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Electrochemistry question

2022 · Shift 1 · Q2
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Electrochemistry question

2022 · Shift 1 · Q2

JEE AdvancedChemistryElectrochemistryNumerical+3 / −1
The reduction potential (E0\left(E^{0}\right.(E0, in V)\left.\mathrm{V}\right)V) of MnO4−(aq)/Mn(s)\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})MnO4−​(aq)/Mn(s) is ‾\underline{\hspace{2cm}}​. [Given: E(MnO4−(aq)/MnO2( s))0=1.68 V;E(MnO2( s)/Mn2+(aq))0=1.21 V;E(Mn2+(aq)/Mn(s))0=−1.03 VE_{\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}E(MnO4−​(aq)/MnO2​( s))0​=1.68 V;E(MnO2​( s)/Mn2+(aq))0​=1.21 V;E(Mn2+(aq)/Mn(s))0​=−1.03 V ]
Numerical answer
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