CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Nernst equation
Question Bank - Set 2
Liberty University
Question 1
Question
A voltaic cell consists of a silver electrode in a 0.10 M AgNO3solution and
a copper electrode in a 0.20 M Cu(NO3)2solution. If the standard reduction
potential of Ag+to Ag(s) is +0.80 V and the standard reduction potential of
Cu2+ to Cu(s) is +0.34 V, calculate the cell potential at 25
°
C using the Nernst
equation.
Solution
Step 1: Write the balanced redox reaction for the cell. Since the cell consists
of a silver electrode and a copper electrode, the balanced redox reaction is:
2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+ (aq)
Step 2: Write the expression for the cell potential (Ecell). The cell potential
(Ecell) is given by the formula: Ecell =Ecathode −Eanode Ecell =E◦
cathode −
E◦
anode
where E◦
cathode and E◦
anode are the standard reduction potentials of the cath-
ode and anode respectively.
Step 3: Calculate the standard cell potential (E◦
cell). E◦
cell =E◦
cathode −
E◦
anode E◦
cell = (+0.80 V) −(+0.34 V) E◦
cell = +0.46 V
Step 4: Calculate the reaction quotient (Q). Q=[Ag]2
[Cu2+]Given: [Ag] =
0.10 M and [Cu2+]=0.20 M Q=(0.10)2
0.20 Q= 0.05 M
Step 5: Calculate the cell potential at 25
°
C using the Nernst equation. The
Nernst equation is given by: E=E◦−0.0592
nlog Qwhere nis the number of
moles of electrons transferred in the balanced redox reaction.
For the given reaction: n= 2 E= 0.46 V −0.0592
2log(0.05) E= 0.46 V −
0.0592
2×(−1.3) E= 0.46 V −0.0385 E= 0.4215 V
Therefore, the cell potential at 25
°
C using the Nernst equation is 0.4215 V.
Question 2
Question
Calculate the cell potential for a galvanic cell with the following half-reactions
at 25
°
C:
Ni2+(0.10M)+2e−→Ni(s)E◦=−0.25 V
MnO−
4(0.20M)+8H+(1.0M)+5e−→Mn2+(1.0M)+4H2O E◦= 1.23 V
Solution
Step 1: Write the overall cell reaction.
Ni2+ + MnO−
4+ 8H+→Ni + Mn2+ + 4H2O
Step 2: Calculate the standard cell potential (E◦) using the given half-
reactions.
E◦
cell =E◦
cathode −E◦
anode = 1.23 V −(−0.25 V) = 1.48 V
Step 3: Write the Nernst equation.
E=E◦−RT
nF ln Q
where: - Eis the cell potential, - E◦is the standard cell potential, - Ris the
gas constant (8.314 J mol−1K−1), - Tis the temperature in Kelvin (25
°
C = 298
K), - nis the number of moles of electrons transferred in the balanced equation
(5 in this case), - Fis the Faraday constant (96485 C mol−1), - Qis the reaction
quotient.
Step 4: Calculate the reaction quotient, Q.
Q=[Ni][Mn2+]
[MnO−
4][
Step 5: Substitute the known values into the Nernst equation and solve for
E.
E= 1.48 V −(8.314 J mol−1K−1)(298 K)
5(96485 C mol−1)ln5.0×10−3
E≈1.48 V −0.059 V ln5.0×10−3
E≈1.48 V −0.059 V ×(−5.30)
E≈1.48 V + 0.313 V = 1.79 V
Therefore, the cell potential for the galvanic cell at 25
°
C is 1.79 V.
2
Question 3
Question
Calculate the cell potential at 25
°
C for the following redox reaction:
Pb2+(aq) + 2Fe3+(aq)→Pb(s) + 2Fe2+(aq)
Given the following half-reactions and standard reduction potentials:
Pb2+(aq)+2e−→Pb(s)E◦=−0.13 V
Fe3+(aq) + e−→Fe2+(aq)E◦= 0.77 V
Also, the concentration of Pb2+ is 0.10 M, the concentration of Fe3+ is 0.50
M, and the concentration of Fe2+ is 1.00 M.
Solution
Step 1: Write the balanced overall redox reaction:
Pb2+(aq) + 2Fe3+(aq)→Pb(s) + 2Fe2+(aq)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
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 in the balanced
redox reaction, - Fis the Faraday constant (96485 C/mol), - Qis the reaction
quotient.
Step 3: Calculate Q, the reaction quotient:
Q=[Fe2+]2
[Pb2+][Fe3+]2=(1.00)2
(0.10)(0.50)2= 40
Step 4: Calculate the cell potential using the Nernst equation: Plugging
values into the Nernst equation:
E= 0.77 V−(8.314)(298)
2(96485) ln(40) = 0.77 V−0.059 ln(40) = 0.77 V−0.59 ≈0.18 V
Therefore, the cell potential at 25
°
C for the given redox reaction is approxi-
mately 0.18 V.
3
Question 4
Question
The concentration of Cu2+ ions in a cell is 0.010 M while the concentration of
Fe2+ ions is 0.100 M. The cell potential at 25◦C is 0.23 V. Determine the value
of the equilibrium constant Kfor the following reaction:
Cu2+(aq) + Fe(s)→Cu(s) + Fe2+(aq)
Solution
Step 1: Write the half-reactions for the given redox reaction:
Oxidation: Cu2+(aq)+2e−→Cu(s) E◦
Cu = 0.34 V
Reduction: Fe2+(aq)+2e−→Fe(s) E◦
Fe =−0.44 V
Step 2: Write the overall cell reaction and find the standard cell potential,
E◦
cell:
Cu2+(aq) + Fe(s)→Cu(s) + Fe2+(aq)
E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.44 V) = 0.78 V
Step 3: Calculate the reaction quotient, Q, using the Nernst equation:
E=E◦−0.0592
nlog [Cu2+]
[Fe2+]
0.23 = 0.78−0.0592
2log 0.01
0.1= 0.78−0.0296 log 0.1=0.78−0.0296(−1) = 0.78+0.0296 = 0.8096 V
Step 4: Calculate the equilibrium constant, K, using the equation:
E=RT
nF ln K
0.8096 = (8.314 J/mol ·K)(298 K)
2(96485 C/mol) ln K
ln K= 0.8096 ·2(96485 C/mol)
(8.314 J/mol ·K)(298 K) = 0.1619
K=e0.1619 ≈1.175
Therefore, the equilibrium constant Kfor the given reaction is approximately
1.175.
4
Question 5
Question
A voltaic cell is constructed with a standard hydrogen electrode as the anode
and a copper electrode as the cathode. The initial concentrations of H+and
Cu2+ are both 0.10 M. The cell potential is measured to be 0.45 V at 25◦C.
Determine the equilibrium constant (K) for the following cell reaction:
2H++Cu2+ →H2+Cu
Solution
Step 1: Write the half-reactions for the anode and cathode: Anode (oxidation):
2H+(aq)+2e−→H2(g) Cathode (reduction): Cu2+(aq)+2e−→Cu(s)
Step 2: Determine the cell potential using the Nernst equation:
E=E◦−0.0592
nlog Q
where, E= cell potential = 0.45 V (given) E◦= standard cell potential n
= number of electrons transferred = 2 in this case Q= reaction quotient =
[H2][Cu]
[H+]2[Cu2+ ]
Step 3: Calculate the standard cell potential (E◦) using standard reduction
potentials: E◦=E◦
cathode −E◦
anode =E◦
reduction of Cu2+ −E◦
oxidation of H+
Step 4: Calculate the reaction quotient (Q) at equilibrium: Given initial
concentrations are both 0.10 M.
Step 5: Substitute the known values into the Nernst equation and solve for
K:
K=enE◦
0.0592
Now, you can proceed with solving for K.
Question 6
Question
Calculate the cell potential at 25
°
C for a cell in which the cathode reaction is:
Cr3+(aq) + 3e−→Cr(s) and the anode reaction is: MnO−
4(aq) + 8H+(aq) +
5e−→Mn2+(aq) + 4H2O(l). Given [Cr3+] = 0.10 M, [MnO−
4] = 0.20 M,
[H+] = 1.0 M.
Solution
Step 1: Write the overall cell reaction.
Cr3+(aq) + MnO−
4(aq) + 9H+(aq)→Cr(s) + Mn2+(aq) + 5H2O(l)
5
Step 2: Break down the overall reaction into the half-reactions. Cathode
half-reaction:
Cr3+(aq) + 3e−→Cr(s)E◦=−0.74 V
Anode half-reaction:
MnO−
4(aq) + 8H+(aq) + 5e−→Mn2+(aq) + 4H2O(l)E◦= 1.51 V
Step 3: Calculate the cell potential, E◦
cell, using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cr3+][MnO−
4]8
[Cr][Mn2+][H+]9
where n is the number of electrons transferred in the balanced redox reaction.
Step 4: Calculate the cell potential.
Ecell = (1.51 V −(−0.74 V)) −0.0592
5log (0.10 M)(0.20 M)8
(1)(1) (1.0 M)9
Ecell = 2.25 V −0.0592
5log1.6×10−10
Ecell = 2.25 V - 0.0237 log1.6×10−10
Ecell ≈2.25 V
Question 7
Question
The concentration of lead (Pb2+) ions in a voltaic cell is measured to be 1.5×
10−3M. If the standard reduction potential for the Pb2+/Pb half-cell is −0.13
V, calculate the cell potential at 25
°
C. (Given: E◦
Ag+/Ag = 0.80 V.)
Solution
Step 1: Write the half-reaction for the Pb2+/Pb half-cell.
Pb2+ + 2e−→Pb
Step 2: Write the Nernst equation for the cell potential, E, which relates
the standard cell potential, E◦, the actual cell potential, the reaction quotient,
Q, and the gas constant, R, and temperature, T.
E=E◦−0.0592
nlog(Q)
Step 3: Determine the number of electrons transferred, n, in the half-
reaction. The number of electrons transferred is the coefficient of the electron
in the balanced half-reaction, which is 2.
6
Step 4: Calculate the reaction quotient, Q, using the concentrations of the
reactants and products.
Q=[Pb]
[Pb2+]=1
1.5×10−3= 666.67
Step 5: Substitute the given values and calculated Qinto the Nernst equation
to solve for the cell potential, E.
E=−0.13 V −0.0592
2log(666.67)
E=−0.13 V −0.0592
2×2.824
E=−0.13 V −0.1057
E=−0.2357 V
Therefore, the cell potential at 25
°
C is -0.2357 V.
Question 8
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) as the
anode and a silver-silver chloride electrode as the cathode. The half-reaction at
the cathode is AgCl(s) + e−−> Ag(s)+Cl−(aq)withastandardreductionpotentialof0.22V.T hehalf −
reactionattheanodeis2H+(aq)+2e−−> H2(g)withastandardreductionpotentialof0.00V.If theconcentrationsofthemetalionsinsolutionare[Ag+] =
1.0Mand[Cl−]=1.0M, andthepHofthesolutionis2.0, calculatethecellpotentialat25C.Given :R
= 8.314 J/(mol ·K), F= 96485 C/mol.
Solution
Step 1: Write the balanced overall reaction for the cell. The overall cell reaction
can be obtained by summing the half-reactions for the anode and cathode. The
half-reaction at the anode (oxidation) will be multiplied by 2 to balance the
electrons in the overall reaction. The overall reaction is:
2H+2H+2H+2H+(aq) + 2AgCl(s)→2Ag(s)+2
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the standard reduction potentials for the half-
reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.22 V −0.00 V = 0.22 V
Step 3: Determine the concentrations of ions involved in the cell reaction.
Given:
[Ag+]=1.0 M
7
[Cl−]=1.0 M
pH = 2.0
The pH of the solution indicates that the concentration of H+ions is 10−2M.
In this case, the concentrations of the ions in the cell reaction are:
[H+] = 10−2M
[Ag+]=1.0 M
[Cl−]=1.0 M
Step 4: Calculate the non-standard cell potential (Ecell). The Nernst equa-
tion can be used to calculate the cell potential under non-standard conditions:
Ecell =E◦
cell −0.0592
nlog [Ag+]2[H+]2
[AgCl]2
For this reaction, n= 2 (from the balanced overall reaction), and the concen-
tration of solid AgCl does not affect the cell potential. Thus, the cell potential
can be calculated as:
Ecell = 0.22 V −0.0592
2log (1.0 M)2×(10−2M)2
(1.0 M)2
Ecell = 0.22 V −0.0298 log10−6
Ecell = 0.22 V + 0.0298 ×6
Ecell = 0.394 V
Therefore, the cell potential at 25
°
C is 0.394 V.
Question 9
Question
A voltaic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and an unknown metal electrode as the cathode. Given that the standard
reduction potential for the unknown metal electrode is −0.20 V, and the [H+]
concentration is 0.10 M, calculate the cell potential at 25◦C. (Use R= 8.31
J/(mol·K), T= 298 K, and F= 96485 C/mol)
8
Solution
Step 1: Write the half-reaction for the reduction occurring at the unknown
metal electrode: The reduction half-reaction for the unknown metal electrode
is:
Unknown metal2+(aq)+2e−→Unknown metal(s)
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potential: The standard cell potential, E◦
cell, can be calculated using
the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Given that the standard reduction potential for the unknown metal electrode is
−0.20 V, we have:
E◦
anode =E◦
SHE = 0 V
Thus,
E◦
cell =−0.20 V −0 V = −0.20 V
Step 3: Calculate the cell potential at non-standard conditions: The Nernst
equation is given by:
Ecell =E◦
cell −RT
nF ln(Q)
where Qis the reaction quotient and can be calculated using the concentrations
of products and reactants.
Step 4: Calculate Qfor the cell reaction: Since the unknown metal is in a
2+ oxidation state and the [H+] concentration is 0.10 M, the reaction quotient
Qis:
Q= (0.10)2= 0.01
Step 5: Plug in the values to calculate the cell potential at 25◦C: Substitute
the given values into the Nernst equation:
Ecell =−0.20 V −(8.31 J/(mol ·K)) ×298 K
2×96485 C/mol ln(0.01)
Step 6: Calculate Ecell:
Ecell =−0.20 V −2480.38
192970 ln(0.01)
Ecell =−0.20 V −0.012857
−4.605
Ecell =−0.20 V + 0.002794
Ecell ≈ −0.197 V
9
Question 10
Question
A reaction is taking place in an electrochemical cell where the following infor-
mation is known:
E◦
cell = 0.75 V
T= 298 K
n= 2
Calculate the cell potential at 25 degrees Celsius (298 K) for the same reaction
when the concentration of one of the ions is ten times greater than the initial
concentration.
Solution
Step 1: Calculate the change in cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Red]final
[Ox]final
Step 2: Since one ion concentration is ten times greater, we can represent
the new concentrations as:
[Red]final
[Ox]final
=10[Red]initial
[Ox]initial
Step 3: Substitute the given values into the Nernst equation:
Ecell = 0.75 V −0.0592
2log 10[Red]initial
[Ox]initial
Step 4: Simplify and solve for the new cell potential:
Ecell = 0.75 V −0.0296 log(10)
Step 5: Calculate the logarithm:
Ecell = 0.75 V - 0.0296 ×1V - 0.0296 ×1V - 0.0296 ×1V - 0.0296 ×1
Step 6: Final result:
Ecell = 0.7204 V
Question 11
Question
A concentration cell is set up where one half-cell has a silver electrode in a 0.10
MAgNO3solution and the other half-cell has a silver electrode in a 0.50 M
AgNO3solution. Calculate the cell potential at 25
°
C. The standard reduction
potential for the Ag+/Ag half-cell is 0.80 V.
10
Solution
Step 1: Write the half-reactions for the two half-cells. The half-cell reactions
for the silver electrode in the two solutions are:
Anode: Ag(s)→Ag+(aq) + e−
Cathode: Ag+(aq) + e−→Ag(s)
Step 2: Calculate the cell potential at standard conditions. The standard
cell potential E◦
cell can be calculated by the difference in standard reduction
potentials for the two half-cells:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80V−0.80V
E◦
cell = 0V
Step 3: Determine the reaction quotient, Q. The Nernst equation relates cell
potential to reaction quotient Q:
Ecell =E◦
cell −0.0592
nlog Q
where n is the number of moles of electrons transferred in the balanced equation.
Since the reaction quotient Q is the ratio of the concentrations of the products
to the concentrations of the reactants, and the reaction is at equilibrium, we
have:
Q=[Ag+]0.50M
[Ag+]0.10M
= 5
Step 4: Calculate the cell potential. Substitute the values into the Nernst
equation:
Ecell = 0V−0.0592
1log 5
Ecell = 0V−0.086V
Ecell =−0.086V
Therefore, the cell potential at 25
°
C for the concentration cell with the two
different silver electrode solutions is -0.086 V.
Question 12
Question
For a galvanic cell using the half-reactions below, calculate the cell potential at
25
°
C when the concentrations of [F e3+]=0.10 Mand [F e2+]=1.0M.
Cathode: F e3+(aq)+3e−→F e2+(aq)E◦= 0.771 V
Anode: Sn2+(aq)→Sn4+(aq)+2e−E◦= 0.150 V
11
Solution
Step 1: Write the overall cell reaction by reversing the anode reaction and
adding the cathode reaction:
Overall cell reaction: Sn4+(aq) + F e2+(aq)→Sn2+(aq) + F e3+(aq)
Step 2: Determine E◦
cell by using the standard cell potential formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.771 V−0.150 V= 0.621 V
Step 3: Write the Nernst equation for this cell:
E=E◦−0.0592
nlog [Fe3+]
[Sn4+][Fe2+]3
Step 4: Calculate the cell potential (E) using the Nernst equation with the
given concentrations:
E= 0.621 V−0.0592
2log 0.10
[Sn4+](1.0)3
E= 0.621 V−0.0296 log 0.10
[Sn4+]
Step 5: Determine the cell potential at 25
°
C by simplifying the expression
and substituting the known values for [F e3+]=0.10 Mand [F e2+] = 1.0M:
E= 0.621 V−0.0296 log 0.10
[Sn4+]
Question 13
Question
Calculate the cell potential at 25
°
C for a hydrogen-oxygen fuel cell in which
the hydrogen electrode is at a pressure of 1.0 atm and the oxygen electrode
is at a pressure of 0.10 atm. The standard cell potential for this reaction is
1.23 V. Given that the Faraday constant is 96485 C/mol, the gas constant is
8.314 J/(mol ·K), and the temperature is 298 K.
Solution
Step 1: Write the balanced half-reactions and the overall redox reaction. The
half-reactions for a hydrogen-oxygen fuel cell are:
Anode: 2H2(g)→4H+(aq)+4e−
12
Cathode: O2(g) + 4H+(aq)+4e−→2H2O(l)
The overall redox reaction is the sum of these two half-reactions:
2H2(g) + O2(g)→2H2O(l)
Step 2: Calculate the standard cell potential using the Nernst equation. The
Nernst equation relates the standard cell potential (E◦), the cell potential under
nonstandard conditions (E), the reaction quotient (Q), the Faraday constant
(F), and the temperature (T):
E=E◦−RT
nF ln(Q)
where nis the number of moles of electrons transferred in the cell reaction, R
is the gas constant, and Qis the reaction quotient.
Given: - Standard cell potential (E◦) = 1.23 V - Pressure of hydrogen = 1.0
atm - Pressure of oxygen = 0.10 atm - Temperature (T) = 298 K - Gas constant
(R) = 8.314 J/(mol ·K) - Faraday constant (F) = 96485 C/mol
Step 3: Calculate the reaction quotient (Q). Since the partial pressures of
the gases are given, we can use the ideal gas law to calculate the concentration
of H+:
0.10 atm = 1.0atm
1.0atm2
= [H+]
Since Qis the product of the concentrations of the products divided by the
product of the concentrations of the reactants, and H2O(l) does not appear in
the expression for Q, the value of Qis (1)2= 1.
Step 4: Substitute the values of E◦,R,T,n,F, and Qinto the Nernst
equation to calculate the cell potential (E) at 25
°
C.
E= 1.23 V −(8.314 J/(mol ·K))(298 K)
4(96485 C/mol) ln(1)
Step 5: Solve for Eto find the cell potential for the hydrogen-oxygen fuel
cell at 25
°
C.
Question 14
Question
Calculate the cell potential for the following reaction at 25
°
C:
Al(s) + Fe3+(aq)→Al3+(aq) + Fe(s)
Given the standard reduction potentials:
E◦
cell(Al3+/Al) = −1.66 V
E◦
cell(Fe3+/Fe) = 0.77 V
and the concentration of Fe3+ is 0.010 M.
13
Solution
Step 1: Write the overall cell reaction:
Al(s) + Fe3+(aq)→Al3+(aq) + Fe(s)
Step 2: Identify the oxidation and reduction half-reactions: Oxidation half-
reaction: Al(s)→Al3+(aq)+3e−
Reduction half-reaction: Fe3+(aq)+3e−→Fe(s)
Step 3: Calculate the standard cell potential, E◦
cell, using the formula:
E◦
cell =E◦
cell,reduction −E◦
cell,oxidation
E◦
cell = 0.77 V −(−1.66 V) = 2.43 V
Step 4: Write the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
where Qis the reaction quotient, and Kis the equilibrium constant.
Step 5: Calculate Qusing the concentrations provided:
Q=[Fe](Al3+)
[Fe3+]
Q=1·1
0.010 = 100
Step 6: Calculate the cell potential at 25
°
C using the Nernst equation:
Ecell = 2.43 V −0.0592
6log(100) = 2.43 V −0.0987 = 2.33 V
Therefore, the cell potential for the given reaction at 25
°
C is 2.33 V.
Question 15
Question
For a galvanic cell with a silver electrode in a 0.10 M Ag+solution and a copper
electrode in a 0.50 M Cu2+ solution, the cell potential is measured to be 0.90 V
at 25
°
C. Calculate the standard cell potential, E, for this cell at 25
°
C. (Given:
Ecell for the Ag+—Ag and Cu2+—Cu cells are 0.80 V and 0.34 V, respectively)
14
Solution
Step 1: Write the half-reactions for each electrode.
The half-reaction for the silver electrode is: Ag+(aq) + e−→Ag(s)
The half-reaction for the copper electrode is: Cu2+(aq) + 2e−→Cu(s)
Step 2: Write the overall cell reaction.
Overall: Ag+(aq)+ Cu2+(aq) →Ag(s) + Cu(s)
Step 3: Calculate Ecell using the Nernst equation.
The Nernst equation is: E=E−0.0592
nlog Q
K
The number of electrons transferred, n, for this cell is 1.
Qis the reaction quotient, which for this cell is [Ag][Cu2+ ]
[Ag+][Cu] =0.10×0.50
1= 0.05
Substitute the given values into the Nernst equation:
0.90 = E−0.0592
1log(0.05)
Step 4: Solve for E.
0.90 = E−0.0592 log(0.05)
E= 0.90 + 0.0592 log(0.05)
E= 0.90 + 0.0592 ×(−1.30)
E≈0.818 V
Therefore, the standard cell potential for this cell at 25
°
C is approximately
0.818 V.
Question 16
Question
A zinc electrode is immersed in a solution containing Zn2+ ions at a concentra-
tion of 0.10 M. If the standard reduction potential for the Zn2+ + 2e−→Zn
half-reaction is -0.76 V, calculate the cell potential at 25
°
C when the concentra-
tion of Zn2+ ions is actually 0.20 M.
Solution
Step 1: Write the half-reaction and Nernst equation. The half-reaction is Zn2+ +
2e−→Zn. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
K
where: E= cell potential at non-standard conditions E◦= standard cell po-
tential n= number of electrons transferred in the balanced half-reaction Q=
reaction quotient K= equilibrium constant
Step 2: Determine the values given in the question. E◦=−0.76 V T= 298
K [Zn2+] = 0.10 M [Zn2+]actual = 0.20 M
Step 3: Calculate the equilibrium constant. The equilibrium constant, K, is
related to the standard cell potential by the equation:
E◦=0.0592
nlog K
15
K= 10n·(E◦/0.0592)
For the given half-reaction, n= 2:
K= 102·(−0.76/0.0592)
Step 4: Calculate Qfor the non-standard conditions.
Q=[Zn2+]actual
[Zn2+]◦=0.20
0.10
Step 5: Substitute the values into the Nernst equation and calculate the cell
potential.
E=−0.76 −0.0592
2log 0.20
0.10
Step 6: Perform the calculations to find the cell potential.
Question 17
Question
Calculate the cell potential for the following electrochemical cell at 25
°
C:
Ag|AgCl(0.10 M) || Cl−(0.0020 M)|Ag
Given:
E◦
AgCl(s)/Ag+(aq)= 0.22 V
E◦
Ag+(aq)/Ag(s)= 0.80 V
R= 8.314 J ·mol−1·K−1
F= 96485 C ·mol−1
Solution
Step 1: Write the overall cell reaction. The overall cell reaction can be written
as:
AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Calculate the standard cell potential, E◦
cell, using the given standard
reduction potentials. The standard cell potential can be calculated using the
formula:
E◦
cell =E◦
cathode −E◦
anode
Given that the cathode is Ag+(aq) + e−→Ag(s) and the anode is AgCl(s) +
e−→Ag(s) + Cl−(aq), we have:
E◦
cell =E◦
Ag+(aq)/Ag(s)−E◦
AgCl(s)/Ag+(aq)
16
E◦
cell = 0.80 V −0.22 V = 0.58 V
Step 3: Calculate the reaction quotient, Q, for the cell at non-standard
conditions.
Q=[Ag+]final ·[Cl−]final
[AgCl]final
Given that [Ag+]final = 0.0020 M, [Cl−]final = 0.0020 M, and [AgCl]final =
0.10 M, we have:
Q=(0.0020)(0.0020)
0.10 = 4.0×10−5
Question 18
Question
A voltaic cell is constructed with a silver-silver chloride electrode (half-cell re-
action: AgCl(s) + e−−→ Ag(s) + Cl−(aq))andazincelectrode(Zn(s)−→ Zn2+
(aq)+2e−).T heinitialconcentrationsof Ag+, Zn2+, andCl−ionsare1.0M, 0.1M, and0.01M, respectively.Ifthestandardreductionpotentialof theAgCl(s)electrodeis0.222V andthestandardreductionpotentialoftheZn(s)electrodeis−
0.76V, whatisthecellpotentialat25C?
Solution
Step 1: Write the half-reactions at each electrode. The half-reactions are:
AgCl(s) + e−−→ Ag(s) + Cl−(aq)withastandardreductionpotentialof0.222V(cathode)Zn(s)−→
Zn2+ (aq)+2e−withastandardreductionpotentialof −0.76V(anode)
Step 2: Calculate the cell potential using the Nernst equation. The cell
potential (Ecell) can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Ag+][Cl−]
Step 3: Determine the number of electrons transferred. Since AgCl has a
1:1 stoichiometry and Zn forms 2 electrons, the number of electrons transferred
(n) is 2.
Step 4: Substitute values into the Nernst equation.
Ecell = 0.222 V −0.0592
2log 0.1
1.0×0.01
Step 5: Solve for the cell potential.
Ecell = 0.222 V −0.0592
2log(10)
Ecell = 0.222 V −0.0296 ×1
Ecell = 0.1924 V
Therefore, the cell potential at 25
°
C is 0.1924 V.
17
Question 19
Question
A concentration cell is set up with two silver-silver chloride electrodes. One
half-cell has a silver electrode dipped in a 0.10 M Ag+solution, and the other
half-cell has a silver electrode dipped in a 0.0010 M Ag+solution. Calculate
the cell potential at 25
°
C for this concentration cell given that E◦
cell = 0.80 V
and F= 96,485 C/mol.
Solution
Step 1: Write the half-reactions:
AgCl(s) + e−→Ag(s) + Cl−(aq)
This half-reaction occurs at both electrodes.
Step 2: Write the Nernst equation for the concentration cell:
Ecell =E◦
cell −0.0592
nlog [Ag+]low
[Ag+]high
where nis the number of electrons transferred in the balanced overall cell reac-
tion.
Step 3: Calculate the number of electrons transferred, n: In this case, the
overall cell reaction is:
Ag+
(0.10 M) + e−→Ag+
(0.0010 M)
which involves the transfer of 1 electron.
Step 4: Solve for Ecell using the Nernst equation:
Ecell = 0.80 V −0.0592
1log 0.0010
0.10 = 0.80 −0.057 = 0.743 V
Therefore, the cell potential at 25
°
C for this concentration cell is 0.743 V.
Question 20
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a copper strip submerged in a 0.10 M Cu2+ solution. Given that the standard
reduction potential for the Cu2+/Cu half-reaction is +0.34 V, calculate the cell
potential when the concentration of Cu2+ is reduced to 0.050 M at 25◦C. (Hint:
Use the Nernst equation)
18
Solution
Step 1: Write the half-reactions for the system: The overall cell reaction can be
broken down into two half-reactions:
Anode (oxidation): Cu(s) →Cu2+(aq) + 2e−
Cathode (reduction): 2H+(aq) + 2e−→H2(g)
Step 2: Calculate the standard cell potential (E◦) using the given standard
reduction potential for the Cu2+/Cu half-reaction (+0.34 V):
E◦
cell =E◦
cathode −E◦
anode = 0 −(+0.34) = −0.34 V
Step 3: Calculate the reaction quotient (Q) using the concentrations given:
Q=[Cu2+]
[H+]2=0.050
12= 0.050
Step 4: Calculate the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the balanced half-
reaction. Since 2 moles of electrons are transferred in the reduction of Cu2+, n
= 2.
Step 5: Substitute the values into the Nernst equation:
Ecell =−0.34 −0.0592
2log(0.050)
Ecell =−0.34 −0.0296 log(0.050)
Ecell =−0.34 −0.0296 ×(−1.30)
Ecell =−0.34 + 0.03848
Ecell =−0.3015 V
Therefore, the cell potential when the concentration of Cu2+ is reduced to
0.050 M at 25◦C is -0.3015 V.
Question 21
Question
A concentration cell is set up using two half-cells, one with a silver electrode
in a 0.10 M AgNO3solution and the other with a silver electrode in a 1.0 M
AgNO3solution. If the standard reduction potential of the Ag+/Ag half-cell is
0.80 V at 25
°
C, calculate the cell potential at 25
°
C using the Nernst equation.
19
Solution
Step 1: Write the half-cell reactions and the overall cell reaction. The half-cell
reactions are given by:
Ag+(aq) + e−→Ag(s)
The cell reaction is:
Ag+(0.10 M, Ag)→Ag+(1.0M, Ag)
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the difference in standard reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cell =E◦
cathode −E◦
anode = 0.80 V−0.80 V= 0 V
Step 3: Write the Nernst equation. The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog [Ag+]right
[Ag+]lef t
where: - Ecell is the cell potential - E◦
cell is the standard cell potential - nis the
number of electrons transferred (1 in this case) - [Ag+]right is the concentration
of Ag+on the right side (1.0 M) - [Ag+]lef t is the concentration of Ag+on the
left side (0.10 M)
Step 4: Calculate the cell potential using the Nernst equation. Plugging in
the values:
Ecell = 0 V−0.0592
1log 1.0
0.10
Ecell =−0.0592 ×log(10)
Ecell =−0.0592 ×1
Ecell =−0.0592 V
Therefore, the cell potential at 25
°
C using the Nernst equation is -0.0592 V.
Question 22
Question
A redox reaction is carried out in a cell with a silver electrode. The half-reaction
occurring at the electrode is:
Ag++ e−→Ag(s)
Given that the standard reduction potential for this reaction is E◦= 0.80
V, calculate the cell potential under the following conditions: [Ag+]=1.0 M
and [Ag] = 0.1 M. (Assume T= 298 K)
20
Solution
Step 1: Write the Nernst equation, which relates the cell potential (E) to the
standard cell potential (E◦), the reaction quotient (Q), the gas constant (R),
the temperature (T), and the number of electrons transferred (n):
E=E◦−RT
nF ln(Q)
Step 2: Determine the reaction quotient Qusing the concentrations provided:
Since the reaction is:
Ag++ e−→Ag(s)
The reaction quotient is:
Q=[Ag]
[Ag+]
Given that [Ag] = 0.1 M and [Ag+] = 1.0 M, we have:
Q=0.1
1.0= 0.1
Step 3: Calculate the cell potential using the Nernst equation. Given that
E◦= 0.80 V, R= 8.314 J·mol−1·K−1,T= 298 K, n= 1 (1 electron transferred),
and Faraday’s constant F= 96485 C ·mol−1:
Plugging in the values:
E= 0.80 −(8.314)(298)
1(96485) ln(0.1)
Calculating the cell potential:
E= 0.80 −2476.472
96485 ln(0.1)
E≈0.80 −0.643 ln(0.1)
E≈0.80 −0.643(−2.303)
E≈1.47 V
Therefore, the cell potential under the given conditions is approximately 1.47
V.
21
Question 23
Question
A voltaic cell consists of a Ag—AgCl half-cell and a Pt—HCl half-cell. The
half-cell reactions are:
AgCl(s) + e−→Ag(s) + Cl−(aq)
2H2(g)+2Cl−(aq)→2HCl(aq)+2e−
If the concentration of H+ions in the HCl half-cell is 1.0×10−3M, what
is the cell potential of this voltaic cell at 25
°
C? Given E◦
Ag+/Ag = 0.80 V and
E◦
H+/H2= 0.00 V.
Solution
Step 1: Write the overall cell reaction and determine the cell potential. The
overall cell reaction is the sum of the two half-cell reactions. We’ll need to flip
the second half-cell reaction and multiply it by 1/2 to balance the electrons:
AgCl(s) + 2H+(aq)→Ag(s) + Cl−(aq)+H2(g)
The cell potential, Ecell, can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
where Qis the reaction quotient and nis the number of electrons transferred
in the balanced cell reaction.
Step 2: Determine the reaction quotient, Q. Since we are given the con-
centrations of H+ions, we can use the Nernst equation to calculate the cell
potential. The reaction quotient, Q, can be calculated as follows:
Q=[Ag+][H+]
[Cl−]
Plugging in the given values:
Q=1.0×10−3×1.0
1.0= 1.0×10−3
Step 3: Calculate the cell potential, Ecell. Substitute the known values into
the Nernst equation:
Ecell =E◦
Ag+/Ag −0.0592
nlog Q
Ecell = 0.80 −0.0592
1log1.0×10−3
Ecell = 0.80 −0.0592 ×(−3) = 0.98 V
Therefore, the cell potential of this voltaic cell at 25
°
C is 0.98 V.
22
Question 24
Question
A voltaic cell is constructed with a standard hydrogen electrode (Pt — H+(aq, 1
M) —— H+(aq, xM) — MnO4(aq, yM), Mn2+(aq)). The cell potential is found
to be 1.00 V at 25
°
C. Given that E◦= 1.49 V, determine the concentrations of
H+and MnO4in the cell. Assume the cell reaction is:
MnO4(aq) + 8H+(aq) + 5e −→ Mn2+(aq) + 4H2O(l).
Solution
Step 1: Write the half-reaction for the standard hydrogen electrode:
2H+(aq)+2e−−→ H2(g)
Step 2: Calculate the cell potential at nonstandard conditions using the
Nernst equation:
E=E◦−0.0592
nlog [H+]2
PH2
where nis the number of moles of electrons transferred, PH2is the partial
pressure of hydrogen gas (1 atm for standard conditions), and Eis the cell
potential at nonstandard conditions.
Step 3: Given that E◦= 1.49 V, E= 1.00 V, and n= 2, solve for [H+]:
1.00 V = 1.49 V −0.0592
2log [H+]2
1
Step 4: Rearrange the equation and solve for [H+]:
−0.49 V = −0.0296 log[H+]2
16.55 = log[H+]2
[H+]2= 1016.55
[H+] = 108.275 = 7.49 ×108M
Step 5: Using the stoichiometry of the cell reaction, calculate the concentra-
tion of MnO4(y) in the cell: Since the stoichiometry is 1:8 for MnO4:H+, the
concentration of MnO4can be determined as:
8[H+] = 8(7.49 ×108) M = 5.99 ×109M
Therefore, the concentration of H+is 7.49 ×108M and the concentration of
MnO4is 5.99 ×109M.
23
Question 25
Question
A galvanic cell is constructed with a silver electrode in a 1.0 M Ag+solution and
a copper electrode in a 0.10 M Cu2+ solution. At 25
°
C, the standard reduction
potentials for the half-reactions are as follows:
Ag++ e−→Ag, E◦= 0.80 V
Cu2+ + 2e−→Cu, E◦= 0.34 V
Calculate the cell potential at 25
°
C, given that the number of electrons trans-
ferred in the balanced equation for the cell reaction is 2.
Solution
Step 1: Write the cell reaction by combining the two half-reactions. The cell
reaction is obtained by summing the reduction half-reactions after multiplying
the equations by the necessary coefficients to balance the electrons. Since 2
electrons are involved, we need to multiply the first half-reaction by 2 before
adding them together:
2 Ag++ Cu2+ →2 Ag + Cu
Step 2: Determine the standard cell potential (E◦
cell). The standard cell
potential is calculated by subtracting the standard reduction potentials of the
two half-reactions involved in the cell reaction.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Ag −E◦
Cu2+
E◦
cell = 0.80 V −0.34 V = 0.46 V
Step 3: Use the Nernst equation to find the cell potential at 25
°
C. The
Nernst equation relates the cell potential under non-standard conditions to the
cell potential under standard conditions:
E=E◦−0.0592
nlog(Q)
where - Eis the cell potential under non-standard conditions, - E◦is the cell
potential under standard conditions, - nis the number of moles of electrons
transferred in the balanced equation, and - Qis the reaction quotient, which is
the ratio of products to reactants at any point during the reaction.
Step 4: Calculate the concentration of Ag+and Cu2+ ions to determine
Q. Since the reaction is at equilibrium, Qis equal to the equilibrium constant,
K. To calculate Q, we need to consider the concentration of the products and
reactants in the cell reaction.
24
Q=[Ag]2
[Cu2+]
Given that the initial concentration of Cu2+ is 0.10 M and the initial concen-
tration of Ag+is 1.0 M, we can substitute these values into the equation.
Step 5: Plug in the values to find the cell potential under non-standard
conditions. Substitute the known values into the Nernst equation to find the
cell potential under the given conditions.
E= 0.46 V −0.0592
2log(Q)
E= 0.46 V −0.0592
2log 1.0 M2
0.10 M
Step 6: Calculate the cell potential at 25
°
C. Plug in the values to find the
cell potential under the given conditions.
E= 0.46 V −0.0592
2log(10) = 0.46 V −0.0296 = 0.43 V
Therefore, the cell potential at 25
°
C is 0.43 V.
Question 26
Question
A voltaic cell is constructed with a zinc electrode in a 1.0 M solution of Zn2+
ions, and a hydrogen electrode in an acidic solution having a concentration of
[H+]=0.10 M. Calculate the cell potential at 25
°
C given that the standard
reduction potential for the Zn2+/Zn half-reaction is −0.76 V and the standard
reduction potential for the H+/H2half-reaction is 0.00 V.
Solution
Step 1: Write the two half-reactions involved in the cell: Zn2+ + 2e−→Zn
2H++ 2e−→H2
Step 2: Calculate the cell potential at standard conditions, E◦
cell:E◦
cell =
E◦
cathode −E◦
anode E◦
cell = 0.00 V −(−0.76 V) E◦
cell = 0.76 V
Step 3: Calculate the reaction quotient Q:Q=[Zn2+]
[H+]=1.0
0.10 = 10
Step 4: Calculate the Nernst equation to find the cell potential at 25
°
C:
Ecell =E◦
cell −0.0592 V
nlog(Q)Ecell = 0.76 V −0.0592 V
2log(10) Ecell = 0.76 V −
0.0296 V ×1Ecell = 0.7304 V
25
Question 27
Question
A voltaic cell consists of a copper electrode in a 1.0 M copper(II) sulfate solution
and a silver electrode in a 0.10 M silver nitrate solution at 25
°
C. Calculate the
cell potential at this temperature. Given E◦
Cu2+/Cu = 0.34 V and E◦
Ag+/Ag =
0.80 V.
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction can be written as:
Cu2+(aq) + 2e−→Cu(s)
2Ag+(aq) + 2e−→2Ag(s)
Combining these two half-reactions:
Cu2+(aq) + 2Ag(s)→Cu(s) + 2Ag+(aq)
Step 2: Calculate the cell potential using the Nernst equation.
The Nernst equation is given by:
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 Faraday’s constant (96485 C/mol) - Qis the reaction quotient, which can be
calculated using the concentrations of reactants and products.
First, calculate Qbased on the concentrations given:
Q=[Cu2+]
[Ag+]2=1.0
(0.10)2= 100
Step 3: Substitute the values into the Nernst equation and solve for E:
E= 0.34 −(8.314)(298)
(2)(96485) ln(100)
E= 0.34 −2465.572
192970 ln(100)
E= 0.34 −0.0127 ln(100)
E0.23 V
Therefore, the cell potential at 25
°
C is approximately 0.23 V.
26
Question 28
Question
An electrochemical cell consists of a zinc electrode in a 1.0 M Zn2+ solution and
a copper electrode in a 1.0 M Cu2+ solution. If the standard reduction potential
for Zn2+ + 2e−→Zn is -0.76 V and for Cu2+ + 2e−→Cu is +0.34 V, calculate
the cell potential at 25
°
C.
Given: R= 8.314 J/(mol·K), T= 298 K, F= 96,485 C/mol
Solution
Step 1: Write the half-cell reactions:
Zn2+ + 2e−→Zn E◦
cell 1 =−0.76 V
Cu2+ + 2e−→Cu E◦
cell 2 = +0.34 V
Step 2: Calculate the cell potential at standard conditions:
E◦
cell =E◦
cell 2 −E◦
cell 1 = 0.34 V −(−0.76 V) = 1.10 V
Step 3: Write down the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Cu2+]
Step 4: Calculate the cell potential using the Nernst equation:
Ecell = 1.10 V −0.0592
2log 1.0
1.0
= 1.10 V −0
= 1.10 V
Therefore, the cell potential at 25
°
C is 1.10 V.
Question 29
Question
A voltaic cell consists of a standard hydrogen electrode (SHE) and a silver
electrode in a solution that is 0.10 M in Ag+ions. The cell potential at 25
°
C
is found to be 0.80 V. Calculate the concentration of Ag+ions that would be
necessary to achieve a cell potential of 1.00 V.
27
Solution
The Nernst equation relates the cell potential of a voltaic cell to the concentra-
tions of the reactants and products involved. It is given by:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential at any given point, - E◦is the standard cell
potential, - nis the number of moles of electrons transferred in the cell reaction,
and - Qis the reaction quotient.
Step 1: Calculate the standard cell potential (E◦)
The standard cell potential for the reaction between the silver electrode and
the SHE can be calculated as:
E◦=ESHE −EAg+/Ag
Given:
ESHE = 0 V
EAg+/Ag = 0.80 V
Therefore:
E◦= 0 −0.80 V = −0.80 V
Step 2: Calculate the reaction quotient (Q)
The cell reaction involves the reduction of Ag+ions to Ag, and since the cell
potential is 1.00 V when the cell is at equilibrium, we can write:
E=E◦−0.0592
nlog Q
Substitute the given values:
1.00 = −0.80 −0.0592
1log Q
log Q=−0.20
Q= 10−0.20 = 0.63
Step 3: Calculate the concentration of Ag+ions
The reaction quotient is given by:
Q=[Ag]
[Ag+]
Given:
[Ag+] = 0.10 M
Substitute the known values into the equation:
0.63 = [Ag]
0.10
[Ag] = 0.063 M
Therefore, the concentration of Ag+ions required to achieve a cell potential
of 1.00 V is 0.063 M.
28
Question 30
Question
A cell is constructed with a silver electrode dipping into a 0.010 M AgNO3 solu-
tion and a cadmium electrode dipping into a 0.10 M Cd(NO3)2 solution. Given
that the standard reduction potentials are E◦
Ag+/Ag = 0.80 V and E◦
Cd2+/Cd =
−0.40 V, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction.
Cd2+ + 2e−→Cd E◦
cell =−0.40 V
2Ag++ 2e−→2Ag E◦
cell = 0.80 V
Cd2+ + 2Ag →Cd + 2Ag+
Step 2: Calculate the standard cell potential (E◦
cell).
E◦
cell =E◦
Ag+/Ag −E◦
Cd2+/Cd
E◦
cell = 0.80 V −(−0.40) V = 1.20 V
Step 3: Calculate the reaction quotient (Q) at 25
°
C.
Q=[Ag+]2
[Cd2+]
Given [Ag+]=0.010 M and [Cd2+] = 0.10 M, we have
Q=(0.010)2
0.10 = 0.001
Step 4: Calculate the cell potential (Ecell) using the Nernst equation.
Ecell =E◦
cell −0.0592
nlog Q
Since the reaction involves the transfer of 2 electrons:
n= 2
Ecell = 1.20 V −0.0592
2log 0.001
Ecell = 1.20 V + 0.0592 ×3×10 = 1.20 V - 0.1776 = 1.0224 V
Therefore, the cell potential at 25
°
C is 1.0224 V.
29
Question 31
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell contains a solution with [Ag+] = 1.0×10−5M, while the other half-cell
contains a solution with [Ag+]=1.0×10−2M. Calculate the cell potential at
25
°
C. (Given: E◦
cell = 0.8 V)
Solution
Step 1: Write the Nernst equation for the cell potential of the concentration
cell:
Ecell =E◦
cell −0.0592
nlog [Ag+]left
[Ag+]right
Where Ecell is the cell potential, E◦
cell is the standard cell potential, nis the
number of moles of electrons transferred, [Ag+]left is the concentration in the
left half-cell, and [Ag+]right is the concentration in the right half-cell.
Step 2: Calculate the cell potential using the given values:
Ecell = 0.8−0.0592
1log 1.0×10−5
1.0×10−2
Ecell = 0.8−0.0592 log 1.0×10−5
1.0×10−2
Ecell = 0.8−0.0592 log 10−3
Ecell = 0.8−0.0592 ×(−3)
Ecell = 0.8+0.1776
Ecell = 0.9776 V
Therefore, the cell potential at 25
°
C for the concentration cell using two
silver-silver chloride electrodes with different concentrations is 0.9776 V.
Question 32
Question
A concentration cell is set up with two half-cells, each containing a solution of
iron(II) ions (Fe2+). One half-cell has a standard Fe2+ concentration of 1.0 M,
while the other half-cell has a Fe2+ concentration of 0.10 M. Calculate the cell
potential at 25◦C for this concentration cell. The standard reduction potential
for the Fe2+/Fe half-cell is +0.77 V.
30
Solution
Step 1: Write the balanced redox half-reaction for the given cell. The half-
reaction is: Fe2+ →Fe + 2e−
Step 2: Write the Nernst equation for the cell potential. The Nernst equation
is:
E=E◦−0.0592
nlog [Fe2+]
[Fe]
where: E= cell potential E◦= standard cell potential n= number of electrons
transferred in the balanced redox reaction [Fe2+] = concentration of Fe2+ in the
cathode (higher concentration) [Fe] = concentration of Fe in the anode (lower
concentration)
Step 3: Find the number of electrons transferred in the balanced half-
reaction. From the balanced half-reaction, we see that 1 mol of Fe2+ is reduced
to form 1 mol of Fe. Therefore, the number of electrons transferred (n) is 2.
Step 4: Substitute known values into the Nernst equation.
E= 0.77 V −0.0592
2log 0.10
1.0
= 0.77 V −0.0296 log(0.10)
= 0.77 V −0.0296 ×(−1)
= 0.77 V + 0.0296
= 0.7996 V
Therefore, the cell potential at 25◦C for this concentration cell is 0.7996 V.
Question 33
Question
A galvanic cell consists of a standard hydrogen electrode (H2(g)|H+(aq)) and
a copper electrode dipped in a solution of Cu2+ ions with a concentration of
0.10 M. The standard reduction potential of Cu2+ →Cu is 0.34 V and the
standard reduction potential of H+|H2(g) is 0.00 V. Calculate the cell potential
at a temperature where E◦is 0.25 V.
Solution
Step 1: Write the half-reactions and the overall cell reaction. The half-reactions
are: Cu2+ +2e−→Cu (Standard reduction potential: 0.34 V) 2H++2e−→H2
(Standard reduction potential: 0.00 V)
The overall cell reaction is: Cu2+ + 2H+→Cu +H2
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potentials.
E◦
cell =E◦
cathode −E◦
anode
31
E◦
cell = 0.00 V −0.34 V
E◦
cell =−0.34 V
Step 3: Calculate the reaction quotient, Q, and the cell potential, E, under
non-standard conditions using the Nernst equation. The reaction quotient, Q,
is given by:
Q=[Cu][H2]
[Cu2+][H+]= 1
The Nernst equation is:
E=E◦
cell −0.0592
nlog(Q)
where n is the number of electrons transferred in the balanced cell reaction.
Step 4: Calculate the cell potential, E, at the given temperature.
E=−0.34 V −0.0592
2log(1)
E=−0.34 V −0 V
E=−0.34 V
Therefore, at a temperature where E◦is 0.25 V, the cell potential is -0.34
V.
Question 34
Question
Calculate the standard electrode potential (E◦) for the reaction below given the
following information:
2Ag++ 2e−→2Ag(s)E◦= 0.80V
Cu2+ + 2e−→Cu(s)E◦= 0.34V
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction can be written by summing the two half-reactions
and canceling out any electrons that appear on both sides of the reaction. The
overall reaction is:
2Ag++Cu2+ →2Ag(s) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell).
The standard cell potential is the difference in the standard electrode poten-
tials of the two half-reactions. We can calculate it using the formula:
32
E◦
cell =E◦
cathode −E◦
anode
Given E◦
cathode =E◦
Cu2+/Cu = 0.34 V and E◦
anode =E◦
Ag+/Ag = 0.80 V, we
can substitute these values into the formula:
E◦
cell = 0.34 V −0.80 V
E◦
cell =−0.46 V
Step 3: Verify the direction of the cell reaction.
Since E◦
cell is negative, the reaction, as written, is not spontaneous. To make
the reaction spontaneous, we need to reverse the reaction.
The spontaneous reaction should be:
2Ag(s) + Cu2+ →2Ag++Cu(s)
Step 4: Calculate the standard electrode potential for the new reaction.
The standard cell potential for the new reaction will be the negative of E◦
cell
for the original reaction:
E◦
cell (reversed) =−E◦
cell
E◦
cell (reversed) =−(−0.46 V)
E◦
cell (reversed) = 0.46 V
Therefore, the standard electrode potential for the reversed reaction is 0.46 V.
Question 35
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell contains a 0.10 M solution of AgCl, while the other half-cell contains
a 0.0010 M solution of AgCl. If the standard cell potential (E◦
cell) is 0.19 V at
25
°
C, calculate the cell potential of the concentration cell at 25
°
C.
Solution
Step 1: Write the half-reactions and the overall cell reaction. The half-reactions
involved are: 1. Reduction half-reaction: Ag+(aq) + e−→Ag(s) 2. Oxidation
half-reaction: Ag(s)→Ag+(aq) + e−
The overall cell reaction is the combination of these two half-reactions:
Ag(s) + Ag+(aq)→Ag+(aq) + Ag(s)
Step 2: Calculate E◦
cell using the Nernst equation. The Nernst equation is
given by:
Ecell =E◦
cell −0.0592
nlog(Q)
33
For the given reaction: n= 2 E= 0.46 V −0.0592
2log(0.05) E= 0.46 V −
0.0592
2×(−1.3) E= 0.46 V −0.0385 E= 0.4215 V
Therefore, the cell potential at 25
°
C using the Nernst equation is 0.4215 V.
Question 2
Question
Calculate the cell potential for a galvanic cell with the following half-reactions
at 25
°
C:
Ni2+(0.10M)+2e−→Ni(s)E◦=−0.25 V
MnO−
4(0.20M)+8H+(1.0M)+5e−→Mn2+(1.0M)+4H2O E◦= 1.23 V
Solution
Step 1: Write the overall cell reaction.
Ni2+ + MnO−
4+ 8H+→Ni + Mn2+ + 4H2O
Step 2: Calculate the standard cell potential (E◦) using the given half-
reactions.
E◦
cell =E◦
cathode −E◦
anode = 1.23 V −(−0.25 V) = 1.48 V
Step 3: Write the Nernst equation.
E=E◦−RT
nF ln Q
where: - Eis the cell potential, - E◦is the standard cell potential, - Ris the
gas constant (8.314 J mol−1K−1), - Tis the temperature in Kelvin (25
°
C = 298
K), - nis the number of moles of electrons transferred in the balanced equation
(5 in this case), - Fis the Faraday constant (96485 C mol−1), - Qis the reaction
quotient.
Step 4: Calculate the reaction quotient, Q.
Q=[Ni][Mn2+]
[MnO−
4][
Step 5: Substitute the known values into the Nernst equation and solve for
E.
E= 1.48 V −(8.314 J mol−1K−1)(298 K)
5(96485 C mol−1)ln5.0×10−3
E≈1.48 V −0.059 V ln5.0×10−3
E≈1.48 V −0.059 V ×(−5.30)
E≈1.48 V + 0.313 V = 1.79 V
Therefore, the cell potential for the galvanic cell at 25
°
C is 1.79 V.
2
Question 3
Question
Calculate the cell potential at 25
°
C for the following redox reaction:
Pb2+(aq) + 2Fe3+(aq)→Pb(s) + 2Fe2+(aq)
Given the following half-reactions and standard reduction potentials:
Pb2+(aq)+2e−→Pb(s)E◦=−0.13 V
Fe3+(aq) + e−→Fe2+(aq)E◦= 0.77 V
Also, the concentration of Pb2+ is 0.10 M, the concentration of Fe3+ is 0.50
M, and the concentration of Fe2+ is 1.00 M.
Solution
Step 1: Write the balanced overall redox reaction:
Pb2+(aq) + 2Fe3+(aq)→Pb(s) + 2Fe2+(aq)
Step 2: Calculate the cell potential using the Nernst equation: The Nernst
equation is given by:
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 in the balanced
redox reaction, - Fis the Faraday constant (96485 C/mol), - Qis the reaction
quotient.
Step 3: Calculate Q, the reaction quotient:
Q=[Fe2+]2
[Pb2+][Fe3+]2=(1.00)2
(0.10)(0.50)2= 40
Step 4: Calculate the cell potential using the Nernst equation: Plugging
values into the Nernst equation:
E= 0.77 V−(8.314)(298)
2(96485) ln(40) = 0.77 V−0.059 ln(40) = 0.77 V−0.59 ≈0.18 V
Therefore, the cell potential at 25
°
C for the given redox reaction is approxi-
mately 0.18 V.
3
Question 4
Question
The concentration of Cu2+ ions in a cell is 0.010 M while the concentration of
Fe2+ ions is 0.100 M. The cell potential at 25◦C is 0.23 V. Determine the value
of the equilibrium constant Kfor the following reaction:
Cu2+(aq) + Fe(s)→Cu(s) + Fe2+(aq)
Solution
Step 1: Write the half-reactions for the given redox reaction:
Oxidation: Cu2+(aq)+2e−→Cu(s) E◦
Cu = 0.34 V
Reduction: Fe2+(aq)+2e−→Fe(s) E◦
Fe =−0.44 V
Step 2: Write the overall cell reaction and find the standard cell potential,
E◦
cell:
Cu2+(aq) + Fe(s)→Cu(s) + Fe2+(aq)
E◦
cell =E◦
cathode −E◦
anode = 0.34 V −(−0.44 V) = 0.78 V
Step 3: Calculate the reaction quotient, Q, using the Nernst equation:
E=E◦−0.0592
nlog [Cu2+]
[Fe2+]
0.23 = 0.78−0.0592
2log 0.01
0.1= 0.78−0.0296 log 0.1=0.78−0.0296(−1) = 0.78+0.0296 = 0.8096 V
Step 4: Calculate the equilibrium constant, K, using the equation:
E=RT
nF ln K
0.8096 = (8.314 J/mol ·K)(298 K)
2(96485 C/mol) ln K
ln K= 0.8096 ·2(96485 C/mol)
(8.314 J/mol ·K)(298 K) = 0.1619
K=e0.1619 ≈1.175
Therefore, the equilibrium constant Kfor the given reaction is approximately
1.175.
4
Question 5
Question
A voltaic cell is constructed with a standard hydrogen electrode as the anode
and a copper electrode as the cathode. The initial concentrations of H+and
Cu2+ are both 0.10 M. The cell potential is measured to be 0.45 V at 25◦C.
Determine the equilibrium constant (K) for the following cell reaction:
2H++Cu2+ →H2+Cu
Solution
Step 1: Write the half-reactions for the anode and cathode: Anode (oxidation):
2H+(aq)+2e−→H2(g) Cathode (reduction): Cu2+(aq)+2e−→Cu(s)
Step 2: Determine the cell potential using the Nernst equation:
E=E◦−0.0592
nlog Q
where, E= cell potential = 0.45 V (given) E◦= standard cell potential n
= number of electrons transferred = 2 in this case Q= reaction quotient =
[H2][Cu]
[H+]2[Cu2+ ]
Step 3: Calculate the standard cell potential (E◦) using standard reduction
potentials: E◦=E◦
cathode −E◦
anode =E◦
reduction of Cu2+ −E◦
oxidation of H+
Step 4: Calculate the reaction quotient (Q) at equilibrium: Given initial
concentrations are both 0.10 M.
Step 5: Substitute the known values into the Nernst equation and solve for
K:
K=enE◦
0.0592
Now, you can proceed with solving for K.
Question 6
Question
Calculate the cell potential at 25
°
C for a cell in which the cathode reaction is:
Cr3+(aq) + 3e−→Cr(s) and the anode reaction is: MnO−
4(aq) + 8H+(aq) +
5e−→Mn2+(aq) + 4H2O(l). Given [Cr3+] = 0.10 M, [MnO−
4] = 0.20 M,
[H+] = 1.0 M.
Solution
Step 1: Write the overall cell reaction.
Cr3+(aq) + MnO−
4(aq) + 9H+(aq)→Cr(s) + Mn2+(aq) + 5H2O(l)
5
Step 2: Break down the overall reaction into the half-reactions. Cathode
half-reaction:
Cr3+(aq) + 3e−→Cr(s)E◦=−0.74 V
Anode half-reaction:
MnO−
4(aq) + 8H+(aq) + 5e−→Mn2+(aq) + 4H2O(l)E◦= 1.51 V
Step 3: Calculate the cell potential, E◦
cell, using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Cr3+][MnO−
4]8
[Cr][Mn2+][H+]9
where n is the number of electrons transferred in the balanced redox reaction.
Step 4: Calculate the cell potential.
Ecell = (1.51 V −(−0.74 V)) −0.0592
5log (0.10 M)(0.20 M)8
(1)(1) (1.0 M)9
Ecell = 2.25 V −0.0592
5log1.6×10−10
Ecell = 2.25 V - 0.0237 log1.6×10−10
Ecell ≈2.25 V
Question 7
Question
The concentration of lead (Pb2+) ions in a voltaic cell is measured to be 1.5×
10−3M. If the standard reduction potential for the Pb2+/Pb half-cell is −0.13
V, calculate the cell potential at 25
°
C. (Given: E◦
Ag+/Ag = 0.80 V.)
Solution
Step 1: Write the half-reaction for the Pb2+/Pb half-cell.
Pb2+ + 2e−→Pb
Step 2: Write the Nernst equation for the cell potential, E, which relates
the standard cell potential, E◦, the actual cell potential, the reaction quotient,
Q, and the gas constant, R, and temperature, T.
E=E◦−0.0592
nlog(Q)
Step 3: Determine the number of electrons transferred, n, in the half-
reaction. The number of electrons transferred is the coefficient of the electron
in the balanced half-reaction, which is 2.
6
Step 4: Calculate the reaction quotient, Q, using the concentrations of the
reactants and products.
Q=[Pb]
[Pb2+]=1
1.5×10−3= 666.67
Step 5: Substitute the given values and calculated Qinto the Nernst equation
to solve for the cell potential, E.
E=−0.13 V −0.0592
2log(666.67)
E=−0.13 V −0.0592
2×2.824
E=−0.13 V −0.1057
E=−0.2357 V
Therefore, the cell potential at 25
°
C is -0.2357 V.
Question 8
Question
An electrochemical cell consists of a standard hydrogen electrode (SHE) as the
anode and a silver-silver chloride electrode as the cathode. The half-reaction at
the cathode is AgCl(s) + e−−> Ag(s)+Cl−(aq)withastandardreductionpotentialof0.22V.T hehalf −
reactionattheanodeis2H+(aq)+2e−−> H2(g)withastandardreductionpotentialof0.00V.If theconcentrationsofthemetalionsinsolutionare[Ag+] =
1.0Mand[Cl−]=1.0M, andthepHofthesolutionis2.0, calculatethecellpotentialat25C.Given :R
= 8.314 J/(mol ·K), F= 96485 C/mol.
Solution
Step 1: Write the balanced overall reaction for the cell. The overall cell reaction
can be obtained by summing the half-reactions for the anode and cathode. The
half-reaction at the anode (oxidation) will be multiplied by 2 to balance the
electrons in the overall reaction. The overall reaction is:
2H+2H+2H+2H+(aq) + 2AgCl(s)→2Ag(s)+2
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the standard reduction potentials for the half-
reactions:
E◦
cell =E◦
cathode −E◦
anode = 0.22 V −0.00 V = 0.22 V
Step 3: Determine the concentrations of ions involved in the cell reaction.
Given:
[Ag+]=1.0 M
7
[Cl−]=1.0 M
pH = 2.0
The pH of the solution indicates that the concentration of H+ions is 10−2M.
In this case, the concentrations of the ions in the cell reaction are:
[H+] = 10−2M
[Ag+]=1.0 M
[Cl−]=1.0 M
Step 4: Calculate the non-standard cell potential (Ecell). The Nernst equa-
tion can be used to calculate the cell potential under non-standard conditions:
Ecell =E◦
cell −0.0592
nlog [Ag+]2[H+]2
[AgCl]2
For this reaction, n= 2 (from the balanced overall reaction), and the concen-
tration of solid AgCl does not affect the cell potential. Thus, the cell potential
can be calculated as:
Ecell = 0.22 V −0.0592
2log (1.0 M)2×(10−2M)2
(1.0 M)2
Ecell = 0.22 V −0.0298 log10−6
Ecell = 0.22 V + 0.0298 ×6
Ecell = 0.394 V
Therefore, the cell potential at 25
°
C is 0.394 V.
Question 9
Question
A voltaic cell is constructed with a standard hydrogen electrode (SHE) as the
anode and an unknown metal electrode as the cathode. Given that the standard
reduction potential for the unknown metal electrode is −0.20 V, and the [H+]
concentration is 0.10 M, calculate the cell potential at 25◦C. (Use R= 8.31
J/(mol·K), T= 298 K, and F= 96485 C/mol)
8
Solution
Step 1: Write the half-reaction for the reduction occurring at the unknown
metal electrode: The reduction half-reaction for the unknown metal electrode
is:
Unknown metal2+(aq)+2e−→Unknown metal(s)
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potential: The standard cell potential, E◦
cell, can be calculated using
the Nernst equation:
E◦
cell =E◦
cathode −E◦
anode
Given that the standard reduction potential for the unknown metal electrode is
−0.20 V, we have:
E◦
anode =E◦
SHE = 0 V
Thus,
E◦
cell =−0.20 V −0 V = −0.20 V
Step 3: Calculate the cell potential at non-standard conditions: The Nernst
equation is given by:
Ecell =E◦
cell −RT
nF ln(Q)
where Qis the reaction quotient and can be calculated using the concentrations
of products and reactants.
Step 4: Calculate Qfor the cell reaction: Since the unknown metal is in a
2+ oxidation state and the [H+] concentration is 0.10 M, the reaction quotient
Qis:
Q= (0.10)2= 0.01
Step 5: Plug in the values to calculate the cell potential at 25◦C: Substitute
the given values into the Nernst equation:
Ecell =−0.20 V −(8.31 J/(mol ·K)) ×298 K
2×96485 C/mol ln(0.01)
Step 6: Calculate Ecell:
Ecell =−0.20 V −2480.38
192970 ln(0.01)
Ecell =−0.20 V −0.012857
−4.605
Ecell =−0.20 V + 0.002794
Ecell ≈ −0.197 V
9
Question 10
Question
A reaction is taking place in an electrochemical cell where the following infor-
mation is known:
E◦
cell = 0.75 V
T= 298 K
n= 2
Calculate the cell potential at 25 degrees Celsius (298 K) for the same reaction
when the concentration of one of the ions is ten times greater than the initial
concentration.
Solution
Step 1: Calculate the change in cell potential using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Red]final
[Ox]final
Step 2: Since one ion concentration is ten times greater, we can represent
the new concentrations as:
[Red]final
[Ox]final
=10[Red]initial
[Ox]initial
Step 3: Substitute the given values into the Nernst equation:
Ecell = 0.75 V −0.0592
2log 10[Red]initial
[Ox]initial
Step 4: Simplify and solve for the new cell potential:
Ecell = 0.75 V −0.0296 log(10)
Step 5: Calculate the logarithm:
Ecell = 0.75 V - 0.0296 ×1V - 0.0296 ×1V - 0.0296 ×1V - 0.0296 ×1
Step 6: Final result:
Ecell = 0.7204 V
Question 11
Question
A concentration cell is set up where one half-cell has a silver electrode in a 0.10
MAgNO3solution and the other half-cell has a silver electrode in a 0.50 M
AgNO3solution. Calculate the cell potential at 25
°
C. The standard reduction
potential for the Ag+/Ag half-cell is 0.80 V.
10
Solution
Step 1: Write the half-reactions for the two half-cells. The half-cell reactions
for the silver electrode in the two solutions are:
Anode: Ag(s)→Ag+(aq) + e−
Cathode: Ag+(aq) + e−→Ag(s)
Step 2: Calculate the cell potential at standard conditions. The standard
cell potential E◦
cell can be calculated by the difference in standard reduction
potentials for the two half-cells:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.80V−0.80V
E◦
cell = 0V
Step 3: Determine the reaction quotient, Q. The Nernst equation relates cell
potential to reaction quotient Q:
Ecell =E◦
cell −0.0592
nlog Q
where n is the number of moles of electrons transferred in the balanced equation.
Since the reaction quotient Q is the ratio of the concentrations of the products
to the concentrations of the reactants, and the reaction is at equilibrium, we
have:
Q=[Ag+]0.50M
[Ag+]0.10M
= 5
Step 4: Calculate the cell potential. Substitute the values into the Nernst
equation:
Ecell = 0V−0.0592
1log 5
Ecell = 0V−0.086V
Ecell =−0.086V
Therefore, the cell potential at 25
°
C for the concentration cell with the two
different silver electrode solutions is -0.086 V.
Question 12
Question
For a galvanic cell using the half-reactions below, calculate the cell potential at
25
°
C when the concentrations of [F e3+]=0.10 Mand [F e2+]=1.0M.
Cathode: F e3+(aq)+3e−→F e2+(aq)E◦= 0.771 V
Anode: Sn2+(aq)→Sn4+(aq)+2e−E◦= 0.150 V
11
Solution
Step 1: Write the overall cell reaction by reversing the anode reaction and
adding the cathode reaction:
Overall cell reaction: Sn4+(aq) + F e2+(aq)→Sn2+(aq) + F e3+(aq)
Step 2: Determine E◦
cell by using the standard cell potential formula:
E◦
cell =E◦
cathode −E◦
anode
E◦
cell = 0.771 V−0.150 V= 0.621 V
Step 3: Write the Nernst equation for this cell:
E=E◦−0.0592
nlog [Fe3+]
[Sn4+][Fe2+]3
Step 4: Calculate the cell potential (E) using the Nernst equation with the
given concentrations:
E= 0.621 V−0.0592
2log 0.10
[Sn4+](1.0)3
E= 0.621 V−0.0296 log 0.10
[Sn4+]
Step 5: Determine the cell potential at 25
°
C by simplifying the expression
and substituting the known values for [F e3+]=0.10 Mand [F e2+] = 1.0M:
E= 0.621 V−0.0296 log 0.10
[Sn4+]
Question 13
Question
Calculate the cell potential at 25
°
C for a hydrogen-oxygen fuel cell in which
the hydrogen electrode is at a pressure of 1.0 atm and the oxygen electrode
is at a pressure of 0.10 atm. The standard cell potential for this reaction is
1.23 V. Given that the Faraday constant is 96485 C/mol, the gas constant is
8.314 J/(mol ·K), and the temperature is 298 K.
Solution
Step 1: Write the balanced half-reactions and the overall redox reaction. The
half-reactions for a hydrogen-oxygen fuel cell are:
Anode: 2H2(g)→4H+(aq)+4e−
12
Cathode: O2(g) + 4H+(aq)+4e−→2H2O(l)
The overall redox reaction is the sum of these two half-reactions:
2H2(g) + O2(g)→2H2O(l)
Step 2: Calculate the standard cell potential using the Nernst equation. The
Nernst equation relates the standard cell potential (E◦), the cell potential under
nonstandard conditions (E), the reaction quotient (Q), the Faraday constant
(F), and the temperature (T):
E=E◦−RT
nF ln(Q)
where nis the number of moles of electrons transferred in the cell reaction, R
is the gas constant, and Qis the reaction quotient.
Given: - Standard cell potential (E◦) = 1.23 V - Pressure of hydrogen = 1.0
atm - Pressure of oxygen = 0.10 atm - Temperature (T) = 298 K - Gas constant
(R) = 8.314 J/(mol ·K) - Faraday constant (F) = 96485 C/mol
Step 3: Calculate the reaction quotient (Q). Since the partial pressures of
the gases are given, we can use the ideal gas law to calculate the concentration
of H+:
0.10 atm = 1.0atm
1.0atm2
= [H+]
Since Qis the product of the concentrations of the products divided by the
product of the concentrations of the reactants, and H2O(l) does not appear in
the expression for Q, the value of Qis (1)2= 1.
Step 4: Substitute the values of E◦,R,T,n,F, and Qinto the Nernst
equation to calculate the cell potential (E) at 25
°
C.
E= 1.23 V −(8.314 J/(mol ·K))(298 K)
4(96485 C/mol) ln(1)
Step 5: Solve for Eto find the cell potential for the hydrogen-oxygen fuel
cell at 25
°
C.
Question 14
Question
Calculate the cell potential for the following reaction at 25
°
C:
Al(s) + Fe3+(aq)→Al3+(aq) + Fe(s)
Given the standard reduction potentials:
E◦
cell(Al3+/Al) = −1.66 V
E◦
cell(Fe3+/Fe) = 0.77 V
and the concentration of Fe3+ is 0.010 M.
13
Solution
Step 1: Write the overall cell reaction:
Al(s) + Fe3+(aq)→Al3+(aq) + Fe(s)
Step 2: Identify the oxidation and reduction half-reactions: Oxidation half-
reaction: Al(s)→Al3+(aq)+3e−
Reduction half-reaction: Fe3+(aq)+3e−→Fe(s)
Step 3: Calculate the standard cell potential, E◦
cell, using the formula:
E◦
cell =E◦
cell,reduction −E◦
cell,oxidation
E◦
cell = 0.77 V −(−1.66 V) = 2.43 V
Step 4: Write the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
K
where Qis the reaction quotient, and Kis the equilibrium constant.
Step 5: Calculate Qusing the concentrations provided:
Q=[Fe](Al3+)
[Fe3+]
Q=1·1
0.010 = 100
Step 6: Calculate the cell potential at 25
°
C using the Nernst equation:
Ecell = 2.43 V −0.0592
6log(100) = 2.43 V −0.0987 = 2.33 V
Therefore, the cell potential for the given reaction at 25
°
C is 2.33 V.
Question 15
Question
For a galvanic cell with a silver electrode in a 0.10 M Ag+solution and a copper
electrode in a 0.50 M Cu2+ solution, the cell potential is measured to be 0.90 V
at 25
°
C. Calculate the standard cell potential, E, for this cell at 25
°
C. (Given:
Ecell for the Ag+—Ag and Cu2+—Cu cells are 0.80 V and 0.34 V, respectively)
14
Solution
Step 1: Write the half-reactions for each electrode.
The half-reaction for the silver electrode is: Ag+(aq) + e−→Ag(s)
The half-reaction for the copper electrode is: Cu2+(aq) + 2e−→Cu(s)
Step 2: Write the overall cell reaction.
Overall: Ag+(aq)+ Cu2+(aq) →Ag(s) + Cu(s)
Step 3: Calculate Ecell using the Nernst equation.
The Nernst equation is: E=E−0.0592
nlog Q
K
The number of electrons transferred, n, for this cell is 1.
Qis the reaction quotient, which for this cell is [Ag][Cu2+ ]
[Ag+][Cu] =0.10×0.50
1= 0.05
Substitute the given values into the Nernst equation:
0.90 = E−0.0592
1log(0.05)
Step 4: Solve for E.
0.90 = E−0.0592 log(0.05)
E= 0.90 + 0.0592 log(0.05)
E= 0.90 + 0.0592 ×(−1.30)
E≈0.818 V
Therefore, the standard cell potential for this cell at 25
°
C is approximately
0.818 V.
Question 16
Question
A zinc electrode is immersed in a solution containing Zn2+ ions at a concentra-
tion of 0.10 M. If the standard reduction potential for the Zn2+ + 2e−→Zn
half-reaction is -0.76 V, calculate the cell potential at 25
°
C when the concentra-
tion of Zn2+ ions is actually 0.20 M.
Solution
Step 1: Write the half-reaction and Nernst equation. The half-reaction is Zn2+ +
2e−→Zn. The Nernst equation is given by:
E=E◦−0.0592
nlog Q
K
where: E= cell potential at non-standard conditions E◦= standard cell po-
tential n= number of electrons transferred in the balanced half-reaction Q=
reaction quotient K= equilibrium constant
Step 2: Determine the values given in the question. E◦=−0.76 V T= 298
K [Zn2+] = 0.10 M [Zn2+]actual = 0.20 M
Step 3: Calculate the equilibrium constant. The equilibrium constant, K, is
related to the standard cell potential by the equation:
E◦=0.0592
nlog K
15
K= 10n·(E◦/0.0592)
For the given half-reaction, n= 2:
K= 102·(−0.76/0.0592)
Step 4: Calculate Qfor the non-standard conditions.
Q=[Zn2+]actual
[Zn2+]◦=0.20
0.10
Step 5: Substitute the values into the Nernst equation and calculate the cell
potential.
E=−0.76 −0.0592
2log 0.20
0.10
Step 6: Perform the calculations to find the cell potential.
Question 17
Question
Calculate the cell potential for the following electrochemical cell at 25
°
C:
Ag|AgCl(0.10 M) || Cl−(0.0020 M)|Ag
Given:
E◦
AgCl(s)/Ag+(aq)= 0.22 V
E◦
Ag+(aq)/Ag(s)= 0.80 V
R= 8.314 J ·mol−1·K−1
F= 96485 C ·mol−1
Solution
Step 1: Write the overall cell reaction. The overall cell reaction can be written
as:
AgCl(s)+e−→Ag(s) + Cl−(aq)
Step 2: Calculate the standard cell potential, E◦
cell, using the given standard
reduction potentials. The standard cell potential can be calculated using the
formula:
E◦
cell =E◦
cathode −E◦
anode
Given that the cathode is Ag+(aq) + e−→Ag(s) and the anode is AgCl(s) +
e−→Ag(s) + Cl−(aq), we have:
E◦
cell =E◦
Ag+(aq)/Ag(s)−E◦
AgCl(s)/Ag+(aq)
16
E◦
cell = 0.80 V −0.22 V = 0.58 V
Step 3: Calculate the reaction quotient, Q, for the cell at non-standard
conditions.
Q=[Ag+]final ·[Cl−]final
[AgCl]final
Given that [Ag+]final = 0.0020 M, [Cl−]final = 0.0020 M, and [AgCl]final =
0.10 M, we have:
Q=(0.0020)(0.0020)
0.10 = 4.0×10−5
Question 18
Question
A voltaic cell is constructed with a silver-silver chloride electrode (half-cell re-
action: AgCl(s) + e−−→ Ag(s) + Cl−(aq))andazincelectrode(Zn(s)−→ Zn2+
(aq)+2e−).T heinitialconcentrationsof Ag+, Zn2+, andCl−ionsare1.0M, 0.1M, and0.01M, respectively.Ifthestandardreductionpotentialof theAgCl(s)electrodeis0.222V andthestandardreductionpotentialoftheZn(s)electrodeis−
0.76V, whatisthecellpotentialat25C?
Solution
Step 1: Write the half-reactions at each electrode. The half-reactions are:
AgCl(s) + e−−→ Ag(s) + Cl−(aq)withastandardreductionpotentialof0.222V(cathode)Zn(s)−→
Zn2+ (aq)+2e−withastandardreductionpotentialof −0.76V(anode)
Step 2: Calculate the cell potential using the Nernst equation. The cell
potential (Ecell) can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Ag+][Cl−]
Step 3: Determine the number of electrons transferred. Since AgCl has a
1:1 stoichiometry and Zn forms 2 electrons, the number of electrons transferred
(n) is 2.
Step 4: Substitute values into the Nernst equation.
Ecell = 0.222 V −0.0592
2log 0.1
1.0×0.01
Step 5: Solve for the cell potential.
Ecell = 0.222 V −0.0592
2log(10)
Ecell = 0.222 V −0.0296 ×1
Ecell = 0.1924 V
Therefore, the cell potential at 25
°
C is 0.1924 V.
17
Question 19
Question
A concentration cell is set up with two silver-silver chloride electrodes. One
half-cell has a silver electrode dipped in a 0.10 M Ag+solution, and the other
half-cell has a silver electrode dipped in a 0.0010 M Ag+solution. Calculate
the cell potential at 25
°
C for this concentration cell given that E◦
cell = 0.80 V
and F= 96,485 C/mol.
Solution
Step 1: Write the half-reactions:
AgCl(s) + e−→Ag(s) + Cl−(aq)
This half-reaction occurs at both electrodes.
Step 2: Write the Nernst equation for the concentration cell:
Ecell =E◦
cell −0.0592
nlog [Ag+]low
[Ag+]high
where nis the number of electrons transferred in the balanced overall cell reac-
tion.
Step 3: Calculate the number of electrons transferred, n: In this case, the
overall cell reaction is:
Ag+
(0.10 M) + e−→Ag+
(0.0010 M)
which involves the transfer of 1 electron.
Step 4: Solve for Ecell using the Nernst equation:
Ecell = 0.80 V −0.0592
1log 0.0010
0.10 = 0.80 −0.057 = 0.743 V
Therefore, the cell potential at 25
°
C for this concentration cell is 0.743 V.
Question 20
Question
A galvanic cell consists of a standard hydrogen electrode (SHE) connected to
a copper strip submerged in a 0.10 M Cu2+ solution. Given that the standard
reduction potential for the Cu2+/Cu half-reaction is +0.34 V, calculate the cell
potential when the concentration of Cu2+ is reduced to 0.050 M at 25◦C. (Hint:
Use the Nernst equation)
18
Solution
Step 1: Write the half-reactions for the system: The overall cell reaction can be
broken down into two half-reactions:
Anode (oxidation): Cu(s) →Cu2+(aq) + 2e−
Cathode (reduction): 2H+(aq) + 2e−→H2(g)
Step 2: Calculate the standard cell potential (E◦) using the given standard
reduction potential for the Cu2+/Cu half-reaction (+0.34 V):
E◦
cell =E◦
cathode −E◦
anode = 0 −(+0.34) = −0.34 V
Step 3: Calculate the reaction quotient (Q) using the concentrations given:
Q=[Cu2+]
[H+]2=0.050
12= 0.050
Step 4: Calculate the Nernst equation:
Ecell =E◦
cell −0.0592
nlog(Q)
where n is the number of moles of electrons transferred in the balanced half-
reaction. Since 2 moles of electrons are transferred in the reduction of Cu2+, n
= 2.
Step 5: Substitute the values into the Nernst equation:
Ecell =−0.34 −0.0592
2log(0.050)
Ecell =−0.34 −0.0296 log(0.050)
Ecell =−0.34 −0.0296 ×(−1.30)
Ecell =−0.34 + 0.03848
Ecell =−0.3015 V
Therefore, the cell potential when the concentration of Cu2+ is reduced to
0.050 M at 25◦C is -0.3015 V.
Question 21
Question
A concentration cell is set up using two half-cells, one with a silver electrode
in a 0.10 M AgNO3solution and the other with a silver electrode in a 1.0 M
AgNO3solution. If the standard reduction potential of the Ag+/Ag half-cell is
0.80 V at 25
°
C, calculate the cell potential at 25
°
C using the Nernst equation.
19
Solution
Step 1: Write the half-cell reactions and the overall cell reaction. The half-cell
reactions are given by:
Ag+(aq) + e−→Ag(s)
The cell reaction is:
Ag+(0.10 M, Ag)→Ag+(1.0M, Ag)
Step 2: Calculate the standard cell potential (E◦
cell). The standard cell
potential can be calculated using the difference in standard reduction potentials:
E◦
cell =E◦
cathode −E◦
anode
Given that E◦
cell =E◦
cathode −E◦
anode = 0.80 V−0.80 V= 0 V
Step 3: Write the Nernst equation. The Nernst equation is given by:
Ecell =E◦
cell −0.0592
nlog [Ag+]right
[Ag+]lef t
where: - Ecell is the cell potential - E◦
cell is the standard cell potential - nis the
number of electrons transferred (1 in this case) - [Ag+]right is the concentration
of Ag+on the right side (1.0 M) - [Ag+]lef t is the concentration of Ag+on the
left side (0.10 M)
Step 4: Calculate the cell potential using the Nernst equation. Plugging in
the values:
Ecell = 0 V−0.0592
1log 1.0
0.10
Ecell =−0.0592 ×log(10)
Ecell =−0.0592 ×1
Ecell =−0.0592 V
Therefore, the cell potential at 25
°
C using the Nernst equation is -0.0592 V.
Question 22
Question
A redox reaction is carried out in a cell with a silver electrode. The half-reaction
occurring at the electrode is:
Ag++ e−→Ag(s)
Given that the standard reduction potential for this reaction is E◦= 0.80
V, calculate the cell potential under the following conditions: [Ag+]=1.0 M
and [Ag] = 0.1 M. (Assume T= 298 K)
20
Solution
Step 1: Write the Nernst equation, which relates the cell potential (E) to the
standard cell potential (E◦), the reaction quotient (Q), the gas constant (R),
the temperature (T), and the number of electrons transferred (n):
E=E◦−RT
nF ln(Q)
Step 2: Determine the reaction quotient Qusing the concentrations provided:
Since the reaction is:
Ag++ e−→Ag(s)
The reaction quotient is:
Q=[Ag]
[Ag+]
Given that [Ag] = 0.1 M and [Ag+] = 1.0 M, we have:
Q=0.1
1.0= 0.1
Step 3: Calculate the cell potential using the Nernst equation. Given that
E◦= 0.80 V, R= 8.314 J·mol−1·K−1,T= 298 K, n= 1 (1 electron transferred),
and Faraday’s constant F= 96485 C ·mol−1:
Plugging in the values:
E= 0.80 −(8.314)(298)
1(96485) ln(0.1)
Calculating the cell potential:
E= 0.80 −2476.472
96485 ln(0.1)
E≈0.80 −0.643 ln(0.1)
E≈0.80 −0.643(−2.303)
E≈1.47 V
Therefore, the cell potential under the given conditions is approximately 1.47
V.
21
Question 23
Question
A voltaic cell consists of a Ag—AgCl half-cell and a Pt—HCl half-cell. The
half-cell reactions are:
AgCl(s) + e−→Ag(s) + Cl−(aq)
2H2(g)+2Cl−(aq)→2HCl(aq)+2e−
If the concentration of H+ions in the HCl half-cell is 1.0×10−3M, what
is the cell potential of this voltaic cell at 25
°
C? Given E◦
Ag+/Ag = 0.80 V and
E◦
H+/H2= 0.00 V.
Solution
Step 1: Write the overall cell reaction and determine the cell potential. The
overall cell reaction is the sum of the two half-cell reactions. We’ll need to flip
the second half-cell reaction and multiply it by 1/2 to balance the electrons:
AgCl(s) + 2H+(aq)→Ag(s) + Cl−(aq)+H2(g)
The cell potential, Ecell, can be calculated using the Nernst equation:
Ecell =E◦
cell −0.0592
nlog Q
where Qis the reaction quotient and nis the number of electrons transferred
in the balanced cell reaction.
Step 2: Determine the reaction quotient, Q. Since we are given the con-
centrations of H+ions, we can use the Nernst equation to calculate the cell
potential. The reaction quotient, Q, can be calculated as follows:
Q=[Ag+][H+]
[Cl−]
Plugging in the given values:
Q=1.0×10−3×1.0
1.0= 1.0×10−3
Step 3: Calculate the cell potential, Ecell. Substitute the known values into
the Nernst equation:
Ecell =E◦
Ag+/Ag −0.0592
nlog Q
Ecell = 0.80 −0.0592
1log1.0×10−3
Ecell = 0.80 −0.0592 ×(−3) = 0.98 V
Therefore, the cell potential of this voltaic cell at 25
°
C is 0.98 V.
22
Question 24
Question
A voltaic cell is constructed with a standard hydrogen electrode (Pt — H+(aq, 1
M) —— H+(aq, xM) — MnO4(aq, yM), Mn2+(aq)). The cell potential is found
to be 1.00 V at 25
°
C. Given that E◦= 1.49 V, determine the concentrations of
H+and MnO4in the cell. Assume the cell reaction is:
MnO4(aq) + 8H+(aq) + 5e −→ Mn2+(aq) + 4H2O(l).
Solution
Step 1: Write the half-reaction for the standard hydrogen electrode:
2H+(aq)+2e−−→ H2(g)
Step 2: Calculate the cell potential at nonstandard conditions using the
Nernst equation:
E=E◦−0.0592
nlog [H+]2
PH2
where nis the number of moles of electrons transferred, PH2is the partial
pressure of hydrogen gas (1 atm for standard conditions), and Eis the cell
potential at nonstandard conditions.
Step 3: Given that E◦= 1.49 V, E= 1.00 V, and n= 2, solve for [H+]:
1.00 V = 1.49 V −0.0592
2log [H+]2
1
Step 4: Rearrange the equation and solve for [H+]:
−0.49 V = −0.0296 log[H+]2
16.55 = log[H+]2
[H+]2= 1016.55
[H+] = 108.275 = 7.49 ×108M
Step 5: Using the stoichiometry of the cell reaction, calculate the concentra-
tion of MnO4(y) in the cell: Since the stoichiometry is 1:8 for MnO4:H+, the
concentration of MnO4can be determined as:
8[H+] = 8(7.49 ×108) M = 5.99 ×109M
Therefore, the concentration of H+is 7.49 ×108M and the concentration of
MnO4is 5.99 ×109M.
23
Question 25
Question
A galvanic cell is constructed with a silver electrode in a 1.0 M Ag+solution and
a copper electrode in a 0.10 M Cu2+ solution. At 25
°
C, the standard reduction
potentials for the half-reactions are as follows:
Ag++ e−→Ag, E◦= 0.80 V
Cu2+ + 2e−→Cu, E◦= 0.34 V
Calculate the cell potential at 25
°
C, given that the number of electrons trans-
ferred in the balanced equation for the cell reaction is 2.
Solution
Step 1: Write the cell reaction by combining the two half-reactions. The cell
reaction is obtained by summing the reduction half-reactions after multiplying
the equations by the necessary coefficients to balance the electrons. Since 2
electrons are involved, we need to multiply the first half-reaction by 2 before
adding them together:
2 Ag++ Cu2+ →2 Ag + Cu
Step 2: Determine the standard cell potential (E◦
cell). The standard cell
potential is calculated by subtracting the standard reduction potentials of the
two half-reactions involved in the cell reaction.
E◦
cell =E◦
cathode −E◦
anode
E◦
cell =E◦
Ag −E◦
Cu2+
E◦
cell = 0.80 V −0.34 V = 0.46 V
Step 3: Use the Nernst equation to find the cell potential at 25
°
C. The
Nernst equation relates the cell potential under non-standard conditions to the
cell potential under standard conditions:
E=E◦−0.0592
nlog(Q)
where - Eis the cell potential under non-standard conditions, - E◦is the cell
potential under standard conditions, - nis the number of moles of electrons
transferred in the balanced equation, and - Qis the reaction quotient, which is
the ratio of products to reactants at any point during the reaction.
Step 4: Calculate the concentration of Ag+and Cu2+ ions to determine
Q. Since the reaction is at equilibrium, Qis equal to the equilibrium constant,
K. To calculate Q, we need to consider the concentration of the products and
reactants in the cell reaction.
24
Q=[Ag]2
[Cu2+]
Given that the initial concentration of Cu2+ is 0.10 M and the initial concen-
tration of Ag+is 1.0 M, we can substitute these values into the equation.
Step 5: Plug in the values to find the cell potential under non-standard
conditions. Substitute the known values into the Nernst equation to find the
cell potential under the given conditions.
E= 0.46 V −0.0592
2log(Q)
E= 0.46 V −0.0592
2log 1.0 M2
0.10 M
Step 6: Calculate the cell potential at 25
°
C. Plug in the values to find the
cell potential under the given conditions.
E= 0.46 V −0.0592
2log(10) = 0.46 V −0.0296 = 0.43 V
Therefore, the cell potential at 25
°
C is 0.43 V.
Question 26
Question
A voltaic cell is constructed with a zinc electrode in a 1.0 M solution of Zn2+
ions, and a hydrogen electrode in an acidic solution having a concentration of
[H+]=0.10 M. Calculate the cell potential at 25
°
C given that the standard
reduction potential for the Zn2+/Zn half-reaction is −0.76 V and the standard
reduction potential for the H+/H2half-reaction is 0.00 V.
Solution
Step 1: Write the two half-reactions involved in the cell: Zn2+ + 2e−→Zn
2H++ 2e−→H2
Step 2: Calculate the cell potential at standard conditions, E◦
cell:E◦
cell =
E◦
cathode −E◦
anode E◦
cell = 0.00 V −(−0.76 V) E◦
cell = 0.76 V
Step 3: Calculate the reaction quotient Q:Q=[Zn2+]
[H+]=1.0
0.10 = 10
Step 4: Calculate the Nernst equation to find the cell potential at 25
°
C:
Ecell =E◦
cell −0.0592 V
nlog(Q)Ecell = 0.76 V −0.0592 V
2log(10) Ecell = 0.76 V −
0.0296 V ×1Ecell = 0.7304 V
25
Question 27
Question
A voltaic cell consists of a copper electrode in a 1.0 M copper(II) sulfate solution
and a silver electrode in a 0.10 M silver nitrate solution at 25
°
C. Calculate the
cell potential at this temperature. Given E◦
Cu2+/Cu = 0.34 V and E◦
Ag+/Ag =
0.80 V.
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction can be written as:
Cu2+(aq) + 2e−→Cu(s)
2Ag+(aq) + 2e−→2Ag(s)
Combining these two half-reactions:
Cu2+(aq) + 2Ag(s)→Cu(s) + 2Ag+(aq)
Step 2: Calculate the cell potential using the Nernst equation.
The Nernst equation is given by:
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 Faraday’s constant (96485 C/mol) - Qis the reaction quotient, which can be
calculated using the concentrations of reactants and products.
First, calculate Qbased on the concentrations given:
Q=[Cu2+]
[Ag+]2=1.0
(0.10)2= 100
Step 3: Substitute the values into the Nernst equation and solve for E:
E= 0.34 −(8.314)(298)
(2)(96485) ln(100)
E= 0.34 −2465.572
192970 ln(100)
E= 0.34 −0.0127 ln(100)
E0.23 V
Therefore, the cell potential at 25
°
C is approximately 0.23 V.
26
Question 28
Question
An electrochemical cell consists of a zinc electrode in a 1.0 M Zn2+ solution and
a copper electrode in a 1.0 M Cu2+ solution. If the standard reduction potential
for Zn2+ + 2e−→Zn is -0.76 V and for Cu2+ + 2e−→Cu is +0.34 V, calculate
the cell potential at 25
°
C.
Given: R= 8.314 J/(mol·K), T= 298 K, F= 96,485 C/mol
Solution
Step 1: Write the half-cell reactions:
Zn2+ + 2e−→Zn E◦
cell 1 =−0.76 V
Cu2+ + 2e−→Cu E◦
cell 2 = +0.34 V
Step 2: Calculate the cell potential at standard conditions:
E◦
cell =E◦
cell 2 −E◦
cell 1 = 0.34 V −(−0.76 V) = 1.10 V
Step 3: Write down the Nernst equation:
Ecell =E◦
cell −0.0592
nlog [Zn2+]
[Cu2+]
Step 4: Calculate the cell potential using the Nernst equation:
Ecell = 1.10 V −0.0592
2log 1.0
1.0
= 1.10 V −0
= 1.10 V
Therefore, the cell potential at 25
°
C is 1.10 V.
Question 29
Question
A voltaic cell consists of a standard hydrogen electrode (SHE) and a silver
electrode in a solution that is 0.10 M in Ag+ions. The cell potential at 25
°
C
is found to be 0.80 V. Calculate the concentration of Ag+ions that would be
necessary to achieve a cell potential of 1.00 V.
27
Solution
The Nernst equation relates the cell potential of a voltaic cell to the concentra-
tions of the reactants and products involved. It is given by:
E=E◦−0.0592
nlog Q
where: - Eis the cell potential at any given point, - E◦is the standard cell
potential, - nis the number of moles of electrons transferred in the cell reaction,
and - Qis the reaction quotient.
Step 1: Calculate the standard cell potential (E◦)
The standard cell potential for the reaction between the silver electrode and
the SHE can be calculated as:
E◦=ESHE −EAg+/Ag
Given:
ESHE = 0 V
EAg+/Ag = 0.80 V
Therefore:
E◦= 0 −0.80 V = −0.80 V
Step 2: Calculate the reaction quotient (Q)
The cell reaction involves the reduction of Ag+ions to Ag, and since the cell
potential is 1.00 V when the cell is at equilibrium, we can write:
E=E◦−0.0592
nlog Q
Substitute the given values:
1.00 = −0.80 −0.0592
1log Q
log Q=−0.20
Q= 10−0.20 = 0.63
Step 3: Calculate the concentration of Ag+ions
The reaction quotient is given by:
Q=[Ag]
[Ag+]
Given:
[Ag+] = 0.10 M
Substitute the known values into the equation:
0.63 = [Ag]
0.10
[Ag] = 0.063 M
Therefore, the concentration of Ag+ions required to achieve a cell potential
of 1.00 V is 0.063 M.
28
Question 30
Question
A cell is constructed with a silver electrode dipping into a 0.010 M AgNO3 solu-
tion and a cadmium electrode dipping into a 0.10 M Cd(NO3)2 solution. Given
that the standard reduction potentials are E◦
Ag+/Ag = 0.80 V and E◦
Cd2+/Cd =
−0.40 V, calculate the cell potential at 25
°
C.
Solution
Step 1: Write the overall cell reaction.
Cd2+ + 2e−→Cd E◦
cell =−0.40 V
2Ag++ 2e−→2Ag E◦
cell = 0.80 V
Cd2+ + 2Ag →Cd + 2Ag+
Step 2: Calculate the standard cell potential (E◦
cell).
E◦
cell =E◦
Ag+/Ag −E◦
Cd2+/Cd
E◦
cell = 0.80 V −(−0.40) V = 1.20 V
Step 3: Calculate the reaction quotient (Q) at 25
°
C.
Q=[Ag+]2
[Cd2+]
Given [Ag+]=0.010 M and [Cd2+] = 0.10 M, we have
Q=(0.010)2
0.10 = 0.001
Step 4: Calculate the cell potential (Ecell) using the Nernst equation.
Ecell =E◦
cell −0.0592
nlog Q
Since the reaction involves the transfer of 2 electrons:
n= 2
Ecell = 1.20 V −0.0592
2log 0.001
Ecell = 1.20 V + 0.0592 ×3×10 = 1.20 V - 0.1776 = 1.0224 V
Therefore, the cell potential at 25
°
C is 1.0224 V.
29
Question 31
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell contains a solution with [Ag+] = 1.0×10−5M, while the other half-cell
contains a solution with [Ag+]=1.0×10−2M. Calculate the cell potential at
25
°
C. (Given: E◦
cell = 0.8 V)
Solution
Step 1: Write the Nernst equation for the cell potential of the concentration
cell:
Ecell =E◦
cell −0.0592
nlog [Ag+]left
[Ag+]right
Where Ecell is the cell potential, E◦
cell is the standard cell potential, nis the
number of moles of electrons transferred, [Ag+]left is the concentration in the
left half-cell, and [Ag+]right is the concentration in the right half-cell.
Step 2: Calculate the cell potential using the given values:
Ecell = 0.8−0.0592
1log 1.0×10−5
1.0×10−2
Ecell = 0.8−0.0592 log 1.0×10−5
1.0×10−2
Ecell = 0.8−0.0592 log 10−3
Ecell = 0.8−0.0592 ×(−3)
Ecell = 0.8+0.1776
Ecell = 0.9776 V
Therefore, the cell potential at 25
°
C for the concentration cell using two
silver-silver chloride electrodes with different concentrations is 0.9776 V.
Question 32
Question
A concentration cell is set up with two half-cells, each containing a solution of
iron(II) ions (Fe2+). One half-cell has a standard Fe2+ concentration of 1.0 M,
while the other half-cell has a Fe2+ concentration of 0.10 M. Calculate the cell
potential at 25◦C for this concentration cell. The standard reduction potential
for the Fe2+/Fe half-cell is +0.77 V.
30
Solution
Step 1: Write the balanced redox half-reaction for the given cell. The half-
reaction is: Fe2+ →Fe + 2e−
Step 2: Write the Nernst equation for the cell potential. The Nernst equation
is:
E=E◦−0.0592
nlog [Fe2+]
[Fe]
where: E= cell potential E◦= standard cell potential n= number of electrons
transferred in the balanced redox reaction [Fe2+] = concentration of Fe2+ in the
cathode (higher concentration) [Fe] = concentration of Fe in the anode (lower
concentration)
Step 3: Find the number of electrons transferred in the balanced half-
reaction. From the balanced half-reaction, we see that 1 mol of Fe2+ is reduced
to form 1 mol of Fe. Therefore, the number of electrons transferred (n) is 2.
Step 4: Substitute known values into the Nernst equation.
E= 0.77 V −0.0592
2log 0.10
1.0
= 0.77 V −0.0296 log(0.10)
= 0.77 V −0.0296 ×(−1)
= 0.77 V + 0.0296
= 0.7996 V
Therefore, the cell potential at 25◦C for this concentration cell is 0.7996 V.
Question 33
Question
A galvanic cell consists of a standard hydrogen electrode (H2(g)|H+(aq)) and
a copper electrode dipped in a solution of Cu2+ ions with a concentration of
0.10 M. The standard reduction potential of Cu2+ →Cu is 0.34 V and the
standard reduction potential of H+|H2(g) is 0.00 V. Calculate the cell potential
at a temperature where E◦is 0.25 V.
Solution
Step 1: Write the half-reactions and the overall cell reaction. The half-reactions
are: Cu2+ +2e−→Cu (Standard reduction potential: 0.34 V) 2H++2e−→H2
(Standard reduction potential: 0.00 V)
The overall cell reaction is: Cu2+ + 2H+→Cu +H2
Step 2: Calculate the cell potential at standard conditions using the standard
reduction potentials.
E◦
cell =E◦
cathode −E◦
anode
31
E◦
cell = 0.00 V −0.34 V
E◦
cell =−0.34 V
Step 3: Calculate the reaction quotient, Q, and the cell potential, E, under
non-standard conditions using the Nernst equation. The reaction quotient, Q,
is given by:
Q=[Cu][H2]
[Cu2+][H+]= 1
The Nernst equation is:
E=E◦
cell −0.0592
nlog(Q)
where n is the number of electrons transferred in the balanced cell reaction.
Step 4: Calculate the cell potential, E, at the given temperature.
E=−0.34 V −0.0592
2log(1)
E=−0.34 V −0 V
E=−0.34 V
Therefore, at a temperature where E◦is 0.25 V, the cell potential is -0.34
V.
Question 34
Question
Calculate the standard electrode potential (E◦) for the reaction below given the
following information:
2Ag++ 2e−→2Ag(s)E◦= 0.80V
Cu2+ + 2e−→Cu(s)E◦= 0.34V
Solution
Step 1: Write the overall cell reaction.
The overall cell reaction can be written by summing the two half-reactions
and canceling out any electrons that appear on both sides of the reaction. The
overall reaction is:
2Ag++Cu2+ →2Ag(s) + Cu(s)
Step 2: Calculate the standard cell potential (E◦
cell).
The standard cell potential is the difference in the standard electrode poten-
tials of the two half-reactions. We can calculate it using the formula:
32
E◦
cell =E◦
cathode −E◦
anode
Given E◦
cathode =E◦
Cu2+/Cu = 0.34 V and E◦
anode =E◦
Ag+/Ag = 0.80 V, we
can substitute these values into the formula:
E◦
cell = 0.34 V −0.80 V
E◦
cell =−0.46 V
Step 3: Verify the direction of the cell reaction.
Since E◦
cell is negative, the reaction, as written, is not spontaneous. To make
the reaction spontaneous, we need to reverse the reaction.
The spontaneous reaction should be:
2Ag(s) + Cu2+ →2Ag++Cu(s)
Step 4: Calculate the standard electrode potential for the new reaction.
The standard cell potential for the new reaction will be the negative of E◦
cell
for the original reaction:
E◦
cell (reversed) =−E◦
cell
E◦
cell (reversed) =−(−0.46 V)
E◦
cell (reversed) = 0.46 V
Therefore, the standard electrode potential for the reversed reaction is 0.46 V.
Question 35
Question
A concentration cell is set up using two silver-silver chloride electrodes. One
half-cell contains a 0.10 M solution of AgCl, while the other half-cell contains
a 0.0010 M solution of AgCl. If the standard cell potential (E◦
cell) is 0.19 V at
25
°
C, calculate the cell potential of the concentration cell at 25
°
C.
Solution
Step 1: Write the half-reactions and the overall cell reaction. The half-reactions
involved are: 1. Reduction half-reaction: Ag+(aq) + e−→Ag(s) 2. Oxidation
half-reaction: Ag(s)→Ag+(aq) + e−
The overall cell reaction is the combination of these two half-reactions:
Ag(s) + Ag+(aq)→Ag+(aq) + Ag(s)
Step 2: Calculate E◦
cell using the Nernst equation. The Nernst equation is
given by:
Ecell =E◦
cell −0.0592
nlog(Q)
33
where - Ecell is the cell potential, - E◦
cell is the standard cell potential, - nis the
number of moles of electrons transferred in the balanced cell reaction, and - Q
is the reaction quotient.
The reaction quotient Qis given by:
Q=[Ag+]cell1
[Ag+]cell2
where - [Ag+]cell1 = 0.10 M and - [Ag+]cell2 = 0.0010 M
Since there is a 1:1 molar ratio of Ag to AgCl in the half-reactions, n= 1.
Now substitute these values into the Nernst equation:
Ecell = 0.19V−0.0592
1log 0.10
0.0010
Step 3: Calculate Ecell by solving the equation.
Ecell = 0.19V−(0.0592) log(100)
Ecell = 0.19V−(0.0592)(2)
Ecell = 0.19V−0.1184V
Ecell = 0.0716V
Therefore, the cell potential of the concentration cell at 25
°
C is 0.0716 V.
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