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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minPYQ_2022
ChemistryMediumnumerical

The reduction potential E0, in V of MnO4-aq/Mns is

[Given: EMnO4-aq/MnO2so=1.68 V; EMnO2s/Mn2+aqo=1.21 V;  EMn2+aq/Mnso=-1.03 V]

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

Answer:
0.77
Solution:

Given

(1) MnO4-aq+4H++3e⟶MnO2s+2H2O; E°=1.68 V

ΔG1°=-3 F1.68=-5.04 F

(2) MnO2s+4H++2e⟶Mn2+(aq) +2H2O; E°=1.21 V

ΔG2°=-2 F1.21=-2.42 F

(3) Mn2+aq+2e⟶Mns; E°=-1.03 V

ΔG3°=-2 F-1.03=+2.06 F

Adding 1,2 and 3,

MnO4-aq+8H++7e⟶Mns+4H2O

ΔGo=ΔG1°+ΔG2°+ΔG3°

=-5.04-2.42+2.06F

-7 FE°=-5.40 F

E°=0.77 V

Stream:JEE_ADVSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells
⏱ 2mℹ️ Source: PYQ_2022

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