Calculate the emf of the cell.
`Zn|Zn^(2+)(0.001M)||Ag^(+)(0.1M)|Ag`
The standard potential of `Ag//Ag^(+)` half - cell is `+0.80 V` and `Zn//Zn^(2+)` is `-0.76V.`

Explanation

Step 1 : Write the half-cell reactions of the anode and the cathode.
Then add the anode and cathode half reactions to obtain the cell reaction and the value of `E_("cell")^(@)`
`{:("Cathode",:,2Ag^(+)+2e^(-),rarr,2Ag,E^(@)=+0.80),("Anode",:," "Zn,rarr,Zn^(2+)+2e^(-),E^(@)=-0.76V),("Cell",:,Zn+"2Ag"^(+),rarr,Zn^(2+)+2Ag,E^(@)=1.56V):}`
Step 2. K for the cell reaction `=([Zn^(2+)])/([Ag^(+)]^(2))`
Substituting the given values in the Nernst equation and solving for `E_("cell")`, we have
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK`
`=1.56-(0.0591)/(2)log""([Zn^(2+)])/([Ag^(+)]^(2))`
`=1.56-(0.0591)/(2)log""([10^(-3)])/([10^(-1)]^(2))`
`=1.56-0.02955`
`=1.58955V`
Calculation of Equilibrium constant for the cell reaction
The Nernst equation for a cell is
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK`
`"or "logK=(E_("cell")^(@))/(0.0591)`