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The $E_{cell}$ in which the reaction :$MnO_4^-+Fe^{+2}+H^+\rightarrow Mn^{+2}+Fe^{+3}+H_2O$ occurs is 0.59V at $25^{\large\circ}C$.The equilibrium constant for the reaction is

$\begin{array}{1 1}(a)\;50\\(b)\;10\\(c)\;10^{50}\\(d)\;10^5\end{array}$

1 Answer

For the given half-cell reaction
$-nFE^0_{cell}=-RTln K_{eq}$
$-5\times 96500\times 0.59=-2.303\times 8.314\times 298 \log K_{eq}$
$K_{eq}=7.8\times 10^{49}\approx 10^{50}$
Hence (c) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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