Question

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

Answer 1

Correct Answer - D
For the reaction,
`2Fe^(3+)+2I^(-)rarr 2Fe^(2+)+I_(2)`
`E_("cell")^(@)=E_(Fe^(3+)//Fe^(2+))^(@)-E_(I_(2)//I^(-))^(@)=0.77-(0.54)=+0.23V`
Here, `E_("cell")^(@)` is +ve so, reaction is feasible.
For the reaction,
`Cu+2Ag^(+)rarr Cu^(2+)+2Ag`
`E_("cell")^(@)=E_(Ag^(+)//Ag)^(@)-E_(Cu^(2+)//Cu)^(@)=0.80-(0.34=+0.46 V`
Here, `E_("cell")^(@)` is +ve so, the reaction is feasible.
For the reaction, `2Fe^(3+)+Cu rarr 2 Fe^(2+)+Cu^(2+)`
`E_("cell")^(@)=E_(Fe^(3+)//Fe^(2+))^(@)-E_(Cu^(2+)//Cu)^(@)=0.77-(0.34)=+0.43 V`
Here, `E_("cell")^(@)` is +ve so, the reaction is feasible.
For the reaction, `Ag+Fe^(3+)rarr Ag^(+)+Fe^(2+)`
`E_("cell")^(@)=E_(Fe^(3+)//Fe^(2+))^(@)-E_(Ag^(+)//Ag)^(@)=0.77-(0.80)=-0.03 V`
Here, `E_("cell")^(@) ` is negative so, the reaction is not feasible.