Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V` Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will be

Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(

1.	Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V` Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will be
Answer» `DeltaG^(@)=-nFE_(cell)^(@)` `DeltaG_(3)^(@)=DeltaG_(1)^(@)-DeltaG_(2)^(@)` `(-n_(3)FE_(3)^(@))=(-n_(1)FE_(1)^(@))-(-n_(2)FE_(2)^(@))` or `" " (-E_(3)^(@))=(-2E_(1)^(@))-(-E_(2)^(@))` or `" " E_(3)^(@)=2E_(1)^(@)-E_(2)^(@)=(2xx0.337)-(0.153)` =(0.674-0.153)=0.521 V

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Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V` Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will be

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