Consider the reaction `2Ag^(+) +Cd to 2Ag to 2Ag +Cd^(2+)`.
The standard reduction potentials of `Ag^(+)//Ag and Cd^(2+)//Cd` are + 0.80 volt and - 0.40, respectively.
(i) Give the cell representation.
(ii) What is the standard cell emf. `E^(Theta)` ?
(iii) What will the emf of the cell if concentration of `Cd^(2+)` is 0.1 M and `Ag^(+)` is 0.2 M?
(iv) Will the cell work spontaneously for the condition given in (ii) above ?

Explanation

(i) From the values of electrode potentials, it is clear that cadmium is acting as anode and silver electrode as cathode.
Thus, the celll is represented as:
`Cd//Cd^(2+) (aq)"||" Ag^(+) (aq) //Ag`
(ii) `E_("cell")^(Theta) =E_("cathode")^(Theta)-E_("anode")^(Theta)=0.80-(-0.40)=1.20V`
(iii) `E_("cell")=E_("cell")^(Theta) -(0.0591)/(2)"log"([Cd^(2+)])/([Ag^(+)]^(2))`
`=1.20V-(0.0591)/(2)"log"([0.1])/([0.2]^(2))`
`=1.20 -(0.0591)/(2) xx 0.3979=1.20 -0.01 =1.19V`
(iv) The cell reaction is spontaneous because `E_("cell") ` is positive. Therefore, free enregy change for this spontaneous reaction will be negative.