The standard reduction potential at $25^{o}$ C for the reduction of water $2 H_{2} O + 2 e ⇌ H_{2} + 2 O H$ is $- 0.8277$ volt. Calculate the equilibrium constant for the reaction $2 H_{2} O ⇌ H_{3} O^{+} + O H$ at $25^{o}$ C.(Answer up to 2 decimal places.)

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Solution

Two half reactions
Oxidation: $H_{2} O + 1 / 2 H_{2} ⟶ H_{3} O^{+} + e^{-}$ , $E_{c e l l}^{o} = 0$
Reduction: $H_{2} O + e^{-} ⟶ {1 / 2 H_{2}}_{(g)} + O H^{-}$ , $E_{c e l l}^{o} = - 0.8277 V$
$n = 1$
$log K_{C} = \frac{n E_{c e l l}^{o}}{0.059} = \frac{1 \times (- 0.8277)}{0.059} = 15.972$
$∴ K_{C} = 9.88 \times 10^{- 15}$

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Similar questions

Q. 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^{-}

25^{0}

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^{-}

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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 $25^{o} C$ .

E_{c e l l}^{o}

for the reaction,

2 H_{2} O \to H_{3} O^{+} + O H^{-}

25^{o}

C is

- 0.8277 V .

The equilibrium constant for the reaction is:

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The standard reduction potential at 25oC for the reduction of water 2H2O+2e⇌H2+2OH is −0.8277 volt. Calculate the equilibrium constant for the reaction 2H2O⇌H3O++OH at 25oC.(Answer up to 2 decimal places.)

Two half reactionsOxidation: H2O+1/2H2⟶H3O++e−, Eocell=0Reduction: H2O+e−⟶1/2H2(g)+OH−, Eocell=−0.8277Vn=1logKC=nEocell0.059=1×(−0.8277)0.059=15.972∴KC=9.88×10−15

The standard reduction potential at $25^{o}$ C for the reduction of water $2 H_{2} O + 2 e ⇌ H_{2} + 2 O H$ is $- 0.8277$ volt. Calculate the equilibrium constant for the reaction $2 H_{2} O ⇌ H_{3} O^{+} + O H$ at $25^{o}$ C.(Answer up to 2 decimal places.)