$\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
પ્રમાણિત રિડકશન પોટેન્શિયલ નીચે આપેલા છે. $\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)$
$\mathrm{E}_{\mathrm{MmO}_4^{-} / \mathrm{Mm}^{2+}}^{\circ}=+1.51 \mathrm{~V}$
$\mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}$
જો ઉપરની પ્રક્રિયાને સંતુલન અચળાંક $K_{e q}=10^x$, તરીકે આપેલ હોય તો, $x$ નું મૂલ્ય = ___________. (નજીકનો પૂણુાંક)
\(\mathrm{E}_{\text {coll }}^{\mathrm{c}}=\mathrm{E}_{\mathrm{op}}^{\circ} \text { of anode }+\mathrm{E}_{\mathrm{RP}}^{\circ} \text { of cathode }\)
\(=0.49+1.51=2.00 \mathrm{~V}\)
At equilibrium
\(\mathrm{E}_{\text {cell }}=0,\)
\(\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.059}{\mathrm{n}} \log \mathrm{K}\)
\(\text { (As per NCERT } \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \text { But } \frac{\mathrm{RT}}{\mathrm{F}}=0.0591\)can also be taken.)
\(2=\frac{0.059}{10} \log \mathrm{K}\)
\(\log \mathrm{K}=338.98\)