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

Options

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

Answer

B

Explanation

`Cu^(2+) +2e^(-) to Cu`
`E^(@) =0.337V `
`DeltaG=-nFE_("cell")^(@)`
`=-2xx F xx 0.337`
`=-0.674`
`Cu^(2+) to Cu^(2+) +e^(-)`
`E^(@) =-0.153V `
`Delta G=+1 xx F xx 0.153`
Final
` Cu^(+) +e^(-) to Cu`
`DeltaG=-0.52 V`
`DeltaG=-nFE_("cell")^(@)`
`E_("cell")^(@)=0.52V`