Question

The standard cell potential of the following cell Zn|Zn²⁺ (aq)|Fe²⁺(aq)|Fe is 0.32 V. Calculate the standard Gibbs energy change for the reaction:Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s)(Given: 1 F = 96487 C)

• A. −61.75 kJ mol⁻¹
• B. +5.006 kJ mol⁻¹
• C. −5.006 kJ mol⁻¹
• D. +61.75 kJ mol⁻¹

Answer 1

The Correct Option is A

To calculate the standard Gibbs energy change for the reaction:

Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s) 

we use the formula that relates the Gibbs energy change (ΔG°) with the cell potential (E°_cell):

\(ΔG° = -nFE°_{cell}\)

where:

• \(n\) is the number of moles of electrons transferred in the balanced equation,
• \(F\) is Faraday's constant (96487 C mol⁻¹),
• \(E°_{cell}\) is the standard cell potential (0.32 V).

 

For the given reaction, the balanced equation shows that 2 moles of electrons are transferred because both zinc and iron change oxidation states by 2:

Zn(s) → Zn²⁺(aq) + 2e⁻
Fe²⁺(aq) + 2e⁻ → Fe(s)

Therefore, \(n = 2\).

Substituting the given values:

\(ΔG° = - (2)(96487 \ \text{C mol}^{-1})(0.32 \ \text{V})\)

Calculating this gives:

\(ΔG° = - (2 \times 96487 \times 0.32) = -61775.68 \ \text{J mol}^{-1}\)

To convert joules to kilojoules, divide by 1000:

\(ΔG° = -61.77568 \ \text{kJ mol}^{-1}\)

Rounding appropriately, we have:

\(ΔG° = -61.75 \ \text{kJ mol}^{-1}\)

Thus, the standard Gibbs energy change for the reaction is −61.75 kJ mol⁻¹.

This corresponds to the correct option.