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Equilibrium Constant from Nernst Equation

If the equili...

Question

If the equilibrium constant for the reaction : $H^{+} (a q .) + O H^{-} (a q .) ⇌ H_{2} O (l)$ is $10^{13}$ at certain temperature then what is the $E^{\circ}$ 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 $- 0.858 V$
As we know,
$E^{0} = \frac{2.303 R T}{n F} \times l o g K = \frac{0.066}{2} \times l o g K = 0.033 \times (- 26) = - 0.858 V$ .
(as $K$ for $2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q .)$ is $\frac{1}{K^{2}} = \frac{1}{{10^{13}}^{2}} = \frac{1}{10^{26}}$ )

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Similar questions

H^{+} (a q) + O H^{-} (a q) ⇌ H_{2} O (l)

10^{- 26}

E^{0}

2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)

\frac{2.303 R T}{F} = 0.066

Z n (s) + A g_{2} O (s) + H_{2} O (l) \to 2 A g (s) + Z n^{2 +} (a q .) + 2 O H^{-} (a q .)

E_{c e l l}^{\circ} = 1.11

298 K

△ G^{\circ}

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

E_{A g^{+} / A g}^{\circ} = + 0.80 V

E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V

C u

F e

E_{c e l l}^{\circ}

(K_{f})

Z n^{2 +} (a q) + 4 O H^{-} (a q) ⇌ Z n (O H)_{4}^{2 -} (a q)

p \times 10^{q},

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

Z n (O H)_{4}^{2 -} (a q) + 2 e^{-} \to Z n (s) + 4 O H^{-} (a q); E^{0} = - 1.36 V

2.303 \frac{R T}{F} = 0.06

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