Question

For the cell reaction:

\[\ce{2Fe^{3+}_{ (aq)} + 2l^-_{ (aq)} -> 2Fe^{2+}_{ (aq)} + l2_{(aq)}}\]

\[\ce{E{^{\circ}_{cell}}}\] = 0.24 V at 298 K. The standard Gibbs energy (Δ_rG°) of the cell reaction is:

[Given that Faraday constant, F = 96500 C mol⁻¹]

Options

• −23.16 kJ mol⁻¹

• 46.32 kJ mol⁻¹

• 23.16 kJ mol⁻¹

• −46.32 kJ mol⁻¹

Answer 1

−46.32 kJ mol⁻¹

Explanation:

Given:

n = 2 electrons (from the balanced equation)

F = 96500 C mol⁻¹

\[\ce{E{^{\circ}_{cell}}}\] = 0.24 V

Use the formula:

\[\ce{\Delta G{^{\circ}} = -nFE{^{\circ}_{cell}}}\]

= −2 × 96500 × 0.24

= −46320 J mol⁻¹

= −46.320 kJ mol⁻¹