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
A. 0.52 V
B. 0.90 V
C. 0.30 V
D. 0.38 V

Answer 1

Correct Answer - A
(a) `DeltaG^(@)=-nFE^(@)`
For reaction, `Cu^(2+)+2e^(-) to Cu, ` …(i)
`DeltaG^(@)=-2xxFxx0.337`
For reaction, `Cu^(+) to Cu^(2+)+e^(-)`, ….(ii)
`DeltaG^(@)=-1xxFxx(-0.153)`
`=+0.153 F`
Adding Eqs. (i) and (ii) , we get
`Cu^(+)+e^(-) to Cu, DeltaG^(@)=-0.521 F`
`DeltaG^(@)=-nFE^(@)`
`:. -0.521 F=-nFE^(@)`
`:. E^(@)=0.52 V`