Question

Given below are the half-cell reactions:

\[\ce{Mn^2+ + 2e- -> Mn}\], E° = −1.18 V

\[\ce{2(Mn^3+ + e- -> Mn^2+)}\], E° = +1.51 V

The E° for \[\ce{3Mn^2+ -> Mn + 2Mn^3+}\] will be ______.

Options

• −2.69 V, the reaction will not occur.

• −2.69 V, the reaction will occur.

• −0.33 V, the reaction will not occur.

• −0.33 V, the reaction will occur.

Answer 1

The E° for \[\ce{3Mn^2+ -> Mn + 2Mn^3+}\] will be −2.69 V, the reaction will not occur.

Explanation:

The overall reaction is:

\[\ce{3Mn^2+ -> Mn + 2Mn^3+}\]

This involves

Oxidation: \[\ce{Mn^2+ −> Mn^3+}\], reverse of \[\ce{Mn^3+ −> Mn^2+}\]; so E° = −1.51 V

Reduction: \[\ce{Mn^2+ + 2e- −> Mn}\]; E° = −1.18 V

Now, 

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{reduction}} - E{^{\circ}_{oxidation}}}\]

= −1.18 − 1.51

= −2.69 V

Since E° < 0, the Gibbs free energy ΔG° > 0, meaning the reaction is non-spontaneous under standard conditions.