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

Answer 1

Correct Answer - D
The desired half-reaction is
`Cu^(+)(aq.)+e^(-) rarr Cu(s)`
We are given
(i) `Cu^(2+)(aq.)+2e^(-) rarr Cu(s)`
(ii) `Cu^(2+)(aq.)+e^(-) rarr Cu^(+)(s)`
We can obtain the desired half-reaction by subtracting second equation from first equation. Thus `E_(T)^(@)=(n_(1)E_(1)^(@)+n_(2)E_(2)^(@))/(n_(T))`
`= ((2)(0.337 V)(-0.153 V))/(1)`
`= 0.521 V`