Std reduction potential for calomel half cell is 0.28 V AT 25 C. Calculate half potential when 0.1 N KCl solution is used.

Asked by Kalita Padmanath | 23rd Jul, 2017, 11:42: PM

Expert Answer:

 
The half-cell is represented as Hg/ Hg2Cl2 (salt)/ KCl (sat)
Calomel electrode can act either as anode or cathode depending on the other electrode used 
For Anode:- 2 Hg + 2Cl - Hg2Cl2 + 2e-

For Cathode:- Hg2Cl2 + 2e-  → 2 Hg + 2Cl-

Net reaction is:- Hg2Cl2 (s) + 2e- → 2 Hg + 2Cl-

The electrode potential may be represented by the Nernst equation as
E = Eo – 0.059/n log [Cl -]2 
E = Eo – 0.0591/2 log [Cl-]2
E = Eo – 0.0591 log [Cl ] at 298 K

The potential of the calomel electrode depends on the concentration of KCl used.
The electrode potential decreases with increase in the concentration of chloride ions.
For 0.1N KCl E = 0.33V
1N KCl E = 0.28V
Saturated KCl E = 0.24V
 

Answered by Prachi Sawant | 24th Jul, 2017, 11:17: AM