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Equilibrium Constant and Standard Free Energy Change

Consider the ...

Question

Consider the standard reduction potential (in volts) as shown in figure. Find $E^{o}$ .

- 0.326

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- 0.425

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- 0.756

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- 0.512

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Solution

The correct option is C $- 0.756$
$S O_{4}^{2 -} + 2 e ⟶ S O_{3}^{2 -} Δ G_{1} = - n f E_{1}^{o}$
$S O_{3}^{2 -} + 2 e ⟶ \frac{1}{2} S_{2} O_{3}^{2 -} Δ G_{2} = - n f E_{2}^{o}$
------------------------------------------------
$S O_{4}^{2 -} + 4 e ⟶ \frac{1}{2} S_{2} O_{3}^{2 -} Δ G_{3} = - n f E^{o} = Δ G_{1} + Δ G_{2}$
So, $- n F E^{o} = - n f E_{1}^{o} - n f E_{2}^{o}$
$\Rightarrow n E^{o} = n E_{1}^{o} + n E_{2}^{o}$
$\Rightarrow 4 E^{o} = 2 E_{1}^{o} + 2 E_{2}^{o}$
$\Rightarrow E^{o} = \frac{- 2 \times 0.936 - 2 \times 0.576}{4} = - 0.756$

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