Equinormal solution of two weak acids , HA(pKa= 3) and HB (pKb = 5) are each placed in contact with standard hydrogen electrode at 25oC. When a cell is construsted by interconnecting them through a salt bridge , find the emf of the cell.
Answer : E = 0.059
Take HA as the right hand electrode and HB as the left hand electrode.
For Hydrogen electrode at 1 atm and 25°C
E red = -0.059pH
pH of weak acid = 1/2 pKa - 1/2 log C
so for the electrode HA E red = -0.059/2 [ pKa - log C] = - 0.0295 [3 - log C]
For electrode HB E red = -0.059/2 [ pKa - log C] = -0.0295 [5 - log C]
Ecell = E red of HA - E red of HB = - 0.0295 [3 - log C] + 0.0295 [5 - log C] = 0.059 V