Electrochemistry Question 3
Question 3 - 2024 (01 Feb Shift 2)
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 $\left(\mathrm{E}_{\text {red }}^{\circ}\right)$
$\mathrm{E}{\mathrm{MnO}{4}^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V}$
$\mathrm{E}{\mathrm{CO}{2} / \mathrm{H}{2} \mathrm{C}{2} \mathrm{O}_{4}}^{\circ}=-0.49 \mathrm{~V}$
If the equilibrium constant of the above reaction is given as $\mathrm{K}_{\mathrm{eq}}=10^{\mathrm{x}}$, then the value of $\mathrm{x}=$ _________ (nearest integer)
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Answer (339)
Solution
CellRx ${ }^{\mathrm{n}} ; \mathrm{MnO}{4}^{-}+\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$
$\mathrm{E}{\text {cell }}^{\circ}=\mathrm{E}{\mathrm{op}}^{\circ}$ of anode $+\mathrm{E}_{\mathrm{RP} \text { of cathode }}^{\circ}$
$=0.49+1.51=2.00 \mathrm{~V}$
At equilibrium
$\mathrm{E}_{\text {cell }}=0$,
$\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.059}{\mathrm{n}} \log \mathrm{K}$
(As per NCERT $\frac{R T}{F}=0.059$ But $\frac{R T}{F}=0.0591$
can also be taken.)
$2=\frac{0.059}{10} \log \mathrm{K}$
$\log \mathrm{K}=338.98$
