The reduction potential (E^o, in V) of MnO₄⁻(aq) / Mn(s) is ________ .

[Given: $E_{{\mathrm{MnO}}_4^{-}(\mathrm{aq})/{\mathrm{MnO}}_2(s)}^0=1.68V;$ $E_{{\mathrm{MnO}}_2(s)/{\mathrm{Mn}}^{2+}(\mathrm{aq})}^0=1.21V;$ $E_{{\mathrm{Mn}}^{2+}(\mathrm{aq})/\mathrm{Mn}(s)}^0=-1.03V$ ]

Correct Answer: The correct answer is 0.77 V (or simply 0.77).

Step-by-Step Explanation:

To find the standard reduction potential $E_{{\mathrm{MnO}}_4^{-}/\mathrm{Mn}}^0$ for the overall reaction where ${\mathrm{MnO}}_4^{-}$ is reduced to solid $\mathrm{Mn}$, we must combine the given half-cell reactions using standard Gibbs free energy changes ($\Delta G^0=-nF{E}^0$). Standard potential values cannot be added directly, but Gibbs free energy changes are additive.

Step 1: Write down the half-cell reactions and determine the number of electrons involved in each step.

1st Reaction: Reduction of ${\mathrm{MnO}}_4^{-}(\mathrm{aq})$ to ${\mathrm{MnO}}_2(s)$

Manganese oxidation state changes from +7 in ${\mathrm{MnO}}_4^{-}$ to +4 in ${\mathrm{MnO}}_2$.

Number of electrons transferred, $n_1=3$

Given standard reduction potential, $E_1^0=1.68\text{V}$

2nd Reaction: Reduction of ${\mathrm{MnO}}_2(s)$ to ${\mathrm{Mn}}^{2+}(\mathrm{aq})$

Manganese oxidation state changes from +4 in ${\mathrm{MnO}}_2$ to +2 in ${\mathrm{Mn}}^{2+}$.

Number of electrons transferred, $n_2=2$

Given standard reduction potential, $E_2^0=1.21\text{V}$

3rd Reaction: Reduction of ${\mathrm{Mn}}^{2+}(\mathrm{aq})$ to $\mathrm{Mn}(s)$

Manganese oxidation state changes from +2 in ${\mathrm{Mn}}^{2+}$ to 0 in $\mathrm{Mn}(s)$.

Number of electrons transferred, $n_3=2$

Given standard reduction potential, $E_3^0=-1.03\text{V}$

Target Reaction: Overall reduction of ${\mathrm{MnO}}_4^{-}(\mathrm{aq})$ to $\mathrm{Mn}(s)$

Manganese oxidation state changes from +7 in ${\mathrm{MnO}}_4^{-}$ to 0 in $\mathrm{Mn}(s)$.

Total electrons transferred in the target reaction, $n_{\text{total}}=n_1+n_2+n_3=3+2+2=7$

Step 2: Relate Gibbs free energies of the individual steps to the overall process.

$\Delta G_{\text{overall}}^0=\Delta G_1^0+\Delta G_2^0+\Delta G_3^0$

Substituting $\Delta G^0=-nF{E}^0$ into the equation:

$-n_{\text{total}}F{E}_{\text{overall}}^0=-n_{1}F{E}_1^0-n_{2}F{E}_2^0-n_{3}F{E}_3^0$

Dividing both sides by $-F$:

$n_{\text{total}}{E}_{\text{overall}}^0=n_1{E}_1^0+n_2{E}_2^0+n_3{E}_3^0$

Step 3: Substitute the known values and solve for $E_{\text{overall}}^0$.

$7\times E_{\text{overall}}^0=(3\times 1.68)+(2\times 1.21)+[2\times (-1.03)]$

$7\times E_{\text{overall}}^0=5.04+2.42-2.06$

$7\times E_{\text{overall}}^0=5.40$

$E_{\text{overall}}^0=\frac{5.40}{7}\approx 0.7714\text{V}$

Rounding to two decimal places gives 0.77 V.