Question

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 1

`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