in two vessel containing 500ml water , 0.5 mol of aniline (Kb = 10-9) and 25 mmol of HCl are added seperately . Two hydrogen electrodes are constructed using these solutions. Calculate the emf of cell made by connecting them appropriately.
Answer :0.395V
Please explain briefly .
C6H5NH2 + H2O -------> C6H5NH3+ + OH-
Given conc. of Aniline = 0.5 mol in 500 ml i.e. 1mol/lit
Kb = 10-9
Therefore,
Kb = [C6H5NH3+] [OH-]/[C6H5NH2] = 10-9
[OH-] = [C6H5NH3+] = √10-9
= 3.1 x 10-5
Conc. of HCl = 0.25 mmol in 500 ml i.e 0.5 mmol/ lit = 5 x 10-3
[H+] = 50 x 10-3
From nernst eqn:
Ecell = E0cell - 0.0592/n log10 [OH-] / [H+]
As it is hydrogen electrode E0cell = 0
Therefore,
Ecell = - 0.0592/n log10 3.1 x 10-5 / 50 x 10-3 as n= 1
Ecell = - 0.0592 log 3.98 x 10-7
= -0.0592 x -6.42
= 0.0387 V
=