Given: Mn++ne−→M E1∘ and Mm++me−→M E2∘If n>m, the standard potential E3∘ for the reaction Mn++(n−m)e−→Mm+ is given by the expression

Mn++ne−→M ΔG10=−nFE1∘ Mm++me−→M ΔG20=−mFE2oMn++(n−m)e−→Mm+ ΔG3o=−FnE1o−mE2o Also ΔG3∘=−(n−m)FE3∘ −(n−m)FE3∘=−FnE1∘−mE2∘ or E3∘=nE1∘−mE2∘/(n−m)

Given: Mn++ne−→M E1∘ and Mm++me−→M E2∘If n>m, the standard potential E3∘ for the reaction Mn++(n−m)e−→Mm+ is given by the expression

Mn++ne−→M ΔG10=−nFE1∘ Mm++me−→M ΔG20=−mFE2oMn++(n−m)e−→Mm+ ΔG3o=−FnE1o−mE2o Also ΔG3∘=−(n−m)FE3∘ −(n−m)FE3∘=−FnE1∘−mE2∘ or E3∘=nE1∘−mE2∘/(n−m)

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Given: $M^{n +} + n e^{-} \to M E_{1}^{\circ}$ and $M^{m +} + m e^{-} \to M E_{2}^{\circ}$
If $n > m,$ the standard potential $(E_{3}^{\circ})$ for the reaction $M^{n +} + (n - m) e^{-} \to M^{m +}$ is given by the expression

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$E_{3}^{\circ} = \frac{n E_{1}^{\circ} - m E_{2}^{\circ}}{n - m}$

$E_{3}^{\circ} = \frac{m E_{1}^{\circ} + n E_{2}^{\circ}}{n + m}$

$E_{3}^{\circ} = \frac{m E_{1}^{\circ} - n E_{2}^{\circ}}{n - m}$

$E_{3}^{\circ} = \frac{n E_{1}^{\circ} + m E_{2}^{\circ}}{n + m}$

answer is B.

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$\frac{M^{n +} + n e^{-} \to M Δ G_{1}^{0} = - n F E_{1}^{\circ} M^{m +} + m e^{-} \to M Δ G_{2}^{0} = - m F E_{2}^{o}}{M^{n +} + (n - m) e^{-} \to M^{m +} {ΔG}_{3}^{o} = - F ({nE}_{1}^{o} - {mE}_{2}^{o})}$

Also $Δ G_{3}^{\circ} = - (n - m) F E_{3}^{\circ}$

$- (n - m) F E_{3}^{\circ} = - F (n E_{1}^{\circ} - m E_{2}^{\circ}) or E_{3}^{\circ} = (n E_{1}^{\circ} - m E_{2}^{\circ}) / (n - m)$

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Given: Mn++ne−→M E1∘ and Mm++me−→M E2∘If n>m, the standard potential E3∘ for the reaction Mn++(n−m)e−→Mm+ is given by the expression

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Given: $M^{n +} + n e^{-} \to M E_{1}^{\circ}$ and $M^{m +} + m e^{-} \to M E_{2}^{\circ}$
If $n > m,$ the standard potential $(E_{3}^{\circ})$ for the reaction $M^{n +} + (n - m) e^{-} \to M^{m +}$ is given by the expression