CHEM 121 - GENERAL CHEMISTRY
I - Galvanic and Electrolytic Cells
Question Bank - Set 2
Liberty University
Question 1
Question
Consider a galvanic cell with a standard electrode potential of E◦
cell = 1.02 V.
Determine the standard cell potential for the cell when the concentration of Fe2+
is 0.10 M and the concentration of Mn2+ is 1.0 M. The half-reactions involved
are:
Fe2+(aq)→Fe3+(aq) + e−E◦
cell = 0.77 V
Mn2+(aq)+2e−→Mn(s)E◦
cell =−1.18 V
Solution
Step 1: Write out the overall cell reaction and find the standard cell potential.
The overall cell reaction is obtained by adding the two half-reactions to-
gether:
Fe2+(aq) + Mn(s)→Fe3+(aq) + Mn2+(aq)
The standard cell potential is given by the difference in standard electrode
potentials of the two half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Fe3+/Fe2+ −E◦
Mn2+/Mn
E◦
cell = 0.77 −(−1.18)
E◦
cell = 0.77 + 1.18
E◦
cell = 1.95 V
Step 2: Calculate the cell potential at the given concentrations.
Using the Nernst equation, the cell potential at non-standard conditions is
given by:
Ecell =E◦
cell −0.0592
nlog [Fe3+]
[Fe2+]
Where nis the number of electrons transferred in the balanced equation. In
this case, n= 1.
Substitute the given concentrations into the equation:
Ecell = 1.95 −0.0592
1log 1.0
0.10
Ecell = 1.95 −0.0592 log 10
Ecell = 1.95 −0.0592(1)
Ecell = 1.95 −0.0592
Ecell = 1.89 V
Therefore, the standard cell potential for the cell at the given concentrations
is 1.89 V.
Question 2
Question
Explain the key differences between Galvanic and Electrolytic Cells. Provide
examples of each type of cell.
Solution
Galvanic cells and electrolytic cells are two types of electrochemical cells that
involve redox reactions. They have distinct differences in their operation and
purpose.
Galvanic Cells:
Galvanic cells convert chemical energy to electrical energy spontaneously.
The redox reaction in a galvanic cell is spontaneous and generates an
electric current.
Electrons flow from the anode (where oxidation occurs) to the cathode
(where reduction occurs).
Examples of galvanic cells include the Daniell cell (Zn-Cu cell) and the
Voltaic pile.
Electrolytic Cells:
Electrolytic cells require an external electrical energy source to drive a
non-spontaneous redox reaction.
The redox reaction in an electrolytic cell is non-spontaneous and consumes
electric current.
Electrons flow from the external power source to the anode, where oxida-
tion occurs, and to the cathode, where reduction occurs.
Examples of electrolytic cells include the electrolysis of water to produce
hydrogen and oxygen gas, and the electroplating of metals.
2
In summary, galvanic cells produce electrical energy from chemical reactions
that occur spontaneously, while electrolytic cells require external energy input
to drive non-spontaneous reactions.
Question 3
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 0.70 V. If
the concentration of Fe2+ ions in the cell is 0.10 M and the concentration of
Cu2+ ions is 1.0 M, determine the cell potential at 25◦C. Assume that the cell
reaction is:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
Solution
Step 1: Write the cell reaction for the galvanic cell and identify the half-
reactions. The cell reaction is:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
The half-reactions are: Oxidation half-reaction:
Fe2+(aq)→Fe3+(aq) + e−
Reduction half-reaction:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the expressions for the equilibrium constant Kand the cell
potential Ecell. The equilibrium constant Kis given by:
K=[Fe3+][Cu]
[Fe2+][Cu2+]
The cell potential Ecell can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(K)
Step 3: Calculate the equilibrium constant K. Substitute the given concen-
trations into the expression for K:
K=(x)(1)
(0.10)(1.0)
Step 4: Calculate the cell potential Ecell. Since the reaction involves the
transfer of one electron, n= 1. Substitute Kinto the Nernst equation:
Ecell = 0.70 −0.0592
1log x
0.10
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Step 5: Solve for Ecell. To find Ecell, we need to solve for xin the expression
for K. Solving for xgives us:
x= 0.10
Now substitute xback into the Nernst equation:
Ecell = 0.70 −0.0592
1log 0.10
0.10
Ecell = 0.70 −0.0592
1log(1)
Ecell = 0.70 V
Therefore, the cell potential at 25◦C is 0.70 V.
Question 4
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦
cell =−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦
cell = 0.34 V
Calculate the standard cell potential when the cell is reacting under non-standard
conditions with Zn2+= 0.10 M, Cu2+= 1.0 M at 25◦C.
Solution
Step 1: Write the Nernst equation for the cell reaction:
Ecell =E◦
cell −0.0592
nlog Q
K
Where: - Ecell is the cell potential under non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of moles of electrons transferred in the
balanced equation, - Qis the reaction quotient, - Kis the equilibrium constant.
Step 2: Calculate n, the number of moles of electrons transferred in the
balanced equation. Since both half-reactions involve the transfer of 2 moles of
electrons, n= 2.
Step 3: Calculate the reaction quotient Qusing the concentrations provided:
Q=[Zn2+]
[Cu2+]=0.10
1.0= 0.10
Step 4: Determine the equilibrium constant Kusing the standard cell po-
tential:
E◦
cell = 0.34 −(−0.76) = 1.10 V
4
K= exp nE◦
cell
0.0592= exp 2×1.10
0.0592 ≈1536.30
Step 5: Substitute the values of E◦
cell,n,Q, and Kinto the Nernst equation:
Ecell = 1.10 −0.0592
2log 0.10
1536.30≈1.044 V
Therefore, the standard cell potential when the cell is reacting under non-
standard conditions is approximately 1.044 V.
Question 5
Question
A galvanic cell is constructed with a standard hydrogen electrode (E◦
cell = 0.00
V) and a copper electrode (E◦
Cu = 0.34 V). If the concentration of Cu2+ ions
is 1.0 M, what is the half-cell potential (E◦
Cu2+/Cu) of the copper electrode at
25
°
C?
Solution
Step 1: Write the cell reaction for the galvanic cell.
Cell reaction: H+(aq)+e−→1
2H2(g)E◦= 0.00 V
Cu2+(aq) + 2e−→Cu(s) E◦= 0.34 V
Step 2: Determine the overall cell reaction and the cell potential. The overall
cell reaction is given by adding the two half reactions:
Overall cell reaction: 2H+(aq) + Cu2+(aq)→1
2H2(g) + Cu(s)
The overall standard cell potential can be calculated using the standard
reduction potentials:
E◦
cell =E◦
Cu2+/Cu −E◦
H+/H2= 0.34 V −0.00 V = 0.34 V
Step 3: Use the Nernst equation to calculate the actual cell potential. The
Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Qis the reaction quotient.
5
Step 4: Calculate the actual cell potential for the given concentration of
Cu2+ ions. Since the cell reaction involves the transfer of 2 moles of electrons,
n= 2. The reaction quotient can be calculated as:
Q=[Cu]
[H+]2=1.0
(1.0)2= 1.0
Therefore, the actual cell potential (Ecell) can be calculated as:
Ecell = 0.34 V −0.0592
2log(1.0) = 0.34 V
Therefore, the half-cell potential (E◦
Cu2+/Cu) of the copper electrode at 25
°
C
is 0.34 V.
Question 6
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
If the concentration of Zn2+ in the cathode compartment is 0.10 M, while
the concentration of Cu2+ in the anode compartment is 1.0 M. Calculate the
cell potential at standard conditions and determine the direction of electron
flow.
Solution
Step 1: Write the overall reaction and calculate the standard cell potential. The
overall cell reaction is the sum of the two half-reactions:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 2: Determine the direction of electron flow. Since E◦
cell is positive, the
reaction is spontaneous as written. This means electrons flow from the anode
(where oxidation occurs) to the cathode (where reduction occurs).
Therefore, in this galvanic cell, electrons flow from Zn (anode) to Cu (cath-
ode).
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Question 7
Question
Consider a galvanic cell composed of a standard hydrogen electrode (SHE) and
a silver electrode. The reduction half-reaction for the silver electrode is the
following:
Ag++e−→Ag(s)
Given that the standard reduction potential for this half-reaction is E0=
0.80 V, calculate the cell potential when the concentration of Ag+ions is 0.1 M
and the pressure of hydrogen gas at the SHE is 1 atm. Is the cell spontaneous
under these conditions?
Solution
Step 1: Write out the full cell reaction: The overall cell reaction can be obtained
by combining the given reduction half-reaction with the standard hydrogen elec-
trode (SHE) reaction, which is the reduction of hydrogen ion:
2H++ 2e−→H2(g)
The full cell reaction is:
2Ag++ 2H+→2Ag + H2
Step 2: Calculate the standard cell potential, E0
cell: Using the standard
reduction potentials provided:
E0
cell =E0
cathode −E0
anode =E0
Ag −E0
H= 0.80V−0V= 0.80V
Step 3: Calculate the cell potential under the given conditions: The Nernst
equation for cell potential is:
Ecell =E0
cell −0.0592
nlog Q
K
where Qis the reaction quotient, Kis the equilibrium constant, and nis the
number of moles of electrons transferred.
Step 4: Calculate the reaction quotient, Q: For the given cell reaction:
Q=[Ag]2
[H+]2
Given [Ag+]=0.1 M and [H+] = 1Min this case:
Q=(0.1)2
12= 0.01
7
Step 5: Calculate the cell potential: Substitute the values of E0
cell,Q, and
n= 2 into the Nernst equation:
Ecell = 0.80 −0.0592
2log(0.01)
Ecell = 0.80 + 0.0296 ×2 log(0.01)
Ecell = 0.80 + 0.0296 ×2×(−2) = 0.80 −0.1184 = 0.6816 V
Step 6: Determine if the cell is spontaneous: Since the calculated cell po-
tential Ecell = 0.6816 V is positive, the cell reaction is spontaneous under the
given conditions.
Question 8
Question
Consider a galvanic cell with a standard cell potential of 1.13 V that involves
the following half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
MnO−
4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦= 1.51 V
Calculate the standard cell potential for the reaction when the following
concentrations are used: Zn2+= 0.10 M, MnO−
4= 0.20 M, H+= 1.0 M.
Solution
Step 1: Write the overall cell reaction using the given half-reactions and their
standard reduction potentials.
Zn(s) + MnO−
4(aq)+8H+(aq)→Zn2+(aq) + Mn2+(aq)+4H2O(l)
Step 2: Determine the standard cell potential (E◦
cell) for this reaction by
adding the standard reduction potentials of the half-reactions.
E◦
cell =E◦
cathode −E◦
anode
= (1.51 V) −(−0.76 V)
= 2.27 V
Step 3: Apply the Nernst equation to calculate the cell potential under
non-standard conditions. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
K
Where: - Eis the cell potential under non-standard conditions - E◦is the
standard cell potential - nis the number of moles of electrons transferred in the
8
cell reaction - Qis the reaction quotient - Kis the equilibrium constant for the
cell reaction
Step 4: Calculate the reaction quotient Qusing the concentrations provided.
Q=Zn2+Mn2+
MnO−
4H+8
Q=(0.10)(1.0)
(0.20)(1.0)8= 0.10
Step 5: Substitute the values into the Nernst equation to find the cell po-
tential under the given concentrations.
E= 2.27 V −0.0592
5log(0.10) ≈2.249 V
Therefore, the standard cell potential for the reaction under the given con-
centrations is approximately 2.249 V.
Question 9
Question
Consider a galvanic cell that consists of a zinc electrode in a 1.0 M Zn(NO3)2
solution and a copper electrode in a 1.0 M Cu(NO3)2 solution. The standard
reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu = 0.34 V. Calculate
the cell potential at 25
°
C for this galvanic cell.
Solution
Step 1: The cell potential (E◦
cell) for a galvanic cell can be calculated using the
formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode is
the standard reduction potential of the anode.
Step 2: In this case, the zinc electrode is the anode and the copper electrode
is the cathode. Therefore, we have:
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential at 25
°
C for this galvanic cell is 1.10 V.
9
Question 10
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
E◦
Zn =−0.76 V
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
E◦
Cu = 0.34 V
a) Calculate the cell potential for this reaction at standard conditions.
b) Determine the equilibrium constant, K, for this cell reaction.
Solution
a) The cell potential for a galvanic cell can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Step 1: Identify the given standard electrode potentials for the cathode and
anode:
E◦
Zn =−0.76 V
E◦
Cu = 0.34 V
Step 2: Substitute the values into the formula to find the cell potential:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential for this reaction at standard conditions is 1.10
V.
b) The equilibrium constant, K, for this cell reaction can be determined
using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(K)
where n is the number of moles of electrons transferred in the balanced
reaction.
Step 1: Determine the number of moles of electrons transferred in the
balanced reaction: The balanced reaction is:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
From this, we see that n = 2 (two moles of electrons are transferred).
Step 2: Find K using the Nernst equation: Substitute the known values
into the equation:
1.10 V = 1.10 V −0.0592
2log(K)
10
0 = −0.0592
2log(K)
log(K) = 0
K= 1
Therefore, the equilibrium constant, K, for this cell reaction is 1.
Question 11
Question
Consider the following Galvanic Cell reaction involving the reduction of copper
ions:
Cu2+(aq)+2e−→Cu(s)
If a Galvanic Cell is constructed with a standard cell potential of E◦
cell = 0.34
V at 25◦C, determine the standard reduction potential, E◦, for the reduction
of copper ions.
Solution
Step 1: Write the Nernst equation for the Galvanic Cell:
Ecell =E◦
cell −0.0592
2log [Cu(s)]
[Cu2+(aq)]2
Step 2: Substitute the given values into the Nernst equation:
0.34 = E◦
cell −0.0592
2log 1
[Cu2+]2
Step 3: Simplify the equation:
0.34 = E◦
cell + 0.0296 log [Cu2+]2
0.34 = E◦
cell + 0.0592 log [Cu2+]
Step 4: Rearrange the equation to solve for the standard reduction potential,
E◦:
0.0592 log [Cu2+]= 0.34 −E◦
cell
log [Cu2+]=0.34 −E◦
cell
0.0592
Step 5: Solve for [Cu2+]:
[Cu2+] = 10
0.34−E◦
cell
0.0592
Therefore, the standard reduction potential, E◦, for the reduction of copper
ions is 100.34−E◦
cell
0.0592 .
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Question 12
Question
Consider a galvanic cell with a standard cell potential of 1.23 V at 298 K. The
half-reactions involved are:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the concentrations are [Zn2+] = 1.0 M and [Cu2+] = 0.1 M, calculate the
cell potential at 298 K.
Solution
Step 1: Write the overall balanced cell reaction:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Write the cell reaction with half-cell potentials and concentrations:
E◦
cell =E◦
cathode −E◦
anode
Step 3: Calculate the cell potential by plugging in the given values:
E◦
cell = 0.34 V
Question 13
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a silver-silver chloride electrode (Ag/AgCl) in half-cells. The standard
reduction potential for the Ag/AgCl electrode is Eo= 0.222 V. If the measured
cell potential is Ecell = 0.48 V, calculate the standard reduction potential of the
hydrogen electrode.
Solution
To find the standard reduction potential of the hydrogen electrode, we can use
the Nernst equation. The Nernst equation relates the measured cell potential
to the standard reduction potentials of the half-reactions involved:
Ecell =Eo
cathode −Eo
anode
where Ecell is the measured cell potential, Eo
cathode is the standard reduction
potential of the cathode (Ag/AgCl in this case), and Eo
anode is the standard
reduction potential of the anode (hydrogen electrode).
12
Given: Ecell = 0.48 V Eo
cathode = 0.222 V
Substitute these values into the Nernst equation to solve for Eo
anode:
0.48 V = 0.222 V −Eo
anode
Eo
anode = 0.222 V −0.48 V = −0.258 V
Therefore, the standard reduction potential of the hydrogen electrode is
−0.258 V.
Question 14
Question
Consider a galvanic cell with the following half-reactions and standard reduction
potentials:
Mn2+(aq)+2e−→Mn(s)E◦=−1.18 V
Ag+(aq) + e−→Ag(s)E◦= 0.80 V
(a) Write the overall cell reaction for this galvanic cell. Indicate the direction
of electron flow and identify the anode and cathode.
(b) Calculate the standard cell potential (E◦
cell) for the galvanic cell.
(c) If the initial concentrations of Mn2+ and Ag+are both 0.10 M, determine
the cell potential when the cell reaches equilibrium.
Solution
(a) The overall cell reaction for the galvanic cell can be obtained by adding the
two half-reactions:
Mn2+(aq) + 2Ag+(aq)→Mn(s) + 2Ag(s)
In this reaction, electrons flow from Mn2+ to Ag+, so the anode is the side where
oxidation occurs (Mn2+) and the cathode is where reduction occurs (Ag+).
(b) The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Where E◦
cathode = 0.80 V and E◦
anode =−1.18 V. Therefore,
E◦
cell = 0.80 V −(−1.18 V) = 1.98 V
(c) At equilibrium, the cell potential is determined by the Nernst equation:
Ecell =E◦
cell −0.0592 V
nlog [Ag+]2
[Mn2+]
13
Since the reaction involves the transfer of 2 moles of electrons, n= 2. At
equilibrium, [Mn2+] = 0.10 M, [Ag+] = 0.10 M. Substituting these values, we
get:
Ecell = 1.98 V −0.0592 V
2log 0.102
0.10
Ecell = 1.98 V −0.0296 V log(1) = 1.98 V
Question 15
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
2Ag(s)→2Ag+(aq) + 2e−E◦= 0.80 V
If the concentration of Zn2+ in the compartment with the zinc electrode is
1.0 M and the concentration of Ag+in the compartment with the silver electrode
is 0.1 M, what is the cell potential at 25 degrees Celsius?
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is obtained by
adding the two half-reactions together after multiplying the first half-reaction
by 2:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Step 2: Calculate the standard cell potential, E◦
cell. The standard cell po-
tential is the difference between the standard reduction potentials of the two
half-reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −(−0.76 V) = 1.56 V
Step 3: Calculate the cell potential, E, at 25 degrees Celsius using the Nernst
equation. The Nernst equation is given by:
E=E◦
cell −0.0592
nlog [Zn2+]
[Ag+]2
where n is the number of moles of electrons transferred. In this case, n = 2.
Substitute the values into the equation:
E= 1.56 V−0.0592
2log 1.0
0.12= 1.56 V−0.0592 log(100) = 1.56 V−0.0592×2=1.4416 V
Therefore, the cell potential at 25 degrees Celsius is 1.4416 V.
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Question 16
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) con-
nected to a copper electrode in one compartment, and a standard hydrogen elec-
trode connected to a silver electrode in the other compartment. The cell operates
under standard conditions (T= 298 K, P= 1 atm). The standard reduction
potential for copper is E◦
Cu2+/Cu = 0.34 V and for silver is E◦
Ag+/Ag = 0.80 V.
Calculate E◦
cell, ∆G◦, ∆S◦, and ∆H◦for the cell reaction.
Solution
Step 1: Calculate E◦
cell using the standard cell potential formula:
E◦
cell =E◦
cathode −E◦
anode
Given E◦
Cu2+/Cu = 0.34 V and E◦
Ag+/Ag = 0.80 V, the overall cell reaction is:
Cu2+(aq) + 2e−→Cu(s)
2H+(aq) + 2e−→H2(g)
Ag+(aq)+e−→Ag(s)
The cell reaction is:
Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)
E◦
cell =E◦
cathode −E◦
anode = +0.80 V −(+0.34 V) = +0.46 V
Step 2: Calculate ∆G◦using the relationship between ∆G◦and E◦
cell:
∆G◦=−nF ×E◦
cell
Given n= 2 (since 2 electrons are involved in the cell reaction) and Faraday
constant F= 96485 C/mol:
∆G◦=−2×96485 ×0.46 J/mol = −88860 J/mol = −88.86 kJ/mol
Step 3: Calculate ∆S◦using the relationship between ∆G◦and ∆S◦:
∆G◦= ∆H◦−T∆S◦
Given T= 298 K, we can rearrange the equation to solve for ∆S◦:
∆S◦=∆H◦−∆G◦
T
Substitute the values of ∆G◦and T:
∆S◦=∆H◦−(−88860 J/mol)
298 K
Step 4: Calculate ∆H◦using the relationship between ∆H◦and ∆G◦:
∆H◦= ∆G◦+T∆S◦
Substitute the values of ∆G◦, ∆S◦, and Tto solve for ∆H◦.
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Question 17
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 0.82 V. If the
cell reaction is:
Cd(s) + 2Ag+(aq)→Cd2+(aq) + 2Ag(s)
Calculate the standard reduction potentials for the half-reactions involved
in this cell reaction.
Solution
Step 1: Write the half-reactions for the cell reaction. The half-reactions for the
cell reaction are: Cathode: 2Ag+(aq)+2e−→2Ag(s) Anode: Cd2+(aq)+2e−→
Cd(s)
Step 2: Use the standard cell potential to find the standard reduction po-
tential for the cathode half-reaction. Given: E◦
cell = 0.82 V The standard
cell potential is equal to the difference in standard reduction potentials for
the cathode and anode half-reactions: E◦
cell =E◦
cathode −E◦
anode 0.82 V =
E◦
cathode −E◦
Cd2+(aq)+2e−→Cd(s)
Step 3: Calculate the standard reduction potential for the cathode half-
reaction. Solving for E◦
cathode:E◦
cathode = 0.82 V + E◦
Cd2+(aq)+2e−→Cd(s)
Step 4: Substitute standard reduction potentials to find E◦
cathode. The stan-
dard reduction potential for the reduction of Cd2+(aq) to Cd(s) is −0.40 V.
E◦
cathode = 0.82 V + (−0.40 V) = 0.42 V
Step 5: Calculate the standard reduction potential for the anode half-reaction.
Using the standard cell potential equation: E◦
anode =E◦
cathode −E◦
cell E◦
anode =
0.42 V - 0.82 V = −0.40 V
Therefore, the standard reduction potential for the cathode half-reaction is
0.42 V, while the standard reduction potential for the anode half-reaction is
−0.40 V.
Question 18
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−E◦=−0.76 V
Cathode: Cl2(g) + 2 e−−−→ 2 Cl−(aq) E◦= 1.36 V
Determine the cell potential (E◦
cell) and if the reaction is spontaneous or not.
16
Solution
Step 1: Write the overall cell reaction by combining the half-reactions.
The cell reaction is the sum of the half-reactions at the anode and cathode.
Cell reaction: Zn(s) + Cl2(g) −−→ Zn2+(aq) + 2 Cl−(aq)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials.
The standard cell potential is given by the formula
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = (1.36 V) −(−0.76 V)
E◦
cell = 2.12 V
Step 3: Determine if the reaction is spontaneous.
For a reaction to be spontaneous, E◦
cell must be positive. Since E◦
cell =
2.12 V >0, the reaction is spontaneous.
Question 19
Question
Consider a galvanic cell that consists of a copper electrode in a 1.0 M Cu2+
solution and a zinc electrode in a 1.0 M Zn2+ solution. The standard reduction
potentials for the half-reactions are as follows:
Cu2+(aq)+2e−→Cu(s) E◦= 0.34 V
Zn2+(aq)+2e−→Zn(s) E◦=−0.76 V
Determine the cell potential for this galvanic cell at 25
°
C.
Solution
Step 1: Determine the overall cell reaction. The overall cell reaction can be
determined by adding the half-reactions together:
Cu2+(aq) + Zn(s) →Cu(s) + Zn2+(aq)
Step 2: Determine the standard cell potential. The standard cell poten-
tial can be calculated by taking the difference between the standard reduction
potentials of the cathode and anode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential for this galvanic cell at 25
°
C is 1.10 V.
17
Question 20
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.85 V. If the
concentration of Fe3+ ions in the cathode compartment is 0.01 M and the con-
centration of Fe2+ ions in the anode compartment is 0.1 M, determine the cell
potential for the reaction:
Fe3+(aq) + Fe(s)→Fe2+(aq) + Fe3+(aq)
Solution
Step 1: Write the half-reactions for the galvanic cell: Cathode (Reduction):
Fe3+(aq) + e−→Fe2+(aq)
Anode (Oxidation): Fe(s)→Fe3+(aq)+3e−
Step 2: Calculate the standard cell potential based on the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
cathode −E◦
anode = (0.85 V) −(3 ×0.04 V) = 0.73 V
Step 3: Use the Nernst equation to determine the cell potential under non-
standard conditions: E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 4: Calculate the cell potential under non-standard conditions: E=
0.73 V −0.0592
1log 0.1
0.01= 0.83 V
Therefore, the cell potential for the given reaction is 0.83 V.
Question 21
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
as the anode and a copper electrode immersed in a 1.0 M CuSO4solution as the
cathode. The standard reduction potential for copper is E◦= 0.34 V. Calculate
the cell potential at 25
°
C assuming ideal behavior.
Solution
Step 1: The half-reaction occurring at the anode is the oxidation of hydrogen
gas:
Anode: 2H2(g)→4H+(aq) + 4e−
Step 2: The cell potential E◦
cell can be calculated using the Nernst equation:
Ecell =Ecathode −Eanode =E◦
cathode −E◦
anode −RT
nF ln [Cu2+]
P2
H2
18
Step 3: Calculate the cathode potential using the standard reduction poten-
tial for copper:
E◦
cathode =E◦
Cu = 0.34 V
Step 4: The number of electrons transferred in the cell reaction is 2 (from
the overall balanced reaction). Step 5: Substitute the known values into the
Nernst equation:
Ecell = 0.34 V −0 V −(8.314 J/K ·mol)(298 K)
2(96485 C/mol) ln 1.0
(1 atm)2
Step 6: Calculate the cell potential at 25
°
C:
Ecell = 0.34 V −0.0592 V ln102
Ecell = 0.34 V −0.0592 V ×2
Ecell = 0.34 V −0.1184 V
Ecell = 0.2216 V
Question 22
Question
Consider a galvanic cell that consists of a silver electrode immersed in a 1.0 M
solution of Ag+ions and a platinum electrode immersed in a 1.0 M solution of
Pt2+ ions. The standard reduction potentials for the half-reactions are:
Ag++ e−→Ag E◦
red = 0.80 V
Pt2+ + 2e−→Pt E◦
red = 1.20 V
Calculate the standard cell potential for this galvanic cell.
Solution
Step 1: Write the overall cell reaction using the reduction half-reactions.
Ag++ e−→Ag E◦
red = 0.80 V
Pt2+ + 2e−→Pt E◦
red = 1.20 V
Overall reaction: Ag++ Pt2+ →Ag + Pt2+
The standard cell potential, E◦
cell, is given by:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the reduction potential of the cathode and E◦
anode is the re-
duction potential of the anode.
Step 2: Identify the cathode and anode.
Cathode: Pt2+ + 2e−→Pt E◦
red = 1.20 V
Anode: Ag++ e−→Ag E◦
red = 0.80 V
19
Step 3: Substitute the values into the formula to calculate the standard cell
potential.
E◦
cell = 1.20 V −0.80 V = 0.40 V
Therefore, the standard cell potential for this galvanic cell is 0.40 V.
Question 23
Question
Suppose you have a galvanic cell constructed with a copper electrode immersed
in a 1.0 M CuSO4solution and a silver electrode immersed in a 1.0 M AgNO3
solution. Calculate the standard cell potential for this galvanic cell at 25◦C.
Given: E◦
cell(Cu|Cu2+) = 0.34 V and E◦
cell(Ag|Ag+) = 0.80 V.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction can be written
as:
Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)
Step 2: Identify the half-reactions. The half-reactions involved in this cell
are: Oxidation half-reaction: Cu(s) →Cu2+(aq)+2e−
Reduction half-reaction: 2Ag+(aq)+2e−→2Ag(s)
Step 3: Determine E◦
cell for the galvanic cell. The standard cell potential,
E◦
cell, can be calculated using the formula:
E◦
cell =E◦
red,cathode −E◦
red,anode
where E◦
red,cathode and E◦
red,anode are the standard reduction potentials for the
cathode and anode, respectively.
Given that E◦
cell(Cu|Cu2+) = 0.34 V and E◦
cell(Ag|Ag+) = 0.80 V, we can
substitute these values into the formula:
E◦
cell = 0.80 V −0.34 V = 0.46 V
Therefore, the standard cell potential for this galvanic cell at 25◦C is 0.46
V.
Question 24
Question
Consider a Galvanic Cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦
red =−0.76 V
20
Cu2+(aq) + 2e−→Cu(s)E◦
red = +0.34 V
Calculate the cell potential when the concentration of Zn2+ is 0.10 M and the
concentration of Cu2+ is 1.00 M. Is this cell Galvanic or Electrolytic?
Solution
Step 1: Calculate the cell potential using the Nernst Equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
where E◦
cell is the standard cell potential, nis the number of electrons trans-
ferred, and [Cu2+] and [Zn2+] are the concentrations of Cu2+ and Zn2+, respec-
tively.
Step 2: Determine the number of electrons transferred in the cell reaction:
Since the half-reactions involve the transfer of 2 electrons each, the total number
of electrons transferred in the cell reaction is 2.
Step 3: Substitute the given values into the Nernst Equation:
Ecell = (0.34 V −(−0.76 V)) −0.0592
2log 1.00
0.10
Ecell = 1.10 V −0.0296 log(10)
Ecell = 1.10 V −0.0296 ×1
Ecell = 1.10 V −0.0296
Ecell = 1.0704 V
Step 4: Determine if the cell is Galvanic or Electrolytic: Since the calculated
cell potential is positive (Ecell = 1.0704V), the cell is Galvanic.
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe2+(aq)→Fe3+(aq) + e−E◦= 0.77 V
a) Write the overall cell reaction for this galvanic cell.
b) Calculate the cell potential at standard conditions for this galvanic cell.
c) If the initial concentrations of Cu2+ and Fe2+ are both 0.1 M, which metal
will be plated out first as the cell operates?
21
Solution
a) The overall cell reaction for this galvanic cell can be determined by adding
the two half-reactions after multiplying the first reaction by 2:
Cu2+(aq) + Fe(s)→Cu(s) + Fe3+(aq)
b) The cell potential at standard conditions (E◦
cell) is given by the difference
in standard reduction potentials of the two half reactions:
E◦
cell =E◦
reduction, cathode −E◦
reduction, anode
E◦
cell =E◦
Fe3+/Fe −E◦
Cu2+/Cu
E◦
cell = 0.77 V −0.34 V = 0.43 V
c) To determine which metal will be plated out first, we compare the stan-
dard reduction potentials of the two half-reactions. The larger the reduction
potential, the more likely the reduction reaction will occur. Since Fe3+/Fe2+
has a higher reduction potential than Cu2+/Cu, iron will be plated out first as
the cell operates.
Question 26
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential for the Zn2+|Zn half-reaction is −0.76 V
and for the Cu2+|Cu half-reaction is 0.34 V, determine the cell potential (E◦
cell)
for this galvanic cell.
Solution
Step 1: Write the overall cell reaction by adding the half-reactions together.
Make sure to balance the number of electrons.
The overall cell reaction is:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions.
The standard cell potential is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
22
Given: E◦
cathode = 0.34 V E◦
anode =−0.76 V
Substitute the values into the formula:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the cell potential for this galvanic cell is 1.10 V .
Question 27
Question
Consider a galvanic cell that consists of a standard hydrogen electrode as the
anode and a copper electrode as the cathode. The standard reduction potential
for the copper electrode is E◦
Cu2+/Cu = 0.34 V. Calculate the cell potential at
25
°
C when the concentration of Cu2+ is 0.10 M. What will happen to the cell
potential if the concentration of Cu2+ is increased to 1.0 M?
Solution
Step 1: Write the balanced redox reaction for the cell:
2H++ 2e−→H2
Cu2+ + 2e−→Cu
The overall cell reaction is:
2H++ Cu2+ →H2+ Cu
Step 2: Calculate the standard cell potential (E◦
cell) using the given standard
reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
H+/H2
E◦
cell = 0.34 V −0.00 V
E◦
cell = 0.34 V
Step 3: Calculate the cell potential (Ecell) at 25
°
C using the Nernst equa-
tion:
Ecell =E◦
cell −RT
nF ln Q
Since the reaction quotient (Q) for the cell is equal to the concentration of
products over reactants, when Cu2+ is 0.10 M, Q=[H2][Cu]
[H+][Cu2+]= 1
Plugging in the values:
Ecell = 0.34 V −(8.314J/(mol ·K)·298K)
2·96485C/mol ln 1
23
Ecell = 0.34 V
Step 4: Now, when the concentration of Cu2+ is increased to 1.0 M, Q=
[H2][Cu]
[H+][Cu2+]>1
This will shift the equilibrium to the left, increasing the concentration of
reactants and decreasing the concentration of products, resulting in a decrease
in cell potential.
Therefore, increasing the concentration of Cu2+ will decrease the cell poten-
tial of the galvanic cell.
Question 28
Question
Consider a galvanic cell with the following half-reactions:
Anode: Pb2+(aq) + 2e−→Pb(s)
Cathode: 2Ag+(aq) + 2e−→2Ag(s)
If the standard reduction potentials are E◦(Pb2+/Pb) = −0.13 V and
E◦(Ag+/Ag) = 0.80 V, calculate the standard cell potential, E◦
cell, for this
galvanic cell and determine if the cell reaction is spontaneous.
Solution
Step 1: Write the overall cell reaction by combining the half-reactions of the
anode and cathode:
Pb2+(aq) + 2Ag+(aq)→Pb(s) + 2Ag(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials:
E◦
cell =E◦(cathode) −E◦(anode)
E◦
cell = 0.80 V −(−0.13 V)
E◦
cell = 0.93 V
Step 3: Determine if the cell reaction is spontaneous by checking if E◦
cell is
positive (>0). Since E◦
cell = 0.93 V, the cell reaction is spontaneous because
the standard cell potential is positive.
24
Question 29
Question
Consider a galvanic cell with the following cell notation: Zn|Zn2+||Cu2+|Cu.
If the standard reduction potential for Cu2+ + 2e−→Cu is +0.34 V and the
standard reduction potential for Zn2+ + 2e−→Zn is −0.76 V:
1. Calculate the cell potential under standard conditions.
2. Determine the reaction that occurs at the cathode and at the anode.
3. Is the cell acting as a galvanic cell or an electrolytic cell?
Solution
1. To calculate the cell potential under standard conditions, we use the formula
E◦
cell =E◦
cathode −E◦
anode, where E◦
cathode and E◦
anode are the standard reduction
potentials of the cathode and anode, respectively.
E◦
cell =E◦
cathode −E◦
anode = (+0.34 V) −(−0.76 V) = 1.10 V
Therefore, the cell potential under standard conditions is 1.10 V.
2. The reactions that occur at the cathode and anode can be determined
from the cell notation:
Cathode: Cu2+ + 2e−→Cu
Anode: Zn →Zn2+ + 2e−
3. A galvanic cell generates electrical energy from spontaneous redox reac-
tions. Since the cell potential is positive (1.10 V), the cell is acting as a galvanic
cell.
Question 30
Question
An electrochemical cell consists of a copper electrode immersed in a 1.0 M
Cu(NO3)2solution and a zinc electrode immersed in a 1.0 M Zn(NO3)2solution.
The standard reduction potentials are E◦
cell = 1.10 V for the following reaction:
Cu2+(aq)+2e−→Cu(s)
Zn2+(aq)+2e−→Zn(s)
What is the standard cell potential for the galvanic cell at 25◦C formed by
connecting these two half-cells?
25
Solution
Step 1: Identify the half-reactions and their standard reduction potentials. The
given half-reactions are:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Step 2: Write the overall cell reaction. The overall cell reaction can be
obtained by adding the two half-reactions, and the standard cell potential is the
sum of the standard reduction potentials of the two half-cells.
Cu2+(aq) + Zn(s)→Cu(s) + Zn2+(aq)
Step 3: Calculate the standard cell potential. Given E◦
cell = 1.10 V, and
E◦
cell =E◦
cathode −E◦
anode.
Therefore, E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Thus, the standard cell potential for the galvanic cell is 1.10 V.
Question 31
Question
Consider a galvanic cell with the standard reduction potentials of E◦
cathode =
0.50 V and E◦
anode =−0.70 V. Calculate the cell potential at 25◦C. Next,
describe the differences between a galvanic cell and an electrolytic cell.
Solution
Step 1: Calculate the cell potential of the galvanic cell at 25◦C using the Nernst
equation:
∆G=−nF E
E=E◦−RT ln Q
nF
Given E◦
cathode = 0.50 V, E◦
anode =−0.70 V, and the number of electrons
transferred (n) is 1, we have:
E= (0.50 V) −(8.314 J/mol·K)(298 K) ln(1)
1(96485 C/mol)
E= 0.50 −(2470.92) ln(1)
96485
E= 0.50 V
Step 2: Describe the differences between a galvanic cell and an electrolytic
cell: - Galvanic Cell: 1. Spontaneous redox reaction occurs. 2. Electrons flow
26
from anode to cathode through the external circuit. 3. Energy is released from
the redox reaction and converted into electrical energy. 4. Anode is negative
and cathode is positive. 5. Salt bridge is used to maintain electrical neutrality.
- Electrolytic Cell: 1. Non-spontaneous redox reaction occurs. 2. External
electrical source is needed to drive the reaction. 3. Electrons are forced to flow
from cathode to anode through the external circuit. 4. Anode is positive and
cathode is negative. 5. Electrons flow in the opposite direction compared to a
galvanic cell. 6. Used for processes like electroplating.
Question 32
Question
Consider a galvanic cell with the following half-reactions:
Anode: Ni(s) →Ni2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Given that the standard reduction potentials are Cu2+/Cu : E◦= 0.34 V and
Ni2+/Ni : E◦=−0.25 V, calculate the cell potential at 25◦C.
Solution
Step 1: Write the cell reaction by adding the half-reactions together:
Ni(s) + Cu2+(aq)→Ni2+(aq) + Cu(s)
Step 2: Write the cell potential equation using the standard reduction po-
tentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.25) V
E◦
cell = 0.59 V
Step 3: Calculate the effect of temperature on the cell potential using the
Nernst equation:
E=E◦−0.0592 V
n·log [Ni2+][Cu]
[Cu2+][Ni]
Step 4: Since we are under standard conditions, Q= 1 and E=E◦
cell.
0.59 = E◦−0.0592 V
2·log(1)
Step 5: Thus, the cell potential at 25◦C is 0.59 V .
27
Question 33
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.80 V that is
initially at equilibrium. If the concentration of the oxidizing agent is doubled
while the concentration of the reducing agent remains the same, calculate the
cell potential at this new equilibrium.
Solution
Step 1: Write the half-reactions for the cell.
Anode: Oxidation (reducing agent) →Oxidation product + e−
Cathode: e−+ Reduction agent →Reduction product
Step 2: Write the cell reaction using the two half-reactions.
Overall Reaction: Oxidizing agent + Reducing agent →Oxidation product+Reduction product
Step 3: Calculate the cell potential using the Nernst equation:
E=E◦−RT
nF ln Q
K
where Eis the cell potential, E◦is the standard cell potential, Ris the gas
constant, Tis the temperature in Kelvin, nis the number of moles of electrons
transferred in the balanced equation, Fis Faraday’s constant, Qis the reaction
quotient, and Kis the equilibrium constant.
Step 4: Calculate the new cell potential after the concentration change.
Given that the concentrations of the oxidizing agent was doubled, Qwill increase
by a factor of 2. Therefore, Q= 2 and the new cell potential (E′) can be
calculated as:
E′= 0.80 −RT
nF ln 2
1
E′= 0.80 −RT
nF ln(2)
E′= 0.80 −0.0257 V
n×ln(2)
Step 5: Determine nfrom the balanced chemical equation. The number of
moles of electrons transferred in the balanced equation is equal to the coefficient
of e−in the balanced equation for the overall cell reaction. Determine this value
and substitute into the equation above to find the final cell potential at the new
equilibrium.
28
Question 34
Question
Consider a galvanic cell that consists of a copper electrode immersed in a 1.0 M
Cu2+ solution and a platinum electrode immersed in a 1.0 M Cl−solution. The
cell diagram is as follows: Cu(s) — Cu2+ (1.0 M) —— Cl−(1.0 M) — Pt(s).
Determine the standard cell potential for this galvanic cell.
Solution
Step 1: Write down the half-reactions and their standard reduction potentials.
Cathode (Reduction): Cu2+(aq)+2e−→Cu(s)E◦
cathode = 0.34 V
Anode (Oxidation): 2Cl−(aq)→Cl2(g)+2e−E◦
anode = 1.36 V
Step 2: Calculate the standard cell potential using the equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −1.36 V
E◦
cell =−1.02 V
Therefore, the standard cell potential for this galvanic cell is -1.02 V.
Question 35
Question
Consider a galvanic cell with a standard cell potential of 1.10 V. Suppose the cell
is operating at a temperature of 298 K and the concentration of Zn2+ in the cell
is 0.050 M. If the cell voltage drops to 0.90 V, calculate the new concentration
of Zn2+ in the cell assuming all other conditions remain constant.
Solution
Step 1: Determine the initial reaction occurring in the cell. The initial reac-
tion in the cell involves the oxidation of Zn(s) to Zn2+(aq) and reduction of
MnO−
4(aq) to Mn2+(aq). This can be represented as:
Zn(s) + MnO−
4(aq)+H+(aq)→Zn2+(aq) + Mn2+(aq)+H2O(l)
Step 2: Determine the initial cell potential (E◦
cell) using the Nernst equation.
The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Mn2+]
29
Where nis the number of electrons transferred in the balanced equation. In
this case, n= 1.
Substitute the given concentrations into the equation:
Ecell = 1.95 −0.0592
1log 1.0
0.10
Ecell = 1.95 −0.0592 log 10
Ecell = 1.95 −0.0592(1)
Ecell = 1.95 −0.0592
Ecell = 1.89 V
Therefore, the standard cell potential for the cell at the given concentrations
is 1.89 V.
Question 2
Question
Explain the key differences between Galvanic and Electrolytic Cells. Provide
examples of each type of cell.
Solution
Galvanic cells and electrolytic cells are two types of electrochemical cells that
involve redox reactions. They have distinct differences in their operation and
purpose.
Galvanic Cells:
Galvanic cells convert chemical energy to electrical energy spontaneously.
The redox reaction in a galvanic cell is spontaneous and generates an
electric current.
Electrons flow from the anode (where oxidation occurs) to the cathode
(where reduction occurs).
Examples of galvanic cells include the Daniell cell (Zn-Cu cell) and the
Voltaic pile.
Electrolytic Cells:
Electrolytic cells require an external electrical energy source to drive a
non-spontaneous redox reaction.
The redox reaction in an electrolytic cell is non-spontaneous and consumes
electric current.
Electrons flow from the external power source to the anode, where oxida-
tion occurs, and to the cathode, where reduction occurs.
Examples of electrolytic cells include the electrolysis of water to produce
hydrogen and oxygen gas, and the electroplating of metals.
2
In summary, galvanic cells produce electrical energy from chemical reactions
that occur spontaneously, while electrolytic cells require external energy input
to drive non-spontaneous reactions.
Question 3
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 0.70 V. If
the concentration of Fe2+ ions in the cell is 0.10 M and the concentration of
Cu2+ ions is 1.0 M, determine the cell potential at 25◦C. Assume that the cell
reaction is:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
Solution
Step 1: Write the cell reaction for the galvanic cell and identify the half-
reactions. The cell reaction is:
Fe2+(aq) + Cu2+(aq)→Fe3+(aq) + Cu(s)
The half-reactions are: Oxidation half-reaction:
Fe2+(aq)→Fe3+(aq) + e−
Reduction half-reaction:
Cu2+(aq)+2e−→Cu(s)
Step 2: Write the expressions for the equilibrium constant Kand the cell
potential Ecell. The equilibrium constant Kis given by:
K=[Fe3+][Cu]
[Fe2+][Cu2+]
The cell potential Ecell can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(K)
Step 3: Calculate the equilibrium constant K. Substitute the given concen-
trations into the expression for K:
K=(x)(1)
(0.10)(1.0)
Step 4: Calculate the cell potential Ecell. Since the reaction involves the
transfer of one electron, n= 1. Substitute Kinto the Nernst equation:
Ecell = 0.70 −0.0592
1log x
0.10
3
Step 5: Solve for Ecell. To find Ecell, we need to solve for xin the expression
for K. Solving for xgives us:
x= 0.10
Now substitute xback into the Nernst equation:
Ecell = 0.70 −0.0592
1log 0.10
0.10
Ecell = 0.70 −0.0592
1log(1)
Ecell = 0.70 V
Therefore, the cell potential at 25◦C is 0.70 V.
Question 4
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦
cell =−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦
cell = 0.34 V
Calculate the standard cell potential when the cell is reacting under non-standard
conditions with Zn2+= 0.10 M, Cu2+= 1.0 M at 25◦C.
Solution
Step 1: Write the Nernst equation for the cell reaction:
Ecell =E◦
cell −0.0592
nlog Q
K
Where: - Ecell is the cell potential under non-standard conditions, - E◦
cell is the
standard cell potential, - nis the number of moles of electrons transferred in the
balanced equation, - Qis the reaction quotient, - Kis the equilibrium constant.
Step 2: Calculate n, the number of moles of electrons transferred in the
balanced equation. Since both half-reactions involve the transfer of 2 moles of
electrons, n= 2.
Step 3: Calculate the reaction quotient Qusing the concentrations provided:
Q=[Zn2+]
[Cu2+]=0.10
1.0= 0.10
Step 4: Determine the equilibrium constant Kusing the standard cell po-
tential:
E◦
cell = 0.34 −(−0.76) = 1.10 V
4
K= exp nE◦
cell
0.0592= exp 2×1.10
0.0592 ≈1536.30
Step 5: Substitute the values of E◦
cell,n,Q, and Kinto the Nernst equation:
Ecell = 1.10 −0.0592
2log 0.10
1536.30≈1.044 V
Therefore, the standard cell potential when the cell is reacting under non-
standard conditions is approximately 1.044 V.
Question 5
Question
A galvanic cell is constructed with a standard hydrogen electrode (E◦
cell = 0.00
V) and a copper electrode (E◦
Cu = 0.34 V). If the concentration of Cu2+ ions
is 1.0 M, what is the half-cell potential (E◦
Cu2+/Cu) of the copper electrode at
25
°
C?
Solution
Step 1: Write the cell reaction for the galvanic cell.
Cell reaction: H+(aq)+e−→1
2H2(g)E◦= 0.00 V
Cu2+(aq) + 2e−→Cu(s) E◦= 0.34 V
Step 2: Determine the overall cell reaction and the cell potential. The overall
cell reaction is given by adding the two half reactions:
Overall cell reaction: 2H+(aq) + Cu2+(aq)→1
2H2(g) + Cu(s)
The overall standard cell potential can be calculated using the standard
reduction potentials:
E◦
cell =E◦
Cu2+/Cu −E◦
H+/H2= 0.34 V −0.00 V = 0.34 V
Step 3: Use the Nernst equation to calculate the actual cell potential. The
Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog Q
where nis the number of moles of electrons transferred in the balanced cell
reaction, and Qis the reaction quotient.
5
Step 4: Calculate the actual cell potential for the given concentration of
Cu2+ ions. Since the cell reaction involves the transfer of 2 moles of electrons,
n= 2. The reaction quotient can be calculated as:
Q=[Cu]
[H+]2=1.0
(1.0)2= 1.0
Therefore, the actual cell potential (Ecell) can be calculated as:
Ecell = 0.34 V −0.0592
2log(1.0) = 0.34 V
Therefore, the half-cell potential (E◦
Cu2+/Cu) of the copper electrode at 25
°
C
is 0.34 V.
Question 6
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
Cu2+(aq) + 2e−→Cu(s)E◦= 0.34 V
If the concentration of Zn2+ in the cathode compartment is 0.10 M, while
the concentration of Cu2+ in the anode compartment is 1.0 M. Calculate the
cell potential at standard conditions and determine the direction of electron
flow.
Solution
Step 1: Write the overall reaction and calculate the standard cell potential. The
overall cell reaction is the sum of the two half-reactions:
Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)
The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Step 2: Determine the direction of electron flow. Since E◦
cell is positive, the
reaction is spontaneous as written. This means electrons flow from the anode
(where oxidation occurs) to the cathode (where reduction occurs).
Therefore, in this galvanic cell, electrons flow from Zn (anode) to Cu (cath-
ode).
6
Question 7
Question
Consider a galvanic cell composed of a standard hydrogen electrode (SHE) and
a silver electrode. The reduction half-reaction for the silver electrode is the
following:
Ag++e−→Ag(s)
Given that the standard reduction potential for this half-reaction is E0=
0.80 V, calculate the cell potential when the concentration of Ag+ions is 0.1 M
and the pressure of hydrogen gas at the SHE is 1 atm. Is the cell spontaneous
under these conditions?
Solution
Step 1: Write out the full cell reaction: The overall cell reaction can be obtained
by combining the given reduction half-reaction with the standard hydrogen elec-
trode (SHE) reaction, which is the reduction of hydrogen ion:
2H++ 2e−→H2(g)
The full cell reaction is:
2Ag++ 2H+→2Ag + H2
Step 2: Calculate the standard cell potential, E0
cell: Using the standard
reduction potentials provided:
E0
cell =E0
cathode −E0
anode =E0
Ag −E0
H= 0.80V−0V= 0.80V
Step 3: Calculate the cell potential under the given conditions: The Nernst
equation for cell potential is:
Ecell =E0
cell −0.0592
nlog Q
K
where Qis the reaction quotient, Kis the equilibrium constant, and nis the
number of moles of electrons transferred.
Step 4: Calculate the reaction quotient, Q: For the given cell reaction:
Q=[Ag]2
[H+]2
Given [Ag+]=0.1 M and [H+] = 1Min this case:
Q=(0.1)2
12= 0.01
7
Step 5: Calculate the cell potential: Substitute the values of E0
cell,Q, and
n= 2 into the Nernst equation:
Ecell = 0.80 −0.0592
2log(0.01)
Ecell = 0.80 + 0.0296 ×2 log(0.01)
Ecell = 0.80 + 0.0296 ×2×(−2) = 0.80 −0.1184 = 0.6816 V
Step 6: Determine if the cell is spontaneous: Since the calculated cell po-
tential Ecell = 0.6816 V is positive, the cell reaction is spontaneous under the
given conditions.
Question 8
Question
Consider a galvanic cell with a standard cell potential of 1.13 V that involves
the following half-reactions:
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
MnO−
4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦= 1.51 V
Calculate the standard cell potential for the reaction when the following
concentrations are used: Zn2+= 0.10 M, MnO−
4= 0.20 M, H+= 1.0 M.
Solution
Step 1: Write the overall cell reaction using the given half-reactions and their
standard reduction potentials.
Zn(s) + MnO−
4(aq)+8H+(aq)→Zn2+(aq) + Mn2+(aq)+4H2O(l)
Step 2: Determine the standard cell potential (E◦
cell) for this reaction by
adding the standard reduction potentials of the half-reactions.
E◦
cell =E◦
cathode −E◦
anode
= (1.51 V) −(−0.76 V)
= 2.27 V
Step 3: Apply the Nernst equation to calculate the cell potential under
non-standard conditions. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
K
Where: - Eis the cell potential under non-standard conditions - E◦is the
standard cell potential - nis the number of moles of electrons transferred in the
8
cell reaction - Qis the reaction quotient - Kis the equilibrium constant for the
cell reaction
Step 4: Calculate the reaction quotient Qusing the concentrations provided.
Q=Zn2+Mn2+
MnO−
4H+8
Q=(0.10)(1.0)
(0.20)(1.0)8= 0.10
Step 5: Substitute the values into the Nernst equation to find the cell po-
tential under the given concentrations.
E= 2.27 V −0.0592
5log(0.10) ≈2.249 V
Therefore, the standard cell potential for the reaction under the given con-
centrations is approximately 2.249 V.
Question 9
Question
Consider a galvanic cell that consists of a zinc electrode in a 1.0 M Zn(NO3)2
solution and a copper electrode in a 1.0 M Cu(NO3)2 solution. The standard
reduction potentials are E◦
Zn2+/Zn =−0.76 V and E◦
Cu2+/Cu = 0.34 V. Calculate
the cell potential at 25
°
C for this galvanic cell.
Solution
Step 1: The cell potential (E◦
cell) for a galvanic cell can be calculated using the
formula:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the standard reduction potential of the cathode and E◦
anode is
the standard reduction potential of the anode.
Step 2: In this case, the zinc electrode is the anode and the copper electrode
is the cathode. Therefore, we have:
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential at 25
°
C for this galvanic cell is 1.10 V.
9
Question 10
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
E◦
Zn =−0.76 V
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
E◦
Cu = 0.34 V
a) Calculate the cell potential for this reaction at standard conditions.
b) Determine the equilibrium constant, K, for this cell reaction.
Solution
a) The cell potential for a galvanic cell can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Step 1: Identify the given standard electrode potentials for the cathode and
anode:
E◦
Zn =−0.76 V
E◦
Cu = 0.34 V
Step 2: Substitute the values into the formula to find the cell potential:
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential for this reaction at standard conditions is 1.10
V.
b) The equilibrium constant, K, for this cell reaction can be determined
using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(K)
where n is the number of moles of electrons transferred in the balanced
reaction.
Step 1: Determine the number of moles of electrons transferred in the
balanced reaction: The balanced reaction is:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
From this, we see that n = 2 (two moles of electrons are transferred).
Step 2: Find K using the Nernst equation: Substitute the known values
into the equation:
1.10 V = 1.10 V −0.0592
2log(K)
10
0 = −0.0592
2log(K)
log(K) = 0
K= 1
Therefore, the equilibrium constant, K, for this cell reaction is 1.
Question 11
Question
Consider the following Galvanic Cell reaction involving the reduction of copper
ions:
Cu2+(aq)+2e−→Cu(s)
If a Galvanic Cell is constructed with a standard cell potential of E◦
cell = 0.34
V at 25◦C, determine the standard reduction potential, E◦, for the reduction
of copper ions.
Solution
Step 1: Write the Nernst equation for the Galvanic Cell:
Ecell =E◦
cell −0.0592
2log [Cu(s)]
[Cu2+(aq)]2
Step 2: Substitute the given values into the Nernst equation:
0.34 = E◦
cell −0.0592
2log 1
[Cu2+]2
Step 3: Simplify the equation:
0.34 = E◦
cell + 0.0296 log [Cu2+]2
0.34 = E◦
cell + 0.0592 log [Cu2+]
Step 4: Rearrange the equation to solve for the standard reduction potential,
E◦:
0.0592 log [Cu2+]= 0.34 −E◦
cell
log [Cu2+]=0.34 −E◦
cell
0.0592
Step 5: Solve for [Cu2+]:
[Cu2+] = 10
0.34−E◦
cell
0.0592
Therefore, the standard reduction potential, E◦, for the reduction of copper
ions is 100.34−E◦
cell
0.0592 .
11
Question 12
Question
Consider a galvanic cell with a standard cell potential of 1.23 V at 298 K. The
half-reactions involved are:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the concentrations are [Zn2+] = 1.0 M and [Cu2+] = 0.1 M, calculate the
cell potential at 298 K.
Solution
Step 1: Write the overall balanced cell reaction:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Write the cell reaction with half-cell potentials and concentrations:
E◦
cell =E◦
cathode −E◦
anode
Step 3: Calculate the cell potential by plugging in the given values:
E◦
cell = 0.34 V
Question 13
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
and a silver-silver chloride electrode (Ag/AgCl) in half-cells. The standard
reduction potential for the Ag/AgCl electrode is Eo= 0.222 V. If the measured
cell potential is Ecell = 0.48 V, calculate the standard reduction potential of the
hydrogen electrode.
Solution
To find the standard reduction potential of the hydrogen electrode, we can use
the Nernst equation. The Nernst equation relates the measured cell potential
to the standard reduction potentials of the half-reactions involved:
Ecell =Eo
cathode −Eo
anode
where Ecell is the measured cell potential, Eo
cathode is the standard reduction
potential of the cathode (Ag/AgCl in this case), and Eo
anode is the standard
reduction potential of the anode (hydrogen electrode).
12
Given: Ecell = 0.48 V Eo
cathode = 0.222 V
Substitute these values into the Nernst equation to solve for Eo
anode:
0.48 V = 0.222 V −Eo
anode
Eo
anode = 0.222 V −0.48 V = −0.258 V
Therefore, the standard reduction potential of the hydrogen electrode is
−0.258 V.
Question 14
Question
Consider a galvanic cell with the following half-reactions and standard reduction
potentials:
Mn2+(aq)+2e−→Mn(s)E◦=−1.18 V
Ag+(aq) + e−→Ag(s)E◦= 0.80 V
(a) Write the overall cell reaction for this galvanic cell. Indicate the direction
of electron flow and identify the anode and cathode.
(b) Calculate the standard cell potential (E◦
cell) for the galvanic cell.
(c) If the initial concentrations of Mn2+ and Ag+are both 0.10 M, determine
the cell potential when the cell reaches equilibrium.
Solution
(a) The overall cell reaction for the galvanic cell can be obtained by adding the
two half-reactions:
Mn2+(aq) + 2Ag+(aq)→Mn(s) + 2Ag(s)
In this reaction, electrons flow from Mn2+ to Ag+, so the anode is the side where
oxidation occurs (Mn2+) and the cathode is where reduction occurs (Ag+).
(b) The standard cell potential can be calculated using the formula:
E◦
cell =E◦
cathode −E◦
anode
Where E◦
cathode = 0.80 V and E◦
anode =−1.18 V. Therefore,
E◦
cell = 0.80 V −(−1.18 V) = 1.98 V
(c) At equilibrium, the cell potential is determined by the Nernst equation:
Ecell =E◦
cell −0.0592 V
nlog [Ag+]2
[Mn2+]
13
Since the reaction involves the transfer of 2 moles of electrons, n= 2. At
equilibrium, [Mn2+] = 0.10 M, [Ag+] = 0.10 M. Substituting these values, we
get:
Ecell = 1.98 V −0.0592 V
2log 0.102
0.10
Ecell = 1.98 V −0.0296 V log(1) = 1.98 V
Question 15
Question
Consider a galvanic cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V
2Ag(s)→2Ag+(aq) + 2e−E◦= 0.80 V
If the concentration of Zn2+ in the compartment with the zinc electrode is
1.0 M and the concentration of Ag+in the compartment with the silver electrode
is 0.1 M, what is the cell potential at 25 degrees Celsius?
Solution
Step 1: Write the overall cell reaction. The overall cell reaction is obtained by
adding the two half-reactions together after multiplying the first half-reaction
by 2:
Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)
Step 2: Calculate the standard cell potential, E◦
cell. The standard cell po-
tential is the difference between the standard reduction potentials of the two
half-reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.80 V −(−0.76 V) = 1.56 V
Step 3: Calculate the cell potential, E, at 25 degrees Celsius using the Nernst
equation. The Nernst equation is given by:
E=E◦
cell −0.0592
nlog [Zn2+]
[Ag+]2
where n is the number of moles of electrons transferred. In this case, n = 2.
Substitute the values into the equation:
E= 1.56 V−0.0592
2log 1.0
0.12= 1.56 V−0.0592 log(100) = 1.56 V−0.0592×2=1.4416 V
Therefore, the cell potential at 25 degrees Celsius is 1.4416 V.
14
Question 16
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) con-
nected to a copper electrode in one compartment, and a standard hydrogen elec-
trode connected to a silver electrode in the other compartment. The cell operates
under standard conditions (T= 298 K, P= 1 atm). The standard reduction
potential for copper is E◦
Cu2+/Cu = 0.34 V and for silver is E◦
Ag+/Ag = 0.80 V.
Calculate E◦
cell, ∆G◦, ∆S◦, and ∆H◦for the cell reaction.
Solution
Step 1: Calculate E◦
cell using the standard cell potential formula:
E◦
cell =E◦
cathode −E◦
anode
Given E◦
Cu2+/Cu = 0.34 V and E◦
Ag+/Ag = 0.80 V, the overall cell reaction is:
Cu2+(aq) + 2e−→Cu(s)
2H+(aq) + 2e−→H2(g)
Ag+(aq)+e−→Ag(s)
The cell reaction is:
Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)
E◦
cell =E◦
cathode −E◦
anode = +0.80 V −(+0.34 V) = +0.46 V
Step 2: Calculate ∆G◦using the relationship between ∆G◦and E◦
cell:
∆G◦=−nF ×E◦
cell
Given n= 2 (since 2 electrons are involved in the cell reaction) and Faraday
constant F= 96485 C/mol:
∆G◦=−2×96485 ×0.46 J/mol = −88860 J/mol = −88.86 kJ/mol
Step 3: Calculate ∆S◦using the relationship between ∆G◦and ∆S◦:
∆G◦= ∆H◦−T∆S◦
Given T= 298 K, we can rearrange the equation to solve for ∆S◦:
∆S◦=∆H◦−∆G◦
T
Substitute the values of ∆G◦and T:
∆S◦=∆H◦−(−88860 J/mol)
298 K
Step 4: Calculate ∆H◦using the relationship between ∆H◦and ∆G◦:
∆H◦= ∆G◦+T∆S◦
Substitute the values of ∆G◦, ∆S◦, and Tto solve for ∆H◦.
15
Question 17
Question
Consider a galvanic cell with a standard cell potential of E◦
cell = 0.82 V. If the
cell reaction is:
Cd(s) + 2Ag+(aq)→Cd2+(aq) + 2Ag(s)
Calculate the standard reduction potentials for the half-reactions involved
in this cell reaction.
Solution
Step 1: Write the half-reactions for the cell reaction. The half-reactions for the
cell reaction are: Cathode: 2Ag+(aq)+2e−→2Ag(s) Anode: Cd2+(aq)+2e−→
Cd(s)
Step 2: Use the standard cell potential to find the standard reduction po-
tential for the cathode half-reaction. Given: E◦
cell = 0.82 V The standard
cell potential is equal to the difference in standard reduction potentials for
the cathode and anode half-reactions: E◦
cell =E◦
cathode −E◦
anode 0.82 V =
E◦
cathode −E◦
Cd2+(aq)+2e−→Cd(s)
Step 3: Calculate the standard reduction potential for the cathode half-
reaction. Solving for E◦
cathode:E◦
cathode = 0.82 V + E◦
Cd2+(aq)+2e−→Cd(s)
Step 4: Substitute standard reduction potentials to find E◦
cathode. The stan-
dard reduction potential for the reduction of Cd2+(aq) to Cd(s) is −0.40 V.
E◦
cathode = 0.82 V + (−0.40 V) = 0.42 V
Step 5: Calculate the standard reduction potential for the anode half-reaction.
Using the standard cell potential equation: E◦
anode =E◦
cathode −E◦
cell E◦
anode =
0.42 V - 0.82 V = −0.40 V
Therefore, the standard reduction potential for the cathode half-reaction is
0.42 V, while the standard reduction potential for the anode half-reaction is
−0.40 V.
Question 18
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−E◦=−0.76 V
Cathode: Cl2(g) + 2 e−−−→ 2 Cl−(aq) E◦= 1.36 V
Determine the cell potential (E◦
cell) and if the reaction is spontaneous or not.
16
Solution
Step 1: Write the overall cell reaction by combining the half-reactions.
The cell reaction is the sum of the half-reactions at the anode and cathode.
Cell reaction: Zn(s) + Cl2(g) −−→ Zn2+(aq) + 2 Cl−(aq)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials.
The standard cell potential is given by the formula
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = (1.36 V) −(−0.76 V)
E◦
cell = 2.12 V
Step 3: Determine if the reaction is spontaneous.
For a reaction to be spontaneous, E◦
cell must be positive. Since E◦
cell =
2.12 V >0, the reaction is spontaneous.
Question 19
Question
Consider a galvanic cell that consists of a copper electrode in a 1.0 M Cu2+
solution and a zinc electrode in a 1.0 M Zn2+ solution. The standard reduction
potentials for the half-reactions are as follows:
Cu2+(aq)+2e−→Cu(s) E◦= 0.34 V
Zn2+(aq)+2e−→Zn(s) E◦=−0.76 V
Determine the cell potential for this galvanic cell at 25
°
C.
Solution
Step 1: Determine the overall cell reaction. The overall cell reaction can be
determined by adding the half-reactions together:
Cu2+(aq) + Zn(s) →Cu(s) + Zn2+(aq)
Step 2: Determine the standard cell potential. The standard cell poten-
tial can be calculated by taking the difference between the standard reduction
potentials of the cathode and anode:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
Zn2+/Zn
E◦
cell = 0.34 V −(−0.76 V)
E◦
cell = 1.10 V
Therefore, the cell potential for this galvanic cell at 25
°
C is 1.10 V.
17
Question 20
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.85 V. If the
concentration of Fe3+ ions in the cathode compartment is 0.01 M and the con-
centration of Fe2+ ions in the anode compartment is 0.1 M, determine the cell
potential for the reaction:
Fe3+(aq) + Fe(s)→Fe2+(aq) + Fe3+(aq)
Solution
Step 1: Write the half-reactions for the galvanic cell: Cathode (Reduction):
Fe3+(aq) + e−→Fe2+(aq)
Anode (Oxidation): Fe(s)→Fe3+(aq)+3e−
Step 2: Calculate the standard cell potential based on the half-reactions:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
cathode −E◦
anode = (0.85 V) −(3 ×0.04 V) = 0.73 V
Step 3: Use the Nernst equation to determine the cell potential under non-
standard conditions: E=E◦−0.0592
nlog [Fe2+]
[Fe3+]
Step 4: Calculate the cell potential under non-standard conditions: E=
0.73 V −0.0592
1log 0.1
0.01= 0.83 V
Therefore, the cell potential for the given reaction is 0.83 V.
Question 21
Question
Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)
as the anode and a copper electrode immersed in a 1.0 M CuSO4solution as the
cathode. The standard reduction potential for copper is E◦= 0.34 V. Calculate
the cell potential at 25
°
C assuming ideal behavior.
Solution
Step 1: The half-reaction occurring at the anode is the oxidation of hydrogen
gas:
Anode: 2H2(g)→4H+(aq) + 4e−
Step 2: The cell potential E◦
cell can be calculated using the Nernst equation:
Ecell =Ecathode −Eanode =E◦
cathode −E◦
anode −RT
nF ln [Cu2+]
P2
H2
18
Step 3: Calculate the cathode potential using the standard reduction poten-
tial for copper:
E◦
cathode =E◦
Cu = 0.34 V
Step 4: The number of electrons transferred in the cell reaction is 2 (from
the overall balanced reaction). Step 5: Substitute the known values into the
Nernst equation:
Ecell = 0.34 V −0 V −(8.314 J/K ·mol)(298 K)
2(96485 C/mol) ln 1.0
(1 atm)2
Step 6: Calculate the cell potential at 25
°
C:
Ecell = 0.34 V −0.0592 V ln102
Ecell = 0.34 V −0.0592 V ×2
Ecell = 0.34 V −0.1184 V
Ecell = 0.2216 V
Question 22
Question
Consider a galvanic cell that consists of a silver electrode immersed in a 1.0 M
solution of Ag+ions and a platinum electrode immersed in a 1.0 M solution of
Pt2+ ions. The standard reduction potentials for the half-reactions are:
Ag++ e−→Ag E◦
red = 0.80 V
Pt2+ + 2e−→Pt E◦
red = 1.20 V
Calculate the standard cell potential for this galvanic cell.
Solution
Step 1: Write the overall cell reaction using the reduction half-reactions.
Ag++ e−→Ag E◦
red = 0.80 V
Pt2+ + 2e−→Pt E◦
red = 1.20 V
Overall reaction: Ag++ Pt2+ →Ag + Pt2+
The standard cell potential, E◦
cell, is given by:
E◦
cell =E◦
cathode −E◦
anode
where E◦
cathode is the reduction potential of the cathode and E◦
anode is the re-
duction potential of the anode.
Step 2: Identify the cathode and anode.
Cathode: Pt2+ + 2e−→Pt E◦
red = 1.20 V
Anode: Ag++ e−→Ag E◦
red = 0.80 V
19
Step 3: Substitute the values into the formula to calculate the standard cell
potential.
E◦
cell = 1.20 V −0.80 V = 0.40 V
Therefore, the standard cell potential for this galvanic cell is 0.40 V.
Question 23
Question
Suppose you have a galvanic cell constructed with a copper electrode immersed
in a 1.0 M CuSO4solution and a silver electrode immersed in a 1.0 M AgNO3
solution. Calculate the standard cell potential for this galvanic cell at 25◦C.
Given: E◦
cell(Cu|Cu2+) = 0.34 V and E◦
cell(Ag|Ag+) = 0.80 V.
Solution
Step 1: Write the overall cell reaction. The overall cell reaction can be written
as:
Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)
Step 2: Identify the half-reactions. The half-reactions involved in this cell
are: Oxidation half-reaction: Cu(s) →Cu2+(aq)+2e−
Reduction half-reaction: 2Ag+(aq)+2e−→2Ag(s)
Step 3: Determine E◦
cell for the galvanic cell. The standard cell potential,
E◦
cell, can be calculated using the formula:
E◦
cell =E◦
red,cathode −E◦
red,anode
where E◦
red,cathode and E◦
red,anode are the standard reduction potentials for the
cathode and anode, respectively.
Given that E◦
cell(Cu|Cu2+) = 0.34 V and E◦
cell(Ag|Ag+) = 0.80 V, we can
substitute these values into the formula:
E◦
cell = 0.80 V −0.34 V = 0.46 V
Therefore, the standard cell potential for this galvanic cell at 25◦C is 0.46
V.
Question 24
Question
Consider a Galvanic Cell with the following half-reactions:
Zn2+(aq) + 2e−→Zn(s)E◦
red =−0.76 V
20
Cu2+(aq) + 2e−→Cu(s)E◦
red = +0.34 V
Calculate the cell potential when the concentration of Zn2+ is 0.10 M and the
concentration of Cu2+ is 1.00 M. Is this cell Galvanic or Electrolytic?
Solution
Step 1: Calculate the cell potential using the Nernst Equation:
Ecell =E◦
cell −0.0592
nlog [Cu2+]
[Zn2+]
where E◦
cell is the standard cell potential, nis the number of electrons trans-
ferred, and [Cu2+] and [Zn2+] are the concentrations of Cu2+ and Zn2+, respec-
tively.
Step 2: Determine the number of electrons transferred in the cell reaction:
Since the half-reactions involve the transfer of 2 electrons each, the total number
of electrons transferred in the cell reaction is 2.
Step 3: Substitute the given values into the Nernst Equation:
Ecell = (0.34 V −(−0.76 V)) −0.0592
2log 1.00
0.10
Ecell = 1.10 V −0.0296 log(10)
Ecell = 1.10 V −0.0296 ×1
Ecell = 1.10 V −0.0296
Ecell = 1.0704 V
Step 4: Determine if the cell is Galvanic or Electrolytic: Since the calculated
cell potential is positive (Ecell = 1.0704V), the cell is Galvanic.
Question 25
Question
Consider a galvanic cell with the following half-reactions:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Fe2+(aq)→Fe3+(aq) + e−E◦= 0.77 V
a) Write the overall cell reaction for this galvanic cell.
b) Calculate the cell potential at standard conditions for this galvanic cell.
c) If the initial concentrations of Cu2+ and Fe2+ are both 0.1 M, which metal
will be plated out first as the cell operates?
21
Solution
a) The overall cell reaction for this galvanic cell can be determined by adding
the two half-reactions after multiplying the first reaction by 2:
Cu2+(aq) + Fe(s)→Cu(s) + Fe3+(aq)
b) The cell potential at standard conditions (E◦
cell) is given by the difference
in standard reduction potentials of the two half reactions:
E◦
cell =E◦
reduction, cathode −E◦
reduction, anode
E◦
cell =E◦
Fe3+/Fe −E◦
Cu2+/Cu
E◦
cell = 0.77 V −0.34 V = 0.43 V
c) To determine which metal will be plated out first, we compare the stan-
dard reduction potentials of the two half-reactions. The larger the reduction
potential, the more likely the reduction reaction will occur. Since Fe3+/Fe2+
has a higher reduction potential than Cu2+/Cu, iron will be plated out first as
the cell operates.
Question 26
Question
Consider a galvanic cell with the following half-reactions:
Anode: Zn(s) −−→ Zn2+(aq) + 2 e−
Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)
If the standard reduction potential for the Zn2+|Zn half-reaction is −0.76 V
and for the Cu2+|Cu half-reaction is 0.34 V, determine the cell potential (E◦
cell)
for this galvanic cell.
Solution
Step 1: Write the overall cell reaction by adding the half-reactions together.
Make sure to balance the number of electrons.
The overall cell reaction is:
Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell) using the standard re-
duction potentials of the half-reactions.
The standard cell potential is given by the formula:
E◦
cell =E◦
cathode −E◦
anode
22
Given: E◦
cathode = 0.34 V E◦
anode =−0.76 V
Substitute the values into the formula:
E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Therefore, the cell potential for this galvanic cell is 1.10 V .
Question 27
Question
Consider a galvanic cell that consists of a standard hydrogen electrode as the
anode and a copper electrode as the cathode. The standard reduction potential
for the copper electrode is E◦
Cu2+/Cu = 0.34 V. Calculate the cell potential at
25
°
C when the concentration of Cu2+ is 0.10 M. What will happen to the cell
potential if the concentration of Cu2+ is increased to 1.0 M?
Solution
Step 1: Write the balanced redox reaction for the cell:
2H++ 2e−→H2
Cu2+ + 2e−→Cu
The overall cell reaction is:
2H++ Cu2+ →H2+ Cu
Step 2: Calculate the standard cell potential (E◦
cell) using the given standard
reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Cu2+/Cu −E◦
H+/H2
E◦
cell = 0.34 V −0.00 V
E◦
cell = 0.34 V
Step 3: Calculate the cell potential (Ecell) at 25
°
C using the Nernst equa-
tion:
Ecell =E◦
cell −RT
nF ln Q
Since the reaction quotient (Q) for the cell is equal to the concentration of
products over reactants, when Cu2+ is 0.10 M, Q=[H2][Cu]
[H+][Cu2+]= 1
Plugging in the values:
Ecell = 0.34 V −(8.314J/(mol ·K)·298K)
2·96485C/mol ln 1
23
Ecell = 0.34 V
Step 4: Now, when the concentration of Cu2+ is increased to 1.0 M, Q=
[H2][Cu]
[H+][Cu2+]>1
This will shift the equilibrium to the left, increasing the concentration of
reactants and decreasing the concentration of products, resulting in a decrease
in cell potential.
Therefore, increasing the concentration of Cu2+ will decrease the cell poten-
tial of the galvanic cell.
Question 28
Question
Consider a galvanic cell with the following half-reactions:
Anode: Pb2+(aq) + 2e−→Pb(s)
Cathode: 2Ag+(aq) + 2e−→2Ag(s)
If the standard reduction potentials are E◦(Pb2+/Pb) = −0.13 V and
E◦(Ag+/Ag) = 0.80 V, calculate the standard cell potential, E◦
cell, for this
galvanic cell and determine if the cell reaction is spontaneous.
Solution
Step 1: Write the overall cell reaction by combining the half-reactions of the
anode and cathode:
Pb2+(aq) + 2Ag+(aq)→Pb(s) + 2Ag(s)
Step 2: Calculate the standard cell potential using the standard reduction
potentials:
E◦
cell =E◦(cathode) −E◦(anode)
E◦
cell = 0.80 V −(−0.13 V)
E◦
cell = 0.93 V
Step 3: Determine if the cell reaction is spontaneous by checking if E◦
cell is
positive (>0). Since E◦
cell = 0.93 V, the cell reaction is spontaneous because
the standard cell potential is positive.
24
Question 29
Question
Consider a galvanic cell with the following cell notation: Zn|Zn2+||Cu2+|Cu.
If the standard reduction potential for Cu2+ + 2e−→Cu is +0.34 V and the
standard reduction potential for Zn2+ + 2e−→Zn is −0.76 V:
1. Calculate the cell potential under standard conditions.
2. Determine the reaction that occurs at the cathode and at the anode.
3. Is the cell acting as a galvanic cell or an electrolytic cell?
Solution
1. To calculate the cell potential under standard conditions, we use the formula
E◦
cell =E◦
cathode −E◦
anode, where E◦
cathode and E◦
anode are the standard reduction
potentials of the cathode and anode, respectively.
E◦
cell =E◦
cathode −E◦
anode = (+0.34 V) −(−0.76 V) = 1.10 V
Therefore, the cell potential under standard conditions is 1.10 V.
2. The reactions that occur at the cathode and anode can be determined
from the cell notation:
Cathode: Cu2+ + 2e−→Cu
Anode: Zn →Zn2+ + 2e−
3. A galvanic cell generates electrical energy from spontaneous redox reac-
tions. Since the cell potential is positive (1.10 V), the cell is acting as a galvanic
cell.
Question 30
Question
An electrochemical cell consists of a copper electrode immersed in a 1.0 M
Cu(NO3)2solution and a zinc electrode immersed in a 1.0 M Zn(NO3)2solution.
The standard reduction potentials are E◦
cell = 1.10 V for the following reaction:
Cu2+(aq)+2e−→Cu(s)
Zn2+(aq)+2e−→Zn(s)
What is the standard cell potential for the galvanic cell at 25◦C formed by
connecting these two half-cells?
25
Solution
Step 1: Identify the half-reactions and their standard reduction potentials. The
given half-reactions are:
Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V
Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V
Step 2: Write the overall cell reaction. The overall cell reaction can be
obtained by adding the two half-reactions, and the standard cell potential is the
sum of the standard reduction potentials of the two half-cells.
Cu2+(aq) + Zn(s)→Cu(s) + Zn2+(aq)
Step 3: Calculate the standard cell potential. Given E◦
cell = 1.10 V, and
E◦
cell =E◦
cathode −E◦
anode.
Therefore, E◦
cell = 0.34 V −(−0.76 V) = 1.10 V
Thus, the standard cell potential for the galvanic cell is 1.10 V.
Question 31
Question
Consider a galvanic cell with the standard reduction potentials of E◦
cathode =
0.50 V and E◦
anode =−0.70 V. Calculate the cell potential at 25◦C. Next,
describe the differences between a galvanic cell and an electrolytic cell.
Solution
Step 1: Calculate the cell potential of the galvanic cell at 25◦C using the Nernst
equation:
∆G=−nF E
E=E◦−RT ln Q
nF
Given E◦
cathode = 0.50 V, E◦
anode =−0.70 V, and the number of electrons
transferred (n) is 1, we have:
E= (0.50 V) −(8.314 J/mol·K)(298 K) ln(1)
1(96485 C/mol)
E= 0.50 −(2470.92) ln(1)
96485
E= 0.50 V
Step 2: Describe the differences between a galvanic cell and an electrolytic
cell: - Galvanic Cell: 1. Spontaneous redox reaction occurs. 2. Electrons flow
26
from anode to cathode through the external circuit. 3. Energy is released from
the redox reaction and converted into electrical energy. 4. Anode is negative
and cathode is positive. 5. Salt bridge is used to maintain electrical neutrality.
- Electrolytic Cell: 1. Non-spontaneous redox reaction occurs. 2. External
electrical source is needed to drive the reaction. 3. Electrons are forced to flow
from cathode to anode through the external circuit. 4. Anode is positive and
cathode is negative. 5. Electrons flow in the opposite direction compared to a
galvanic cell. 6. Used for processes like electroplating.
Question 32
Question
Consider a galvanic cell with the following half-reactions:
Anode: Ni(s) →Ni2+(aq)+2e−
Cathode: Cu2+(aq)+2e−→Cu(s)
Given that the standard reduction potentials are Cu2+/Cu : E◦= 0.34 V and
Ni2+/Ni : E◦=−0.25 V, calculate the cell potential at 25◦C.
Solution
Step 1: Write the cell reaction by adding the half-reactions together:
Ni(s) + Cu2+(aq)→Ni2+(aq) + Cu(s)
Step 2: Write the cell potential equation using the standard reduction po-
tentials:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −(−0.25) V
E◦
cell = 0.59 V
Step 3: Calculate the effect of temperature on the cell potential using the
Nernst equation:
E=E◦−0.0592 V
n·log [Ni2+][Cu]
[Cu2+][Ni]
Step 4: Since we are under standard conditions, Q= 1 and E=E◦
cell.
0.59 = E◦−0.0592 V
2·log(1)
Step 5: Thus, the cell potential at 25◦C is 0.59 V .
27
Question 33
Question
Consider a galvanic cell with a standard cell potential of E◦= 0.80 V that is
initially at equilibrium. If the concentration of the oxidizing agent is doubled
while the concentration of the reducing agent remains the same, calculate the
cell potential at this new equilibrium.
Solution
Step 1: Write the half-reactions for the cell.
Anode: Oxidation (reducing agent) →Oxidation product + e−
Cathode: e−+ Reduction agent →Reduction product
Step 2: Write the cell reaction using the two half-reactions.
Overall Reaction: Oxidizing agent + Reducing agent →Oxidation product+Reduction product
Step 3: Calculate the cell potential using the Nernst equation:
E=E◦−RT
nF ln Q
K
where Eis the cell potential, E◦is the standard cell potential, Ris the gas
constant, Tis the temperature in Kelvin, nis the number of moles of electrons
transferred in the balanced equation, Fis Faraday’s constant, Qis the reaction
quotient, and Kis the equilibrium constant.
Step 4: Calculate the new cell potential after the concentration change.
Given that the concentrations of the oxidizing agent was doubled, Qwill increase
by a factor of 2. Therefore, Q= 2 and the new cell potential (E′) can be
calculated as:
E′= 0.80 −RT
nF ln 2
1
E′= 0.80 −RT
nF ln(2)
E′= 0.80 −0.0257 V
n×ln(2)
Step 5: Determine nfrom the balanced chemical equation. The number of
moles of electrons transferred in the balanced equation is equal to the coefficient
of e−in the balanced equation for the overall cell reaction. Determine this value
and substitute into the equation above to find the final cell potential at the new
equilibrium.
28
Question 34
Question
Consider a galvanic cell that consists of a copper electrode immersed in a 1.0 M
Cu2+ solution and a platinum electrode immersed in a 1.0 M Cl−solution. The
cell diagram is as follows: Cu(s) — Cu2+ (1.0 M) —— Cl−(1.0 M) — Pt(s).
Determine the standard cell potential for this galvanic cell.
Solution
Step 1: Write down the half-reactions and their standard reduction potentials.
Cathode (Reduction): Cu2+(aq)+2e−→Cu(s)E◦
cathode = 0.34 V
Anode (Oxidation): 2Cl−(aq)→Cl2(g)+2e−E◦
anode = 1.36 V
Step 2: Calculate the standard cell potential using the equation:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.34 V −1.36 V
E◦
cell =−1.02 V
Therefore, the standard cell potential for this galvanic cell is -1.02 V.
Question 35
Question
Consider a galvanic cell with a standard cell potential of 1.10 V. Suppose the cell
is operating at a temperature of 298 K and the concentration of Zn2+ in the cell
is 0.050 M. If the cell voltage drops to 0.90 V, calculate the new concentration
of Zn2+ in the cell assuming all other conditions remain constant.
Solution
Step 1: Determine the initial reaction occurring in the cell. The initial reac-
tion in the cell involves the oxidation of Zn(s) to Zn2+(aq) and reduction of
MnO−
4(aq) to Mn2+(aq). This can be represented as:
Zn(s) + MnO−
4(aq)+H+(aq)→Zn2+(aq) + Mn2+(aq)+H2O(l)
Step 2: Determine the initial cell potential (E◦
cell) using the Nernst equation.
The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Mn2+]
29
Given that Ecell = 1.10 V and Ecell =E◦
cell−0.0592
1log 0.050
[Mn2+], we can calculate
[Mn2+].
Thus 1.10 = 1.10 −0.0592 log 0.050
[Mn2+]
0 = −0.0592 log 0.050
[Mn2+]
log 0.050
[Mn2+]= 0
0.050
[Mn2+]= 1
[Mn2+] = 0.050 M
Step 3: Calculate the new concentration of Zn2+ in the cell. Using Ecell =
E◦
cell −0.0592
nlog 0.050
[Mn2+], and the new cell potential Enew = 0.90 V,
0.90 = 1.10 −0.0592 log [Zn2+]
0.050
0.20 = −0.0592 log [Zn2+]
0.050
log [Zn2+]
0.050 =0.20
−0.0592
[Zn2+]
0.050 = 3.3784
[Zn2+]=3.3784 ×0.050 M
[Zn2+] = 0.169 M
Therefore, the new concentration of Zn2+ in the cell is 0.169 M.
30
