ElectrochemistryHard

Question

The standard reduction potential at 25°C of the reaction 2H2O + 2e $\rightleftharpoons$H2 + 2OH is −0.84 V. Calculate equilibrium constant for the reaction: 2H2O $\rightleftharpoons$H3O+ + OH at 25°C. (2.303 RT/F = 0.06)

Options

A.10−14
B.1014
C.107
D.10−7

Solution

For the given reaction, $E_{cell}^{o} = E_{\text{given}}^{o} - E_{\text{hydrogen}}^{o}$

$\therefore( - 0.84) - 0 = \frac{0.06}{1}.\log K_{eq} \Rightarrow K_{eq} = 10^{- 14}$

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