Given: (i) `Cu^(2+)+2e^(-)toCu," "E^(o)=0.337V` (ii) `Cu^(2+)+e^(-)toCu^(+)," "E^(o)=0.153V` Electrode potential, `E^(o)` for the reaction, `Cu^(+)+e^(-)toCu`, will beA. 0.52 VB. 0.90 VC. 0.30 VD. 0.38 V

Given: (i) `Cu^(2+)+2e^(-)toCu," "E^(o)=0.337V` (ii) `Cu^(2+)+e^(-)toC

1.	Given: (i) `Cu^(2+)+2e^(-)toCu," "E^(o)=0.337V` (ii) `Cu^(2+)+e^(-)toCu^(+)," "E^(o)=0.153V` Electrode potential, `E^(o)` for the reaction, `Cu^(+)+e^(-)toCu`, will beA. 0.52 VB. 0.90 VC. 0.30 VD. 0.38 V
Answer» Correct Answer - A `DeltaG^(@)=nFE^(@)` For, `Cu^(2+)+2e^(-)toCu,DeltaG_(1)^(@)=-2F(0.337)` . . . . (i) `Cu^(2+)+e^(-)toCu^(+),DeltaG_(2)^(@)=-F(0.153)` . . . (ii) Subtracting (ii) from (i), `Cu^(+)+e^(-)toCu` `DeltaG^(@)=-0.674F+0.153F` . . . (iii) `nFE^(@)=-0.521F` `because`For (iii), n=1, `thereforeE^(@)=0.521V`.

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Given: (i) `Cu^(2+)+2e^(-)toCu," "E^(o)=0.337V` (ii) `Cu^(2+)+e^(-)toCu^(+)," "E^(o)=0.153V` Electrode potential, `E^(o)` for the reaction, `Cu^(+)+e^(-)toCu`, will beA. 0.52 VB. 0.90 VC. 0.30 VD. 0.38 V

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