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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