Question

7. Consider the following electrode potentials: Mg2+ + 2e--> Mg E* = -2.37 V V2+ + 2e -> V E = -1.19 V Cu2+ + --> Cut E* = 0.16 V Which one of the reactions below will proceed spontaneously from left to right? A) Mg2+ + V -> V2+ + Mg B) Mg2+ + 2Cu* -> 2Cu2+ + Mg C) V2+ + 2Cut -> V + 2Cu2+ D) V + 2Cu2+ -> V2+ + 2Cut E) none of these

Please explain the steps on how to solve this problem, I already have the answer.

Answer 1

Spontainity is judged by the gibbs free energy of a reaction. If the value is positive then the reaction is non spontaneous and if its value is negative then the reaction is spontaneous in nature. And ΔG° = -nFE° where E is the emf of the cell reaction at standard conditions.

If the value of EMF of the cell is positive only then the Gibbs free energy is negative therefore we need to find the combination which provides the EMF to be a positive value.

A. Mg2+ is reduced and V is oxidised.

E = E of cathode - E of anode = E (reduction) _E(oxidation)

= -2.37 - (-1.19)= -1.18 V

B. Mg2+ is reduced and Cu+ is oxidised.

E = -2.37 - ( 0.16) = - 2.53 V

C. Cu + is oxidised and V2+ is reduced.

E = -1.19 -(0.16) = - 1.35 V

D. Cu2+ is reduced and V is oxidised

E = 0.16 - ( -1.19) = 1.35 V

So only part D has positive value of Emf so it will give a negative value of ΔG so that the reaction can be spontaneous.

So answer is D only.

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