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

 

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