CHEM 121 - GENERAL CHEMISTRY I Galvanic and Electrolytic Cells Question Bank Set 5

CHEM 121 - GENERAL CHEMISTRY

I - Galvanic and Electrolytic Cells

Question Bank - Set 5

Liberty University

Question 1

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (E◦

H+/H2=

0 V) and a copper electrode in a 1 M copper(II) sulfate solution (E◦

Cu2+/Cu =

0.34 V). Calculate the cell potential at 25

Solution

Step 1: Write the overall cell reaction by combining the half-reactions of the

two electrodes:

Cu2+(aq)+2e−→Cu(s) (Reduction half-reaction)

2H+(aq)+2e−→H2(g) (Oxidation half-reaction)

Cu2+(aq) + 2H+(aq)→Cu(s)+H2(g) (Overall cell reaction)

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

potentials for the half-reactions:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

Cu2+/Cu −E◦

H+/H2

E◦

cell = 0.34 V −0 V = 0.34 V

Step 3: Calculate the actual cell potential at 25

C using the Nernst equation:

Ecell =E◦

cell −RT

nF ln(Q)

where R= 8.314 J ·K−1·mol−1(gas constant), T= 298 K (temperature),

n= 2 (number of moles of electrons exchanged), F= 96485 C/mol (Faraday

constant), and Q=[Cu2+ ][H+]

[H2]=1

1= 1 (since the concentration of Cu2+ and

H+are both 1 M).

Ecell = 0.34 V −(8.314 J ·K−1·mol−1)(298 K)

2(96485 C/mol) ×ln(1)

Ecell = 0.34 V −(8.314 J ·K−1·mol−1)(298 K)

2(96485 C/mol) ×0

Ecell = 0.34 V

Therefore, the cell potential for the galvanic cell at 25

C is 0.34 V.

Question 2

Question

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

cell = 1.10 V. If

the cell operates until the concentration of Cu2+ ions in the copper half-cell

is reduced to 0.025 M, calculate the cell potential at this point assuming ideal

behavior under standard conditions (T= 25◦C).

Solution

Step 1: Write the cell reaction and Nernst equation for the given scenario. The

cell reaction for the given galvanic cell is:

Cu2+(aq)+2e−→Cu(s)

The Nernst equation is:

Ecell =E◦

cell −0.0592

nlog [Cu2+]

[Cu2+]◦

where: - Ecell is the cell potential at any given point, - E◦

cell is the standard cell

potential, - nis the number of moles of electrons transferred in the balanced

cell reaction, - [Cu2+] is the concentration of Cu2+ ions at any given point, -

[Cu2+]◦is the standard concentration of Cu2+ ions (usually 1 M), - 0.0592 is

the value of RT

Fat 25◦C, and - log denotes the logarithm to the base 10.

Step 2: Calculate the cell potential at the given point. Given that the initial

concentration of Cu2+ ions is 1 M and the final concentration is 0.025 M:

Ecell = 1.10 V −0.0592

2log 0.025

1

= 1.10 V −0.0296 ×log(0.025)

= 1.10 V −0.0296 ×(−1.6)

= 1.10 V −(−0.0474)

= 1.1474 V

Therefore, the cell potential at this point is 1.1474 V.

Question 3

Question

A galvanic cell is constructed with a copper electrode immersed in a 1.0 M

Cu2+ solution and a silver electrode immersed in a 1.0 M Ag+solution. If the

standard reduction potentials are E◦

Cu2+ /Cu = +0.34 V and E◦

Ag+/Ag = +0.80

V, determine the cell potential at 25

Solution

To determine the cell potential, we can use the formula:

Ecell =E◦

cathode −E◦

anode

where E◦

cathode is the standard reduction potential of the half-reaction at the

cathode and E◦

anode is the standard reduction potential of the half-reaction at

the anode.

Step 1: Write the overall cell reaction, with the cathode and anode half-

reactions:

At the cathode (Cu2+ reduction):

Cu2+(aq)+2e−→Cu(s)

At the anode (Ag+reduction):

Ag+(aq) + e−→Ag(s)

Overall cell reaction:

2Ag+(aq) + Cu(s)→2Ag(s) + Cu2+(aq)

Step 2: Calculate the cell potential by using the standard reduction poten-

tials:

Ecell =E◦

cathode −E◦

anode

Ecell =E◦

Ag+/Ag −E◦

Cu2+ /Cu

Ecell = (+0.80 V) −(+0.34 V)

Ecell = +0.46 V

Therefore, the cell potential at 25

C is +0.46 V.

Question 4

Question

Consider the following half-reactions:

Zn2+(aq) + 2e−→Zn(s)E◦=−0.76 V

2H+(aq) + 2e−→H2(g)E◦= 0.00 V

(a) Write the overall cell reaction for a galvanic cell constructed using these

half-reactions. Indicate the anode and cathode.

(b) Calculate the standard cell potential, E

cell, for the galvanic cell.

Solution

(a) The overall cell reaction for a galvanic cell constructed using these half-

reactions can be obtained by adding the two half-reactions. The half-reaction

with the more positive standard reduction potential will be reduced and the

other oxidized.

The overall cell reaction is:

Zn2+(aq) + 2H+(aq)→Zn(s)+H2(g)

In this cell reaction, Zn2+(aq)isreducedtoZn(s)atthecathode, while2H+(aq)isoxidizedtoH2(g)attheanode.

(b) The standard cell potential, E

cell, can be determined by adding the

standard reduction potentials for the cathode and anode half-reactions. The

standard cell potential, E◦

cell, is given by:

E◦

cell =E◦

cathode −E◦

anode

Substitute the given standard reduction potentials:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.00 V −(−0.76 V)

E◦

cell = 0.76 V

Therefore, the standard cell potential for the galvanic cell is 0.76 V.

Question 5

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 cell potential (E◦

cell) at 25

C for this galvanic cell using stan-

dard reduction potentials.

Solution

To calculate the cell potential, we will use the formula:

E◦

cell =E◦

cathode −E◦

anode

Step 1: Find the standard reduction potentials for each half-reaction. The

standard reduction potentials (E◦

red) for the given half-reactions are as follows:

E◦

red,Zn2++2 e−−−→Zn =−0.76 V

E◦

red,Cu2++2 e−−−→Cu = 0.34 V

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

cell.

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the cell potential (E◦

cell) at 25

C for this galvanic cell is 1.10 V.

Question 6

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 potentials are E◦

Zn2+/Zn =−0.76 V and E◦

Cu2+/Cu =

0.34 V, calculate the cell potential (E◦

cell) at 25◦C and comment on the spon-

taneity of the redox reaction.

Solution

1. Calculate E◦

cell:

The cell potential (E◦

cell) can be calculated using the formula:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode is the standard reduction potential of the cathode and

E◦

anode is the standard reduction potential of the anode.

Given data: E◦

Zn2+/Zn =−0.76 V E◦

Cu2+/Cu = 0.34 V

Therefore, E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Hence, the cell potential (E◦

cell) is 1.10 V.

2. Comment on spontaneity:

The redox reaction is spontaneous when E◦

cell >0. In this case, E◦

cell =

1.10 V, which is greater than 0. Therefore, the redox reaction is sponta-

neous at 25◦C. The positive cell potential indicates that the reaction will

proceed spontaneously from the anode (Zn) to the cathode (Cu).

Question 7

Question

Consider a galvanic cell with the following half-reactions:

Cathode: Cu2+(aq)+2e−→Cu(s)

Anode: Zn(s) →Zn2+(aq)+2e−

Calculate the cell potential at standard conditions (T = 298K, P = 1 atm, and

[Cu2+]=1.0M, [Zn2+] = 1.0M).

Solution

Step 1: Write the overall cell reaction by combining the two half-reactions:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 2: Write the cell potential equation using the standard reduction po-

tentials of the half-reactions. The standard reduction potentials (E◦) for the

given half-reactions are:

E◦

Cu2+/Cu = 0.34 V

E◦

Zn2+/Zn =−0.76 V

The cell potential (E◦

cell) can be calculated using the equation:

E◦

cell =E◦

cathode −E◦

anode

Plug in the values:

E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Step 3: Calculate the cell potential under the given standard conditions.

The Nernst equation relates the standard cell potential to the reaction quotient

(Q), temperature, and standard cell potential:

E=E◦

cell −0.0592

nlog(Q)

where nis the number of electrons transferred.

In this case, n= 2 because 2 electrons are transferred in both half-reactions.

Since the reaction is at standard conditions, Q= 1 and the Nernst equation

simplifies to:

E=E◦

cell

Substitute the value of E◦

cell into the equation:

E= 1.10 V

Therefore, the cell potential at standard conditions is 1.10 V.

Question 8

Question

Consider a Galvanic Cell with a standard cell potential of 1.10 V that operates

under standard conditions. If the cell potential drops to 0.95 V due to a con-

centration change in one of the half-cells, calculate the change in concentration

of the reactant. Assume the temperature and pressure remain constant. Given:



Standard reduction potential of the half-cell reaction: Cu2+(aq)+2e−→

Cu(s): E◦= 0.34 V



Concentration of Cu2+ ions in the original cell: 1.0 M

Solution

Step 1: Determine the half-cell reactions and their standard reduction poten-

tials. The two half-cell reactions involved are:

Zn(s) →Zn2+(aq)+2e−

Standard reduction potential for Zn half-cell: E◦

Zn =−0.76 V

Cu2+(aq)+2e−→Cu(s)

Given standard reduction potential for Cu half-cell: E◦

Cu = 0.34 V

Step 2: Calculate the standard cell potential based on the standard reduction

potentials of the two half-cell reactions. The standard cell potential (E◦

cell) is

given by:

E◦

cell =E◦

Cu −E◦

E◦

cell = 0.34 −(−0.76) = 1.10 V

Step 3: Calculate the reaction quotient Qat the lower cell potential of 0.95

V. At equilibrium, the cell potential is equal to zero. Therefore, at 0.95 V, the

reaction quotient Qis given by:

0.95 = E◦

cell −0.0592

2log Q

Substitute known values into the equation and solve for Q:

0.95 = 1.10 −0.0296 log Q

0.15 = 0.0296 log Q

log Q= 5.0676

Q= 105.0676 = 112260

Step 4: Calculate the new concentration of Cu2+ ions in the lower potential

cell. The reaction quotient Qis related to the concentrations of the products

and reactants in the cell. For the copper half-cell:

Q=[Cu](solid)

[Cu2+(aq)]2

Since the concentration of Cu solid does not change significantly, we can ap-

proximate the concentration of Cu2+ in the lower potential cell to be:

[Cu2+] = r1.0

Q=r1.0

112260

[Cu2+]≈0.01 M

Step 5: Calculate the change in concentration of the reactant. The change

in concentration of the reactant Cu2+ is given by:

∆[Cu2+] = [Cu2+]initial −[Cu2+]final

∆[Cu2+]=1.0−0.01

∆[Cu2+] = 0.99 M

Therefore, the change in concentration of the Cu2+ reactant is 0.99 M.

Question 9

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

Calculate the cell potential at 25◦C when the concentrations are [Zn2+] = 1.0

M and [H+]=0.10 M.

Solution

Step 1: Write the overall cell reaction based on the given half-reactions. Identify

the oxidizing and reducing agents. The overall cell reaction is the sum of the

two half-reactions:

Zn2+(aq) + 2H+(aq)→Zn(s)+H2(g)

Zinc is the reducing agent (it is oxidized) and hydrogen ions are the oxidizing

agent (they are reduced).

Step 2: Calculate the standard cell potential (E◦) using the standard reduc-

tion potentials given. The standard cell potential is given by:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the standard reduction potentials:

E◦

cell = 0.00 V −(−0.76 V) = 0.76 V

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

standard conditions. The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog Q

K

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

quotient.

Step 4: Calculate the cell potential with the given concentrations. First,

calculate the reaction quotient:

Q=[products]

[reactants] =PZn ·PH2

PZn2+ ·P2

H+

Substitute the concentrations and partial pressures:

Q=1·1 atm

1·(0.1)2= 10 atm

Step 5: Calculate the cell potential using the Nernst equation. Substitute

the values into the Nernst equation:

Ecell = 0.76 V −0.0592

2log(10) ≈0.716 V

Therefore, the cell potential at 25◦C when the concentrations are [Zn2+] =

1.0 M and [H+]=0.10 M is approximately 0.716 V.

Question 10

Question

Consider a galvanic cell with a standard voltage of 1.23 V. If the cell potential

drops to 1.10 V when the cell is operating, calculate the reaction quotient Qfor

the cell and determine if the reaction at the anode or cathode is favored.

Solution

Step 1: Write the Nernst equation to relate the cell potential (E), standard

cell potential (E◦), the reaction quotient (Q), the gas constant (R), absolute

temperature (T), and the Faraday constant (F).

E=E◦−RT

nF ln(Q)

Step 2: As the cell potential drops to 1.10 V, we have:

1.10 V = 1.23 V −(0.0257 V) ·T

nln(Q)

Step 3: Rearrange the equation to solve for ln(Q).

ln(Q) = (1.23 −1.10) V ·n

0.0257 V

ln(Q) = 0.13 V ·n

0.0257 V

ln(Q) = 5.05n

Step 4: Exponentiate both sides of the equation to solve for Q.

Q=e5.05n

Step 5: Since Qis based on the reaction quotient, we can determine that if

Q > 1, the reaction is product-favored (occurs spontaneously), and if Q < 1,

the reaction is reactant-favored (does not occur spontaneously).

Therefore, if Q > 1, the reaction at the anode is favored; if Q < 1, the

reaction at the cathode is favored.

Question 11

Question

Consider a galvanic cell with a standard cell potential of 0.98 V. If the con-

centration of Cu2+ ions is 0.10 M and the concentration of Zn2+ ions is 2.0 M,

determine the standard cell potential when the cell operates under non-standard

conditions with Cu2+ at 0.025 M and Zn2+ at 1.5 M.

Solution

Let’s label the half-reactions for the galvanic cell:

Zn(s)→Zn2+(aq)+2e−E◦=−0.76 V

Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V

The standard cell potential, E◦

cell, is calculated as follows:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Next, we will determine the reaction quotient Qunder the new concentra-

tions of Cu2+ and Zn2+:

Q=[Cu2+]

[Zn2+]=0.025

1.5= 0.0167

The Nernst equation is used to find the cell potential, Ecell, under non-

standard conditions:

Ecell =E◦

cell −0.0592

2log(Q)

Ecell = 1.10 V −(0.0296) log(0.0167)

Ecell = 1.10 V −(0.0296)(−1.78)

Ecell = 1.10 V + 0.0529

Ecell = 1.15 V

Therefore, the standard cell potential under non-standard conditions is 1.15 V.

Question 12

Question

A student is given a cell with a nickel electrode and a copper electrode. The

standard reduction potentials are E◦

Ni2+/Ni =−0.25 V and E◦

Cu2+/Cu = 0.34 V.

The student is asked to determine if a spontaneous redox reaction will occur

between these two electrodes.

Solution

Step 1: Write the overall cell reaction and calculate the standard cell potential,

E◦

cell. The overall cell reaction is: Ni(s) + Cu2+(aq) →Ni2+(aq) + Cu(s)

Using the standard reduction potentials given: E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

Cu2+/Cu −E◦

Ni2+/Ni E◦

cell = 0.34 V −(−0.25 V) E◦

cell = 0.59 V

Step 2: Determine the spontaneity of the redox reaction. Since E◦

cell is

positive (0.59 V), the cell reaction is spontaneous as written. This means that a

spontaneous redox reaction will occur between the nickel and copper electrodes.

Question 13

Question

Consider a galvanic cell where the standard cell potential (E◦

cell) is 1.02 V. If the

cell reaction is 2H+(aq) + Pb(s)→Pb2+(aq)+H2(g), determine the standard

reduction potential of the Pb half-reaction. Additionally, explain the difference

between a galvanic cell and an electrolytic cell.

Solution

Step 1: Calculate the standard reduction potential of the Pb half-reaction.

Given the standard cell potential E◦

cell = 1.02 V and the half-reaction:

2H+(aq) + Pb(s)→Pb2+(aq)+H2(g)

We know that the overall standard cell potential is the difference in the

standard reduction potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode is the reduction potential of the cathode (Pb in this case) and

E◦

anode is the reduction potential of the anode (H+in this case).

Given E◦

cell = 1.02 V, the reduction potential of the Pb half-reaction can be

calculated as:

E◦

Pb =E◦

cell −E◦

H+

E◦

Pb = 1.02 V −0 V

E◦

Pb = 1.02 V

Therefore, the standard reduction potential of the Pb half-reaction is 1.02

Step 2: Explain the difference between a galvanic cell and an electrolytic

cell. A galvanic cell is an electrochemical cell that produces electrical energy

from a spontaneous chemical reaction. It involves the conversion of chemical

energy into electrical energy. The cell reaction is driven by a negative Gibbs

free energy change (∆G < 0). Electrons flow from the anode (where oxidation

occurs) to the cathode (where reduction occurs).

On the other hand, an electrolytic cell is an electrochemical cell that con-

sumes electrical energy to drive a non-spontaneous chemical reaction. It in-

volves the conversion of electrical energy into chemical energy. The cell reaction

is driven by a positive Gibbs free energy change (∆G > 0). Electrons are forced

to flow from the external electrical source to the cathode (where reduction oc-

curs), and from the anode (where oxidation occurs) back to the electrical source.

In summary, the key difference between galvanic and electrolytic cells lies in

their energy source and the spontaneity of the cell reaction. Galvanic cells op-

erate spontaneously, producing electrical energy, while electrolytic cells require

external electrical energy input to drive non-spontaneous reactions.

Question 14

Question

Consider a galvanic cell with the following half-reactions:

Zn →Zn2+ + 2e−E◦=−0.76 V

Cu2+ + 2e−→Cu E◦= 0.34 V

If the concentrations of Zn2+ and Cu2+ are both 1.0 M, calculate the cell

potential at 25

C. Determine whether the cell is functioning as a galvanic cell

or as an electrolytic cell.

Solution

Step 1: Write the equation for the overall cell reaction. The overall cell reaction

is the combination of the two half-reactions. To do this, we reverse the first

half-reaction and add the two reactions together:

Cu2+ + Zn →Zn2+ + Cu

Step 2: Calculate the standard cell potential (E◦

cell). The standard cell

potential is the difference in standard reduction potentials for the two half-

reactions:

E◦

cell =E◦

cathode −E◦

anode E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

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

ratio of the concentrations of the products to the concentrations of the reactants

raised to the power of their stoichiometric coefficients:

Q=[Zn2+][Cu]

[Cu2+][Zn] = 1.0

Step 4: Calculate the cell potential (Ecell) at 25

C using the Nernst equation.

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

the reaction quotient:

Ecell =E◦

cell −0.0592

nlog(Q)

where nis the number of electrons transferred in the overall balanced equation.

For the given reaction, n= 2, so:

Ecell = 1.10 V −0.0592

2log(1.0)

Ecell = 1.10 V

Step 5: Conclusion. The cell potential is 1.10 V, which is positive. Therefore,

the cell is functioning as a galvanic cell since a positive cell potential indicates

a spontaneous redox reaction occurring in the cell.

Question 15

Question

Consider a galvanic cell with the following half-reactions:

Anode: Zn(s)→Zn2+(aq)+2e−

Cathode: 2H+(aq)+2e−→H2(g)

Determine the cell potential at standard conditions for this galvanic cell.

Solution

Step 1: Write the overall cell reaction.

Zn(s) + 2H+(aq)→Zn2+(aq)+H2(g)

Step 2: Find the standard reduction potentials for each half-reaction from a

table.

E◦

Zn2+/Zn =−0.76 V

E◦

H+/H2= 0 V

Step 3: Calculate the cell potential at standard conditions.

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0 V −(−0.76 V)

E◦

cell = 0.76 V

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

0.76 V.

Question 16

Question

A voltaic cell is constructed using a standard hydrogen electrode (SHE) as

the anode and an unknown metal electrode as the cathode. The measured cell

potential is 0.67 V. Determine the standard reduction potential for the unknown

metal electrode.

Solution

Step 1: Write the cell reaction.

The overall cell reaction for the voltaic cell is:

H+

(aq)+ e−→1

2H2(g)

Let the unknown metal electrode be represented as M:

Mn+

(aq)+ ne−→M(s)

The overall cell reaction can be written as:

Mn+

(aq)+n

2H2(g)→M(s)+ nH+

(aq)

Step 2: Determine the standard cell potential.

Given: Measured cell potential, Ecell = 0.67V

T hestandardcellpotential, Ecell, canbecalculatedusingtheNernstequation :Ecell =

Ecell −RT

nF ln QSince the cell is at standard conditions, Q = 1 and the equation

simplifies to:

Ecell =Ecell

Therefore, the standard cell potential is 0.67 V.

Step 3: Apply the standard cell potential.

By comparing with the standard reduction potential table, we know that the

standard reduction potential for the standard hydrogen electrode (SHE) is 0

V. The standard reduction potential for the unknown metal electrode can be

calculated as:

Ecell =Ecathode −Eanode

0.67 = Ecathode −0

Ecathode = 0.67 V

Therefore, the standard reduction potential for the unknown metal electrode

is 0.67 V.

Question 17

Question

Consider a galvanic cell with a standard cell potential of E◦= 0.87 V. The cell

reaction is given by:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Determine the standard cell potential for the corresponding electrolytic cell

involving the same half-reactions.

Solution

Step 1: Write the cell reaction for the electrolytic cell by reversing the reaction

for the galvanic cell.

Zn2+(aq) + Cu(s)→Zn(s) + Cu2+(aq)

Step 2: Determine the standard cell potential for the electrolytic cell. Since

the cell reaction has been reversed, the standard cell potential for the electrolytic

cell is the negative of the standard cell potential for the galvanic cell.

E◦

electrolytic =−E◦=−0.87 V

Therefore, the standard cell potential for the corresponding electrolytic cell

involving the same half-reactions is E◦

electrolytic =−0.87 V.

Question 18

Question

Consider a galvanic cell with a standard cell potential of E◦= 1.10 V. The

half-reactions in the cell are:

Anode: Zn(s) −−→ Zn2+(aq) + 2 e−

Cathode: Cu2+(aq) + 2 e−−−→ Cu(s)

Calculate the equilibrium constant Kfor the cell reaction at 25

Solution

Step 1: Write the cell reaction equation. The cell reaction equation is the sum

of the half-reactions at the anode and cathode:

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 2: Write expression for the equilibrium constant K. The equilibrium

constant Kfor the cell reaction is related to the standard cell potential E◦by

the Nernst equation:

E=E◦−RT

nF ln(K)

where nis the number of electrons transferred in the cell reaction, Ris the gas

constant, Tis the temperature in kelvin, and Fis the Faraday constant.

Step 3: Determine the number of electrons transferred n. From the balanced

cell reaction, we see that n= 2, as 2 electrons are transferred in the reaction.

Step 4: Plug in the known values and solve for K. Given: E◦= 1.10 V

R= 8.314 J/mol ·KT= 298 K F= 96485 C/mol

Substitute the values into the Nernst equation to solve for K:

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

(2)(96485 C/mol) ln(K)

Solving for ln(K):

ln(K) = −(8.314 J/mol ·K)(298 K)

(2)(96485 C/mol)

ln(K) = −0.0109

K=e−0.0109

K≈0.989

Therefore, the equilibrium constant Kfor the cell reaction at 25

C is ap-

proximately 0.989.

Question 19

Question

Consider a galvanic cell with the following half-cell 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 standard cell potential (E◦

cell) for this galvanic cell.

Solution

Step 1: Write the overall cell reaction.

Zn(s) + Cu2+(aq) −−→ Zn2+(aq) + Cu(s)

Step 2: Calculate the standard cell potential (E◦

cell) using the standard re-

duction potentials.

E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

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

Question 20

Question

Describe the differences between galvanic cells and electrolytic cells. Provide an

example of each and explain how they operate.

Solution

A galvanic cell is an electrochemical cell that converts chemical energy into

electrical energy through a spontaneous redox reaction, while an electrolytic

cell is an electrochemical cell that requires an external electrical energy source

to drive a non-spontaneous redox reaction.

Differences:



Energy Conversion: Galvanic cells convert chemical energy into elec-

trical energy, while electrolytic cells use electrical energy to drive a non-

spontaneous reaction.



Spontaneity: Galvanic cells have spontaneous redox reactions, while

electrolytic cells have non-spontaneous reactions.



Electrodes: In galvanic cells, the anode is negative and the cathode

is positive; in electrolytic cells, the anode is positive and the cathode is

negative.

Example of a Galvanic Cell (Voltaic Cell): The Daniell cell is a com-

mon example of a galvanic cell. In this cell, a zinc electrode is placed in a

solution of zinc sulfate, connected by a salt bridge to a copper electrode in a

solution of copper sulfate. The zinc undergoes oxidation at the anode, releasing

electrons, which flow through the external circuit to the copper cathode, where

reduction occurs.

Operation of a Galvanic Cell:



Step 1: Zinc metal atoms undergo oxidation at the anode, releasing elec-

trons:

Zn(s) →Zn2+(aq) + 2e−



Step 2: Electrons flow through the external circuit to the copper cathode.



Step 3: Copper ions in solution gain electrons and are reduced at the

cathode:

Cu2+(aq) + 2e−→Cu(s)



Overall Cell Reaction:

Zn(s) + Cu2+(aq) →Zn2+(aq) + Cu(s)

Example of an Electrolytic Cell: An example of an electrolytic cell is

the electrolysis of water. In this cell, an electric current is passed through water

to split it into hydrogen and oxygen gases.

Operation of an Electrolytic Cell:



Step 1: Water molecules undergo electrolysis to form oxygen gas at the

anode and hydrogen gas at the cathode:

2H2O(l) →2H2(g) + O2(g)

Question 21

Question

Consider the following electrochemical cell reaction:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Given that the standard reduction potentials are E◦

Zn2+/Zn =−0.76 V and

E◦

Cu2+/Cu = 0.34 V, determine if the reaction will be spontaneous under stan-

dard conditions. If not, calculate the minimum potential at which the reaction

will be spontaneous.

Solution

Step 1: Calculate the cell potential (E◦

cell) under standard conditions.

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76) V

E◦

cell = 1.10 V

Step 2: Determine if the reaction is spontaneous under standard conditions.

Since E◦

cell is positive, the reaction is spontaneous under standard conditions.

Step 3: Calculate the minimum potential to make the reaction spontaneous.

For a reaction to be spontaneous, the cell potential (Ecell) must be positive.

Ecell =E◦

cell −0.0592

nlog Q

where Qis the reaction quotient and nis the number of electrons transferred.

Step 4: Calculate the reaction quotient (Q) at equilibrium. Since the reac-

tion is at equilibrium when Ecell = 0, we have:

0=1.10 −0.0592

2log Q

log Q=1.10 ×2

0.0592

log Q= 37.1622

Q= 1037.1622

Step 5: Calculate the minimum potential required for the reaction to be

spontaneous.

Ecell = 0 = 1.10 −0.0592

2log Q′

Substitute Qwith Q′:

0=1.10 −0.0592

2log 1037.1622

0=1.10 −0.0592 ×37.1622

0 = 1.10 −1.0968

1.0968 = 1.10

1.0968 >1.10

Therefore, the minimum potential required for the reaction to be sponta-

neous is greater than 1.10 V.

Question 22

Question

Consider a galvanic cell with a standard cell potential of 0.78 V. If the cell

potential drops to 0.70 V at 25

C, calculate the change in standard Gibbs free

energy (∆G◦) for the reaction occurring in the cell.

Solution

Step 1: Calculate the change in cell potential (∆E◦): Given: Standard cell

potential, E◦= 0.78 V Cell potential at 25

C, E= 0.70 V

The change in cell potential can be calculated using the equation:

∆E◦=E◦−E

∆E◦= 0.78 −0.70 = 0.08 V

Step 2: Calculate the Faraday constant (F): The Faraday constant, F, is

given by:

F= 96485 C/mol

Step 3: Using the equation relating ∆G◦, ∆E◦, and n(the number of moles

of electrons transferred in the reaction), we have:

∆G◦=−nF ∆E◦

Step 4: Determine the number of moles of electrons transferred, n: Since the

cell potential dropped, we know the reaction is running in the reverse direction.

Therefore, the number of moles of electrons transferred, n, is:

n=−2

Step 5: Calculate the change in standard Gibbs free energy (∆G◦): Substi-

tute the values of n,F, and ∆E◦into the equation:

∆G◦=−(−2)(96485)(0.08)

∆G◦= 15435.76 J/mol

Therefore, the change in standard Gibbs free energy for the reaction occur-

ring in the cell is 15435.76 J/mol.

Question 23

Question

Consider a galvanic cell with a standard cell potential of 1.13 V that involves

the reaction:

Zn(s) + 2Fe3+(aq)→Zn2+(aq) + 2Fe2+(aq)

(a) Write the cell notation for this galvanic cell.

(b) Calculate the standard cell potential of the galvanic cell if [Fe3+]=0.20

M and [Fe2+]=2.00 M.

Solution

(a) The cell notation for the galvanic cell can be written as:

Zn(s)|Zn2+(aq)|| Fe3+(aq),Fe2+(aq)|Pt(s)

(b) The standard cell potential can be calculated using 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 K)), Tis the temperature in Kelvin, nis the number

of moles of electrons transferred in the balanced equation, Fis the Faraday

constant (96485 C/mol), and Qis the reaction quotient.

In this case, n= 2 because two moles of electrons are transferred in the

balanced equation. The reaction quotient Qcan be expressed as:

Q=[Zn2+]

[Fe3+][Fe2+]=1

(0.20)(2.00)2= 1.25

Substitute the given values into the Nernst equation:

E= 1.13 V −(8.314)(298)

2(96485) ln 1.25 ≈1.08 V

Question 24

Question

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

cell = 1.20 V. The cell

reaction is:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Determine the standard reduction potentials for the Zn and Cu half-reactions

and write the half-reactions. Also, calculate the standard cell potential for the

reverse reaction.

Solution

Step 1: Write the half-reactions and determine their standard reduction poten-

tials.

Zn2+ + 2e−→Zn E◦

Zn =−0.76 V

Cu2+ + 2e−→Cu E◦

Cu = 0.34 V

Step 2: Write the balanced overall cell reaction for the voltaic cell.

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 3: Calculate the standard cell potential for the reverse reaction. Given

the standard cell potential for the forward reaction is E◦

cell = 1.20 V, the stan-

dard cell potential for the reverse reaction is −E◦

cell.

E◦

cell, reverse =−1.20 V

Question 25

Question

Consider the following reaction in an electrochemical cell:

Pb2+(aq) + 2Ag(s)→Pb(s) + 2Ag+(aq)

Determine which electrode acts as the anode and which as the cathode. Explain

your reasoning.

Solution

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

The standard reduction potentials for the half-reactions involved are:

Ag+(aq) + e−→Ag(s)E◦= +0.80 V

Pb2+(aq)+2e−→Pb(s)E◦=−0.13 V

Step 2: Identify the half-reaction with the more positive standard reduction

potential as the reduction half-reaction (cathode). Since the reduction potential

for the reduction of Ag+to Ag is higher (+0.80 V), this is the reduction half-

reaction. Therefore, the Ag electrode acts as the cathode.

Step 3: Deduce the anode based on the cathode. The other electrode, the

Pb electrode, must therefore act as the anode.

Therefore, in the given cell setup: - The Ag electrode acts as the cathode. -

The Pb electrode acts as the anode.

Question 26

Question

Consider a galvanic cell and an electrolytic cell, both involving the same redox

reaction. The galvanic cell produces an electric current when the reaction occurs

spontaneously, while the electrolytic cell requires an external electric current to

drive the non-spontaneous reaction. Explain the key differences between these

two types of cells, focusing on their operating principles and applications.

Solution

Step 1: Galvanic Cell A galvanic cell, also known as a voltaic cell, converts

chemical energy into electrical energy. It consists of two half-cells connected by

a salt bridge or porous barrier. In the spontaneous redox reaction, electrons

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

occurs), generating an electric current. This current can be used to power

external devices.

Step 2: Electrolytic Cell An electrolytic cell uses an external electric current

to drive a non-spontaneous redox reaction. It consists of two electrodes (anode

and cathode) immersed in an electrolyte solution. When an electric current

is applied, electrons are forced to move against their natural direction. This

process allows for the reduction of cations at the cathode and the oxidation of

anions at the anode.

Step 3: Key Differences 1. Energy Conversion: Galvanic cells convert chem-

ical energy into electrical energy, while electrolytic cells require electrical energy

to drive a non-spontaneous reaction. 2. Spontaneity: Galvanic cells run spon-

taneously, while electrolytic cells need an external power source to operate. 3.

Electrode Reactions: In galvanic cells, oxidation occurs at the anode and reduc-

tion at the cathode. In electrolytic cells, the reactions are reversed due to the

external electric current. 4. Applications: Galvanic cells are used in batteries to

power devices, while electrolytic cells are used in processes like electroplating,

metal refining, and electrolysis of water.

In summary, galvanic cells produce electrical energy from chemical reactions,

while electrolytic cells use electrical energy to drive non-spontaneous reactions.

They have different applications and operate based on opposite principles of

redox reactions.

Question 27

Question

Consider a galvanic cell that consists of a standard hydrogen electrode (E◦

cell =

0.00 V) and a silver electrode in a 1.0 M solution of silver nitrate.

Part a: Write the overall cell reaction for this galvanic cell.

Part b: If the standard reduction potential of silver is E◦

Ag+/Ag = 0.80 V,

calculate the cell potential at standard conditions.

Part c: Will the cell potential increase or decrease if the concentration of

silver nitrate is increased? Justify your answer.

Solution

Part a:

The overall cell reaction for the galvanic cell involving a standard hydrogen

electrode and a silver electrode is given by:

2H+(aq) + 2e−→H2(g)E◦= 0.00 V

Ag+(aq)+e−→Ag(s)E◦= 0.80 V

Adding these two half-reactions together, we get the overall cell reaction:

2H+(aq) + 2Ag+(aq)→H2(g) + 2Ag(s)

Part b:

The cell potential E◦

cell at standard conditions can be calculated by sub-

tracting the reduction potential of the anode from the reduction potential of

the cathode:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.80 V −0.00 V = 0.80 V

Therefore, the cell potential at standard conditions is 0.80 V.

Part c:

If the concentration of silver nitrate is increased, the concentration of Ag+

in the cathode compartment increases. According to the Nernst equation, an

increase in the concentration of a reactant shifts the equilibrium towards prod-

ucts resulting in a higher cell potential. Therefore, increasing the concentration

of silver nitrate will increase the cell potential.

Question 28

Question

Consider a galvanic cell with the half-reactions:

Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V

Fe2+(aq)+2e−→Fe(s)E◦=−0.44 V

Suppose the standard cell potential is measured to be 0.78 V. Determine the

concentrations of Cu2+ and Fe2+ if the cell operates under standard conditions.

Solution

Step 1: Write the cell reaction. The cell reaction is obtained by adding the

half-reactions together:

Cu2+(aq) + Fe(s)→Cu(s) + Fe2+(aq)

Step 2: Calculate the standard cell potential. The standard cell potential,

E◦, is the difference between the standard reduction potentials of the two half-

reactions: E◦=E◦(cathode) −E◦(anode)

E◦= 0.34 V −(−0.44 V)

E◦= 0.78 V

Step 3: Determine the reaction quotient. Since the reaction is at equilibrium,

the reaction quotient, Q, is equal to 1.

Step 4: Use the Nernst equation to find the ratio of concentrations. The

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

the concentrations of species involved in the cell reaction:

E=E◦−0.0592

nlog Q

where n is the number of moles of electrons transferred in the cell reaction.

Step 5: Solve for the concentrations of Cu2+ and Fe2+. Given that Q = 1

in this case and n = 2, we can write:

0.78 V = 0.34 V −0.0592

2log 1

0.44 = 0.0592 log 1

0 = log 1

Thus, at standard conditions, the concentrations of Cu2+ and Fe2+ do not

change.

Question 29

Question

A student is studying galvanic and electrolytic cells in their electrochemistry

course. They are given a cell diagram but are unsure whether the cell is gal-

vanic or electrolytic. The diagram shows a zinc metal electrode dipping into

a solution of zinc sulfate and a silver metal electrode dipping into a solution

of silver nitrate. The two electrodes are connected by a wire. The student

must determine whether the cell is galvanic or electrolytic, and explain their

reasoning.

Solution

To determine whether the given cell is galvanic or electrolytic, we need to con-

sider the standard electrode potentials of the half-reactions involved.

Step 1: Write the half-reactions

The half-reaction at the zinc electrode is the oxidation of zinc metal:

Zn(s)→Zn2+(aq)+2e−

The half-reaction at the silver electrode is the reduction of silver ions:

Ag+(aq) + e−→Ag(s)

Step 2: Determine the standard electrode potentials

Given standard reduction potentials:

E◦

Zn2+/Zn =−0.76 V

E◦

Ag+/Ag = 0.80 V

Step 3: Determine if the cell is galvanic or electrolytic

To determine if the cell is galvanic or electrolytic, we can calculate the cell

potential using the standard reduction potentials.

The overall cell reaction is the sum of the two half-reactions:

Zn(s) + Ag+(aq)→Zn2+(aq) + Ag(s)

The overall standard cell potential (E◦

cell) can be calculated as follows:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

Ag+/Ag −E◦

Zn2+/Zn

E◦

cell = 0.80 V −(−0.76 V) = 1.56 V

Since the cell potential is positive (1.56 V), the cell is galvanic. In a galvanic

cell, the cell potential is positive, indicating that the spontaneous electrochem-

ical reaction occurs.

Question 30

Question

Consider a galvanic cell involving the reaction:

Cr3+(aq) + Zn(s)→Zn2+(aq) + Cr2+(aq)

The standard reduction potentials are:

E◦

Cr3+/Cr2+ =−0.74 V

E◦

Zn2+/Zn =−0.76 V

(a) Calculate the standard cell potential for this galvanic cell.

(b) Will the reaction in this galvanic cell proceed spontaneously at standard

conditions?

reactions will take place at the electrodes?

Solution

(a) To calculate the standard cell potential, we use the formula:

E◦

cell =E◦

cathode −E◦

anode

Given that reduction occurs at the cathode and oxidation occurs at the

anode, we have:

E◦

cell =E◦

Cr2+/Cr3+ −E◦

Zn2+/Zn

E◦

cell = (−0.74 V) −(−0.76 V)

E◦

cell = 0.02 V

Therefore, the standard cell potential for this galvanic cell is 0.02 V.

(b) For a reaction to proceed spontaneously, the standard cell potential

must be positive. Since the standard cell potential for this galvanic cell is

0.02 V (which is positive), the reaction will proceed spontaneously at standard

conditions.

the reactions at the electrodes will also be reversed. This means that Zn will be

plated at the cathode and Cr will be oxidized at the anode. The reactions will

be:

At the cathode (Zn):

Zn2+(aq)+2e−→Zn(s)

At the anode (Cr):

Cr2+(aq)→Cr3+(aq) + e−

Question 31

Question

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

cell = 1.10 V. If the

concentration of Fe3+ in one of the half-cells is 0.10 M and the concentration of

Fe2+ in the other half-cell is 1.0 M at 25◦C, determine the cell potential when

the concentration of Fe3+ in the first half-cell is reduced to 0.050 M and the

concentration of Fe2+ in the second half-cell is increased to 4.0 M.

Solution

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

for the cell are:

Fe3+ + 3e−→Fe2+ E◦=−0.77 V

Fe2+ →Fe + 2e−E◦=−0.44 V

The overall cell reaction is the sum of the two half-reactions:

Fe3+ + Fe2+ →Fe + Fe2+

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

standard cell potential is E◦= 1.10 V. The initial concentrations of Fe3+ and

Fe2+ are 0.10 M and 1.0 M, respectively. The Nernst equation is:

Ecell =E◦

cell −0.0592

nlog [Fe2+]

[Fe3+]

Substitute the values and calculate the initial cell potential:

Ecell = 1.10 −0.0592

2log 1.0

0.10

Ecell = 1.10 −0.0296 ×1≈1.0704 V

Step 3: Calculate the final cell potential after changing the concentrations:

The new concentrations are 0.050 M for Fe3+ and 4.0 M for Fe2+. Use the

Nernst equation again to calculate the new cell potential:

Ecell = 1.10 −0.0592

2log 4.0

0.050

Ecell = 1.10 −0.0296 ×3.322 ≈1.0068 V

Therefore, the cell potential after changing the concentrations is approxi-

mately 1.0068 V.

Question 32

Question

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

cell = 0.80 V. If the

reaction in the cell is spontaneous, what can you conclude about the standard

Gibbs free energy change for the cell reaction? Explain your answer.

Solution

Step 1: Recall the relationship between standard Gibbs free energy change

(∆G◦) and standard cell potential (E◦

cell):

∆G◦=−nF E◦

cell

where nis the number of electrons transferred in the cell reaction and Fis the

Faraday constant (96,485 C/mol).

Step 2: Since the cell reaction is spontaneous, it means that the standard

Gibbs free energy change for the reaction is negative:

∆G◦<0

Step 3: Substituting the given standard cell potential into the equation, we

have:

∆G◦=−nF (0.80 V)

Step 4: To determine the sign of ∆G◦, we need to consider the signs of n

and F. Since E◦

cell is positive, nmust also be positive to ensure a negative value

for ∆G◦.

Step 5: Therefore, we can conclude that the standard Gibbs free energy

change for the cell reaction is negative, indicating that the reaction is sponta-

neous and capable of performing work.

Question 33

Question

Consider a galvanic cell constructed with a copper electrode in a 1.0 M Cu2+

solution and a silver electrode in a 1.0 M Ag+solution. The standard reduction

potentials are E◦

Cu2+/Cu = 0.34 V and E◦

Ag+/Ag = 0.80 V. Determine the cell

potential when the concentration of Cu2+ is 0.10 M and Ag+is 0.50 M.

Solution

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

reactions that occur in this cell are: Cathode (Reduction at Ag+): Ag+(aq) +

e−→Ag(s) (E◦

Ag+/Ag = 0.80 V) Anode (Oxidation at Cu2+): Cu(s) →Cu2+(aq)+

2e−(E◦

Cu2+/Cu = 0.34 V)

Step 2: Calculate the cell potential under standard conditions: The cell

potential under standard conditions (E◦

cell) is given by:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.80 V −0.34 V = 0.46 V

Step 3: Use the Nernst equation to calculate the cell potential when the

concentrations are not standard: The Nernst equation is:

Ecell =E◦

cell −0.0592

nlog [Ag+]

[Cu2+]

where n is the number of electrons transferred in the balanced chemical equation.

In this case, n = 1.

Plugging in the values:

Ecell = 0.46 V −0.0592

1log 0.50

0.10

Ecell = 0.46 V −0.0592 log 5

Ecell = 0.46 V −0.11 V = 0.35 V

Therefore, the cell potential when the concentration of Cu2+ is 0.10 M and

Ag+is 0.50 M is 0.35 V.

Question 34

Question

Consider a galvanic cell with a standard cell potential of E◦= 1.20 V at 25◦C

that consists of a standard hydrogen electrode (Pt—H2(1 atm)—H+) and a

silver electrode (Ag). Calculate the cell potential when the concentration of

Ag+is 0.10 M and the pressure of hydrogen gas is 0.5 atm at 25◦C.

Solution

Step 1: Write the cell reaction for the galvanic cell:

Pt(s) + 2H+(aq) + 2e−→H2(g)

Ag(s)→Ag+(aq)+e−

Step 2: Write the balanced overall cell reaction:

Pt(s) + 2Ag+(aq)→2Ag(s) + 2H+(aq)

Step 3: Use the Nernst equation to calculate the cell potential:

E=E◦−0.0592

nlog Q

where Eis the cell potential, E◦is the standard cell potential, nis the number

of electrons transferred in the balanced cell reaction, Qis the reaction quotient,

and Kis the equilibrium constant.

Step 4: Calculate the reaction quotient Qand the number of electrons trans-

ferred nbased on the balanced cell reaction:

Q=[Ag+]2

[H+]2=(0.10)2

(1)2

n= 2

Step 5: Substitute the values into the Nernst equation:

E= 1.20 −0.0592

2log 0.01

E= 1.20 −0.0296 log(0.01)

E= 1.20 −0.0296(−2)

E= 1.20 + 0.0592

E= 1.2592 V

Therefore, the cell potential when the concentration of Ag+is 0.10 M and

the pressure of hydrogen gas is 0.5 atm at 25◦C is 1.2592 V.

Question 35

Question

Consider a galvanic cell in which the following reaction occurs:

Zn(s) — Zn2+(aq, 1.00 M)||Cu2+(aq, 1.00 M) — Cu(s)

Calculate the cell potential at 25

C. Given: E◦

cell(Cu2+/Cu) = +0.34 V and

E◦

cell(Zn2+/Zn) = −0.76 V.

Solution

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

by summing the half-reactions for each electrode. The standard cell potential,

E◦

cell, is the difference between the standard reduction potentials of the half-

reactions. The two half-reactions and their standard reduction potentials are:

Zn2+(aq)+2e−→Zn(s)E◦=−0.76 V

Question 5

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 cell potential (E◦

cell) at 25

C for this galvanic cell using stan-

dard reduction potentials.

Solution

To calculate the cell potential, we will use the formula:

E◦

cell =E◦

cathode −E◦

anode

Step 1: Find the standard reduction potentials for each half-reaction. The

standard reduction potentials (E◦

red) for the given half-reactions are as follows:

E◦

red,Zn2++2 e−−−→Zn =−0.76 V

E◦

red,Cu2++2 e−−−→Cu = 0.34 V

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

cell.

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Therefore, the cell potential (E◦

cell) at 25

C for this galvanic cell is 1.10 V.

Question 6

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 potentials are E◦

Zn2+/Zn =−0.76 V and E◦

Cu2+/Cu =

0.34 V, calculate the cell potential (E◦

cell) at 25◦C and comment on the spon-

taneity of the redox reaction.

Solution

1. Calculate E◦

cell:

The cell potential (E◦

cell) can be calculated using the formula:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode is the standard reduction potential of the cathode and

E◦

anode is the standard reduction potential of the anode.

Given data: E◦

Zn2+/Zn =−0.76 V E◦

Cu2+/Cu = 0.34 V

Therefore, E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Hence, the cell potential (E◦

cell) is 1.10 V.

2. Comment on spontaneity:

The redox reaction is spontaneous when E◦

cell >0. In this case, E◦

cell =

1.10 V, which is greater than 0. Therefore, the redox reaction is sponta-

neous at 25◦C. The positive cell potential indicates that the reaction will

proceed spontaneously from the anode (Zn) to the cathode (Cu).

Question 7

Question

Consider a galvanic cell with the following half-reactions:

Cathode: Cu2+(aq)+2e−→Cu(s)

Anode: Zn(s) →Zn2+(aq)+2e−

Calculate the cell potential at standard conditions (T = 298K, P = 1 atm, and

[Cu2+]=1.0M, [Zn2+] = 1.0M).

Solution

Step 1: Write the overall cell reaction by combining the two half-reactions:

Zn(s) + Cu2+(aq)→Zn2+(aq) + Cu(s)

Step 2: Write the cell potential equation using the standard reduction po-

tentials of the half-reactions. The standard reduction potentials (E◦) for the

given half-reactions are:

E◦

Cu2+/Cu = 0.34 V

E◦

Zn2+/Zn =−0.76 V

The cell potential (E◦

cell) can be calculated using the equation:

E◦

cell =E◦

cathode −E◦

anode

Plug in the values:

E◦

cell =E◦

Cu2+/Cu −E◦

Zn2+/Zn

E◦

cell = 0.34 V −(−0.76 V)

E◦

cell = 1.10 V

Step 3: Calculate the cell potential under the given standard conditions.

The Nernst equation relates the standard cell potential to the reaction quotient

(Q), temperature, and standard cell potential:

E=E◦

cell −0.0592

nlog(Q)

where nis the number of electrons transferred.

In this case, n= 2 because 2 electrons are transferred in both half-reactions.

Since the reaction is at standard conditions, Q= 1 and the Nernst equation

simplifies to:

E=E◦

cell

Substitute the value of E◦

cell into the equation:

E= 1.10 V

Therefore, the cell potential at standard conditions is 1.10 V.

Question 8

Question

Consider a Galvanic Cell with a standard cell potential of 1.10 V that operates

under standard conditions. If the cell potential drops to 0.95 V due to a con-

centration change in one of the half-cells, calculate the change in concentration

of the reactant. Assume the temperature and pressure remain constant. Given:



Standard reduction potential of the half-cell reaction: Cu2+(aq)+2e−→

Cu(s): E◦= 0.34 V



Concentration of Cu2+ ions in the original cell: 1.0 M

Solution

Step 1: Determine the half-cell reactions and their standard reduction poten-

tials. The two half-cell reactions involved are:

Zn(s) →Zn2+(aq)+2e−

Standard reduction potential for Zn half-cell: E◦

Zn =−0.76 V

Cu2+(aq)+2e−→Cu(s)

Given standard reduction potential for Cu half-cell: E◦

Cu = 0.34 V

Step 2: Calculate the standard cell potential based on the standard reduction

potentials of the two half-cell reactions. The standard cell potential (E◦

cell) is

given by:

E◦

cell =E◦

Cu −E◦

E◦

cell = 0.34 −(−0.76) = 1.10 V

Step 3: Calculate the reaction quotient Qat the lower cell potential of 0.95

V. At equilibrium, the cell potential is equal to zero. Therefore, at 0.95 V, the

reaction quotient Qis given by:

0.95 = E◦

cell −0.0592

2log Q

Substitute known values into the equation and solve for Q:

0.95 = 1.10 −0.0296 log Q

0.15 = 0.0296 log Q

log Q= 5.0676

Q= 105.0676 = 112260

Step 4: Calculate the new concentration of Cu2+ ions in the lower potential

cell. The reaction quotient Qis related to the concentrations of the products

and reactants in the cell. For the copper half-cell:

Q=[Cu](solid)

[Cu2+(aq)]2

Since the concentration of Cu solid does not change significantly, we can ap-

proximate the concentration of Cu2+ in the lower potential cell to be:

[Cu2+] = r1.0

Q=r1.0

112260

[Cu2+]≈0.01 M

Step 5: Calculate the change in concentration of the reactant. The change

in concentration of the reactant Cu2+ is given by:

∆[Cu2+] = [Cu2+]initial −[Cu2+]final

∆[Cu2+]=1.0−0.01

∆[Cu2+] = 0.99 M

Therefore, the change in concentration of the Cu2+ reactant is 0.99 M.

Question 9

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

Calculate the cell potential at 25◦C when the concentrations are [Zn2+] = 1.0

M and [H+]=0.10 M.

Solution

Step 1: Write the overall cell reaction based on the given half-reactions. Identify

the oxidizing and reducing agents. The overall cell reaction is the sum of the

two half-reactions:

Zn2+(aq) + 2H+(aq)→Zn(s)+H2(g)

Zinc is the reducing agent (it is oxidized) and hydrogen ions are the oxidizing

agent (they are reduced).

Step 2: Calculate the standard cell potential (E◦) using the standard reduc-

tion potentials given. The standard cell potential is given by:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the standard reduction potentials:

E◦

cell = 0.00 V −(−0.76 V) = 0.76 V

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

standard conditions. The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog Q

K

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

quotient.

Step 4: Calculate the cell potential with the given concentrations. First,

calculate the reaction quotient:

Q=[products]

[reactants] =PZn ·PH2

PZn2+ ·P2

H+

Substitute the concentrations and partial pressures:

Q=1·1 atm

1·(0.1)2= 10 atm

Step 5: Calculate the cell potential using the Nernst equation. Substitute

the values into the Nernst equation:

Ecell = 0.76 V −0.0592

2log(10) ≈0.716 V

Therefore, the cell potential at 25◦C when the concentrations are [Zn2+] =

1.0 M and [H+]=0.10 M is approximately 0.716 V.

Question 10

Question

Consider a galvanic cell with a standard voltage of 1.23 V. If the cell potential

drops to 1.10 V when the cell is operating, calculate the reaction quotient Qfor

the cell and determine if the reaction at the anode or cathode is favored.

Solution

Step 1: Write the Nernst equation to relate the cell potential (E), standard

cell potential (E◦), the reaction quotient (Q), the gas constant (R), absolute

temperature (T), and the Faraday constant (F).

E=E◦−RT

nF ln(Q)

Step 2: As the cell potential drops to 1.10 V, we have:

1.10 V = 1.23 V −(0.0257 V) ·T

nln(Q)

Step 3: Rearrange the equation to solve for ln(Q).

ln(Q) = (1.23 −1.10) V ·n

0.0257 V

ln(Q) = 0.13 V ·n

0.0257 V

ln(Q) = 5.05n

Step 4: Exponentiate both sides of the equation to solve for Q.

Q=e5.05n

Step 5: Since Qis based on the reaction quotient, we can determine that if

Q > 1, the reaction is product-favored (occurs spontaneously), and if Q < 1,

the reaction is reactant-favored (does not occur spontaneously).

Therefore, if Q > 1, the reaction at the anode is favored; if Q < 1, the

reaction at the cathode is favored.

Question 11

Question

Consider a galvanic cell with a standard cell potential of 0.98 V. If the con-

centration of Cu2+ ions is 0.10 M and the concentration of Zn2+ ions is 2.0 M,

determine the standard cell potential when the cell operates under non-standard

conditions with Cu2+ at 0.025 M and Zn2+ at 1.5 M.

Solution

Let’s label the half-reactions for the galvanic cell:

Zn(s)→Zn2+(aq)+2e−E◦=−0.76 V

Cu2+(aq)+2e−→Cu(s)E◦= 0.34 V

The standard cell potential, E◦

cell, is calculated as follows:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.34 V −(−0.76 V) = 1.10 V

Next, we will determine the reaction quotient Qunder the new concentra-

tions of Cu2+ and Zn2+:

Q=[Cu2+]

[Zn2+]=0.025

1.5= 0.0167

The Nernst equation is used to find the cell potential, Ecell, under non-

standard conditions:

Ecell =E◦

cell −0.0592

2log(Q)

Ecell = 1.10 V −(0.0296) log(0.0167)

Ecell = 1.10 V −(0.0296)(−1.78)

Ecell = 1.10 V + 0.0529

Ecell = 1.15 V

Therefore, the standard cell potential under non-standard conditions is 1.15 V.

Question 12

Question

A student is given a cell with a nickel electrode and a copper electrode. The

standard reduction potentials are E◦

Ni2+/Ni =−0.25 V and E◦

Cu2+/Cu = 0.34 V.

The student is asked to determine if a spontaneous redox reaction will occur

between these two electrodes.

Solution

Step 1: Write the overall cell reaction and calculate the standard cell potential,

E◦

cell. The overall cell reaction is: Ni(s) + Cu2+(aq) →Ni2+(aq) + Cu(s)

Using the standard reduction potentials given: E◦

cell =E◦

cathode −E◦

anode

E◦

cell =E◦

Cu2+/Cu −E◦

Ni2+/Ni E◦

cell = 0.34 V −(−0.25 V) E◦

cell = 0.59 V

Step 2: Determine the spontaneity of the redox reaction. Since E◦

cell is

positive (0.59 V), the cell reaction is spontaneous as written. This means that a

spontaneous redox reaction will occur between the nickel and copper electrodes.

Question 13

Question

Consider a galvanic cell where the standard cell potential (E◦

cell) is 1.02 V. If the

cell reaction is 2H+(aq) + Pb(s)→Pb2+(aq)+H2(g), determine the standard

reduction potential of the Pb half-reaction. Additionally, explain the difference

between a galvanic cell and an electrolytic cell.

Solution

Step 1: Calculate the standard reduction potential of the Pb half-reaction.

Given the standard cell potential E◦

cell = 1.02 V and the half-reaction:

2H+(aq) + Pb(s)→Pb2+(aq)+H2(g)

We know that the overall standard cell potential is the difference in the

standard reduction potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode is the reduction potential of the cathode (Pb in this case) and

E◦

anode is the reduction potential of the anode (H+in this case).

Given E◦

cell = 1.02 V, the reduction potential of the Pb half-reaction can be

calculated as:

E◦

Pb =E◦

cell −E◦

H+

E◦

Pb = 1.02 V −0 V

E◦

Pb = 1.02 V

Therefore, the standard reduction potential of the Pb half-reaction is 1.02

Step 2: Explain the difference between a galvanic cell and an electrolytic

cell. A galvanic cell is an electrochemical cell that produces electrical energy

from a spontaneous chemical reaction. It involves the conversion of chemical

energy into electrical energy. The cell reaction is driven by a negative Gibbs

free energy change (∆G < 0). Electrons flow from the anode (where oxidation

occurs) to the cathode (where reduction occurs).