Given : `E_(Fe^(3+)//Fe)^(@) = -0.036V, E_(FE^(2+)//Fe)^(@)= -0.439V`. The value of electrode potential for the change, `Fe_(aq)^(3+) + e^(-)rightarrow Fe^(2+) (aq)` will be :

Options

A. `0.770V`
B. `-0.27V`
C. `-0.072V`
D. `0.385V`

Answer

A

Explanation

`Fe^(3+) +3e rarr Fe` ...(1)
`E^(@) = 0.036V`
`:. DeltaG_(1)^(@) = nFE^(@)`
`=- 3 xx F(-0.036V)`
`= 0.108F`
Also, `Fe^(2+) +2e rarr Fe` ..(2)
`E^(@) =- 0.439V`
`:. DeltaG_(2)^(@) = nFE^(@)`
`=- 2 xx F(0.439)`
`=08.78 F`
For `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq)` ..(3)
`:.` Assume `E^(@) = xV`
eq.(1)-eq.(2) produces eq.(3)
`:. DeltaG^(@) = DeltaG_(1)^(@) - DeltaG_(2)^(@)`
`-1 xx F xx x = 0.108 F - 0.878F`
`- x =- 0.77`
`x = 0.77V`