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Equilibrium Constant and Standard Free Energy Change

If the equili...

Question

If the equilibrium consatnt for the reaction $H^{+} (a q) + O H^{-} (a q) ⇌ H_{2} O (l)$ at a certain temperature is $10^{- 26}$ , then what is the $E^{0}$ for the reaction, $2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$
(Given: $\frac{2.303 R T}{F} = 0.066$ )

1.230 V

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- 0.858 V

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- 0.80 V

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- 0.8274 V

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Solution

The correct option is B $- 0.858 V$
$2 H_{2} O (l) ⇌ 2 H^{+} (g) + 2 O H^{-} (a q); △ G_{1}^{0}$
$2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g); △ G_{2}^{0}$
$2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q); △ G_{3}^{0}$
Now,
$△ G_{3}^{0} =△ G_{1}^{0} + △ G_{2}^{0}$
$\Rightarrow - n f E_{3}^{o} = - 2.303 R T l o g K_{e q} - n f E_{2}^{o}$
$\Rightarrow - 2 f E_{3}^{o} = - 2.303 R T l o g 10^{- 26} - n f \times 0$
$\Rightarrow E_{3}^{0} = \frac{2.303 R T}{2 f} \times (- 26) + 0 = - 13 \times 0.066 = - 0.858 V$

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