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Calculate ΔrG° for the reaction Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) - Chemistry

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Question

Calculate ΔrG° for the reaction 

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

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

Solution 1

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

Solution 2

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

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Solution Calculate ΔrG° for the reaction Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) Concept: Relation Between Gibbs Energy Change and Emf of a Cell.
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