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Using the standard electrode potentials...

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

Text Solution

Verified by Experts

The reaction is feasible if the EMF of the cell reaction is positive
(i). `Fe^(3+)(aq)+I^(-)(aq)toFe^(2+)(aq)+(1)/(2)I_(2)(g)`
i.e., `Pt//I_(2)//I^(-)(aq)||Fe^(3+)(aq)|Fe^(2+)(aq)|Pt`
`E_(cell)^(@)=E_(Fe^(3+),Fe^(2+))^(@)-E_(1//2I_(1),I^(-))^(@)`
`=0.77-0.54=0.23V` (Feasible)
(ii). `2Ag^(+)(aq)+Cu(s)to2Ag(s)+Cu^(2+)(aq)`
i.e., `Cu|Cu^(2+)(aq)||Ag^(+)(aq)|Ag`
`E_(cell)=E_(Ag^(+),Ag)-E_(Cu^(2+),Cu)^(@)`
`=0.80-0.34=0.46V` (feasible)
(iii). `Fe^(3+)(aq)+Br^(-)(aq)toFe^(2+)(aq)+(1)/(2)Br_(2)(g)`
`E_(cell)^(@)=0.77-1.09=-0.32V` (Not feasible)
(iv). `Ag(s)+Fe^(3+)(aq)toAg^(+)(aq)+Fe^(2+)(aq)`
`E_(Cell)^(@)=0.77-0.80=-0.03`V (Not feasible)
(v). `(1)/(2)Br_(2)(g)+Fe^(+2)(aq)toBr^(-)(aq)+Fe^(3+)(aq)`
`E_(cell)^(@)=1.09-0.77=0.32V` (Feasible).
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