Two weak acid solutions `HA_(1)` and `HA_(2)` with the same concentration and having `pK_(a)` values 3 and 5 are placed in contact with hydrogen electrode `(1 atm ` and `25^(@)C)` and are interconnected through a salt bridge. Find the `EMF` of the cell.

Explanation

The cell may be represented as
`Pt|H_(2)(1atm)|HA_(2)||HA_(1)|H_(2)(1atm)|Pt`
electrode reaction expressed as reduction reaction is `H^(+)+e^(-)to(1)/(2)H_(2)`
At anode: `E_((H^(+)//H_(2))_(2))=E_((H^(+)//H_(2))_(2))^(@)+0.059(pH)_(2)`
At cathode: `E_((H^(+)//H_(2))_(1))=E_((H^(+)//H_(2))_(1))^(@)+0.059(pH)_(1)`
But `[H^(+)]=Calpha=Csqrt((K_(a))/(C))=sqrt(K_(a)C)=(K_(a)C)^(1//2)`
or `-log[H^(+)]=-(1)/(2)logK_(a)-(1)/(2)logC`
`pH_(2)=(1)/(2)pK_(a)-(1)/(2)logC`
`E_(cell)^(@)=E_((H^(+)//H_(2))_(1))^(@)-E_((H^(+)//H_(2))_(2))^(@)=0.059[(1)/(2)pK_(a_(2))-(1)/(2)pK_(a_(1))]=(0.0591)/(2)(5-3)`
=0.059V