Electrochemistry Question 3
Question 3 - 2024 (01 Feb Shift 2)
Consider the following redox reaction :
$MnO _4^{-}+H^{+}+H _2 C _2 O _4 \rightleftharpoons Mn^{2+}+H _2 O+CO _2$
The standard reduction potentials are given as below $\left(E _{\text {red }}^{\circ}\right)$
$E _{MnO _4^{-} / Mn^{2+}}^{\circ}=+1.51 V$
$E _{CO _2 / H _2 C _2 O _4}^{\circ}=-0.49 V$
If the equilibrium constant of the above reaction is given as $K _{eq}=10^{x}$, then the value of $x=$ _________ (nearest integer)
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Answer (339)
Solution
CellRx ${ }^{n} ; MnO _4^{-}+H _2 C _2 O _4 \rightarrow Mn^{2+}+CO _2$
$E _{\text {cell }}^{\circ}=E _{op}^{\circ}$ of anode $+E _{RP \text { of cathode }}^{\circ}$
$=0.49+1.51=2.00 V$
At equilibrium
$E _{\text {cell }}=0$,
$E _{\text {cell }}^{\circ}=\frac{0.059}{n} \log K$
(As per NCERT $\frac{R T}{F}=0.059$ But $\frac{R T}{F}=0.0591$
can also be taken.)
$2=\frac{0.059}{10} \log K$
$\log K=338.98$
