Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
`E^(θ)` values : `Fe^(3+)//Fe^(2+)=0.77, I_(2)//I^(-)=+0.54`,
`Cu^(2+)//Cu=+0.34, Ag^(+)//Ag=+0.80V`

Options

A. `Fe^(3+) and I^(-)`
B. `Ag^(+) and Cu`
C. `Fe^(3+) and` Cu
D. `Ag and Fe^(3+)`

Answer

D

Explanation

`2Fe^(3+) +2e^(-) rarr 2Fe^(2+) ,E^(@) = +0 .77V`
`2I^(-) rarr I_2 +2e^(-) ,E^(@) = -0.54v`
(sign of `E^@` is revered)
`2F^(3+) +2I^(-) rarr 2Fe^(2+) +I_2 , E_("cell")^@ = +0.23 V`
This reaction is feasible sin `E_("cell")^(@) =+0.23V`
(b) `Cu rarr Cu^(2+) +2e^(-), E^@ = -0.34 V`
(sign of `E^@` has been reversed)
`2Ag^(+) + 2e^(-) rarr 2Ag, E^@ = +0.80V`
`Cu+2Ag^(+)rarr2Cu^(2+)+2Ag,E^@ =+046V`
This reaction is feasible since `E_("cell")^@` is positive.
(c) `2Fe^(3+)+ 2e^(-) rarr 2Fe^(2+) ,E^@ = +0.77V`
`(Cu rarrCu^(2+)+2e^(-),E^@=-0.34V)/(2Fe^(3+)+Curarr2Fe^(2+)+Cu^(2+),E^@=+0.43V)`
(sign of `E^@` is reversed)
This reaction is feasible since `E_("cell")^@` is positive.
(d) `Ag rarr Ag^(+) + e^(-) ,E^@ = -0.80V`
`Fe^(3+)+ e^(-)rarr Fe^(2+) ,E^0 =+ 0.77V`
`Ag +Fe^(3+) rarr Ag^(+) +Fe^(2+) , E^@ = -0.03 V`
This reaction is not feasible since `E_("cell")^@` is negative.