Jee Main
Electrochemistry
JEE Main 2024 (Online) 1st February Evening Shift
INTEGER+4 / -12024
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)