Prove that for two half reactions having potentials `E_(1)` and `E_(2)` which are combined to yield a third half reaction, having a potential `E_(3)`,
`E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)`
`DeltaG_(3)=DeltaG_(1)+DeltaG_(2)`
`-n_(3)FE_(3)=-n_(1)FE_(1)-n_(2)FE_(2)`
or `n_(3)E_(3)=n_(1)E_(1)+n_(2)E_(2)`
or `E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)`
(i) The half reaction are :
`2Ag^(+)+underset(("Reduction"),("(Cathode)"))(2e^(-)) rarr 2Ag`,
`E_(Ag^(+)//Ag)^(@)=0.80` volt (Reduction potential)
`Cd rarr underset(("Oxidation"),("(Anode)"))(Cd^(2+))+2e^(-)`,
`E_(Cd^(2+)//Cd)^(@)=-0.40` volt (Reduction potential)
or `E_(Cd//Cd^(2+))^(@)=+0.40` volt
`E^(@)=E_(Cd//Cd^(2+))^(@)+E_(Ag^(+)//Ag)^(@)=0.40+0.80=1.20` volt
(ii) The negative electrode is...