Consider the following electrochemical cell.
(a). Write a balanced net ionic equation for the spontaneous reaction that take place in the cell.
(b). Calculate the standard cell potential `E^(0)` for the cell reaction.
(c). If the cell emf is `1.6V` what is the concentration of `Zn^(2+)`?
(d). How will the cell potential be affected if Kl is added to `Ag^(+)` half-cell?

Explanation

(a) The spontaneous reaction taking place in the cell is :
`Zn(s)+2Ag^(+)(aq) to Zn^(2+)(aq)+2Ag(s)`
(b) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.80-(-0.76)=1.56V`
(c ) According to Nernst equation :
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Ag^(+)]^(2))`
`1.60=1.56-(0.0591)/(2)"log"([Zn^(2+)])/((0.1)^(2))`
`0.04=-0.02955" log"([Zn^(2+)])/((0.01))`
`"log"([Zn^(2+)])/((0.01))=(0.04)/((-0.02955))=-1.356`
`([Zn^(2+)])/(0.01)="Antilog"(-1.356)="Antilog"(overset(-)(2).644)4.4xx10^(-2)`
`[Zn^(2+)]=4.4xx10^(-4)M`
(d) On adding KI solution to `Ag^(+)//Ag` half-cell (cathode half cell) `Ag^(2+)` ions will be precipiated as Agl. Thus, the `[Ag^(+)]` will decrease and `E_(cell)` will reduce according to the Nernst equation.
`Ag^(+)+I to Agl`.