Concept Used

To find the standard reduction potential ($E^0$) for a half-reaction that is a combination of other half-reactions, we cannot simply add the potentials. Instead, we must use the relationship between the standard Gibbs free energy change ($\\\Delta G^0$) and the standard electrode potential ($E^0$), which is given by: $\Delta G^0 = -nFE^0$ where:

• $n$ is the number of moles of electrons transferred in the balanced half-reaction.
• $F$ is the Faraday constant.

Since $\\\Delta G^0$ is an extensive property, the Gibbs free energy changes for the individual reactions can be added to find the Gibbs free energy change for the overall reaction. If a target reaction is the sum of several other reactions, then: $\Delta G^0_{total} = \Delta G^0_1 + \Delta G^0_2 + \Delta G^0_3 + ...$ Substituting the expression for $\\\Delta G^0$: $-n_{total}FE^0_{total} = -n_1FE^0_1 - n_2FE^0_2 - n_3FE^0_3 - ...$ Dividing by $-F$ gives the formula to calculate the combined potential: $n_{total}E^0_{total} = n_1E^0_1 + n_2E^0_2 + n_3E^0_3 + ...$

Step-by-step Solution

1. Identify the target half-reaction and find $n_{total}$ We need to find the reduction potential for the couple $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$. The balanced half-reaction in acidic medium is: $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 8\mathrm{H}^{+} + 7e^{-} \rightarrow \mathrm{Mn}(\mathrm{s}) + 4\mathrm{H}_{2}\mathrm{O}$ The oxidation state of Mn changes from +7 in $\mathrm{MnO}_{4}^{-}$ to 0 in $\mathrm{Mn}(\mathrm{s})$. The number of electrons transferred is $n_{total} = 7$. Let the potential for this reaction be $E^0_{total}$.

2. Write down the given half-reactions and their parameters We are given three half-reactions:

   (i) $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{MnO}_{2}(\mathrm{s})$ Reaction: $\mathrm{MnO}_{4}^{-} + 4\mathrm{H}^{+} + 3e^{-} \rightarrow \mathrm{MnO}_{2} + 2\mathrm{H}_{2}\mathrm{O}$ The oxidation state of Mn changes from +7 to +4. So, $n_1 = 3$. $E^0_1 = 1.68 \mathrm{~V}$

   (ii) $\mathrm{MnO}_{2}(\mathrm{s}) / \mathrm{Mn}^{2+}(\mathrm{aq})$ Reaction: $\mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O}$ The oxidation state of Mn changes from +4 to +2. So, $n_2 = 2$. $E^0_2 = 1.21 \mathrm{~V}$

   (iii) $\mathrm{Mn}^{2+}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$ Reaction: $\mathrm{Mn}^{2+} + 2e^{-} \rightarrow \mathrm{Mn}$ The oxidation state of Mn changes from +2 to 0. So, $n_3 = 2$. $E^0_3 = -1.03 \mathrm{~V}$

3. Combine the given reactions to get the target reaction Adding the three half-reactions (i), (ii), and (iii): ($\mathrm{MnO}_{4}^{-} + 4\mathrm{H}^{+} + 3e^{-} \rightarrow \mathrm{MnO}_{2} + 2\mathrm{H}_{2}\mathrm{O}$)

   • ($\mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O}$)
   • ($\mathrm{Mn}^{2+} + 2e^{-} \rightarrow \mathrm{Mn}$)

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   The net reaction after canceling intermediate species ($\mathrm{MnO}_{2}$ and $\mathrm{Mn}^{2+}$) is: $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + (3+2+2)e^{-} \rightarrow \mathrm{Mn} + 4\mathrm{H}_{2}\mathrm{O}$ $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 7e^{-} \rightarrow \mathrm{Mn} + 4\mathrm{H}_{2}\mathrm{O}$ This is our target reaction. So, we can add the corresponding $\\\Delta G^0$ values.

4. Calculate the total potential $E^0_{total}$ Using the formula $n_{total}E^0_{total} = n_1E^0_1 + n_2E^0_2 + n_3E^0_3$: $7 \times E^0_{total} = (3 \times 1.68) + (2 \times 1.21) + (2 \times -1.03)$ $7 \times E^0_{total} = 5.04 + 2.42 - 2.06$ $7 \times E^0_{total} = 7.46 - 2.06$ $7 \times E^0_{total} = 5.40$ $E^0_{total} = \frac{5.40}{7} \approx 0.7714 \mathrm{~V}$

Rounding to two decimal places, we get $E^0_{total} = 0.77 \mathrm{~V}$.

Conclusion

The reduction potential ($E^0$) of $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$ is 0.77 V.