If $O x_{1} + a e^{-} ⇌ {R e d}_{1}, {E_{1}}^{o}$

$O x_{2} + b e^{-} ⇌ {R e d}_{2}, {E_{2}}^{o}$

The potential at the equivalence point is :

E = \frac{a {E_{1}}^{o} + b {E_{2}}^{o}}{(a + b)}

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E = \frac{a {E_{1}}^{o} - b {E_{2}}^{o}}{(a + b)}

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E = \frac{b {E_{1}}^{o} - a {E_{2}}^{o}}{(a + b)}

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E = \frac{a {E_{1}}^{o} - b {E_{2}}^{o}}{(a - b)}

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Solution

The correct option is A $E = \frac{a {E_{1}}^{o} + b {E_{2}}^{o}}{(a + b)}$
$O x_{1} + a e^{-} ⇌ {R e d}_{1}, {E_{1}}^{o}$

$O x_{2} + b e^{-} ⇌ {R e d}_{2}, {E_{2}}^{o}$

At the equivalent point, all species are in unit concentration and

$Δ G^{o} (n e t) = Δ G_{1}^{o} + Δ G_{2}^{o}$

Total electrons exchanged $= (a + b)$ .

$∴ Δ G^{o} = - n F E^{o}$ .

$- (a + b) F E_{3}^{o} = - a F E_{1}^{o} - b F E_{2}^{o}$

$∴ E_{3}^{o} = \frac{a E_{1}^{o} + b E_{2}^{o}}{a + b}$

Hence, the correct option is $A$

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