The standard reduction potentials for two consecutive reduction steps of a hypothetical metal M are given as: E _ M ³⁺ / M ^ =-0.10 ~V E _ M ³⁺ / M ²⁺ ^ =0.20 ~V What is the standard Gibbs free energy change ( G ^ ) for the following disproportionation reaction? 3 M ²⁺( aq ) 2 M ³⁺( aq )+ M ( s ) (Given: 1 ~F =96500 ~C ~mol ⁻¹ )

Step-by-step solution

First, calculate the standard reduction potential for M ²⁺ + 2 e ⁻ M . G ₁^ for M ³⁺ + 3 e ⁻ M is -3 F (-0.10) = +0.30 F G ₂^ for M ³⁺ + e ⁻ M ²⁺ is -1 F (0.20) = -0.20 F Subtracting the second reaction from the first gives M ²⁺ + 2 e ⁻ M : G ₃^ = G ₁^ - G ₂^ = 0.30 F - (-0.20 F ) = 0.50 F E _ M ²⁺/ M ^ = - G ₃^ 2 F = - 0.50 F 2 F = -0.25 ~V For the disproportionation reaction 3 M ²⁺ 2 M ³⁺ + M : Oxidation half-cell: M ²⁺ M ³⁺ + e ⁻ (Anode, E _ anode ^ = 0.20 ~V ) Reduction half-cell: M ²⁺ + 2 e ⁻ M (Cathode, E _ c