Two weak acid solution HA_(1) and HA_(2) each with the same concentration and having pK_(a) values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25^(@)C) and are interconnected through salt bridge. Calculate the e.m.f. of the cell.

Two weak acid solution HA_(1) and HA_(2) each with the same concentrat

1.	Two weak acid solution HA_(1) and HA_(2) each with the same concentration and having pK_(a) values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25^(@)C) and are interconnected through salt bridge. Calculate the e.m.f. of the cell.
Answer» Solution :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