CHEM 121 - GENERAL CHEMISTRY I Galvanic and Electrolytic Cells Question Bank Set 3

CHEM 121 - GENERAL CHEMISTRY

I - Galvanic and Electrolytic Cells

Question Bank - Set 3

Liberty University

Question 1

Question

Consider a galvanic cell with a standard cell potential of E◦

cell = 0.50 V. If the

cell operates until the concentration of the electrolyte in one of the half-cells

decreases by a factor of 10, calculate the new cell potential.

Solution

Step 1: The Nernst equation relates the cell potential (Ecell) to the standard

cell potential (E◦

cell), the reaction quotient (Q), the gas constant (R), the tem-

perature (T), and the number of electrons transferred in the cell reaction (n).

The Nernst equation is given by:

Ecell =E◦

cell −RT

nF ln Q

Step 2: We can calculate the reaction quotient Qfor the original cell by

considering the concentrations of the electrolytes in the two half-cells. Since

the concentration of one electrolyte decreases by a factor of 10, the reaction

quotient for the new cell will be different.

Step 3: Given that the reaction quotient Qfor the original cell is 1, we can

calculate the new reaction quotient Q′for the cell with changed concentrations.

Since the concentration of the electrolyte in one half-cell decreases by a factor

of 10, Q′will be Q/10.

Step 4: Substituting Q′into the Nernst equation, we can now calculate the

new cell potential E′

cell for the changed concentrations:

E′

cell =E◦

cell −RT

nF ln Q

10

Step 5: Finally, substitute the given values (E◦

cell = 0.50 V, T= 298 K, n=

1, R= 8.314 J/(mol·K), F= 96485 C/mol) into the equation to find the new

cell potential E′

cell. Calculate the natural logarithm term before substituting

into the equation.

Step 6: After performing the calculations, we find the new cell potential

E′

cell. This value will indicate the change in cell potential due to the change in

concentrations of the electrolytes in the half-cells.

Question 2

Question

Consider a galvanic cell with a standard cell potential of 1.15 V. If the cell

operates until the concentration of Zn2+ ions is reduced to 0.05 M from an

initial concentration of 1.0 M, calculate the amount of electrical work done by

the cell during this process. Assume complete reaction of Zn2+ ions and use

Faraday’s constant (F= 96,485 C/mol).

Solution

Step 1: Write the balanced redox reaction for the galvanic cell. The cell reaction

in a galvanic cell involving Zn2+ ions and a standard hydrogen electrode is:

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Given that the standard cell potential is 1.15 V, the standard electromotive

force of the cell (E◦

cell) is 1.15 V.

Step 2: Calculate the number of electrons transferred during the reaction.

From the balanced redox reaction, we see that 2 moles of electrons are trans-

ferred per mole of Zn2+ ion reacted.

Step 3: Calculate the number of moles of Zn2+ ions reacted. The initial

and final concentrations of Zn2+ ions are 1.0 M and 0.05 M, respectively. The

number of moles of Zn2+ ions reacted can be calculated using the formula:

moles of Zn2+ ions reacted = (initial concentration−final concentration)×volume

Assuming a volume of 1 L for simplicity, we find:

moles of Zn2+ ions reacted = (1.0−0.05) ×1 = 0.95 mol

Step 4: Calculate the electrical work done by the cell. The electrical work

done by the cell (Wcell) can be calculated using the formula:

Wcell =−nF ×Ecell

Where: - nis the number of moles of electrons transferred during the reaction,

-Fis Faraday’s constant, - Ecell is the standard cell potential (1.15 V in this

case).

Substitute the values to find:

Wcell =−(2 mol)(96,485 C/mol)(1.15 V) = −221,902.5 J

Therefore, the amount of electrical work done by the cell during this process is

-221,902.5 J.

Question 3

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 potentials are E◦

Zn2+/Zn =−0.76 V and E◦

Cu2+/Cu =

0.34 V, calculate the cell potential at standard conditions.

Solution

Step 1: Identify the species being oxidized and reduced. In a galvanic cell,

the species being oxidized occurs at the anode, while the species being reduced

occurs at the cathode. From the given half-reactions, we can see that zinc is

being oxidized at the anode (Zn −−→ Zn2+ + 2 e−) and copper ions are being

reduced at the cathode (Cu2+ + 2 e−−−→ Cu).

Step 2: Write the overall cell reaction. By adding the two half-reactions

together, we get the overall cell reaction:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 3: Calculate the cell potential at standard conditions. The cell potential

can be calculated using the equation:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the values for the standard reduction potentials, we get:

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Therefore, the cell potential at standard conditions for the galvanic cell is

1.10 V.

Question 4

Question

An electrolytic cell contains a solution of CuSO4and a strip of pure copper as the

cathode. A magnesium strip is also placed in the solution to act as an anode.

The standard reduction potentials are E◦

Cu2+/Cu = 0.34 V and E◦

Mg2+/Mg =

−2.37 V. Determine the cell potential of this electrolytic cell.

Solution

Step 1: Write the balanced redox reactions for the anode and cathode reactions.

The anode reaction involves the oxidation of Mg:

Mg →Mg2+ + 2e−

The cathode reaction involves the reduction of Cu2+:

Cu2+ + 2e−→Cu

Step 2: Calculate the cell potential (Ecell) using the formula:

Ecell =E◦

cathode −E◦

anode =E◦

Cu2+/Cu −E◦

Mg2+/Mg

Ecell = 0.34 V −(−2.37 V) = 2.71 V

Therefore, the cell potential of this electrolytic cell is 2.71 V.

Question 5

Question

Consider a galvanic cell with a standard cell potential of E◦

cell = 1.10 V. The

cell reaction is given by:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

If the concentration of Cu2+ is 0.10 M and the concentration of Zn2+ is 1.0

M, determine the cell potential under these conditions.

Solution

Step 1: Write the half-reactions for the cell reaction:

Anode: Zn(s)→Zn2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential using the standard reduction

potentials for Zn and Cu:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 3: Determine the reaction quotient, Q, using the given concentrations:

Q=[Zn2+][Cu(s)]

[Zn(s)][Cu2+]=(1.0)(1.0)

1= 1.0

Step 4: Use the Nernst equation to calculate the cell potential under the

given conditions:

Ecell =E◦

cell −0.0592 V

2log(Q) = 1.10 V −(0.0296 V) log(1.0) = 1.10 V

Therefore, the cell potential under the given conditions is 1.10 V .

Question 6

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)

Calculate the standard cell potential (E◦

cell) at 25

C for this galvanic cell.

Given E◦

cell values are:

E◦

cell(Zn2+/Zn) = −0.76 V

E◦

cell(Cu2+/Cu) = 0.34 V

Solution

Step 1: Identify the half-reactions and standard cell potential formulas.

In a galvanic cell, the standard cell potential (E◦

cell) can be calculated using

the formula:

E◦

cell =E◦

reduction(cathode) −E◦

reduction(anode)

Given half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−E◦

reduction(anode) = −0.76 V

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s) E◦

reduction(cathode) = 0.34 V

Step 2: Substitute the values into the formula to find E◦

cell.

E◦

cell =E◦

reduction(cathode) −E◦

reduction(anode)

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the standard cell potential (E◦

cell) at 25

C for this galvanic cell is

1.10 V.

Question 7

Question

Consider a galvanic cell with a standard cell potential of E◦

cell = 0.78 V. If the

cell potential drops to 0.66 V when a current of 2.5 A is drawn from the cell

for a certain period of time, calculate the amount of substance that has reacted

during this time.

Solution

Step 1: Identify the given information and the cell reaction. The cell reaction

for a galvanic cell can be written as follows:

Anode (oxidation) : M(s)→Mz+(aq) + ze−

Cathode (reduction) : X(aq) + ze−→X(s)

Given: E◦

cell = 0.78 V Ecell = 0.66 V I= 2.5 A

Step 2: Calculate the reaction quotient and the cell potential at the new

condition. The Nernst equation relates the cell potential to the reaction quotient

Ecell =E◦

cell −0.0592

zlog(Q)

Substitute the given values into the Nernst equation:

0.66 = 0.78 −0.0592

zlog(Q)

Step 3: Calculate the new reaction quotient Q. Solve the equation for Q:

Q= exp (0.78 −0.66) ·z

0.0592 

Step 4: Determine the change in Qand the amount of substance reacted.

Since Qis related to the amount of substance reacted, we can calculate the

change in Q:

∆Q=Qfinal −Qinitial

From the change in Q, the amount of substance reacted can be calculated using

stoichiometry.

Question 8

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)

on the left side and a zinc electrode on the right side. The cell has a standard

cell potential of E◦

cell = 1.10 V. If the concentration of Zn2+ ions in the zinc

half-cell is 0.001 M, determine the concentration of H+ions required in the

hydrogen half-cell to maintain the standard cell potential.

Solution

Step 1: Write the half-cell reactions for the galvanic cell. The overall cell reaction

is the sum of the reduction half-reactions for each electrode.

Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V

2H+(aq)+2e−→H2(g)E◦= 0.00 V

Step 2: Write the overall redox reaction for the cell.

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Step 3: Calculate the cell potential at non-standard conditions using the

Nernst Equation:

Ecell =E◦

cell −0.0592

nlog [Zn2+]

[H+]2

Given: E◦

cell = 1.10 V, [Zn2+] = 0.001 M.

Step 4: Solve for [H+] by setting Ecell equal to E◦

cell.

1.10 = 1.10 −0.0592

2log 0.001

[H+]2

Step 5: Simplify and solve for [H+].

0.0592

2log 0.001

[H+]2= 0

Step 6: Calculate [H+].

[H+] = 0.0316 M

Therefore, the concentration of H+ions required in the hydrogen half-cell

to maintain the standard cell potential is 0.0316 M.

Question 9

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a nickel

electrode. If the standard reduction potential of nickel is −0.25 V and the

standard reduction potential of the hydrogen electrode is 0.00 V, determine the

cell potential at standard conditions. Additionally, if the concentration of [Ni2+]

in the nickel half-cell is 1.0 M, calculate the cell potential when the concentration

of [H+] in the hydrogen half-cell is 0.1 M.

Solution

Step 1: Write the overall cell reaction for the galvanic cell. The overall cell reac-

tion for the galvanic cell can be written by subtracting the reduction potential

of the anode from the reduction potential of the cathode:

Ni2+(aq)+2e−→Ni(s)

2H+(aq)+2e−→H2(g)

Net cell reaction: Ni2+(aq) + 2H+(aq)→Ni(s)+H2(g)

Step 2: Calculate the standard cell potential (E◦) at standard conditions.

Given: Standard reduction potential of nickel, E◦

Ni =−0.25 V Standard reduc-

tion potential of the hydrogen electrode, E◦

H2= 0.00 V

The standard cell potential is calculated by summing the reduction potentials

of the cathode and anode:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

H2−E◦

Ni = 0.00 V −(−0.25 V) = 0.25 V

Therefore, the standard cell potential at standard conditions is 0.25 V.

Step 3: Calculate the cell potential under non-standard conditions. Given:

[Ni2+]=1.0 M [H+]=0.1 M

The Nernst equation relates the cell potential under non-standard conditions

to the standard cell potential:

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 electrons transferred in the cell reaction, Qis

the reaction quotient, and Kis the equilibrium constant.

The reaction quotient for the given cell is:

Q=[H+]2

[Ni2+]=(0.1)2

1.0= 0.01

The number of electrons transferred in the cell reaction is 2.

Using the Nernst equation:

E= 0.25 V −0.0592

2log(0.01)

E= 0.25 V + 0.0296 log(0.01)

E= 0.25 V + 0.0296(−2)

E= 0.25 V −0.0592

E= 0.19 V

Therefore, the cell potential under non-standard conditions is 0.19 V.

Question 10

Question

Consider a galvanic cell with a standard cell potential of E◦= 1.20 V. If the

cell operates until the concentration of Zn2+ ions in the anode compartment

becomes 0.0010 M and the concentration of Cu2+ ions in the cathode compart-

ment becomes 0.010 M, determine the number of electrons transferred during

this process.

Solution

Step 1: Write the half-reactions and their standard reduction potentials.

The half-reactions for this galvanic cell are:

Zn2+ + 2 e−−−→ Zn(s) (E◦=−0.76 V)

Cu2+ + 2 e−−−→ Cu(s) (E◦= 0.34 V)

Step 2: Determine the balanced cell reaction and overall cell potential.

The cell reaction is:

Zn(s) + Cu2+ −−→ Cu(s) + Zn2+

The overall cell potential (E◦

cell) is calculated by subtracting the reduction po-

tential of the anode half-reaction from the reduction potential of the cathode

half-reaction:

E◦

cell =E◦(cathode) −E◦(anode) = 0.34 V −(−0.76 V) = 1.10 V

Step 3: Use the Nernst equation to find the cell potential under non-standard

conditions.

The Nernst equation is:

E=E◦−0.0592

nlog [Cu2+]

[Zn2+]

where nis the number of moles of electrons transferred. Rearranging the equa-

tion to solve for ngives:

n=E◦−E

0.0592 log [Cu2+]

[Zn2+]

Substitute the given values into the equation to find n:

n=1.20 V −1.10 V

0.0592 log 0.010

0.0010

n=0.10 V

0.0592 log(10)

n= 1.68

Therefore, the number of electrons transferred during this process is 2 (rounded

to the nearest whole number).

Question 11

Question

Consider a galvanic cell with the following cell notation: Zn|Zn2+||Cu2+|Cu. If

the standard reduction potentials for Zn2+/Zn and Cu2+/Cu are −0.76 V and

0.34 V, respectively, calculate the cell potential at 25◦C for the galvanic cell.

Solution

Let’s denote the oxidation half-reaction at the anode as:

Zn(s)→Zn2+(aq)+2e−

And the reduction half-reaction at the cathode as:

Cu2+(aq)+2e−→Cu(s)

Given the standard reduction potentials, we have: E◦

Zn2+/Zn =−0.76 V and

E◦

Cu2+/Cu = 0.34 V.

Step 1: Calculate the cell potential at standard conditions using the stan-

dard reduction potentials:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 2: Determine the effect of temperature on the cell potential using the

Nernst equation:

Ecell =E◦

cell −0.0592

nlog [Cu2+]

[Zn2+]

Here, nis the number of electrons transferred (2 in this case) and [Cu2+] and

[Zn2+] are the concentrations of Cu2+ and Zn2+ ions.

Step 3: Since the cell is at standard conditions, [Cu2+] = [Zn2+] = 1 M:

Ecell = 1.10 V −0.0592

2log 1

1= 1.10 V

Therefore, the cell potential at 25◦C for the galvanic cell is 1.10 V .

Question 12

Question

Consider a galvanic cell in which the following reaction occurs:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

If the standard reduction potentials are Cu2+/Cu = +0.34 V and Zn2+/Zn =

-0.76 V, calculate the cell potential at 25

C. Is the reaction spontaneous?

Solution

Step 1: Write the half-reactions for the standard reduction potentials given.

Cu2+(aq)+2e−→Cu(s)E◦= +0.34 V

Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V

Step 2: Write the balanced overall cell reaction.

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 3: Use the standard reduction potentials to find the standard cell po-

tential (E◦

cell).

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Step 4: Calculate the non-standard cell potential (Ecell) using the Nernst

Equation:

Ecell =E◦

cell −0.0592 V

nlog(Q)

where Qis the reaction quotient, and for this reaction Q=[Zn2+]

[Cu2+].

Step 5: Determine the spontaneity of the reaction. Since the calculated cell

potential Ecell is positive, the cell reaction is spontaneous.

Question 13

Question

Consider a galvanic cell with a standard cell potential of 1.13 V. If the reaction

in this cell is spontaneous and the cell operates under standard conditions, what

can you say about the sign of ∆G◦? Explain your answer.

Solution

To determine the sign of ∆G◦, we can use the relationship between the standard

Gibbs free energy change, cell potential, and Faraday’s constant:

∆G◦=−nF E◦

where ∆G◦is the standard Gibbs free energy change, nis the number of

moles of electrons transferred in the reaction, Fis Faraday’s constant (96485

C/mol), and E◦is the standard cell potential.

Step 1: Given E◦= 1.13 V, we know that a positive cell potential indicates

a spontaneous reaction.

Since the reaction is spontaneous, we also know that ∆G◦<0.

Step 2: Substitute the values of n= 1 (assuming a single-electron transfer

process) and E◦= 1.13 V into the equation:

∆G◦=−1×96485 C/mol ×1.13 V

∆G◦=−109097.05 J/mol

Step 3: Since ∆G◦<0, we can conclude that ∆G◦is negative. This

confirms that the reaction is spontaneous under standard conditions.

Question 14

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V

2H+(aq) + 2e−→H2(g)E◦= 0.00 V

Determine the standard cell potential for this galvanic cell and state whether

the cell reaction is spontaneous or non-spontaneous.

Solution

Step 1: Write out the balanced overall cell reaction.

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Step 2: Calculate the standard cell potential by summing the standard re-

duction potentials for the half-reactions involved in the cell reaction.

E◦

cell =E◦

reduction, cathode −E◦

reduction, anode

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.00 V −(−0.76 V) = +0.76 V

Step 3: State whether the cell reaction is spontaneous or non-spontaneous.

Since the standard cell potential is positive (+0.76 V), the cell reaction is spon-

taneous.

Question 15

Question

Consider a galvanic cell constructed with a standard hydrogen electrode and a

platinum electrode immersed in a 1.0 M solution of CuSO4. The cell notation

for this galvanic cell is:

H2(g)|H+(aq)||Cu2+ (aq)|Cu(s)

Calculate the standard cell potential of this galvanic cell at 25

Solution

Step 1: Write the half-reactions for the galvanic cell. The half-reactions for the

given cell are:

Anode: 2H2(g)→4H+(aq)+4e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential (E◦

cell) using the standard reduc-

tion potentials (E◦

red) for each half-reaction. The standard reduction potentials

at 25

C are: E◦

red(Cu2+(aq)+2e−→Cu(s)) = 0.34 V

E◦

red(2H2(g)→4H+(aq)+4e−)=1.23 V

The standard cell potential is given by:

E◦

cell =E◦

red(cathode) −E◦

red(anode)

Therefore,

E◦

cell = 0.34 V −1.23 V = −0.89 V

So, the standard cell potential of the galvanic cell is -0.89 V.

Question 16

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq) + 2e−→Zn(s)E◦

red =−0.76 V

Fe3+(aq) + 3e−→Fe(s)E◦

red =−0.036 V

If the initial concentrations of Zn2+ and Fe3+ are both 1.0 M, determine the

cell potential after the cell has been operating for 1.5 hours. Assume that the

temperature is constant and that the reaction quotient Q remains at a constant

value of 1.

Solution

Step 1: Write the overall cell reaction.

The overall cell reaction is the sum of the reduction half-reactions for the

two species:

Zn2+(aq) + Fe3+(aq)→Zn(s) + Fe(s)

Step 2: Calculate the cell potential at standard state.

We can calculate the cell potential at standard state using the given standard

reduction potentials:

E◦

cell =E◦

red, cathode −E◦

red, anode

E◦

cell =E◦

Fe3+/Fe −E◦

Zn2+/Zn =−0.036 V −(−0.76 V) = 0.724 V

Step 3: Calculate the reaction quotient Q.

Since the reaction quotient Q is constant (equal to 1), the Nernst equation

simplifies to:

Ecell =E◦

cell −0.0592

nlog Q

Where n is the number of moles of electrons transferred in the balanced cell

reaction.

Step 4: Calculate the cell potential after 1.5 hours.

Since Q remains constant at 1, the cell potential after 1.5 hours is the same

as the cell potential at standard state:

Ecell = 0.724 V

Therefore, the cell potential after the cell has been operating for 1.5 hours

is 0.724 V.

Question 17

Question

Consider a galvanic cell with a standard cell potential of E◦= 0.95 V. The

half-reactions involved are:

Cathode: 2H2O(l) + 2e−→2OH−(aq)+H2(g), E◦= 0.83 V

Anode: 2I−(aq)→I2(s) + 2e−,E◦= 0.54 V

Calculate: (a) E◦for the overall reaction of the cell. (b) Equilibrium

constant Kfor the overall reaction of the cell. (c) The cell potential when

[OH−]=0.10 M and [I−]=0.20 M.

Solution

(a) To calculate the standard cell potential E◦for the overall reaction of the

cell, we can use the formula:

E◦

cell =E◦

cathode −E◦

anode

Given: E◦

cathode = 0.83 V, E◦

anode = 0.54 V

E◦

cell = 0.83 V −0.54 V = 0.29 V

(b) The equilibrium constant Kfor the overall reaction of the cell can be

determined using the formula:

E◦

cell =RT

nF ln K

Given: E◦

cell = 0.29 V, R= 8.314 J/(mol

K), T= 298 K, n= 2 (number of

moles of electrons transferred), F= 96485 C/mol

0.29 V = (8.314 J/(mol

K))(298 K)

(2)(96485 C/mol) ln K

ln K=(0.29 V)(2)(96485 C/mol)

(8.314 J/(mol

K))(298 K)

ln K= 6.45

K=e6.45 ≈644.00

E=E◦−RT

nF ln Q

Given: E◦= 0.29 V, T= 298 K, n= 2, F= 96485 C/mol

For the given concentrations, Q=[OH−]2

[I−]2=(0.10 M)2

(0.20 M)2= 0.25

E= 0.29 V −(8.314 J/(mol

K))(298 K)

(2)(96485 C/mol) ln 0.25

E≈0.24 V

Question 18

Question

Consider the following galvanic cell setup at 25

(s)Cu2+(0.01 M)|| Cu(s)

Determine if the cell reaction is spontaneous and if so, calculate the cell

potential. Assume standard reduction potential of E◦= 0.34 Vfor the Cu2+/Cu

redox couple.

Solution

Step 1: Write the overall cell reaction.

Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the cell potential at standard conditions using the standard

reduction potential.

E◦

cell =E◦

cathode −E◦

anode = 0 −0.34 = −0.34 V

Step 3: Determine the reaction quotient, Q.

Q=[Cu]1

[Cu2+]1= 1

Step 4: Calculate the cell potential using the Nernst equation:

Ecell =E◦

cell −0.0592

nlog(Q)

Ecell =−0.34 −0.0592

2log(1)

Ecell =−0.34 V

Step 5: Since Ecell is negative, the cell reaction is not spontaneous.

Question 19

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq)+2e−→Zn(s)E◦=−0.76V

Cu2+(aq)+2e−→Cu(s)E◦= +0.34V

Calculate the cell potential at 25

C when the concentrations of Zn2+ and

Cu2+ are both 0.10 M. Determine if the cell is galvanic or electrolytic.

Solution

Step 1: Write the cell reaction by combining the given half-reactions.

Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)

Step 2: Calculate the cell potential at 25

C using the Nernst equation:

Ecell =E◦−0.0592

nlog Q

Where

Q=[products]

[reactants] =[Zn2+][Cu]

[Zn][Cu2+]=(0.10)2

(1)(1) = 0.01

n= 2 (number of electrons transferred)

Substitute E◦= 0.34V, Q= 0.01, n= 2 into the Nernst equation:

Ecell = 0.34V −0.0592

2log 0.01 = 0.34V −0.0296 log 0.01

Ecell = 0.34V −0.0296(−2) = 0.34V + 0.0592 = 0.3992V

Therefore, the cell potential at 25

C is 0.3992 V.

Step 3: Determine if the cell is galvanic or electrolytic. Since the cell poten-

tial is positive (0.3992 V), the cell is galvanic. Galvanic cells have positive cell

potentials and produce electrical energy from spontaneous redox reactions.

Question 20

Question

Consider a galvanic cell where the half-reactions are:

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 cell potential at 25

C when the concentrations are [Zn2+] =

0.10 M and [MnO−

4]=0.20 M.

Solution

Step 1: Write the overall cell reaction. The overall cell reaction is the sum of

the two half-reactions:

2 Zn2+(aq) + MnO−

4(aq) + 16H+(aq)→2 Zn(s) + Mn2+(aq)+8H2O(l)

Step 2: Calculate the standard cell potential, E◦

cell. Using the standard

reduction potentials given, the standard cell potential is:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 1.51 V −(−0.76 V) = 2.27 V

Step 3: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential to the standard cell potential and the

concentrations of the reactants and products:

Ecell =E◦

cell −0.0592

nlog(Q)

where Qis the reaction quotient, which is given by:

Q=[Mn2+]1

[Zn2+]2[MnO−

4]1[H+]16

Step 4: Calculate Qand the cell potential, Ecell. Substitute the given con-

centrations into the Qexpression:

Q=(0.20)1

(0.10)2(0.20)1(10−16)16

Q= 5.0×10−100

Substitute the values of E◦

cell,n= 5, and Qinto the Nernst equation to find

Ecell:

Ecell = 2.27 V −0.0592

5log5.0×10−100

Ecell = 2.27 V −0.0592

5×(−100)

Ecell = 2.27 V + 0.1184 = 2.3884 V

Therefore, the cell potential at 25

C with the given concentrations is 2.3884 V.

Question 21

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)

and a copper electrode. The standard reduction potential for the copper elec-

trode is +0.34 V. Calculate the cell potential at 25

C when the concentration

of Cu2+ is 1.0 M and the pressure of H2gas is 0.80 atm.

Solution

To calculate the cell potential of the galvanic cell, we can use the Nernst equa-

tion:

E=E◦−0.0592

nlog Q

where: - Eis the cell potential, - E◦is the standard cell potential, - nis the

number of electrons transferred in the cell reaction, - Qis the reaction quotient.

Step 1: Write the cell reaction. Since we have a standard hydrogen electrode

and a copper electrode, the cell reaction will be:

Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential (E◦). Given that the stan-

dard reduction potential for the copper electrode is +0.34 V, the standard cell

potential can be calculated as:

E◦=E◦

cathode −E◦

anode = 0 −(+0.34) = −0.34 V

Step 3: Calculate the reaction quotient (Q). The reaction quotient for the

cell reaction is given by:

Q=[Cu](Cu2+)

[H+]

Substitute the given concentration and pressure values into the equation to find

Step 4: Calculate the cell potential (E). Substitute E◦,Q, and n(which is

2 in this case) into the Nernst equation to find E.

Question 22

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s)→Zn2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

If the standard reduction potential for Zn is −0.76 V and for Cu is +0.34 V,

calculate the standard cell potential, E◦

cell, for this galvanic cell.

Solution

Step 1: The standard cell potential, E◦

cell, can be calculated using the formula:

E◦

cell =E◦

reduction,Cathode −E◦

reduction,Anode

Step 2: First, identify the reduction potentials for the cathode and anode

reactions:

E◦

reduction,Cathode = +0.34 V

E◦

reduction,Anode =−0.76 V

Step 3: Substitute the reduction potentials into the formula to determine

E◦

cell:

E◦

cell = +0.34 V −(−0.76 V)

E◦

cell = 0.34 V + 0.76 V

E◦

cell = 1.10 V

Therefore, the standard cell potential for this galvanic cell is 1.10 V.

Question 23

Question

Consider a galvanic cell where an aqueous solution of silver nitrate, AgNO3, is

in one half-cell with a silver electrode, and a solution of iron(III) chloride, FeCl3,

is in the other half-cell with an iron electrode. Write the overall cell reaction

and calculate the standard cell potential, E◦

cell, for this galvanic cell.

Solution

Step 1: Begin by writing the half-reactions for each half-cell. In the anode half-

cell: Fe(s)→Fe3+(aq)+3e−In the cathode half-cell: Ag+(aq) + e−→Ag(s)

Step 2: Write the overall cell reaction by adding the half-reactions. Fe(s) +

3Ag+(aq)→Fe3+(aq) + 3Ag(s)

Step 3: Determine the standard cell potential, E◦

cell, by considering the

standard reduction potentials for the half-reactions involved. From standard

reduction potential tables, E◦

cell =E◦

cathode −E◦

anode E◦

cell =E◦

Ag+/Ag −E◦

Fe3+/Fe

E◦

cell = 0.80 V −(−0.04 V) E◦

cell = 0.84 V

Therefore, the standard cell potential for this galvanic cell is 0.84 V.

Question 24

Question

Consider a galvanic cell where the following reaction takes place:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Given that the standard reduction potentials are E◦

Cu2+/Cu = 0.34 V and

E◦

Zn2+/Zn =−0.76 V, determine the standard cell potential (E◦

cell) for this cell.

Solution

Step 1: Write the overall cell reaction. The standard cell potential can be de-

termined using the standard reduction potentials of each half-reaction involved.

The cell reactions are written as reduction reactions, so we have the following

overall reaction:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 2: Calculate E◦

cell using the standard reduction potentials. The stan-

dard cell potential, E◦

cell, is given by the difference between the standard reduc-

tion potentials of the two half-reactions involved:

E◦

cell =E◦

cathode −E◦

anode

In this case, Cu is the cathode (reduction occurs) and Zn is the anode (oxi-

dation occurs). Therefore:

E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the standard cell potential for the given galvanic cell is 1.10 V .

Question 25

Question

A voltaic cell is constructed with a standard hydrogen electrode and a chromium

electrode. The standard reduction potential for the chromium electrode is

−0.744 V. Calculate the cell potential at 25

C for the voltaic cell described

above.

Solution

Step 1: Write the cell reaction and the standard cell potential. The cell reaction

for the voltaic cell is:

2H++ Cr →H2+ Cr2+

The standard cell potential, E◦

cell, can be determined using the standard

reduction potentials of the half-reactions involved:

E◦

cell =E◦

cathode −E◦

anode

Given that the standard reduction potential for the chromium electrode is

−0.744 V, the standard cell potential is:

E◦

cell = 0.000 V −(−0.744 V) = 0.744 V

Step 2: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential to the standard cell potential and the

reaction quotient, Q:

Ecell =E◦

cell −RT

nF ln Q

where: - R= 8.314 J mol−1K−1is the gas constant, - T= 25 ◦C = 298 K is the

temperature, - nis the number of electrons transferred in the cell reaction, -

F= 96485 C mol−1is the Faraday constant, and - Qis the reaction quotient.

For the given cell reaction:

2H++ Cr →H2+ Cr2+

n= 2 because 2 electrons are transferred.

The reaction quotient, Q, is given by:

Q=[H2][Cr2+]

[H+]2[Cr]

Since the concentrations are not provided in the question, the calculations

for the cell potential at 25

C cannot be completed.

Question 26

Question

Consider a galvanic cell with a standard cell potential of 0.85 V that uses the

following half-reactions:

Anode: Zn(s)→Zn2+(aq)+2e−E◦

Zn2+/Zn =−0.76 V

Cathode: MnO−

4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦

MnO−

4/Mn2+ = 1.51 V

Calculate the equilibrium constant, K, for the reaction in the cell and de-

termine if the cell reaction is spontaneous at 298 K.

Solution

Step 1: Write the balanced cell reaction and calculate the overall standard cell

potential.

The balanced cell reaction is:

Zn(s) + MnO−

4(aq)+8H+(aq)→Zn2+(aq) + Mn2+(aq)+4H2O(l)

The overall standard cell potential, E◦

cell, is the sum of the standard reduction

potentials of the half-reactions at the cathode and anode:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 1.51 V −(−0.76 V) = 2.27 V

Step 2: Use the Nernst equation to relate the equilibrium constant, K, to

the cell potential.

The Nernst equation relates the cell potential to the equilibrium constant:

Ecell =RT

nF ln K

where R= 8.314 J/(mol·K), T= 298 K, nis the number of electrons transferred

in the balanced cell reaction, and F= 96485 C/mol.

Step 3: Determine if the cell reaction is spontaneous at 298 K.

For a spontaneous reaction, the equilibrium constant, K, must be greater

than 1. Let’s solve the Nernst equation for K:

K= exp nF Ecell

RT = exp 5×96485 C/mol ×2.27 V

8.314 J/(mol ·K) ×298 K 

Calculating this expression gives us the value of K.

Question 27

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)

Calculate the standard cell potential for this galvanic cell at 25

C. Determine if

this cell is spontaneous or non-spontaneous.

Solution

Step 1: Write the overall cell reaction. The overall cell reaction can be obtained

by adding the half-reactions together:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 2: Write the standard cell potential equation. The standard cell po-

tential, Ecell, is given by the formula:

Ecell =Ecathode −Eanode

Step 3: Determine the standard reduction potentials. Lookup the standard

reduction potentials for the half-reactions:

ECu2+/Cu = 0.34 V

EZn2+/Zn =−0.76 V

Step 4: Calculate the standard cell potential. Substitute the values into the

standard cell potential formula:

Ecell = 0.34 V −(−0.76 V) = 1.10 V

Step 5: Determine spontaneity. Since the standard cell potential is positive

(1.10 V), the cell reaction is spontaneous.

Question 28

Question

In an electrochemical cell, a copper electrode is placed in a 1.0 M CuSO4solu-

tion, and a silver electrode is placed in a 1.0 M AgNO3solution. The standard

reduction potentials are as follows: Cu2+ + 2e−→Cu(s) with E◦= 0.34 V and

Ag++ e−→Ag(s) with E◦= 0.80 V. Determine the cell potential and predict

the direction of electron flow.

Solution

Step 1: Write the two half-reactions involving the reduction of Cu2+ and Ag+:

Cu2+ + 2e−→Cu(s) E◦= 0.34 V

Ag++ e−→Ag(s) E◦= 0.80 V

Step 2: Calculate the cell potential (E◦

cell) by subtracting the reduction

potential of the anode from the reduction potential of the cathode:

E◦

cell =E◦

cathode −E◦

anode = 0.80 V −0.34 V = 0.46 V

Step 3: Determine the direction of electron flow based on the cell poten-

tial. Since E◦

cell is positive, the reaction is spontaneous, and electrons will flow

from the anode (where oxidation occurs) to the cathode (where reduction oc-

curs). Therefore, the electrons will flow from the copper electrode to the silver

electrode.

Question 29

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) →Zn2+(aq)+2e−

Cathode: 2Ag+(aq)+2e−→2Ag(s)

If the standard reduction potential for the Zn half-reaction is −0.76 V and the

standard reduction potential for the Ag half-reaction is 0.80 V, calculate the

standard cell potential (E◦

cell) of the galvanic cell.

Solution

Step 1: Write the half-reactions and determine which is the anode and which is

the cathode based on their reduction potentials. The anode is where oxidation

occurs (Zn to Zn2+) and the cathode is where reduction occurs (Ag+to Ag).

Anode: Zn(s) →Zn2+(aq)+2e−

Cathode: 2Ag+(aq)+2e−→2Ag(s)

Step 2: Identify the standard reduction potentials for the half-reactions.

The standard reduction potential for the Zn half-reaction, E◦

Zn, is −0.76 V and

for the Ag half-reaction, E◦

Ag, is 0.80 V.

Step 3: Calculate the standard cell potential, E◦

cell, using the equation:

E◦

cell =E◦

cathode −E◦

anode

Step 4: Substitute the values into the equation to calculate E◦

cell.

E◦

cell = 0.80 V −(−0.76 V)

E◦

cell = 0.80 V + 0.76 V

E◦

cell = 1.56 V

Therefore, the standard cell potential of the galvanic cell is 1.56 V .

Question 30

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 initial concentrations are [Zn2+]=1.0 M and [Cu2+]=0.1 M, calcu-

late the cell potential at standard conditions and when the cell reaches equilib-

rium. Determine whether this cell can function as an electrolytic cell without

an external potential applied.

Solution

Step 1: Calculate the cell potential at standard conditions using the Nernst

equation:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 2: Calculate the cell potential when the cell reaches equilibrium by

using the Nernst equation:

Ecell =E◦

cell −RT

nF ln [Cu2+]

[Zn2+]2

Ecell = 1.10 V −(8.314 J/mol ·K)(298 K)

2(96485 C/mol) ln 0.1 M

(1.0 M)2

Ecell ≈1.0884 V

Step 3: Determine whether this cell can function as an electrolytic cell with-

out an external potential applied. Since the cell potential at equilibrium is

slightly less than the standard cell potential, the cell would not function as an

electrolytic cell without an external potential applied because the reaction is

not favorable to proceed spontaneously.

Question 31

Question

A galvanic cell consists of a standard hydrogen electrode (E◦

H+/H2= 0.00 V)

and a copper electrode in a 0.10 M solution of Cu2+ ions. The cell potential is

measured as 0.34 V at 25

Step 1: Write the overall redox reaction for the galvanic cell.

2H++ Cu2+ →Cu + H2

Step 2: Calculate the cell potential using the Nernst equation. The general

form of the Nernst equation is:

E=E◦−0.0592

nlog(Q)

where: - Eis the cell potential - E◦is the standard cell potential - nis the

number of moles of electrons transferred in the balanced redox reaction - Qis

the reaction quotient

For the given cell, E= 0.34 V, E◦=E◦

Cu2+/Cu −E◦

H+/H2,n= 2 (from the

balanced redox reaction).

Therefore, the Nernst equation becomes:

0.34 = E◦−0.0592

2log [Cu2+]

[H+]2

Step 3: Solve for E◦. Given that E◦

H+/H2= 0.00 V, we can simplify the

equation to:

0.34 = E◦−0.0296 log [Cu2+]

[H+]2

Since the standard concentration for H+is 1 M, the equation further simpli-

fies to:

0.34 = E◦−0.0296 log[Cu2+]

Therefore, the standard reduction potential of the copper electrode at 25

Question

Consider the following cell at 25

Mg(s)|Mg2+(0.10M)|| Fe3+(0.20M)|Fe(s)

Write the cell diagram, the overall cell reaction, and determine the cell potential

at standard conditions. Given: E◦(Fe3+/Fe) = 0.77 V and E◦(Mg2+/Mg) =

−2.37 V.

Solution

Step 1: Write the cell diagram The cell diagram can be written as:

Mg(s)|Mg2+(0.10M)|| Fe3+(0.20M)|Fe(s)

Step 2: Write the overall cell reaction The overall cell reaction can be written

by adding the two half-cell reactions together:

Mg(s) + 2Fe3+(aq)→Mg2+(aq) + 2Fe(s)

Step 3: Determine the cell potential at standard conditions The cell potential

at standard conditions can be calculated using the formula:

E◦

cell =E◦

cathode −E◦

anode

Given that E◦(Fe3+/Fe) = 0.77 V and E◦(Mg2+/Mg) = −2.37 V, we can

substitute these values into the formula:

E◦

cell = 0.77 V −(−2.37 V) = 3.14 V

Therefore, the cell potential at standard conditions for the given cell is 3.14

Question 33

Question

Consider a Galvanic cell consisting of a standard hydrogen electrode (SHE) and

a copper electrode. The standard reduction potential for the copper electrode

is E◦= 0.34 V. Calculate the cell potential when the concentration of Cu2+ is

0.10 M and the pressure of hydrogen gas is 0.50 atm. Is this cell galvanic or

electrolytic? Explain your answer.

Solution

Step 1: Write the cell reaction for the Galvanic cell involving the standard

hydrogen electrode (SHE) and the copper electrode. The two half-reactions are:

Copper Reduction: Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V

Hydrogen Oxidation: 2H+(aq)+2e−→H2(g)E◦= 0 V

The overall cell reaction is:

Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)

Step 2: Calculate the cell potential using the Nernst equation: The cell

potential, Ecell, is given by:

Ecell =E◦−0.0592

nlog Q

K

where Qis the reaction quotient and nis the number of moles of electrons

exchanged.

In this case, since the reaction quotient for the reaction is already at equi-

librium (Q=K), the cell potential simplifies to:

Ecell =E◦

Plugging in the values, we have:

Ecell = 0.34 V

Step 3: Determine if the cell is galvanic or electrolytic. Since the cell po-

tential (Ecell = 0.34 V) is positive, the reaction is spontaneous and the cell is

galvanic. In a galvanic cell, the anode is negative, and the cathode is positive.

Question 34

Question

Consider the following redox reaction:

Pb2+(aq) + 2I−(aq)→PbI2(s)

Part (a)

Explain the difference between a galvanic cell and an electrolytic cell, and pro-

vide an example of each type of cell.

Part (b)

Determine if the reaction given above is more likely to occur in a galvanic cell

or an electrolytic cell, and justify your answer.

Solution

Part (a)

A galvanic cell is a device that converts chemical energy into electrical energy

through a spontaneous redox reaction. In a galvanic cell, electrons flow from

the anode (where oxidation occurs) to the cathode (where reduction occurs)

through an external circuit. An example of a galvanic cell is the Daniell cell,

which consists of a zinc electrode immersed in a zinc sulfate solution and a

copper electrode immersed in a copper sulfate solution.

On the other hand, an electrolytic cell is a device that uses electrical energy

to drive a non-spontaneous redox reaction. In an electrolytic cell, an external

voltage source is used to drive the reaction in the opposite direction than it

would occur spontaneously. An example of an electrolytic cell is the electrolysis

of water, where water is split into hydrogen and oxygen gases using an external

electrical current.

Part (b)

For the given reaction:

Pb2+(aq) + 2I−(aq)→PbI2(s)

This reaction is more likely to occur in a galvanic cell. This is because the

reaction as written is spontaneous (positive standard cell potential) and would

release energy in the form of electricity. In a galvanic cell, the spontaneous

redox reaction occurs without the need for an external power source, allowing

for the conversion of chemical energy into electrical energy.

Question 35

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 both half-reactions occur at 25

C and standard conditions, answer the

following:

1. Write the overall cell reaction.

2. Calculate the standard cell potential for this galvanic cell.

3. Determine the equilibrium constant (K) for the cell reaction.

Solution

1. Step 1: Write the overall cell reaction.

Since the electrons cancel out in the overall cell reaction, we can add the

two half-reactions together:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

This is the overall cell reaction for the galvanic cell.

2. Step 2: Calculate the standard cell potential for this galvanic cell.

The standard cell potential is given by the difference in standard reduction

potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

From standard reduction potential tables, E◦

cathode = +0.34 V and E◦

anode =

−0.76 V. Thus,

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Therefore, the standard cell potential for this galvanic cell is 1.10 V.

Step 5: Finally, substitute the given values (E◦

cell = 0.50 V, T= 298 K, n=

1, R= 8.314 J/(mol·K), F= 96485 C/mol) into the equation to find the new

cell potential E′

cell. Calculate the natural logarithm term before substituting

into the equation.

Step 6: After performing the calculations, we find the new cell potential

E′

cell. This value will indicate the change in cell potential due to the change in

concentrations of the electrolytes in the half-cells.

Question 2

Question

Consider a galvanic cell with a standard cell potential of 1.15 V. If the cell

operates until the concentration of Zn2+ ions is reduced to 0.05 M from an

initial concentration of 1.0 M, calculate the amount of electrical work done by

the cell during this process. Assume complete reaction of Zn2+ ions and use

Faraday’s constant (F= 96,485 C/mol).

Solution

Step 1: Write the balanced redox reaction for the galvanic cell. The cell reaction

in a galvanic cell involving Zn2+ ions and a standard hydrogen electrode is:

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Given that the standard cell potential is 1.15 V, the standard electromotive

force of the cell (E◦

cell) is 1.15 V.

Step 2: Calculate the number of electrons transferred during the reaction.

From the balanced redox reaction, we see that 2 moles of electrons are trans-

ferred per mole of Zn2+ ion reacted.

Step 3: Calculate the number of moles of Zn2+ ions reacted. The initial

and final concentrations of Zn2+ ions are 1.0 M and 0.05 M, respectively. The

number of moles of Zn2+ ions reacted can be calculated using the formula:

moles of Zn2+ ions reacted = (initial concentration−final concentration)×volume

Assuming a volume of 1 L for simplicity, we find:

moles of Zn2+ ions reacted = (1.0−0.05) ×1 = 0.95 mol

Step 4: Calculate the electrical work done by the cell. The electrical work

done by the cell (Wcell) can be calculated using the formula:

Wcell =−nF ×Ecell

Where: - nis the number of moles of electrons transferred during the reaction,

-Fis Faraday’s constant, - Ecell is the standard cell potential (1.15 V in this

case).

Substitute the values to find:

Wcell =−(2 mol)(96,485 C/mol)(1.15 V) = −221,902.5 J

Therefore, the amount of electrical work done by the cell during this process is

-221,902.5 J.

Question 3

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 potentials are E◦

Zn2+/Zn =−0.76 V and E◦

Cu2+/Cu =

0.34 V, calculate the cell potential at standard conditions.

Solution

Step 1: Identify the species being oxidized and reduced. In a galvanic cell,

the species being oxidized occurs at the anode, while the species being reduced

occurs at the cathode. From the given half-reactions, we can see that zinc is

being oxidized at the anode (Zn −−→ Zn2+ + 2 e−) and copper ions are being

reduced at the cathode (Cu2+ + 2 e−−−→ Cu).

Step 2: Write the overall cell reaction. By adding the two half-reactions

together, we get the overall cell reaction:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 3: Calculate the cell potential at standard conditions. The cell potential

can be calculated using the equation:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the values for the standard reduction potentials, we get:

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Therefore, the cell potential at standard conditions for the galvanic cell is

1.10 V.

Question 4

Question

An electrolytic cell contains a solution of CuSO4and a strip of pure copper as the

cathode. A magnesium strip is also placed in the solution to act as an anode.

The standard reduction potentials are E◦

Cu2+/Cu = 0.34 V and E◦

Mg2+/Mg =

−2.37 V. Determine the cell potential of this electrolytic cell.

Solution

Step 1: Write the balanced redox reactions for the anode and cathode reactions.

The anode reaction involves the oxidation of Mg:

Mg →Mg2+ + 2e−

The cathode reaction involves the reduction of Cu2+:

Cu2+ + 2e−→Cu

Step 2: Calculate the cell potential (Ecell) using the formula:

Ecell =E◦

cathode −E◦

anode =E◦

Cu2+/Cu −E◦

Mg2+/Mg

Ecell = 0.34 V −(−2.37 V) = 2.71 V

Therefore, the cell potential of this electrolytic cell is 2.71 V.

Question 5

Question

Consider a galvanic cell with a standard cell potential of E◦

cell = 1.10 V. The

cell reaction is given by:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

If the concentration of Cu2+ is 0.10 M and the concentration of Zn2+ is 1.0

M, determine the cell potential under these conditions.

Solution

Step 1: Write the half-reactions for the cell reaction:

Anode: Zn(s)→Zn2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential using the standard reduction

potentials for Zn and Cu:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 3: Determine the reaction quotient, Q, using the given concentrations:

Q=[Zn2+][Cu(s)]

[Zn(s)][Cu2+]=(1.0)(1.0)

1= 1.0

Step 4: Use the Nernst equation to calculate the cell potential under the

given conditions:

Ecell =E◦

cell −0.0592 V

2log(Q) = 1.10 V −(0.0296 V) log(1.0) = 1.10 V

Therefore, the cell potential under the given conditions is 1.10 V .

Question 6

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)

Calculate the standard cell potential (E◦

cell) at 25

C for this galvanic cell.

Given E◦

cell values are:

E◦

cell(Zn2+/Zn) = −0.76 V

E◦

cell(Cu2+/Cu) = 0.34 V

Solution

Step 1: Identify the half-reactions and standard cell potential formulas.

In a galvanic cell, the standard cell potential (E◦

cell) can be calculated using

the formula:

E◦

cell =E◦

reduction(cathode) −E◦

reduction(anode)

Given half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−E◦

reduction(anode) = −0.76 V

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s) E◦

reduction(cathode) = 0.34 V

Step 2: Substitute the values into the formula to find E◦

cell.

E◦

cell =E◦

reduction(cathode) −E◦

reduction(anode)

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the standard cell potential (E◦

cell) at 25

C for this galvanic cell is

1.10 V.

Question 7

Question

Consider a galvanic cell with a standard cell potential of E◦

cell = 0.78 V. If the

cell potential drops to 0.66 V when a current of 2.5 A is drawn from the cell

for a certain period of time, calculate the amount of substance that has reacted

during this time.

Solution

Step 1: Identify the given information and the cell reaction. The cell reaction

for a galvanic cell can be written as follows:

Anode (oxidation) : M(s)→Mz+(aq) + ze−

Cathode (reduction) : X(aq) + ze−→X(s)

Given: E◦

cell = 0.78 V Ecell = 0.66 V I= 2.5 A

Step 2: Calculate the reaction quotient and the cell potential at the new

condition. The Nernst equation relates the cell potential to the reaction quotient

Ecell =E◦

cell −0.0592

zlog(Q)

Substitute the given values into the Nernst equation:

0.66 = 0.78 −0.0592

zlog(Q)

Step 3: Calculate the new reaction quotient Q. Solve the equation for Q:

Q= exp (0.78 −0.66) ·z

0.0592 

Step 4: Determine the change in Qand the amount of substance reacted.

Since Qis related to the amount of substance reacted, we can calculate the

change in Q:

∆Q=Qfinal −Qinitial

From the change in Q, the amount of substance reacted can be calculated using

stoichiometry.

Question 8

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)

on the left side and a zinc electrode on the right side. The cell has a standard

cell potential of E◦

cell = 1.10 V. If the concentration of Zn2+ ions in the zinc

half-cell is 0.001 M, determine the concentration of H+ions required in the

hydrogen half-cell to maintain the standard cell potential.

Solution

Step 1: Write the half-cell reactions for the galvanic cell. The overall cell reaction

is the sum of the reduction half-reactions for each electrode.

Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V

2H+(aq)+2e−→H2(g)E◦= 0.00 V

Step 2: Write the overall redox reaction for the cell.

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Step 3: Calculate the cell potential at non-standard conditions using the

Nernst Equation:

Ecell =E◦

cell −0.0592

nlog [Zn2+]

[H+]2

Given: E◦

cell = 1.10 V, [Zn2+] = 0.001 M.

Step 4: Solve for [H+] by setting Ecell equal to E◦

cell.

1.10 = 1.10 −0.0592

2log 0.001

[H+]2

Step 5: Simplify and solve for [H+].

0.0592

2log 0.001

[H+]2= 0

Step 6: Calculate [H+].

[H+] = 0.0316 M

Therefore, the concentration of H+ions required in the hydrogen half-cell

to maintain the standard cell potential is 0.0316 M.

Question 9

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a nickel

electrode. If the standard reduction potential of nickel is −0.25 V and the

standard reduction potential of the hydrogen electrode is 0.00 V, determine the

cell potential at standard conditions. Additionally, if the concentration of [Ni2+]

in the nickel half-cell is 1.0 M, calculate the cell potential when the concentration

of [H+] in the hydrogen half-cell is 0.1 M.

Solution

Step 1: Write the overall cell reaction for the galvanic cell. The overall cell reac-

tion for the galvanic cell can be written by subtracting the reduction potential

of the anode from the reduction potential of the cathode:

Ni2+(aq)+2e−→Ni(s)

2H+(aq)+2e−→H2(g)

Net cell reaction: Ni2+(aq) + 2H+(aq)→Ni(s)+H2(g)

Step 2: Calculate the standard cell potential (E◦) at standard conditions.

Given: Standard reduction potential of nickel, E◦

Ni =−0.25 V Standard reduc-

tion potential of the hydrogen electrode, E◦

H2= 0.00 V

The standard cell potential is calculated by summing the reduction potentials

of the cathode and anode:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

H2−E◦

Ni = 0.00 V −(−0.25 V) = 0.25 V

Therefore, the standard cell potential at standard conditions is 0.25 V.

Step 3: Calculate the cell potential under non-standard conditions. Given:

[Ni2+]=1.0 M [H+]=0.1 M

The Nernst equation relates the cell potential under non-standard conditions

to the standard cell potential:

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 electrons transferred in the cell reaction, Qis

the reaction quotient, and Kis the equilibrium constant.

The reaction quotient for the given cell is:

Q=[H+]2

[Ni2+]=(0.1)2

1.0= 0.01

The number of electrons transferred in the cell reaction is 2.

Using the Nernst equation:

E= 0.25 V −0.0592

2log(0.01)

E= 0.25 V + 0.0296 log(0.01)

E= 0.25 V + 0.0296(−2)

E= 0.25 V −0.0592

E= 0.19 V

Therefore, the cell potential under non-standard conditions is 0.19 V.

Question 10

Question

Consider a galvanic cell with a standard cell potential of E◦= 1.20 V. If the

cell operates until the concentration of Zn2+ ions in the anode compartment

becomes 0.0010 M and the concentration of Cu2+ ions in the cathode compart-

ment becomes 0.010 M, determine the number of electrons transferred during

this process.

Solution

Step 1: Write the half-reactions and their standard reduction potentials.

The half-reactions for this galvanic cell are:

Zn2+ + 2 e−−−→ Zn(s) (E◦=−0.76 V)

Cu2+ + 2 e−−−→ Cu(s) (E◦= 0.34 V)

Step 2: Determine the balanced cell reaction and overall cell potential.

The cell reaction is:

Zn(s) + Cu2+ −−→ Cu(s) + Zn2+

The overall cell potential (E◦

cell) is calculated by subtracting the reduction po-

tential of the anode half-reaction from the reduction potential of the cathode

half-reaction:

E◦

cell =E◦(cathode) −E◦(anode) = 0.34 V −(−0.76 V) = 1.10 V

Step 3: Use the Nernst equation to find the cell potential under non-standard

conditions.

The Nernst equation is:

E=E◦−0.0592

nlog [Cu2+]

[Zn2+]

where nis the number of moles of electrons transferred. Rearranging the equa-

tion to solve for ngives:

n=E◦−E

0.0592 log [Cu2+]

[Zn2+]

Substitute the given values into the equation to find n:

n=1.20 V −1.10 V

0.0592 log 0.010

0.0010

n=0.10 V

0.0592 log(10)

n= 1.68

Therefore, the number of electrons transferred during this process is 2 (rounded

to the nearest whole number).

Question 11

Question

Consider a galvanic cell with the following cell notation: Zn|Zn2+||Cu2+|Cu. If

the standard reduction potentials for Zn2+/Zn and Cu2+/Cu are −0.76 V and

0.34 V, respectively, calculate the cell potential at 25◦C for the galvanic cell.

Solution

Let’s denote the oxidation half-reaction at the anode as:

Zn(s)→Zn2+(aq)+2e−

And the reduction half-reaction at the cathode as:

Cu2+(aq)+2e−→Cu(s)

Given the standard reduction potentials, we have: E◦

Zn2+/Zn =−0.76 V and

E◦

Cu2+/Cu = 0.34 V.

Step 1: Calculate the cell potential at standard conditions using the stan-

dard reduction potentials:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 2: Determine the effect of temperature on the cell potential using the

Nernst equation:

Ecell =E◦

cell −0.0592

nlog [Cu2+]

[Zn2+]

Here, nis the number of electrons transferred (2 in this case) and [Cu2+] and

[Zn2+] are the concentrations of Cu2+ and Zn2+ ions.

Step 3: Since the cell is at standard conditions, [Cu2+] = [Zn2+] = 1 M:

Ecell = 1.10 V −0.0592

2log 1

1= 1.10 V

Therefore, the cell potential at 25◦C for the galvanic cell is 1.10 V .

Question 12

Question

Consider a galvanic cell in which the following reaction occurs:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

If the standard reduction potentials are Cu2+/Cu = +0.34 V and Zn2+/Zn =

-0.76 V, calculate the cell potential at 25

C. Is the reaction spontaneous?

Solution

Step 1: Write the half-reactions for the standard reduction potentials given.

Cu2+(aq)+2e−→Cu(s)E◦= +0.34 V

Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V

Step 2: Write the balanced overall cell reaction.

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 3: Use the standard reduction potentials to find the standard cell po-

tential (E◦

cell).

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Step 4: Calculate the non-standard cell potential (Ecell) using the Nernst

Equation:

Ecell =E◦

cell −0.0592 V

nlog(Q)

where Qis the reaction quotient, and for this reaction Q=[Zn2+]

[Cu2+].

Step 5: Determine the spontaneity of the reaction. Since the calculated cell

potential Ecell is positive, the cell reaction is spontaneous.

Question 13

Question

Consider a galvanic cell with a standard cell potential of 1.13 V. If the reaction

in this cell is spontaneous and the cell operates under standard conditions, what

can you say about the sign of ∆G◦? Explain your answer.

Solution

To determine the sign of ∆G◦, we can use the relationship between the standard

Gibbs free energy change, cell potential, and Faraday’s constant:

∆G◦=−nF E◦

where ∆G◦is the standard Gibbs free energy change, nis the number of

moles of electrons transferred in the reaction, Fis Faraday’s constant (96485

C/mol), and E◦is the standard cell potential.

Step 1: Given E◦= 1.13 V, we know that a positive cell potential indicates

a spontaneous reaction.

Since the reaction is spontaneous, we also know that ∆G◦<0.

Step 2: Substitute the values of n= 1 (assuming a single-electron transfer

process) and E◦= 1.13 V into the equation:

∆G◦=−1×96485 C/mol ×1.13 V

∆G◦=−109097.05 J/mol

Step 3: Since ∆G◦<0, we can conclude that ∆G◦is negative. This

confirms that the reaction is spontaneous under standard conditions.

Question 14

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V

2H+(aq) + 2e−→H2(g)E◦= 0.00 V

Determine the standard cell potential for this galvanic cell and state whether

the cell reaction is spontaneous or non-spontaneous.

Solution

Step 1: Write out the balanced overall cell reaction.

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Step 2: Calculate the standard cell potential by summing the standard re-

duction potentials for the half-reactions involved in the cell reaction.

E◦

cell =E◦

reduction, cathode −E◦

reduction, anode

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.00 V −(−0.76 V) = +0.76 V

Step 3: State whether the cell reaction is spontaneous or non-spontaneous.

Since the standard cell potential is positive (+0.76 V), the cell reaction is spon-

taneous.

Question 15

Question

Consider a galvanic cell constructed with a standard hydrogen electrode and a

platinum electrode immersed in a 1.0 M solution of CuSO4. The cell notation

for this galvanic cell is:

H2(g)|H+(aq)||Cu2+ (aq)|Cu(s)

Calculate the standard cell potential of this galvanic cell at 25

Solution

Step 1: Write the half-reactions for the galvanic cell. The half-reactions for the

given cell are:

Anode: 2H2(g)→4H+(aq)+4e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential (E◦

cell) using the standard reduc-

tion potentials (E◦

red) for each half-reaction. The standard reduction potentials

at 25

C are: E◦

red(Cu2+(aq)+2e−→Cu(s)) = 0.34 V

E◦

red(2H2(g)→4H+(aq)+4e−)=1.23 V

The standard cell potential is given by:

E◦

cell =E◦

red(cathode) −E◦

red(anode)

Therefore,

E◦

cell = 0.34 V −1.23 V = −0.89 V

So, the standard cell potential of the galvanic cell is -0.89 V.

Question 16

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq) + 2e−→Zn(s)E◦

red =−0.76 V

Fe3+(aq) + 3e−→Fe(s)E◦

red =−0.036 V

If the initial concentrations of Zn2+ and Fe3+ are both 1.0 M, determine the

cell potential after the cell has been operating for 1.5 hours. Assume that the

temperature is constant and that the reaction quotient Q remains at a constant

value of 1.

Solution

Step 1: Write the overall cell reaction.

The overall cell reaction is the sum of the reduction half-reactions for the

two species:

Zn2+(aq) + Fe3+(aq)→Zn(s) + Fe(s)

Step 2: Calculate the cell potential at standard state.

We can calculate the cell potential at standard state using the given standard

reduction potentials:

E◦

cell =E◦

red, cathode −E◦

red, anode

E◦

cell =E◦

Fe3+/Fe −E◦

Zn2+/Zn =−0.036 V −(−0.76 V) = 0.724 V

Step 3: Calculate the reaction quotient Q.

Since the reaction quotient Q is constant (equal to 1), the Nernst equation

simplifies to:

Ecell =E◦

cell −0.0592

nlog Q

Where n is the number of moles of electrons transferred in the balanced cell

reaction.

Step 4: Calculate the cell potential after 1.5 hours.

Since Q remains constant at 1, the cell potential after 1.5 hours is the same

as the cell potential at standard state:

Ecell = 0.724 V

Therefore, the cell potential after the cell has been operating for 1.5 hours

is 0.724 V.

Question 17

Question

Consider a galvanic cell with a standard cell potential of E◦= 0.95 V. The

half-reactions involved are:

Cathode: 2H2O(l) + 2e−→2OH−(aq)+H2(g), E◦= 0.83 V

Anode: 2I−(aq)→I2(s) + 2e−,E◦= 0.54 V

Calculate: (a) E◦for the overall reaction of the cell. (b) Equilibrium

constant Kfor the overall reaction of the cell. (c) The cell potential when

[OH−]=0.10 M and [I−]=0.20 M.

Solution

(a) To calculate the standard cell potential E◦for the overall reaction of the

cell, we can use the formula:

E◦

cell =E◦

cathode −E◦

anode

Given: E◦

cathode = 0.83 V, E◦

anode = 0.54 V

E◦

cell = 0.83 V −0.54 V = 0.29 V

(b) The equilibrium constant Kfor the overall reaction of the cell can be

determined using the formula:

E◦

cell =RT

nF ln K

Given: E◦

cell = 0.29 V, R= 8.314 J/(mol

K), T= 298 K, n= 2 (number of

moles of electrons transferred), F= 96485 C/mol

0.29 V = (8.314 J/(mol

K))(298 K)

(2)(96485 C/mol) ln K

ln K=(0.29 V)(2)(96485 C/mol)

(8.314 J/(mol

K))(298 K)

ln K= 6.45

K=e6.45 ≈644.00

E=E◦−RT

nF ln Q

Given: E◦= 0.29 V, T= 298 K, n= 2, F= 96485 C/mol

For the given concentrations, Q=[OH−]2

[I−]2=(0.10 M)2

(0.20 M)2= 0.25

E= 0.29 V −(8.314 J/(mol

K))(298 K)

(2)(96485 C/mol) ln 0.25

E≈0.24 V

Question 18

Question

Consider the following galvanic cell setup at 25

(s)Cu2+(0.01 M)|| Cu(s)

Determine if the cell reaction is spontaneous and if so, calculate the cell

potential. Assume standard reduction potential of E◦= 0.34 Vfor the Cu2+/Cu

redox couple.

Solution

Step 1: Write the overall cell reaction.

Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the cell potential at standard conditions using the standard

reduction potential.

E◦

cell =E◦

cathode −E◦

anode = 0 −0.34 = −0.34 V

Step 3: Determine the reaction quotient, Q.

Q=[Cu]1

[Cu2+]1= 1

Step 4: Calculate the cell potential using the Nernst equation:

Ecell =E◦

cell −0.0592

nlog(Q)

Ecell =−0.34 −0.0592

2log(1)

Ecell =−0.34 V

Step 5: Since Ecell is negative, the cell reaction is not spontaneous.

Question 19

Question

Consider a galvanic cell with the following half-reactions:

Zn2+(aq)+2e−→Zn(s)E◦=−0.76V

Cu2+(aq)+2e−→Cu(s)E◦= +0.34V

Calculate the cell potential at 25

C when the concentrations of Zn2+ and

Cu2+ are both 0.10 M. Determine if the cell is galvanic or electrolytic.

Solution

Step 1: Write the cell reaction by combining the given half-reactions.

Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)

Step 2: Calculate the cell potential at 25

C using the Nernst equation:

Ecell =E◦−0.0592

nlog Q

Where

Q=[products]

[reactants] =[Zn2+][Cu]

[Zn][Cu2+]=(0.10)2

(1)(1) = 0.01

n= 2 (number of electrons transferred)

Substitute E◦= 0.34V, Q= 0.01, n= 2 into the Nernst equation:

Ecell = 0.34V −0.0592

2log 0.01 = 0.34V −0.0296 log 0.01

Ecell = 0.34V −0.0296(−2) = 0.34V + 0.0592 = 0.3992V

Therefore, the cell potential at 25

C is 0.3992 V.

Step 3: Determine if the cell is galvanic or electrolytic. Since the cell poten-

tial is positive (0.3992 V), the cell is galvanic. Galvanic cells have positive cell

potentials and produce electrical energy from spontaneous redox reactions.

Question 20

Question

Consider a galvanic cell where the half-reactions are:

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 cell potential at 25

C when the concentrations are [Zn2+] =

0.10 M and [MnO−

4]=0.20 M.

Solution

Step 1: Write the overall cell reaction. The overall cell reaction is the sum of

the two half-reactions:

2 Zn2+(aq) + MnO−

4(aq) + 16H+(aq)→2 Zn(s) + Mn2+(aq)+8H2O(l)

Step 2: Calculate the standard cell potential, E◦

cell. Using the standard

reduction potentials given, the standard cell potential is:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 1.51 V −(−0.76 V) = 2.27 V

Step 3: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential to the standard cell potential and the

concentrations of the reactants and products:

Ecell =E◦

cell −0.0592

nlog(Q)

where Qis the reaction quotient, which is given by:

Q=[Mn2+]1

[Zn2+]2[MnO−

4]1[H+]16

Step 4: Calculate Qand the cell potential, Ecell. Substitute the given con-

centrations into the Qexpression:

Q=(0.20)1

(0.10)2(0.20)1(10−16)16

Q= 5.0×10−100

Substitute the values of E◦

cell,n= 5, and Qinto the Nernst equation to find

Ecell:

Ecell = 2.27 V −0.0592

5log5.0×10−100

Ecell = 2.27 V −0.0592

5×(−100)

Ecell = 2.27 V + 0.1184 = 2.3884 V

Therefore, the cell potential at 25

C with the given concentrations is 2.3884 V.

Question 21

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (SHE)

and a copper electrode. The standard reduction potential for the copper elec-

trode is +0.34 V. Calculate the cell potential at 25

C when the concentration

of Cu2+ is 1.0 M and the pressure of H2gas is 0.80 atm.

Solution

To calculate the cell potential of the galvanic cell, we can use the Nernst equa-

tion:

E=E◦−0.0592

nlog Q

where: - Eis the cell potential, - E◦is the standard cell potential, - nis the

number of electrons transferred in the cell reaction, - Qis the reaction quotient.

Step 1: Write the cell reaction. Since we have a standard hydrogen electrode

and a copper electrode, the cell reaction will be:

Cu2+(aq)+2e−→Cu(s)

Step 2: Calculate the standard cell potential (E◦). Given that the stan-

dard reduction potential for the copper electrode is +0.34 V, the standard cell

potential can be calculated as:

E◦=E◦

cathode −E◦

anode = 0 −(+0.34) = −0.34 V

Step 3: Calculate the reaction quotient (Q). The reaction quotient for the

cell reaction is given by:

Q=[Cu](Cu2+)

[H+]

Substitute the given concentration and pressure values into the equation to find

Step 4: Calculate the cell potential (E). Substitute E◦,Q, and n(which is

2 in this case) into the Nernst equation to find E.

Question 22

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s)→Zn2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

If the standard reduction potential for Zn is −0.76 V and for Cu is +0.34 V,

calculate the standard cell potential, E◦

cell, for this galvanic cell.

Solution

Step 1: The standard cell potential, E◦

cell, can be calculated using the formula:

E◦

cell =E◦

reduction,Cathode −E◦

reduction,Anode

Step 2: First, identify the reduction potentials for the cathode and anode

reactions:

E◦

reduction,Cathode = +0.34 V

E◦

reduction,Anode =−0.76 V

Step 3: Substitute the reduction potentials into the formula to determine

E◦

cell:

E◦

cell = +0.34 V −(−0.76 V)

E◦

cell = 0.34 V + 0.76 V

E◦

cell = 1.10 V

Therefore, the standard cell potential for this galvanic cell is 1.10 V.

Question 23

Question

Consider a galvanic cell where an aqueous solution of silver nitrate, AgNO3, is

in one half-cell with a silver electrode, and a solution of iron(III) chloride, FeCl3,

is in the other half-cell with an iron electrode. Write the overall cell reaction

and calculate the standard cell potential, E◦

cell, for this galvanic cell.

Solution

Step 1: Begin by writing the half-reactions for each half-cell. In the anode half-

cell: Fe(s)→Fe3+(aq)+3e−In the cathode half-cell: Ag+(aq) + e−→Ag(s)

Step 2: Write the overall cell reaction by adding the half-reactions. Fe(s) +

3Ag+(aq)→Fe3+(aq) + 3Ag(s)

Step 3: Determine the standard cell potential, E◦

cell, by considering the

standard reduction potentials for the half-reactions involved. From standard

reduction potential tables, E◦

cell =E◦

cathode −E◦

anode E◦

cell =E◦

Ag+/Ag −E◦

Fe3+/Fe

E◦

cell = 0.80 V −(−0.04 V) E◦

cell = 0.84 V

Therefore, the standard cell potential for this galvanic cell is 0.84 V.

Question 24

Question

Consider a galvanic cell where the following reaction takes place:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Given that the standard reduction potentials are E◦

Cu2+/Cu = 0.34 V and

E◦

Zn2+/Zn =−0.76 V, determine the standard cell potential (E◦

cell) for this cell.

Solution

Step 1: Write the overall cell reaction. The standard cell potential can be de-

termined using the standard reduction potentials of each half-reaction involved.

The cell reactions are written as reduction reactions, so we have the following

overall reaction:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 2: Calculate E◦

cell using the standard reduction potentials. The stan-

dard cell potential, E◦

cell, is given by the difference between the standard reduc-

tion potentials of the two half-reactions involved:

E◦

cell =E◦

cathode −E◦

anode

In this case, Cu is the cathode (reduction occurs) and Zn is the anode (oxi-

dation occurs). Therefore:

E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the standard cell potential for the given galvanic cell is 1.10 V .

Question 25

Question

A voltaic cell is constructed with a standard hydrogen electrode and a chromium

electrode. The standard reduction potential for the chromium electrode is

−0.744 V. Calculate the cell potential at 25

C for the voltaic cell described

above.

Solution

Step 1: Write the cell reaction and the standard cell potential. The cell reaction

for the voltaic cell is:

2H++ Cr →H2+ Cr2+

The standard cell potential, E◦

cell, can be determined using the standard

reduction potentials of the half-reactions involved:

E◦

cell =E◦

cathode −E◦

anode

Given that the standard reduction potential for the chromium electrode is

−0.744 V, the standard cell potential is:

E◦

cell = 0.000 V −(−0.744 V) = 0.744 V

Step 2: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential to the standard cell potential and the

reaction quotient, Q:

Ecell =E◦

cell −RT

nF ln Q

where: - R= 8.314 J mol−1K−1is the gas constant, - T= 25 ◦C = 298 K is the

temperature, - nis the number of electrons transferred in the cell reaction, -

F= 96485 C mol−1is the Faraday constant, and - Qis the reaction quotient.

For the given cell reaction:

2H++ Cr →H2+ Cr2+

n= 2 because 2 electrons are transferred.

The reaction quotient, Q, is given by:

Q=[H2][Cr2+]

[H+]2[Cr]

Since the concentrations are not provided in the question, the calculations

for the cell potential at 25

C cannot be completed.

Question 26

Question

Consider a galvanic cell with a standard cell potential of 0.85 V that uses the

following half-reactions:

Anode: Zn(s)→Zn2+(aq)+2e−E◦

Zn2+/Zn =−0.76 V

Cathode: MnO−

4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E◦

MnO−

4/Mn2+ = 1.51 V

Calculate the equilibrium constant, K, for the reaction in the cell and de-

termine if the cell reaction is spontaneous at 298 K.

Solution

Step 1: Write the balanced cell reaction and calculate the overall standard cell

potential.

The balanced cell reaction is:

Zn(s) + MnO−

4(aq)+8H+(aq)→Zn2+(aq) + Mn2+(aq)+4H2O(l)

The overall standard cell potential, E◦

cell, is the sum of the standard reduction

potentials of the half-reactions at the cathode and anode:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 1.51 V −(−0.76 V) = 2.27 V

Step 2: Use the Nernst equation to relate the equilibrium constant, K, to

the cell potential.

The Nernst equation relates the cell potential to the equilibrium constant:

Ecell =RT

nF ln K

where R= 8.314 J/(mol·K), T= 298 K, nis the number of electrons transferred

in the balanced cell reaction, and F= 96485 C/mol.

Step 3: Determine if the cell reaction is spontaneous at 298 K.

For a spontaneous reaction, the equilibrium constant, K, must be greater

than 1. Let’s solve the Nernst equation for K:

K= exp nF Ecell

RT = exp 5×96485 C/mol ×2.27 V

8.314 J/(mol ·K) ×298 K 

Calculating this expression gives us the value of K.

Question 27

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)

Calculate the standard cell potential for this galvanic cell at 25

C. Determine if

this cell is spontaneous or non-spontaneous.

Solution

Step 1: Write the overall cell reaction. The overall cell reaction can be obtained

by adding the half-reactions together:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 2: Write the standard cell potential equation. The standard cell po-

tential, Ecell, is given by the formula:

Ecell =Ecathode −Eanode

Step 3: Determine the standard reduction potentials. Lookup the standard

reduction potentials for the half-reactions:

ECu2+/Cu = 0.34 V

EZn2+/Zn =−0.76 V

Step 4: Calculate the standard cell potential. Substitute the values into the

standard cell potential formula:

Ecell = 0.34 V −(−0.76 V) = 1.10 V

Step 5: Determine spontaneity. Since the standard cell potential is positive

(1.10 V), the cell reaction is spontaneous.

Question 28

Question

In an electrochemical cell, a copper electrode is placed in a 1.0 M CuSO4solu-

tion, and a silver electrode is placed in a 1.0 M AgNO3solution. The standard

reduction potentials are as follows: Cu2+ + 2e−→Cu(s) with E◦= 0.34 V and

Ag++ e−→Ag(s) with E◦= 0.80 V. Determine the cell potential and predict

the direction of electron flow.

Solution

Step 1: Write the two half-reactions involving the reduction of Cu2+ and Ag+:

Cu2+ + 2e−→Cu(s) E◦= 0.34 V

Ag++ e−→Ag(s) E◦= 0.80 V

Step 2: Calculate the cell potential (E◦

cell) by subtracting the reduction

potential of the anode from the reduction potential of the cathode:

E◦

cell =E◦

cathode −E◦

anode = 0.80 V −0.34 V = 0.46 V

Step 3: Determine the direction of electron flow based on the cell poten-

tial. Since E◦

cell is positive, the reaction is spontaneous, and electrons will flow

from the anode (where oxidation occurs) to the cathode (where reduction oc-

curs). Therefore, the electrons will flow from the copper electrode to the silver

electrode.

Question 29

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s) →Zn2+(aq)+2e−

Cathode: 2Ag+(aq)+2e−→2Ag(s)

If the standard reduction potential for the Zn half-reaction is −0.76 V and the

standard reduction potential for the Ag half-reaction is 0.80 V, calculate the

standard cell potential (E◦

cell) of the galvanic cell.

Solution

Step 1: Write the half-reactions and determine which is the anode and which is

the cathode based on their reduction potentials. The anode is where oxidation

occurs (Zn to Zn2+) and the cathode is where reduction occurs (Ag+to Ag).

Anode: Zn(s) →Zn2+(aq)+2e−

Cathode: 2Ag+(aq)+2e−→2Ag(s)

Step 2: Identify the standard reduction potentials for the half-reactions.

The standard reduction potential for the Zn half-reaction, E◦

Zn, is −0.76 V and

for the Ag half-reaction, E◦

Ag, is 0.80 V.

Step 3: Calculate the standard cell potential, E◦

cell, using the equation:

E◦

cell =E◦

cathode −E◦

anode

Step 4: Substitute the values into the equation to calculate E◦

cell.

E◦

cell = 0.80 V −(−0.76 V)

E◦

cell = 0.80 V + 0.76 V

E◦

cell = 1.56 V

Therefore, the standard cell potential of the galvanic cell is 1.56 V .

Question 30

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 initial concentrations are [Zn2+]=1.0 M and [Cu2+]=0.1 M, calcu-

late the cell potential at standard conditions and when the cell reaches equilib-

rium. Determine whether this cell can function as an electrolytic cell without

an external potential applied.

Solution

Step 1: Calculate the cell potential at standard conditions using the Nernst

equation:

E◦

cell =E◦

cathode −E◦

anode = 0.34 V −(−0.76 V) = 1.10 V

Step 2: Calculate the cell potential when the cell reaches equilibrium by

using the Nernst equation:

Ecell =E◦

cell −RT

nF ln [Cu2+]

[Zn2+]2

Ecell = 1.10 V −(8.314 J/mol ·K)(298 K)

2(96485 C/mol) ln 0.1 M

(1.0 M)2

Ecell ≈1.0884 V

Step 3: Determine whether this cell can function as an electrolytic cell with-

out an external potential applied. Since the cell potential at equilibrium is

slightly less than the standard cell potential, the cell would not function as an

electrolytic cell without an external potential applied because the reaction is

not favorable to proceed spontaneously.

Question 31

Question

A galvanic cell consists of a standard hydrogen electrode (E◦

H+/H2= 0.00 V)

and a copper electrode in a 0.10 M solution of Cu2+ ions. The cell potential is

measured as 0.34 V at 25

Step 1: Write the overall redox reaction for the galvanic cell.

2H++ Cu2+ →Cu + H2

Step 2: Calculate the cell potential using the Nernst equation. The general

form of the Nernst equation is:

E=E◦−0.0592

nlog(Q)

where: - Eis the cell potential - E◦is the standard cell potential - nis the

number of moles of electrons transferred in the balanced redox reaction - Qis

the reaction quotient

For the given cell, E= 0.34 V, E◦=E◦

Cu2+/Cu −E◦

H+/H2,n= 2 (from the

balanced redox reaction).

Therefore, the Nernst equation becomes:

0.34 = E◦−0.0592

2log [Cu2+]

[H+]2

Step 3: Solve for E◦. Given that E◦

H+/H2= 0.00 V, we can simplify the

equation to:

0.34 = E◦−0.0296 log [Cu2+]

[H+]2

Since the standard concentration for H+is 1 M, the equation further simpli-

fies to:

0.34 = E◦−0.0296 log[Cu2+]

Therefore, the standard reduction potential of the copper electrode at 25

Question

Consider the following cell at 25

Mg(s)|Mg2+(0.10M)|| Fe3+(0.20M)|Fe(s)

Write the cell diagram, the overall cell reaction, and determine the cell potential

at standard conditions. Given: E◦(Fe3+/Fe) = 0.77 V and E◦(Mg2+/Mg) =

−2.37 V.

Solution

Step 1: Write the cell diagram The cell diagram can be written as:

Mg(s)|Mg2+(0.10M)|| Fe3+(0.20M)|Fe(s)

Step 2: Write the overall cell reaction The overall cell reaction can be written

by adding the two half-cell reactions together:

Mg(s) + 2Fe3+(aq)→Mg2+(aq) + 2Fe(s)

Step 3: Determine the cell potential at standard conditions The cell potential

at standard conditions can be calculated using the formula:

E◦

cell =E◦

cathode −E◦

anode

Given that E◦(Fe3+/Fe) = 0.77 V and E◦(Mg2+/Mg) = −2.37 V, we can

substitute these values into the formula:

E◦

cell = 0.77 V −(−2.37 V) = 3.14 V

Therefore, the cell potential at standard conditions for the given cell is 3.14

Question 33

Question

Consider a Galvanic cell consisting of a standard hydrogen electrode (SHE) and

a copper electrode. The standard reduction potential for the copper electrode

is E◦= 0.34 V. Calculate the cell potential when the concentration of Cu2+ is

0.10 M and the pressure of hydrogen gas is 0.50 atm. Is this cell galvanic or

electrolytic? Explain your answer.

Solution

Step 1: Write the cell reaction for the Galvanic cell involving the standard

hydrogen electrode (SHE) and the copper electrode. The two half-reactions are:

Copper Reduction: Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V

Hydrogen Oxidation: 2H+(aq)+2e−→H2(g)E◦= 0 V

The overall cell reaction is:

Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g)

Step 2: Calculate the cell potential using the Nernst equation: The cell

potential, Ecell, is given by:

Ecell =E◦−0.0592

nlog Q

K

where Qis the reaction quotient and nis the number of moles of electrons

exchanged.

In this case, since the reaction quotient for the reaction is already at equi-

librium (Q=K), the cell potential simplifies to:

Ecell =E◦

Plugging in the values, we have:

Ecell = 0.34 V

Step 3: Determine if the cell is galvanic or electrolytic. Since the cell po-

tential (Ecell = 0.34 V) is positive, the reaction is spontaneous and the cell is

galvanic. In a galvanic cell, the anode is negative, and the cathode is positive.

Question 34

Question

Consider the following redox reaction:

Pb2+(aq) + 2I−(aq)→PbI2(s)

Part (a)

Explain the difference between a galvanic cell and an electrolytic cell, and pro-

vide an example of each type of cell.

Part (b)

Determine if the reaction given above is more likely to occur in a galvanic cell

or an electrolytic cell, and justify your answer.

Solution

Part (a)

A galvanic cell is a device that converts chemical energy into electrical energy

through a spontaneous redox reaction. In a galvanic cell, electrons flow from

the anode (where oxidation occurs) to the cathode (where reduction occurs)

through an external circuit. An example of a galvanic cell is the Daniell cell,

which consists of a zinc electrode immersed in a zinc sulfate solution and a

copper electrode immersed in a copper sulfate solution.

On the other hand, an electrolytic cell is a device that uses electrical energy

to drive a non-spontaneous redox reaction. In an electrolytic cell, an external

voltage source is used to drive the reaction in the opposite direction than it

would occur spontaneously. An example of an electrolytic cell is the electrolysis

of water, where water is split into hydrogen and oxygen gases using an external

electrical current.

Part (b)

For the given reaction:

Pb2+(aq) + 2I−(aq)→PbI2(s)

This reaction is more likely to occur in a galvanic cell. This is because the

reaction as written is spontaneous (positive standard cell potential) and would

release energy in the form of electricity. In a galvanic cell, the spontaneous

redox reaction occurs without the need for an external power source, allowing

for the conversion of chemical energy into electrical energy.

Question 35

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 both half-reactions occur at 25

C and standard conditions, answer the

following:

1. Write the overall cell reaction.

2. Calculate the standard cell potential for this galvanic cell.

3. Determine the equilibrium constant (K) for the cell reaction.

Solution

1. Step 1: Write the overall cell reaction.

Since the electrons cancel out in the overall cell reaction, we can add the

two half-reactions together:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

This is the overall cell reaction for the galvanic cell.

2. Step 2: Calculate the standard cell potential for this galvanic cell.

The standard cell potential is given by the difference in standard reduction

potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

From standard reduction potential tables, E◦

cathode = +0.34 V and E◦

anode =

−0.76 V. Thus,

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Therefore, the standard cell potential for this galvanic cell is 1.10 V.

3. Step 3: Determine the equilibrium constant (K) for the cell reaction.

The standard change in Gibbs free energy (∆G◦) is related to the equilib-

rium constant (K) through the equation:

∆G◦=−nF E◦

cell =−RT ln K

where: n= total number of electrons transferred = 2,

F= Faraday’s constant = 96485 C/mol,

R= gas constant = 8.314 J/mol K,

T= temperature = 25 + 273 = 298 K.

First, calculate ∆G◦using the formula:

∆G◦=−nF E◦

cell =−2×96485 C/mol ×1.10 V

=−2×96485 C/mol ×1.10 J/C = −212067 J/mol

Next, rearrange the equation to solve for K:

ln K=−∆G◦

Plug in the values to find the equilibrium constant K:

K= exp −212067 J/mol

8.314 J/mol K ×298 K 

K= exp(−28.51) = 4.18 ×10−13

Therefore, the equilibrium constant (K) for the cell reaction is 4.18×10−13.

CHEM 121 - GENERAL CHEMISTRY I Galvanic and Electrolytic Cells Question Bank Set 3

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