Question

Calculate Δ_rG^° for the reaction 

Mg (s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu (s)

Given : E^°_cell = + 2.71 V, 1 F = 96500 C mol^–1

Answer 1

The reaction given is Mg(s)+Cu²⁺(aq)→Mg²⁺(aq)+Cu(s)

In the given reaction, the electron transfer is of 2 electrons, hence n = 2

Given, E⁰cell - +2.71 V and 1 F = 96500 Cmol^-1

 We know that,

-ΔG^° = nFE^° cell

= 2 96500 x 2.71

= 523030 J

ΔG^°  = -523030 J

ΔG^° = -5.23 x 10⁵J=-523.03kJ

Answer 2

For the cell reaction,

Mg (s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu (s)       ;     E°_cell = + 2·71 V

the change in the standard Gibbs free energy is given as

∆_rG° = −nFE°_cell = −2×96500×2.71 = 523030 Jmol⁻¹