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)`.

Explanation

(a) The possible reaction between `Fe^(3+) (aq)` and `I^(-)(aq)` is
`2Fe^(3+)(aq)+2I^(-)(aq)rarr2Fe^(2+)(aq)+I_(2)(s)`
The above redox reaction can be split into the following two half reaction one involving oxidation and the other redcution wiritng electrode potential for each half rection from
overall reaction `2Fe^(3+)(aq) +2I^(-)(aq)rarr2Fe^(2+)+I_(2)(s)`
since the EMF for the above reaction `Ag^(+)` (aq) and Cu (s) is
`Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2ag(s)`
The above redox reaction can be split into the following two half reaction writing electrode potential for each half reaction form we have
oxidation `Cu(s)rarrCu^(2)(aq)+2e^(-), E^(@)=-0.34 V`
Reduction `Ag^(+)(aq)+E^(-)rarrAg(s)xx2, E^(@)=+0.80 V`
overall reaction `Cu(s)+2Ag^(+)(aq)rarrCu^(2+) (aq)+2 (s),E^(@)=+0.46V`
since the EMF of the above reaction comes out to be postive therefore the above reaction is feasible (c ) suppose the reaction between `Fe^(3+)(aq)rarr Cu^(aq)+2Ag(s),E^(@)=+0.46 V`
since the EMF of the above reaction comes oiut to be postive therefore to the following equatin
`Cu(s)+2Fe^(3+)(a)rarr3Cu(aq)+2Fe^(2+)(aq)`
the above reaction can be split into the following two half reaction wirting elecrode potential for each half reaction from we have
oxidation `Cu(s)rarrCu^(2+)(aq)+2e^(-) , E^(@)=-0.34 V`
Reduction `Fe^(3+)(aq)+E^(-)rarrFe^(2+)(aq)xx2 , E^(2)=+0.77V`
overall reaction `Cu(s)_+2Fe^(3+)(aq)rarrCu^(2+)+2Fe^(2+)(aq),E^(@)=+0.43 V`
since the EMF of the reaction is postive therefore the above reaction is feasible
alternatively if the raction between 1 `Fe^(3+)` (aQ) and Cu (s) occurs according to the followng equation
The EMF of the reaction comes out to be -ve -0.36 V(-0.34 V -0.034 V) and hence this rection is not feasible
(d) suppose the reaction between Ag(s) and `Fe^(3+)` (Aq) occurs accoding to the following eqation
The above reaction can be split into the following two half rection writing electrode potential for each half reaction from we have
oxidation `Ag(s)rarrAg^(+)(aq)+e^(-), E^(@)=-0.80 V`
Reduction `Fe^(3+)(aq)+E^(-)rarrFe^(2+)(aq), E^(@)=+0.77 V`
since the EMF of the reaction is negative therefore the above reaction is not feasible
Alternatively the rection between Ag (s) and `Fe^(3+)` (aQ) may occur according to the following equation
`3Ag(s)+Fe^(3+)(aq)rarr3Ag^(+)(aq)+Fe(s)`
on similar lines we can calcualte the EMF of this reaction comes to be even followng equation
`Br_(2)(aq)+2Fe^(2+)(aq)rarr2Br^(-)(aq)+2Fe^(2+)(aq)`
the above reaction can be split in to the followng two half rection writing electrode potential for each half reaction from the we have
oxidation `Fe^(2+)(aq)rarrFe^(3+)(aq)+e^(-)xx2,E^(@)=-0.77 V`
Reduction : `Br_(2)(aq)+2e^(-)rarrBr^(-)rarr2Br^(-)(aq),E^(@)=+1.09 V`
overall reaction `2Fe^(2+)(aq)+Br_(2)(aq)rarr2Fe^(3+)(aq)+2 Br^(-)(aq)'E^(@)=+0.32 V` since the EMF for the above reaction is positive therefore this reaction is feasible