Consider the reaction,
`2Ag^(+)+Cd rarr 2Ag+Cd^(2+)`
The standard electrode potentials for `Ag^(+) rarr Ag` and `Cd^(2+) rarr Cd` couples are 0.80 volt and -0.40 volt respectively.
(i) What is the standard potential `E^(@)` for this reaction ?
(ii) For the electrochemical cell, in which this reaction takes place which electrode is negative electrode ?

Explanation

(i) The half reactions are:
`2Ag^(+)+ underset("Reduction cathode") (2e^(-)) to 2Ag`
`E_(Ag^(+)//Ag)^(@)=0.80 volt ("Reduction potential")`
`Cd to underset("Oxidation Anode") Cd^(2+)+2e^(-)`
`E_(Cd^(+)//Cd)^(@)=-0.40` volt (Reduction potentia)
`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 always the electrode whose reduction potential has smaller value or the electrode where oxidation occurs. Thus, Cd electrode is the negative electrode.