The standard reduction potential at $25^{0}$ C for the reduction of water
$2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 O H^{-}$ is -0.8277 volt. Calculate the equlibrium constant for the reaction $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $25^{0}$ C.

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Solution

$E_{c e l l}^{o} = E_{c e l l} - \frac{- 0.059}{n} log K_{C}$
$E_{c e l l} = 0$ at equilibrium
$\Rightarrow - 0.8277 = \frac{- 0.059}{1} log K_{C}$
$\Rightarrow log K_{C} = 14.02$ .

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Similar questions

Q. The standard reduction potential at

25^{o}

C for the reduction of water

2 H_{2} O + 2 e ⇌ H_{2} + 2 O H

- 0.8277

volt. Calculate the equilibrium constant for the reaction

2 H_{2} O ⇌ H_{3} O^{+} + O H

25^{o}

C.(Answer up to 2 decimal places.)

Q. The standard reduction potential at

25^{\circ} C

of the reaction

2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}

- 0.8277 v o l t

. Calculate the equilibrium constant for the reaction,

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}

25^{\circ} C

Q. The standard reduction potential at

25^{\circ} C

of the reaction

2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 ⊖ O H

is -0.8277 V. The equilibrium constant for the reaction

2 H_{2} O ⇌ H_{3} O^{\oplus} + ⊖ O

25^{\circ} C

is:

The standard reduction potential at $25^{o} C$ of the reaction $H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is $- 0 .8277 V$ . Calculate the equilibrium constant for the reaction $2 H_{2} O \to H_{3} O^{+} + O H^{-}$ at $25^{o} C$ .

Q. Calculate reduction potential of the following half cell reaction at

25^{o} C

2 H_{2} O + 2 e \to H_{2} + 2 {O H}^{-}

(1 a t m)

(10 \times 10^{- 7})

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The standard reduction potential at 250C for the reduction of water 2H2O+2e−⇌H2+2OH− is -0.8277 volt. Calculate the equlibrium constant for the reaction 2H2O⇌H3O++OH− at 250C.

Eocell=Ecell−−0.059nlogKCEcell=0 at equilibrium⇒−0.8277=−0.0591logKC⇒logKC=14.02 .

The standard reduction potential at $25^{0}$ C for the reduction of water
$2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 O H^{-}$ is -0.8277 volt. Calculate the equlibrium constant for the reaction $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $25^{0}$ C.

$E_{c e l l}^{o} = E_{c e l l} - \frac{- 0.059}{n} log K_{C}$
$E_{c e l l} = 0$ at equilibrium
$\Rightarrow - 0.8277 = \frac{- 0.059}{1} log K_{C}$
$\Rightarrow log K_{C} = 14.02$ .