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Standard XII

Chemistry

Application of Electrolysis

Calculate red...

Question

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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Solution

$2 H_{2} O + 2 e^{-} \to H_{2} + 2 {O H}^{(-)}$
$(1 a t m)$ $(10 \times 10^{- 7})$
$E_{c e l l} = E^{0} - \frac{0.059}{n} l o g ⎡ ⎢ ⎢ ⎢ ⎣ \frac{{({O H}^{(-)})}^{2}}{{(P_{H_{2}})}^{1}} ⎤ ⎥ ⎥ ⎥ ⎦$
$E_{c e l l} = 0 - \frac{0.059}{2} l o g ⎡ ⎣ \frac{{(10 \times 10^{- 7})}^{2}}{1} ⎤ ⎦$
$E_{c e l l} = - 0.059 \times l o g [(10 \times 10^{- 7})]$
$E_{c e l l} = - 0.059 \times (- 6)$
$E_{c e l l} = - 0.354 V$

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Similar questions

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. 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^{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^{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 half cell reduction potential for

{A g}^{+} | A g

0.7991 V

25^{o} C

. Given that the experimental value of

K_{s p} = 1.56 \times 10^{- 10}

for

A g C l

, calculate the standard half cell reduction potential for the

A g | A g C l

electrode.

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Calculate reduction potential of the following half cell reaction at 25oC2H2O+2e→H2+2OH− (1atm) (10×10−7)

2H2O+2e−→H2+2OH(−) (1atm) (10×10−7)Ecell=E0−0.059nlog⎡⎢ ⎢ ⎢⎣(OH(−))2(PH2)1⎤⎥ ⎥ ⎥⎦Ecell=0−0.0592log⎡⎣(10×10−7)21⎤⎦Ecell=−0.059×log[(10×10−7)]Ecell=−0.059×(−6)Ecell=−0.354V

Calculate reduction potential of the following half cell reaction at $25^{o} C$
$2 H_{2} O + 2 e \to H_{2} + 2 {O H}^{-}$
$(1 a t m)$ $(10 \times 10^{- 7})$