Question

A galvanic electrochemical cell made of Zn²⁺/Zn and Cu²⁺/Cu half-cells produces 1.10 V at 25°C. The ratio of |Zn²⁺| to |Cu²⁺| is maintained at 1.0. The ΔG° for the reaction when 1.0 mol of Zn gets dissolved is____kJ. (round off to the nearest integer)
[Given: Faraday's constant = 96485 C mol^-1]

Answer 1

Correct Answer: -212

To solve for the standard Gibbs free energy change (ΔG°) of the reaction in the galvanic cell, we start by acknowledging the reaction in question: Zn + Cu²⁺ → Zn²⁺ + Cu. The cell potential (E°_cell) is given as 1.10 V. The standard Gibbs free energy change is calculated using the formula:

ΔG° = -nFE°_cell

where n is the number of moles of electrons transferred in the balanced equation, F is Faraday's constant (96485 C mol^-1), and E°_cell is the standard cell potential (1.10 V).

In the given reaction, the oxidation and reduction half-reactions are:
Zn → Zn²⁺ + 2e^- (oxidation)
Cu²⁺ + 2e^- → Cu (reduction)

Here, n = 2 moles of electrons are transferred per mole of Zn reacted. Substituting these into the Gibbs free energy equation, we get:

ΔG° = -(2 mol)(96485 C/mol)(1.10 V)

Calculating this gives:
ΔG° = -212267 J (or -212.267 kJ, as 1 J = 0.001 kJ).

Rounding this value to the nearest integer, we obtain ΔG° = -212 kJ.