The two half-cell reactions of an electrochemical cell is given as {Ag}^{+}+e^{-} \rightarrow {Ag} ; E_{ {Ag}^{+} \/ {Ag}}^{\circ}=-0.3995 {V} {Fe}^{2+} \rightarrow {Fe}^{3+}+e^{-} ; E_{ {Fe}^{3+}}^{\circ} \/ {Fe}^{2+}=-0.7120 {V} The value of EMF will be

MHT CET 2021 Chemistry Solved Paper

Question : 64 of 100

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The two half-cell reactions of an electrochemical cell is given as
Ag++e−⟶Ag;EAg+∕Ag∘=−0.3995V
Fe2+⟶Fe3++e−;EFe3+∘∕Fe2+=−0.7120V
The value of EMF will be

−0.3125V
0.3125V
1.114V
−1.114V

Validate

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