Solved Examples on Cell Potential

Example 1: An example of cell potential can be demonstrated using the redox reaction between zinc (Zn) and copper (Cu) in an electrochemical cell. The overall reaction can be written as:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

The standard reduction potential of Cu²⁺/Cu is +0.34 V, while the standard reduction potential of Zn²⁺/Zn is -0.76 V. Therefore, the Cu²⁺/Cu half-reaction will act as the cathode half-reaction and the Zn²⁺/Zn half-reaction will act as the anode half-reaction.

Solution:

The cell potential can be calculated using the formula:

E_cell = E°_cathode – E°_anode

E°_cathode = +0.34 V (Cu²⁺/Cu)
E°_anode = -0.76 V (Zn²⁺/Zn)

Ecell = +0.34 V – (-0.76 V) = +1.10 V

The positive value of the cell potential indicates that the reaction is spontaneous in the forward direction, i.e., Zn(s) is oxidized and Cu²⁺(aq) is reduced. The reaction proceeds until the concentrations of the reactants and products reach equilibrium.

To measure the actual cell potential under non-standard conditions, the Nernst equation can be used. For example, if the concentrations of Cu²⁺ and Zn²⁺ are both 1.00 M and the cell is at 25°C, the cell potential can be calculated as,

E_cell = E°_cell – (RT/nF) ln(Q)

Q = [Cu⁺²]/[Zn⁺²] = 1.00/1.00 = 1.00

Plugging in the values, we get

E_cell = +1.10 V – (0.0257 V) ln(1.00) = +1.10 V

Thus, the cell potential is the same as the standard cell potential under these conditions.

Example 2: Consider the following reaction: Cu(s) + Ag⁺(aq) -> Cu²⁺(aq) + Ag(s)

Standard reduction potentials are,

• Cu²⁺(aq) + 2e^– -> Cu(s)                (E° = +0.34 V)
• Ag⁺(aq) + e^– -> Ag(s)                  (E° = +0.80 V)

A) Write the balanced equation for the cell reaction and calculate the standard cell potential. 

B) If the initial concentrations of Cu2+ and Ag+ are both 0.1 M and the cell operates at standard conditions, calculate the cell potential.

C) If the initial concentration of Cu2+ is 1 M and the initial concentration of Ag+ is 0.1 M, calculate the cell potential.

Solution:

A) The balanced equation for the cell reaction is,

Cu(s) + 2Ag⁺(aq) -> Cu²⁺(aq) + 2Ag(s)

Standard cell potential is calculated using the formula:

E°_cell = E°_reduction (reduced species) – E°_reduction (oxidized species)

E°_cell = E°Ag⁺ – E°Cu²⁺

E°_cell = 0.80 V + 0.34 V – 0.34 V – 0.80 V

E°_cell = -0.50 V

B) If the initial concentrations of Cu2+ and Ag+ are both 0.1 M, the cell operates at standard conditions and the Nernst equation can be used:

E_cell = E°_cell – (RT/nF)lnQ

At standard conditions, Q = 1 and lnQ = 0, so:

E_cell = E°_cell = -0.50 V

C) If the initial concentration of Cu²⁺ is 1 M and the initial concentration of Ag⁺ is 0.1 M, the reaction quotient Q is:

Q = [Cu²⁺][Ag⁺]²/ [Ag⁺]²[Cu] = [Cu²⁺] / [Ag⁺]

Q = 1 / 0.1 = 10

Cell Potential can be calculated using the Nernst equation,

E_cell = E°_cell – (RT/nF)lnQ

E_cell = -0.50 V – (0.0257 V/K)(298 K/2)(ln10)

E_cell = -0.50 V – 0.059 V

E_cell = -0.559 V

Example 3: Consider the following redox reaction, 2 Fe³⁺(aq) + 2 I^–(aq) → 2 Fe²⁺(aq) + I₂(s). Calculate the cell potential.

Solution:

To calculate the cell potential, we need to find the reduction potentials for each half-reaction and then subtract the reduction potential of the anode from the reduction potential of the cathode.

Half-reactions for the given reaction are,

Fe³⁺(aq) + e^– → Fe²⁺(aq)               E° = +0.77 V

I₂(s) + 2 e^– → 2 I^–(aq)                     E° = +0.54 V

Reduction Potential of the anode is the reduction potential of the species being oxidized, which is Fe³⁺ in this case. So, the reduction potential of the anode is +0.77 V.

The reduction potential of the cathode is the reduction potential of the species being reduced, which is I₂ in this case. So, the reduction potential of the cathode is +0.54 V.

The cell potential (E_cell) is the difference between the reduction potentials of the cathode and anode:

E_cell = E_cathode – E_anode
E_cell = +0.54 V – (+0.77 V)
E_cell = -0.23 V

Therefore, the cell potential for this reaction is -0.23 V. This means that the reaction is not spontaneous and the electrons will not flow from the anode to the cathode without an external energy source.

Cell Potential is the difference in electrical potential between an electrochemical cell’s two electrodes. An electrochemical cell is a device that uses an electrochemical reaction to transform chemical energy into electrical energy. The differential in electron affinities between the two electrodes and the electrolytes’ reactivity are what cause the cell potential.
Electrons will move from the metal with lower electron affinity to the metal with higher electron affinity when two different metals are in contact, creating an electrical potential difference.