The standard reduction potential at 25-C of the reaction 2H2O - 2e- H2 - 2OH- is -0-8277 volt- Calculate the equilibrium constant for the reaction- 2H2O - H3O- - OH- at 25-C-

≈ 10^14 . H_2O + e^ →12H_2 (Cathode); E^∘ = 0.8277 volt H_2O + 12H_2→ H_3O^+ + e^ (Anode); E^∘ = 0 E^∘ for the cell = 0.8288 volt Ap ...

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Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

The standard ...

Question

The standard reduction potential at $25^{\circ} C$ of the reaction $2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is $- 0.8277 v o l t$ . Calculate the equilibrium constant for the reaction,
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $25^{\circ} C$ .

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Solution

$\approx 10^{14}$ .
$H_{2} O + e^{-} \to \frac{1}{2} H_{2} (C a t h o d e); E^{\circ} = - 0.8277 v o l t$
$H_{2} O + \frac{1}{2} H_{2} \to H_{3} O^{+} + e^{-} (A n o d e); E^{\circ} = 0$
$E^{\circ}$ for the cell $= - 0.8288 v o l t$
Apply now $E^{\circ} = \frac{0.0591}{n} [log K] (n = 1)$ .

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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 $25^{o} C$ .

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The standard reduction potential at 25∘C of the reaction 2H2O+2e−→H2+2OH− is −0.8277 volt. Calculate the equilibrium constant for the reaction,2H2O⇌H3O++OH− at 25∘C.

≈1014.H2O+e−→12H2(Cathode);E∘=−0.8277 voltH2O+12H2→H3O++e−(Anode);E∘=0E∘ for the cell =−0.8288 voltApply now E∘=0.0591n[logK] (n=1).

The standard reduction potential at $25^{\circ} C$ of the reaction $2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is $- 0.8277 v o l t$ . Calculate the equilibrium constant for the reaction,
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $25^{\circ} C$ .

$\approx 10^{14}$ .
$H_{2} O + e^{-} \to \frac{1}{2} H_{2} (C a t h o d e); E^{\circ} = - 0.8277 v o l t$
$H_{2} O + \frac{1}{2} H_{2} \to H_{3} O^{+} + e^{-} (A n o d e); E^{\circ} = 0$
$E^{\circ}$ for the cell $= - 0.8288 v o l t$
Apply now $E^{\circ} = \frac{0.0591}{n} [log K] (n = 1)$ .