CBSE Class 12-science Answered
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 | 23 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
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 | 24 Jul, 2017, 11:17: AM
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