
form [Red] = concentration of reduced form
Step 3: Calculate the cell potential for the anode half-cell: For the anode
half-cell with a 0.1 M solution of NaCl: n= 1 (since one electron is transferred
in the half-reaction) [Ox] = [Ag+]=0.1 M [Red] = [Ag+]=0.01 M Substitute
the values into the Nernst equation:
Eanode = 0.222 −0.0592
1log 0.1
0.01
Eanode = 0.222−0.0592 log(10) = 0.222−0.0592×1=0.222−0.0592 = 0.1628 V
Step 4: Calculate the cell potential by considering the difference in electrode
potentials: The total cell potential is the difference between the cathode and
anode potentials:
Ecell =Ecathode −Eanode
Since the cathode and anode are identical in this case, Ecathode = 0.222 V
Therefore, the cell potential is:
Ecell = 0.222 −0.1628 = 0.0592 V
Therefore, the cell potential of this concentration cell is 0.0592 V.
Question 11
Question
A concentration cell is set up using two half-cells. The first half-cell contains
a copper electrode dipped in a 1.0 M Cu2+ solution at 25
°
C, while the second
half-cell contains a copper electrode dipped in a 0.1 M Cu2+ solution at the
same temperature. Calculate the cell potential of this concentration cell.
Given: E◦
cell = 0.34 V, R= 8.31 J/mol ·K, and T= 298 K.
Solution
Step 1: Write the overall cell reaction and equation for cell potential. The overall
cell reaction for the concentration cell can be represented as: Cu2+(1.0 M,cathode) →
Cu2+(0.1 M,anode).
The cell potential can be calculated using the Nernst equation: Ecell =
E◦
cell −0.0592
nlog [Cu2+]anode
[Cu2+]cathode
Given that n= 2 for this cell.
Step 2: Calculate the cell potential. We can now plug in the values into the
Nernst equation: Ecell = 0.34 V −0.0592
2log 0.1
1.0
Ecell = 0.34 V −0.0296 log(0.1)
10
Ecell = 0.34 V −0.0296 ×(−1)
Ecell = 0.34 V + 0.0296
Ecell = 0.3696 V
Therefore, the cell potential of the concentration cell is 0.3696 V.
Question 12
Question
Calculate the cell potential for the reaction occurring in a galvanic cell with the
following half-reactions at 25
°
C:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Fe3+(aq)+3e−→Fe(s)E◦=−0.04 V
Given that the initial concentrations are Zn2+ = 1.0 M, Fe3+ = 0.1 M, and
[Fe2+]=1.0×10−3M. Assume the temperature is 25
°
C.
Solution
Step 1: Write the overall cell reaction by adding the two half-reactions together.
The overall cell reaction is:
Zn2+(aq) + Fe3+(aq)→Zn(s) + Fe2+(aq)
Step 2: Find the standard cell potential, E◦
cell, using the standard reduction
potentials.
E◦
cell =E◦
cathode −E◦
anode
=E◦
Fe3+ /Fe2+ −E◦
Zn2+/Zn
= (−0.04 V) −(−0.76 V)
= 0.72 V
Step 3: Calculate the reaction quotient, Q, using the initial concentrations
given.
Q=[Fe2+]final
[Zn2+]final[Fe3+]final
Since [Zn2+] is used up in the cell, we can assume that [Zn2+]final = 0. This
simplifies the expression to:
Q=[Fe2+]final
[Fe3+]final
Converting the mole ratio of Fe to Fe2+ from the balanced equation, Fe3+ +
3e−→Fe, we find that the moles of Fe that are produced are equal to the moles
of Fe2+ produced. Therefore,
[Fe2+]final = [Fe3+]initial −[Fe3+]final
11
Substitute the given values into the equation and calculate the final concentra-
tion of Fe2+.
Step 4: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced equation.
Since this cell involves the transfer of 3 moles of electrons, n= 3. Substitute
the values into the Nernst equation, along with the calculated value of Q, to
find the cell potential at 25
°
C.
Question 13
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn2+(0.10M) + Cu(s)→Zn(s) + Cu2+(1.0M)
Given: E◦
cell = 1.10 V, E◦
Zn2+/Zn =−0.76 V, and E◦
Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions for the oxidation and reduction processes:
Zn(s)→Zn2+(aq)+2e−E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Step 2: Write the overall cell reaction equation:
Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
Step 3: Calculate E◦
cell using the standard cell potential values:
E◦
cell =E◦
reduction −E◦
oxidation
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 4: The Nernst equation relates the cell potential to the reaction quotient
Qand the standard cell potential:
E=E◦
cell −0.0592
nlog(Q)
where nis the number of moles of electrons transferred in the balanced redox
reaction.
Step 5: Calculate the reaction quotient Qusing the initial concentrations:
Q=[Zn]
[Cu2+]=0.10
1.0= 0.10
12
Step 6: Substitute the given values and calculated Qinto the Nernst equation
to find the cell potential:
E= 1.10 −0.0592
2log(0.10)
E= 1.10 −0.0296 ×(−1)
E= 1.10 + 0.0296 = 1.13 V
Therefore, the cell potential for the reaction at 25
°
C is 1.13 V.
Question 14
Question
A voltaic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and a silver ion electrode as the cathode. The concentration of Ag+(aq)
is 1.0×10−4M in the cathode compartment. If the measured cell potential is
0.54 V at 25◦C, calculate the concentration of H+(aq) in the anode compart-
ment.
Solution
Step 1: Write the cell reaction that occurs in the voltaic cell. The cell reaction
is:
2H+(aq) + 2e−→H2(g)
Step 2: Determine the standard cell potential (E◦) for the cell reaction.
Given that the standard electrode potential for SHE is 0.00 V and for Ag+(aq)|Ag(s)
is 0.80 V, the standard cell potential is:
E◦=E◦
cathode −E◦
anode = 0.80 V −0.00 V = 0.80 V
Step 3: Use the Nernst equation to relate cell potential to concentrations of
ions. The Nernst equation is given by:
E=E◦−0.0592
nlog [H+(aq)]anode
[H+(aq)]cathode
Where Eis the cell potential, E◦is the standard cell potential, nis the number
of electrons transferred, and [H+(aq)]anode and [H+(aq)]cathode are the concen-
trations of H+(aq) in the anode and cathode compartments, respectively.
Step 4: Substitute the known values into the Nernst equation. Given that
E= 0.54 V, E◦= 0.80 V, n= 2 (since 2 electrons are transferred in the cell
13
reaction), and [H+(aq)]cathode = [H+(aq)]SHE = 1.0×10−7M (standard pH
scale for SHE), we can solve for [H+(aq)]anode.
0.54 = 0.80 −0.0592
2log [H+(aq)]anode
1.0×10−7
0.54 = 0.80 −0.0296 log [H+(aq)]anode
1.0×10−7
0.0296 log [H+(aq)]anode
1.0×10−7= 0.26
log [H+(aq)]anode
1.0×10−7=0.26
0.0296 ≈8.78
[H+(aq)]anode
1.0×10−7= 108.78
[H+(aq)]anode ≈108.78 ×1.0×10−7
[H+(aq)]anode ≈2.51 ×101M
Therefore,
Question 15
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) and
a copper electrode. The concentration of Cu2+ ions in the copper half-cell is
0.10 M. The cell potential is measured to be 0.46 V at 25
°
C. Calculate the
standard reduction potential for the copper half-cell.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction for the electro-
chemical cell is:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the Nernst equation. The Nernst equation relates the cell
potential (Ecell) to the standard cell potential (E◦), the gas constant (R), the
temperature (T), the number of electrons transferred (n), and the concentrations
of the species involved. The Nernst equation is given by:
Ecell =E◦−RT
nF ln (Q)
where Ecell = 0.46 V, R= 8.314 J/mol ·K, T= 298 K, n= 2 (number of
electrons transferred), F= 96,485 C/mol, and Qis the reaction quotient.
Step 3: Calculate the reaction quotient. The reaction quotient Qcan be cal-
culated using the concentrations of products and reactants. Since the standard
14
hydrogen electrode has a hydrogen ion concentration of 1.0 M and a pressure of
1 atm, the reaction quotient is:
Q=[Cu](H+)2
[Cu2+]=1×(1.0)2
0.10 = 10
Step 4: Substitute values into the Nernst equation. Now, substitute the
given values and the reaction quotient into the Nernst equation:
0.46 = E◦−(8.314 ·298)
2·96485 ln(10)
Step 5: Solve for the standard reduction potential. Solve the equation for
E◦to find the standard reduction potential for the copper half-cell:
E◦= 0.46 + (8.314 ·298)
2·96485 ln(10)
E◦= 0.46 + 2471.672
192970 ln(10)
E◦= 0.46 + 0.03173 ln(10)
E◦= 0.46 + 0.03173 ×2.303
E◦≈0.535 V
Therefore, the standard reduction potential for the copper half-cell is ap-
proximately 0.535 V.
Question 16
Question
A redox reaction with an equilibrium constant K= 1.0×10−5occurs in a
galvanic cell at 25◦C. If the concentration of Mn2+ is 0.010 M and the concen-
tration of Mn3+ is 0.040 M, calculate the cell potential at standard conditions
and the cell potential at the given concentrations using the Nernst equation.
Solution
Step 1: Write the half-reactions and calculate the standard cell potential: The
half reactions are:
Mn3+ +e−→Mn2+ E◦=−1.18 V
We start by calculating the standard cell potential, E◦
cell:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 −(−1.18) = 1.18 V
15
Step 2: Calculate the cell potential at non-standard conditions using the
Nernst equation: The Nernst equation is given by:
Ecell =E◦
cell −0.0591
nlog [Mn2+]
[Mn3+]
where nis the number of electrons transferred in the balanced half-reaction. In
this case, n = 1.
Substitute the given values into the Nernst equation:
Ecell = 1.18 −0.0591
1log 0.010
0.040
Ecell = 1.18 −0.0591 log(0.25) = 1.18 −0.0591 ×(−0.602) = 1.215 V
Therefore, the cell potential at standard conditions is 1.18 V and at the given
concentrations, it is 1.215 V.
Question 17
Question
A redox reaction occurs in a cell according to the equation: Cu2+ +P b −→
Cu +P b2+. Given that the standard reduction potentials are E◦
Cu2+/Cu = 0.34
V and E◦
Pb2+/Pb =−0.13 V, calculate the cell potential when the concentrations
of Cu2+ and Pb2+ ions are 1.0 M and 0.1 M, respectively.
Solution
Step 1: Write the half-reactions and determine the cell potential in standard
conditions.
Cathode: Cu2+ + 2e−−→ Cu E◦= 0.34 V
Anode: Pb −→ Pb2+ + 2e−E◦=−0.13 V
The cell potential in standard conditions can be calculated as:
E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.13 V) = 0.47 V
Step 2: Use the Nernst equation to calculate the cell potential under non-
standard conditions. The Nernst equation is given by:
Ecell =E◦
cell −0.0592 V
nlog [Pb2+]
[Cu2+]
where nis the number of electrons involved in the reaction. In this case, n= 2.
16
Substitute the given concentrations into the Nernst equation:
Ecell = 0.47 V −0.0592 V
2log 0.1
1.0
Ecell = 0.47 V −0.0296 V log(0.1)
Ecell = 0.47 V −0.0296 V ×(−1)
Ecell = 0.47 V + 0.0296 V = 0.4996 V
Therefore, the cell potential when the concentrations of Cu2+ and Pb2+ ions
are 1.0 M and 0.1 M, respectively, is 0.4996 V.
Question 18
Question
A galvanic cell is constructed with a standard hydrogen electrode (SHE) on the
left side and a Ag/AgCl electrode on the right side. The standard reduction
potential for the Ag/AgCl electrode is 0.222 V. The cell is operated at 298 K.
Calculate the cell potential when the [Cl-] concentration is 0.01 M and the [H+]
concentration is 1.0 M.
Solution
Step 1: Write the cell reaction representing the given information. The cell
reaction for the galvanic cell can be represented as:
2H+(aq)+2e−→H2(g)
AgCl(s) + e−→Ag(s) + Cl−(aq)
Step 2: Calculate the cell potential at standard conditions. The standard
cell potential (E◦
cell) can be calculated using the formula:
E◦
cell =E◦
right −E◦
left = 0.222 V −0 V = 0.222 V
Step 3: Calculate the Nernst equation. The Nernst equation relates the cell
potential at non-standard conditions to the standard cell potential:
Ecell =E◦
cell −0.0592
nlog Q
Pn
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Qis the reaction quotient.
Step 4: Calculate the reaction quotient, Q. The reaction quotient for the
given cell can be calculated using the concentrations provided:
Q=[H+]2[Cl−]
PH2·PAg
=(1.0)2(0.01)
1·1= 0.01
17
Step 5: Substitute values into the Nernst equation. Substitute the given
values and calculated reaction quotient into the Nernst equation to find the cell
potential under non-standard conditions:
Ecell = 0.222 −0.0592
2log 0.01
Step 6: Calculate the cell potential. Now, solve for the cell potential:
Ecell = 0.222 −0.0592
2× −2=0.222 + 0.0592 = 0.2812 V
Therefore, the cell potential when the [Cl-] concentration is 0.01 M and the
[H+] concentration is 1.0 M is 0.2812 V.
Question 19
Question
Calculate the cell potential at 25
°
C for a cell in which the following reaction
occurs:
Fe3+(aq)+e−⇌Fe2+(aq)
Given that the standard reduction potential for the Fe3+/Fe2+ couple is
+0.77V, the concentration of Fe3+ is 0.10M, and the concentration of Fe2+ is
1.0M. Assume that the temperature is 25
°
C and that the Faraday constant, F,
is 96485 C/mol.
Solution
Step 1: Write the half-reaction equation for the Fe3+/Fe2+ couple:
Fe3+(aq)+e−⇌Fe2+(aq)
Step 2: Write the expression for the cell potential using the Nernst equation:
E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 3: Calculate the cell potential using the given values and the Nernst
equation: Given:
E◦= +0.77 V
[Fe3+] = 0.10 M
[Fe2+]=1.0M
n= 1 (number of electrons transferred in the reaction)
T= 25C= 298K
18
F= 96485 C/mol
Plugging in the values:
E= 0.77V−0.0592
1log 1.0
0.10
E= 0.77V−0.0592 log(10)
E= 0.77V−0.0592 ×1
E= 0.77V−0.0592
E= 0.7108V
Therefore, the cell potential at 25
°
C for the given reaction is 0.7108V.
Question 20
Question
A concentration cell is constructed with two silver-silver chloride electrodes.
Electrode 1 has a [Ag+] concentration of 0.10 M and electrode 2 has a [Ag+]
concentration of 0.0010 M. If the standard reduction potential for the reaction
Ag+(0.10 M) + e−→Ag(s) is 0.80 V, calculate the cell potential for this
concentration cell at 25◦C.
Solution
Step 1: Write the half-reaction involved in the concentration cell: The half-
reaction for the silver-silver chloride electrode is:
AgCl(s) + e−→Ag(s) + Cl−
Step 2: Write the Nernst equation for the cell potential (Ecell): The Nernst
equation is given by:
Ecell =E◦
cell −0.0592
nlog [Ag+]cathode
[Ag+]anode
Step 3: Calculate the number of electrons transferred (n): Since the reaction
involves the transfer of 1 electron, n= 1.
Step 4: Calculate the standard cell potential (E◦
cell): Given that the standard
reduction potential for the reaction is 0.80 V, E◦
cell = 0.80 V.
Step 5: Substitute the given values into the Nernst equation:
Ecell = 0.80V−0.0592
1log 0.0010
0.10
Ecell = 0.80V−0.59 log 0.010
19
Step 6: Solve for Ecell:
Ecell = 0.80V−0.59(−2)
Ecell = 0.80V+ 1.18V
Ecell = 1.98 V
Therefore, the cell potential for this concentration cell is 1.98 V at 25◦C.
Question 21
Question
Calculate the cell potential for a galvanic cell at 25
°
C in which the concentration
of Zn2+ is 0.10 M and the concentration of Cu2+ is 1.0 ×10−3M. The standard
reduction potentials are Zn2+/Zn = -0.76 V and Cu2+/Cu = 0.34 V.
Solution
Step 1: Write the half-reactions involved in the cell reaction. The overall cell
reaction is: Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)
The half-reactions are: Zn2+(aq) + 2e−→Zn(s) (oxidation) Cu2+ + 2e−
→Cu(s) (reduction)
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potentials. E◦
cell = E◦
reduction, cathode - E◦
oxidation, anode
E◦
cell = E◦
reduction, cathode - E◦
oxidation, anode E◦
cell = (0.34 V) - (-0.76 V) E◦
cell
= 1.10 V
Step 3: Calculate the Nernst equation to find the cell potential under non-
standard conditions. E = E◦-0.0592
n×log10 [Cu2+]
[Zn2+]
Where: E = cell potential under non-standard conditions E◦= cell potential
at standard conditions (1.10 V) n = number of electrons transferred in the
balanced chemical equation (2) [Cu2+] = concentrationof Cu2+ ions (1.0 ×
10−3M) [Zn2+] = concentrationof Zn2+ ions (0.10 M)
E = 1.10 V - 0.0592
2×log10 1.0×10−3
0.10 E = 1.10 V - 0.0296 ×log1010 E = 1.10
V - 0.0296 ×1 E = 1.10 V - 0.0296 E = 1.07 V
Therefore, the cell potential under the given conditions is 1.07 V.
Question 22
Question
A concentration cell is set up with two Ag—AgCl electrodes. One half-cell
contains a 10−2M concentration of Cl−ions, while the other half-cell contains
a 10−4M concentration of Cl−ions. Calculate the cell potential at 25
°
C using
the Nernst equation. Given: E◦
cell = 0.22 V.
20
Solution
Step 1: Write the two half-reactions that occur in the concentration cell.
Anode: Ag →Ag++e−
Cathode: Ag++e−→Ag
Step 2: Write the overall cell reaction.
Overall: 2Ag →2Ag+
Step 3: Calculate the cell potential at standard conditions using the standard
reduction potentials for each half-reaction. Given: E◦
cell = 0.22 V
Step 4: Substitute the known information into the Nernst equation:
Ecell =E◦
cell −0.0592
2log [Ag+]2
cathode
[Ag+]2
anode
Step 5: Convert the concentrations to molarities.
[Ag+]cathode = 10−4M
[Ag+]anode = 10−2M
Step 6: Substitute the molarities into the Nernst equation and solve for the
cell potential.
Ecell = 0.22 −0.0592
2log (10−4)2
(10−2)2
Step 7: Calculate the cell potential.
Ecell = 0.22 −0.0592
2log 10−8
Ecell = 0.22 −0.0592 ×4≈0.22 −0.2368 ≈ −0.0168 V
Therefore, the cell potential at 25
°
C using the Nernst equation is approxi-
mately -0.0168 V.
Question 23
Question
A galvanic cell is constructed with an Al/Al3+ half-cell and a Br2/Br−half-cell.
The initial concentrations are [Al3+]=1.0 M, [Br−] = 0.10 M, and [Br2]=0.20
M. Given that the standard reduction potentials for the Al3+/Al and Br2/Br−
half-reactions are E◦
cell,Al =−1.68 V and E◦
cell,Br2= 1.09 V, calculate the cell
potential at t= 0.
21
Solution
Step 1: The Nernst equation is given by:
E=E◦−0.0592
nlog Q
where Eis the cell potential, E◦is the standard cell potential, nis the number
of moles of electrons transferred, and Qis the reaction quotient.
Step 2: The balanced overall cell reaction is:
Al + Br2→Al3+ + 2Br−
Step 3: From the standard reduction potentials given, the standard cell
potential E◦
cell can be calculated as:
E◦
cell =E◦
cell, Br2−E◦
cell, Al
Step 4: Calculating the standard cell potential gives:
E◦
cell = 1.09 V −(−1.68 V) = 2.77 V
Step 5: The initial reaction quotient Qcan be determined using the initial
concentrations:
Q=[Al3+][Br−]2
[Al][Br2]=(1.0)(0.10)2
1(0.20) = 0.50
Step 6: Substituting values into the Nernst equation gives:
E= 2.77 −0.0592
1log 0.50
Step 7: Calculating the cell potential at t= 0:
E= 2.77 −0.0592(−0.3010) = 2.79 V
Therefore, the cell potential at t= 0 is 2.79 V.
Question 24
Question
Calculate the potential of a standard hydrogen electrode at 25
°
C when the
concentration of hydrogen ions in the solution is 1.0×10−2M. The standard
reduction potential of the hydrogen electrode is 0.00 V.
22
Solution
Step 1: Write the Nernst equation. The Nernst equation relates the standard
electrode potential to the actual cell potential under non-standard conditions.
The Nernst equation is given by:
E=E◦−0.0592
nlog(Q)
where: - Eis the cell potential under non-standard conditions, - E◦is the
standard electrode potential, - nis the number of electrons involved in the
reaction, - Qis the reaction quotient.
Step 2: Identify the values given in the question. - E◦= 0.00 V (standard
reduction potential of the hydrogen electrode) - T= 25
°
C (298 K) - [H+] =
1.0×10−2M
Step 3: Determine the number of electrons involved in the reaction. Since
the standard hydrogen electrode involves the reduction of hydrogen ions to
hydrogen gas, the reaction is:
2H++ 2e−→H2
Therefore, n= 2 (2 electrons are involved in the reaction).
Step 4: Calculate the reaction quotient Q.
Q=[products]
[reactants] =1
1.0×10−2= 100
Step 5: Substitute the values into the Nernst equation and solve for E.
E= 0.00−0.0592
2log(100) = 0.00 −0.0592 ×2×2=0.00 −0.2368 = −0.2368 V
Therefore, the potential of a standard hydrogen electrode at 25
°
C with 1.0×
10−2M hydrogen ions is −0.2368 V.
Question 25
Question
A galvanic cell consists of a silver electrode in a 0.10 M solution of AgNO3and
a copper electrode in a 0.20 M solution of CuSO4. The standard reduction
potential for Cu2+(aq)+2e−→Cu(s) is 0.34 V, while the standard reduction
potential for Ag+(aq) + e−→Ag(s) is 0.80 V. Calculate the cell potential at
25
°
C. (Given: R= 8.31 J/(mol ·K), F= 96,485 C/mol)
Solution
Step 1: Write the overall cell reaction and find the cell potential at standard
conditions. The overall reaction for the cell is: Cu2+(aq)+2Ag(s)→Cu(s) +
2Ag+(aq)
23
The standard cell potential, Ecell, is given by: Ecell =Ecathode −Eanode
Given that Ecathode is 0.80 V for Ag+(aq) + e−→Ag(s) and Eanode is 0.34
V for Cu2+(aq)+2e−→Cu(s), we have: Ecell = 0.80V−0.34V= 0.46V
Step 2: Calculate the reaction quotient, Q. The reaction quotient, Q, is given
by: Q=[Ag+]2
[Cu2+ ]
Given that [Ag+]=0.10Mand [Cu2+]=0.20M, we have: Q=(0.10)2
0.20 =
0.05
Step 3: Use the Nernst equation to calculate the cell potential at 25
°
C. The
Nernst equation is given by: Ecell =Ecell −0.0592V
nlog(Q)
Since the reaction involves the transfer of 2 moles of electrons, n= 2. Sub-
stituting the values of Ecell,n, and Qinto the equation: Ecell = 0.46V−
0.0592V
2log(0.05) Ecell = 0.46V−0.0592V
2×(−1.3010) Ecell = 0.46V+ 0.0389V
Ecell = 0.4989V
Therefore, the cell potential at 25
°
C for the given galvanic cell setup is 0.4989
V.
Question 26
Question
A half-cell consists of a copper electrode in a 0.05 M CuSO4solution. The
standard reduction potential for the reaction is E◦= 0.34 V. At what pH will
the potential of this half-cell be 0.18 V?
Solution
Step 1: Write the half-reaction and the balanced redox reaction.
The half-reaction for the reduction of copper is: Cu2+ + 2e−→Cu
The balanced redox reaction is: Cu2+ + 2e−→Cu
Step 2: Write the Nernst equation.
The Nernst equation relates the cell potential to the standard cell potential and
the concentrations of the reactants and products:
E=E◦−0.0592
nlog(Q)
where Eis the cell potential, E◦is the standard cell potential, nis the number
of electrons transferred in the balanced equation, and Q=[products]
[reactants].
Step 3: Identify the values given and calculate Q.
Given: E◦= 0.34 V, E= 0.18 V, [Cu2+]=0.05 M
Since the concentration of Cu is not given, we can assume it to be 1 M. Thus,
Q=1
0.052= 400
24
Step 4: Substitute the values into the Nernst equation and solve for the pH.
Substitute the given values into the Nernst equation:
0.18 = 0.34 −0.0592
2log(400)
0.18 = 0.34 −0.0296 log(400)
Now, solve for log(400):
log(400) = 0.34 −0.18
0.0296
log(400) = 0.16
0.0296
log(400)5.41
Step 5: Find the pH.
To find the pH, we need to use the relationship between [H+] and pH: pH =
−log[H+]
Since the concentration of H+can be calculated from the given Q, we have:
[H+] = pQ=√40020
pH = −log(20)1.70
Therefore, the pH at which the potential of the half-cell is 0.18 V is approx-
imately 1.70.
Question 27
Question
Calculate the cell potential at 25
°
C for a galvanic cell with a standard hydrogen
electrode (SHE) on the left and a copper(II) sulfate solution on the right, where
the concentration of Cu2+ is 0.10 M. Given: E◦
Cu2+/Cu = 0.34 V and E◦
H+/H2=
0 V.
Solution
Step 1: Write the overall balanced cell reaction. The overall cell reaction can
be written as:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the half-reactions for the anode and cathode. The half-
reactions at the anode and cathode are as follows: Anode (oxidation):
Cu(s)→Cu2+(aq)+2e−
25
Cathode (reduction):
2H+(aq)+2e−→H2(g)
Step 3: Write the cell potential at standard conditions. The cell potential
at standard conditions (E◦
cell) is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 V −0.34 V = −0.34 V
Step 4: Determine the reaction quotient, Q. The reaction quotient, Q, is
given by:
Q=[Cu2+]
[H+]
2
Given that [Cu2+]=0.10 M and [H+]=1.0 M (from the SHE), we have:
Q=0.10
1.0
2
= 0.01
Step 5: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
Ecell =E◦
cell −0.0592
nlog(Q)
where nis the number of electrons transferred in the balanced cell reaction.
Since 2 electrons are transferred, n= 2.
Ecell =−0.34 V −0.0592
2log(0.01)
Ecell =−0.34 V −0.0296 log(0.01)
Ecell =−0.34 V −0.0296 ×(−2)
Ecell =−0.34 V + 0.0592
Ecell =−0.2808 V
Therefore, the cell potential at 25
°
C for the given galvanic cell is -0.2808 V.
Question 28
Question
Calculate the cell potential for the following reaction at 25
°
C:
Zn(s)|Zn2+(0.10M)||Cu2+(1.0M)|Cu(s)
Given: E◦
cell = 1.10 V R= 8.314 J/(mol ·K) T= 298 K
26
Solution
Step 1: Write the half-reactions for the redox reaction and determine the stan-
dard cell potential. The half-reactions for the redox reaction are:
Zn2+(aq)+2e−→Zn(s) E◦=−0.76 V
Cu2+(aq)+2e−→Cu(s) E◦= 0.34 V
The standard cell potential is given by:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 2: Write the Nernst equation. The Nernst equation is given by:
Ecell =E◦
cell −RT
nF ln Q
K
Where: Ecell = cell potential at non-standard conditions E◦
cell = standard cell
potential R= gas constant (8.314 J/(mol
·
K)) T= temperature in Kelvin (298
K) n= number of moles of electrons exchanged in the balanced redox reaction
F= Faraday’s constant (96485 C/mol) Q= reaction quotient (ratio of products
to reactant concentrations) K= equilibrium constant
Step 3: Calculate the number of moles of electrons exchanged (n). From the
balanced redox reaction, 2 moles of electrons are exchanged by both Zn and Cu:
n= 2
Step 4: Calculate the reaction quotient (Q). The reaction quotient is calcu-
lated using the concentrations of the ions involved in the redox reaction:
Q=[Cu2+]
[Zn2+]
Q=1.0
0.10 = 10
Step 5: Substitute the given values into the Nernst equation and solve for
Ecell.
Ecell = 1.10 V −(8.314 J/mol ·K)(298 K)
(2)(96485 C/mol) ln 10
K
Ecell = 1.10 V −0.0257 ln 10
KV
Therefore, the cell potential at 25
°
C for the given redox reaction is Ecell =
1.10 V −0.0257 ln 10
KV.
27
Question 29
Question
A half-cell consists of a silver metal electrode in contact with a 0.05 M silver
nitrate solution. If the standard electrode potential for the Ag/Ag+half-cell is
E◦= 0.80 V, calculate the potential of this half-cell at 25◦C when [Ag+] = 0.01
M. (Given: R= 8.314 J/(mol ·K), F= 96485 C/mol)
Solution
Step 1: Write the Nernst equation: The Nernst equation relates the measured
cell potential (E) to the standard cell potential (E◦), the reaction quotient (Q),
the gas constant (R), the temperature (T), and the Faraday constant (F). The
Nernst equation for the given half-cell can be written as:
E=E◦−RT
nF ln(Q)
Step 2: Calculate the reaction quotient (Q): In this case, the reaction quo-
tient (Q) is the ratio of product concentrations to reactant concentrations, raised
to the power of their stoichiometric coefficients. The given half-cell reaction is:
Ag(s)→Ag+(aq) + e−
Thus, Q=[Ag+]
1.
Step 3: Plug in the given values to solve for E: Given: E◦= 0.80 V, T=
25◦C = 298 K, [Ag+] = 0.01 M. Plugging these values into the Nernst equation,
we get:
E= 0.80 V −(8.314 J/(mol ·K)) ×(298 K)
1×(96485 C/mol) ln(0.01)
Step 4: Calculate E: Calculating the Natural Logarithm first:
ln(0.01) = ln1×10−2=−2 ln(10) ≈ −4.605
Now plug this value back into the Nernst equation:
E= 0.80 V −(8.314 J/(mol ·K)) ×(298 K)
1×(96485 C/mol) × −4.605
E≈0.80 V −(−0.056)
E≈0.856 V
Therefore, the potential of the half-cell at 25◦C when [Ag+]=0.01 M is
approximately 0.856 V.
28
Question 30
Question
A voltaic cell consists of a half-cell with a copper electrode in a 1.0 M Cu2+
solution and a half-cell with a silver electrode in a 0.010 M Ag+solution. Cal-
culate the cell potential at 25
°
C for this cell. Given that the standard reduction
potentials are: Cu2+ + 2e−→Cu(s) with E◦= +0.34 V and Ag++ e−→
Ag(s) with E◦= +0.80 V.
Solution
Step 1: Write the two half-reactions with their respective standard reduction
potentials. The two half-reactions are: Cu2+ + 2e−→Cu(s) E◦
red = +0.34
V Ag++ e−→Ag(s) E◦
red = +0.80 V
Step 2: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog [Cu2+]
[Ag+]
Step 3: Determine the number of electrons transferred (n) in the overall
cell reaction. From the two half-reactions provided, we can see that 2 moles
of electrons are transferred for the reduction of Cu2+ and 1 mole of electron is
transferred for the reduction of Ag+. So, the least common multiple of 2 and 1
is 2. Hence, n = 2.
Step 4: Substitute the given values into the Nernst equation.
E= 0.80V−0.0592
2log 1.0
0.010
E= 0.80V−0.0296 log(100)
E= 0.80V−0.0296 ×2
E= 0.80V−0.0592
E= 0.74 V
Therefore, the cell potential at 25
°
C for this cell is 0.74 V.
Question 31
Question
A cell is constructed with a standard hydrogen electrode (SHE) on one side
and a copper electrode on the other side. The standard reduction potential
of hydrogen electrode is 0 V and the standard reduction potential of copper
electrode is +0.34 V. Calculate the cell potential at 25◦C.
29
Solution
Step 1: Write the half-reactions for the electrodes involved. The half-reaction
for the standard hydrogen electrode is:
2H++ 2e−→H2E◦= 0 V
The half-reaction for the copper electrode is:
Cu2+ + 2e−→Cu E◦= +0.34 V
Step 2: Determine the overall cell reaction. To find the overall cell reaction,
we add the two half-reactions together. Since the reduction potential of copper
is higher, the copper half-reaction will be the reduction:
Cu2+ + 2H++ 2e−→Cu + H2
Step 3: Calculate the cell potential using the Nernst equation. The Nernst
equation is given by:
E=E◦−0.0592
nlog(Q)
Where: - Eis the cell potential - E◦is the standard cell potential - nis the
number of electrons transferred in the cell reaction - Qis the reaction quotient,
which is the ratio of product concentrations to reactant concentrations
For this cell, E◦= +0.34 V and n= 2.
Step 4: Calculate the reaction quotient (Q). Since concentrations are not
given, Qcannot be calculated.
Therefore, the cell potential of this cell cannot be calculated without knowing
the concentrations of the reactants and products.
Question 32
Question
A voltaic cell consists of a platinum electrode dipping in a solution of 0.01
M Ag+ions and a silver electrode dipping in a solution of 1.0 M Ag+ions.
Calculate the cell potential at 25
°
C given that the standard reduction potential
for the silver/silver ion half-cell is +0.80 V and that the Nernst constant (E◦)
is 0.0592 V.
Solution
Step 1: Write the half-reactions for the cell.
The anode half-reaction is the oxidation of Ag to Ag+:
Ag(s)→Ag+(aq) + e−
30
The cathode half-reaction is the reduction of Ag+to Ag:
Ag+(aq) + e−→Ag(s)
Step 2: Calculate the cell potential using the Nernst equation:
E=E◦−0.0592
1log [products]
[reactants]
The standard reduction potential (E◦) for the silver/silver ion half-cell is
given as +0.80 V.
Step 3: Calculate the concentrations of Ag+ions at both electrodes.
For the anode, [Ag+] = 0.01MF orthecathode, [Ag+]=1.0M
Step 4: Substitute the values into the Nernst equation and solve for the cell
potential.
E= 0.80V−0.0592
1log 1.0
0.01
E= 0.80V−0.0592 ×log(100)
E= 0.80V−0.0592 ×2
E= 0.80V−0.1184
E= 0.6816 V
Therefore, the cell potential at 25
°
C is 0.6816 V.
Question 33
Question
A cell consists of a standard hydrogen electrode (SHE) and a nickel electrode
in a 1.0 M Ni2+ solution. Given that the standard reduction potential of the
Ni2+/Ni electrode is −0.25 V, calculate the cell potential at 25
°
C when the
concentration of Ni2+ is 0.10 M. (F= 96485 C/mol)
Solution
Step 1: Write the half-reaction for the standard hydrogen electrode (SHE) and
the nickel electrode. The half-reaction for the standard hydrogen electrode is:
2H++ 2e−→H2(g)E◦= 0 V
The half-reaction for the nickel electrode is:
Ni2+ + 2e−→Ni(s) E◦=−0.25 V
31
Step 2: Write the overall cell reaction and find E◦
cell. The overall cell reaction
is the sum of the two half-reactions:
2H++ Ni2+ →H2(g) + Ni(s)
Using the standard reduction potentials, E◦
cell can be calculated as:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0 V −(−0.25 V) = 0.25 V
Step 3: Use the Nernst equation to find the cell potential at non-standard
conditions. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
Fn
Where: - E= cell potential at non-standard conditions - E◦= standard cell
potential - n= number of electrons transferred in the balanced cell reaction -
Q= reaction quotient - F= Faraday’s constant
Step 4: Calculate the reaction quotient Q. Since the concentrations of all
species are 1 M in the standard state, the reaction quotient Qcan be calculated
as:
Q=[H2][Ni(s)]
[H+]2[Ni2+]= 1
Step 5: Substitute the values into the Nernst equation. Substitute the values
into the Nernst equation:
E= 0.25 V −0.0592
2log(1)
E= 0.25 V
Therefore, the cell potential at 25
°
C when the concentration of Ni2+ is 0.10
M is 0.25 V.
Question 34
Question
A redox reaction involving the reduction of copper(II) ions to copper metal is
represented by the equation:
Cu2+ + 2e−→Cu
If the standard reduction potential (E◦) for this reaction is +0.34 V, calculate
the cell potential when the concentration of Cu2+ is 0.10 M and the concentra-
tion of Cu is 1.0 M. Assume the temperature is 298 K.
32
Solution
Step 1: Write the Nernst equation for the cell potential E:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential, - E◦is the standard reduction potential, - nis
the number of moles of electrons transferred, - Qis the reaction quotient.
Step 2: Determine the number of moles of electrons transferred (n) from the
balanced redox reaction. In this case, since 2 moles of electrons are involved in
the reduction of 1 mole of Cu2+ to Cu,n= 2.
Step 3: Calculate the reaction quotient Qusing the given concentrations of
Cu2+ and Cu:
Q=[Cu]1
[Cu2+]2
Q=1.0
(0.10)2
Q= 100
Step 4: Plug the values of E◦,n, and Qinto the Nernst equation to solve
for E:
E= 0.34 −0.0592
2log(100)
E= 0.34 −0.0296 ×2
E= 0.34 −0.0592
E= 0.2808 V
Therefore, the cell potential when the concentration of Cu2+ is 0.10 M and
the concentration of Cu is 1.0 M is 0.2808 V.
Question 35
Question
A voltaic cell is constructed with two half-cells. One half-cell consists of a
standard hydrogen electrode (SHE) with H+(aq) at 1.00Mconcentration and
hydrogen gas at 1.00atm. The other half-cell consists of Cu2+(aq) at 0.010M
and Cu(s). Calculate the cell potential at 25◦Cgiven that E◦
cell = 0.34V.
Assume R= 8.314J/(mol ·K) and F= 96485C/mol.
33
Solution
Step 1: Write the half-reactions for the given cell.
Anode (oxidation): Cu(s)→Cu2+(aq)+2e−
Cathode (reduction): 2H+(aq)+2e−→H2(g)
Step 2: Calculate the cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592V
nlog Q
where nis the total number of electrons transferred and Qis the reaction quo-
tient for the cell.
Step 3: Determine the reaction quotient Qfor the cell.
Q=[H+]2
[Cu2+]
Plugging in the given concentrations:
Q=(1.00)2
0.010 = 100
Step 4: Calculate the cell potential using the Nernst equation.
Ecell = 0.34V−0.0592V
2log 100
Ecell = 0.34V−(0.0296V)×2
Ecell = 0.2808V
Therefore, the cell potential at 25◦Cis 0.2808V.
34
