Problem:
In two vessels each containing 500mL of water, 0.5 mmol of aniline (K
b=10
-9) and 25 mmol of HCl are added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of cell made by connecting them appropriately.
(mmol = milli moles)
Attempt:
This is basically a concentration cell. Since the electrodes are connected properly, I need not worry about which is the anode and the cathode, I have to find the absolute value of emf.
Concentration of H
+ in vessel containing HCl, a=0.05 M
I have K
b for aniline so I can find the concentration of H
+ i.e b. It comes out to be 10
-8.
From the nernst equation,
E
cell=E
ocell-0.059/n * log(b/a)
Here, E
ocell=0 and n=2, hence E
cell=-0.059/2 * log(10
-8/0.05).
Solving, I get E
cell=0.197 V which is just half of the given answer.
Any help is appreciated. Thanks!
PS: Can someone tell me how do you write latex here?