Write the half eaction for the following redox reac tiojns (a)2Fe^(3+)(aQ)+2I^(-)(aq) rarr 2 Fe^(2+)(aq)+I_(2)(aq) (b)Zn(s)++2 H^(+)(aq)rarr Zn (c )AI(s)+3Ag^(+)(aQ)rarrAI^(3)rarr AI^(3+)(aq)+3Ag (s)

Write the half eaction for the following redox reac tiojns (a)2Fe^(3+

1.	Write the half eaction for the following redox reac tiojns (a)2Fe^(3+)(aQ)+2I^(-)(aq) rarr 2 Fe^(2+)(aq)+I_(2)(aq) (b)Zn(s)++2 H^(+)(aq)rarr Zn (c )AI(s)+3Ag^(+)(aQ)rarrAI^(3)rarr AI^(3+)(aq)+3Ag (s)
Answer» Solution :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`