Using the standard electrode potentials given in Table, predict if the reaction between the following is feasible`:`
`a. Fe^(3+)(aq)` and `I^(c-)(aq)`
`b.` `Ag^(o+)(aq)` and `Cu(s)`
`c.` `Fe^(3+)(aq)` and `Br^(c-)(aq)`
`d.` `Ag(s)` and `Fe^(3+)(aq)`
`e. ``Br_(2)(aq)` and `Fe^(2+)(aq)`.

Answer

`{:(A. Feasibl e,b. Feasibl e,c. Not feasibl e,,),(d. Not feasibl e,,e. Feasibl e,,):}`

Explanation

A reaction is feasible if the `EMF` of the cell reaction is `+ve`
`a. Fe^(3+)(aq)+I^(c-)(aq)rarrFe^(2+)(aq)+(1)/(2)I_(2)`
`Pt,I_(2)|I^(c-)(aq)||Fe^(3+)(aq)|Fe^(2+)(aq)|Pt`
`:. E_(cell)^(c-)=E_((Fe^(3+)^(c-)|Fe^(2+)))-E_((I_(2)|I^(c-)))^(c-)`
`=0.77-0.54=0.23V("Feasible ")`
b. `Ag^(o+)=Cu rarr Ag(s)+Cu^(2+)(aq)`
i.e., `Cu|Cu^(2+)(aq)||Ag^(o+)(aq)|Ag`
`=0.80-0.34=0.46V( "Feasible ")`
c. `Fe^(3+)(aq)+Br^(c-)(aq) rarr Fe^(2+)(aq)+ (1)/(2)Br_(2),`
`E_(cell)^(c-)=0.77-1.09=-0.32V("not feasible")`
d.`Ag(s)+Fe^(3+)(aq) rarr Ag^(o+)(aq)+Fe^(2+)(aq),`
`E_(cell)^(c-)=0.77-0.80=-0.30V("not feasible")`
e.`(1)/(2)Br_(2)(aq)+Fe^(3+)(aq)rarr Br^(c-)+Fe^(3+),`
`E_(cell)^(c-)=1.09-0.77=0.32V (feasible)`