Given :
(i) `CU^(2+) + 2e^- rarr Cu, E^@ = 0.337 V`
(ii) `Cu^(2+) +e^- rarr Cu^(+) , E^2=0.1 5 3 V Then E^0 for Cu^(+) +e^- rarr Cu`.

Options

A. ` -0. 49 V`
B. ` 0. 49 V`
C. ` -0. 38 V`
D. ` 0.38 V`

Answer

D

Explanation

The required reactin `(Cu^(++) + Cr =Cu^++) rarr 2Cu^+)` can be obttaine by using the following reactions.
` Cu^(++) + e^- rarr Cu^(+) , E_(Cu^(++)//Cu)^@ =0.15 V`….(i)
`Cu^(++) + 2e^- rarr Cu, E_(Cu^(++)//Cu)^@ =0. 35 V` …(ii)
Multiplying Eq.(i) by `2` we get
`2 Cu^(++) + rarr 2C u^(+) ...(iii)
`DeltaG_1=- 2e^(-) n-FE=- 2 xx F xx 0. 15`
` Cu^(++) +2e^- rarr Cu` ...(iv)
` Delta G_2 =- nFE=- 2xx Fxx 0.34`
Subtract eq. (iv) from (iii)
` Cu^(++) + Cu rarr 2 Cu`
` Delta G_3 =- nFE=- 1 xx Fxx E^@`
Also `Delta g_3 = Delta _1 - Delta G_2`
` 1 Fe^@ = (-2 Fxx0.15)_(2) F xx 0. 34)`
`e^@ =- 0. 38`
This is the value fro the reaction
` Cu^(++) + Cu rarr 2Cu`
But the given reaction si just reverse of it
` :. E_(cell)` for given reaction ` =+ 0.38 V`.