Write the half reaction for the following redox reaction
`(a) 2Fe^(3+)(aq)+2I^(-)(aq)rarr2Fe^(2+)(aq)+I_(2)(aq)`
`(b) Zn(s)+2H^(+)(aq)rarrZn^(2+)(aq)+H_(2)(g)`
(iii) `Al(s)+3Ag^(+)rarrAl^(+3)(aq)+3Ag(s)`

Answer

`E_(cell)=0.360 "volt" ,An+cd^(2+)rarrZn^(2+)+Cd`

Explanation

Reverse the sign of oxidation potential to get the values of the redcution or electrode potentials thus
`therefore E_(Zn,Zn^(2+))=0763 V therefore E_(Zn^(2+)//Zn)^(I@)=-0.763 V "and" therefore E_(Cd,Cd^(2+))^(@)=0.403V therefore E_(Cd(2+,Cd))^(@)=0.403V`
Since `Zn^(2+)//Zn` electrode is at lower potential therefore it acts as the anode while `Cd^(2+)//Cd` electrode with higher potential acts as the cathode
In other owrds Zmn loses electrons and `Cd^(2+)` iion accepts them Therefore cell reaction is `Zn+Cd^(2+)rarrZn^(2+)+Cd`
and `E_(Cell)^(@)=E_(Cd^(2+),Cd)^(-E^(@))(Zn^(2+)Zn)=-0.04030-(-0.763)=+0.360V`