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JEE Advanced 2022Paper 1
Exam
JEE Advanced
Year
2022
Paper
Paper 1
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2022 · Paper 1 · Chemistry · Electrochemistry

The reduction potential ( E ^ 0 ), in V) of MnO _ 4 ^ - (aq) /...

The reduction potential (E0),in\leftV)\left({E}^{0}\right), in \leftV) of MnO4−(aq)/Mn(s){MnO}_{4}^{-}\left(aq\right) / Mn\left(s\right) is

[Given: E(MnO4−(aq)/MnO2(s))o=1.68V;E(MnO2(s)/Mn2+(aq))o=1.21V;E(Mn2+(aq)/Mn(s))o=−1.03V{E}_{\left({MnO}_{4}^{-}\left(aq\right) / {MnO}_{2}\left(s\right)\right)}^{o} = 1 . 68V; {E}_{\left({MnO}_{2}\left(s\right) / {Mn}^{2 +}\left(aq\right)\right)}^{o} = 1 . 21V; {E}_{\left({Mn}^{2 +}\left(aq\right) / Mn\left(s\right)\right)}^{o} = - 1 . 03V]

Truncate/round-off the value to TWO decimal places.

Correct answer

0.77

Step-by-step solution

Given

(1) MnO4−(aq)+4H++3e⟶MnO2(s)+2H2O{MnO}_{4}^{-}\left(aq\right) + 4{H}^{+} + 3e ⟶ {MnO}_{2}\left(s\right) + 2{H}_{2}O; E°=1.68VE ° = 1 . 68V

ΔG1°=−3F(1.68)=−5.04F{\Delta G}_{1} ° = - 3F\left(1 . 68\right) = - 5 . 04F

(2) MnO2(s)+4H++2e⟶Mn2+(aq)+2H2O{MnO}_{2}\left(s\right) + 4{H}^{+} + 2e ⟶ {Mn}^{2 +}( aq ) + 2{H}_{2}O; E°=1.21VE ° = 1 . 21V

ΔG2°=−2F(1.21)=−2.42F{\Delta G}_{2} ° = - 2F\left(1 . 21\right) = - 2 . 42F

(3) Mn2+(aq)+2e⟶Mn(s){Mn}^{2 +}\left(aq\right) + 2e ⟶ Mn\left(s\right); E°=−1.03VE ° = - 1 . 03V

ΔG3°=−2F(−1.03)=+2.06F{\Delta G}_{3} ° = - 2F\left(- 1 . 03\right) = + 2 . 06F

Adding \left(1\right), \left(2\right)and(3)and \left(3\right),

MnO4−(aq)+8H++7e⟶Mn(s)+4H2O{MnO}_{4}^{-}\left(aq\right) + 8{H}^{+} + 7e ⟶ Mn\left(s\right) + 4{H}_{2}O

ΔGo=ΔG1°+ΔG2°+ΔG3°{\Delta G}^{o} = {\Delta G}_{1} ° + {\Delta G}_{2} ° + {\Delta G}_{3} °

=(−5.04−2.42+2.06)F= \left(- 5 . 04 - 2 . 42 + 2 . 06\right)F

−7FE°=−5.40F- 7FE ° = - 5 . 40F

E°=0.77VE ° = 0 . 77V

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