Consider the following redox reaction :
$$
\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2
$$
The standard reduction potentials are given as below $\left(\mathrm{E}_{\text {red }}^0\right)$ :
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\\\
& \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}
\end{aligned}
$$
If the equilibrium constant of the above reaction is given as $\mathrm{K}_{\mathrm{eq}}=10^x$, then the value of $x=$ __________ (nearest integer)
Answer (integer)
338
Solution
Cell $\mathrm{Rx}^{\mathrm{n}} ; \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2$
<br/><br/>$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}$ of anode $+\mathrm{E}_{\mathrm{RP}}^{\circ}$ of cathode
<br/><br/>$=0.49+1.51=2.00 \mathrm{~V}$
<br/><br/>At equilibrium
<br/><br/>$\mathrm{E}_{\text {cell }}=0 \text {, }$
<br/><br/>$\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}$
<br/><br/>$2=\frac{0.0591}{10} \log K$
<br/><br/>$\log K=338$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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