JEE Advance - Chemistry (2022 - Paper 1 Online - No. 2)
The reduction potential $\left(E^{0}\right.$, in $\left.\mathrm{V}\right)$ of $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$ is __________.
[Given: $E_{\left(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{MnO}_{2}(\mathrm{~s})\right)}^{0}=1.68 \mathrm{~V} ; E_{\left(\mathrm{MnO}_{2}(\mathrm{~s}) / \mathrm{Mn}^{2+}(\mathrm{aq})\right)}^{0}=1.21 \mathrm{~V} ; E_{\left(\mathrm{Mn}^{2+}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})\right)}^{0}=-1.03 \mathrm{~V}$ ]
Explanation
(1) $$MnO_4^ - (aq) + 4{H^ + } + 3e\buildrel {} \over \longrightarrow Mn{O_2}(s) + 2{H_2}O;\,E^\circ = 1.68\,V$$
$$\Delta G{^\circ _1} = - 3F(1.68) = - 5.04F$$
(2) $$MnO_2^{}(s) + 4{H^ + } + 2e\buildrel {} \over \longrightarrow M{n^{2 + }} + 2{H_2}O;\,E^\circ = 1.21\,V$$
$$\Delta G{^\circ _2} = - 2F(1.21) = - 2.42F$$
(3) $$Mn_{}^{2 + }(aq) + 2e\buildrel {} \over \longrightarrow Mn(s);\,E^\circ = - 1.03\,V$$
$$\Delta G{^\circ _3} = - 2F( - 1.0.3) = + 2.06F$$
Adding (1), (2) and (3),
$$MnO_4^ - (aq) + 8{H^ + } + 7e\buildrel {} \over \longrightarrow Mn(s) + 4{H_2}O$$
$$\Delta G = \Delta G{^\circ _1} + \Delta G{^\circ _2} + \Delta G{^\circ _3}$$
$$ = ( - 5.04 - 2.42 + 2.06)F$$
$$ - 7F\,E^\circ = - 5.4F$$
$$E$$$$^\circ$$ = $$0.77 \,V$$
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