Eocell for the reaction- 2H2O- H3O-OH- at 25oC is -0-8277V- The equilibrium constant for the reaction is-

The correct option is A 10^ 14 Given reaction, 2H_2O→ H_3O^++OH^ We know a relation, logK=E^0_cell× n0.0591 logK= 0.8277×10.0591 = 14 ...

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Equilibrium Constant from Nernst Equation

Eocell for th...

Question

$E_{c e l l}^{o}$ for the reaction, $2 H_{2} O \to H_{3} O^{+} + O H^{-}$ at $25^{o}$ C is $- 0.8277 V .$ The equilibrium constant for the reaction is:

10^{- 14}

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10^{- 23}

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10^{- 7}

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10^{- 21}

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Solution

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Similar questions

N H_{4}^{+} + 2 H_{2} O ⇌ N H_{4} O H + H_{3} O^{+}

K_{1} = 5.56 \times 10^{- 10}

H_{2} O + H_{2} O ⇌ H_{3} O^{+} + O H^{-}

K_{2} = 1 \times 10^{- 14}

K_{3}

N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}

25^{o}

2 H_{2} O + 2 e ⇌ H_{2} + 2 O H

- 0.8277

2 H_{2} O ⇌ H_{3} O^{+} + O H

25^{o}

K_{w}

2 H_{2} O ⇌ H_{3} O^{+} O H^{-}

10^{- 14}

25^{\circ} C

9.62 \times 10^{- 14}

60^{\circ} C

p H

60^{\circ} C

25^{\circ} C

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

- 0.8277 v o l t

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}

25^{\circ} C

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{w} = 1 \times 10^{- 14}

25^{\circ} C

K_{a}

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