JEE MAIN - Chemistry (2024 - 1st February Evening Shift - No. 28)
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)
$$ \mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $$
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
338
Explanation
Cell $\mathrm{Rx}^{\mathrm{n}} ; \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2$
$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}$ of anode $+\mathrm{E}_{\mathrm{RP}}^{\circ}$ of cathode
$$ =0.49+1.51=2.00 \mathrm{~V} $$
At equilibrium
$$ \mathrm{E}_{\text {cell }}=0 \text {, } $$
$$ \mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} $$
$2=\frac{0.0591}{10} \log K$
$\log K=338$
$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}$ of anode $+\mathrm{E}_{\mathrm{RP}}^{\circ}$ of cathode
$$ =0.49+1.51=2.00 \mathrm{~V} $$
At equilibrium
$$ \mathrm{E}_{\text {cell }}=0 \text {, } $$
$$ \mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} $$
$2=\frac{0.0591}{10} \log K$
$\log K=338$
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