Question

The reduction potential (E° in V) of \[\ce{MnO^-_{4(aq)}/Mn_{(s)}}\] is ______.

Given \[\ce{E^{\circ}_{MnO^-_{4(aq)}/MnO_{2(s)}}}\] = 1.68 V

\[\ce{E^{\circ}_{MnO_{2(s)}/Mn^{+2}_{(aq)}}}\] = 1.21 V

\[\ce{E^{\circ}_{Mn^{+2}_{(aq)}/Mn_{(s)}}}\] = −1.03 V

Answer 1

The reduction potential (E° in V) of \[\ce{MnO^-_{4(aq)}/Mn_{(s)}}\] is 77 × 10⁻² V.

Explanation:

We are given three standard electrode potentials:

1. \[\ce{E^{\circ}_{MnO^-_{4(aq)}/MnO_{2(s)}}}\] = 1.68 V
2. \[\ce{E^{\circ}_{MnO_{2(s)}/Mn^{+2}_{(aq)}}}\] = 1.21 V
3. \[\ce{E^{\circ}_{Mn^{+2}_{(aq)}/Mn_{(s)}}}\] = −1.03 V

We are to find the overall standard reduction potential for \[\ce{MnO^-_{ (aq)} + electrons -> Mn_{(s)}}\]

By using Gibbs free energy relation

ΔG° = −nFE°

We will sum ΔG° for each step to get total ΔG°, then convert back to total E°.

Let’s compute ΔG° for each

n₁​ = 3, E₁​ = 1.68:
\[\ce{\Delta G{^{\circ}_{1}}}\] = −3 F(1.68) = −5.04 F

n₂ ​= 2, E₂​ = 1.21:
\[\ce{\Delta G{^{\circ}_{2}}}\] = −2 F(1.21) = −2.42 F

n_3​ = 2, E₃ ​= −1.03:
\[\ce{\Delta G{^{\circ}_{1}}}\] = −2 F(−1.03) = +2.06 F

∴ Total gibbs free energy ΔG° = −5.04 F + (−2.42 F) + 2.06 F

= −5.40 F

This corresponds to a total of 7 electrons transferred from permanganate to Mn metal:

n_total = 3 + 2 + 2 = 7

We know that ΔG° = −nFE

⇒ −5.40 F = −7F × E°

E° = \[\ce{\frac{5.40}{7}}\]

E° = 0.771 V

E° = 77 × 10⁻² V