Calculate emf of the half cell given below : Pt(s) | H2(g, 2 atm) | HCl(aq, 0.02 M) E H2/H+ = 0 V (Given : 2.303RTF = 0.059, log 2 = 0.3010)
- −0.109 V
- 0.109 V
- 0.035 V
- −0.035 V
Answer
(2) 0.109 V
Solution
Electrode reaction (oxidation, H2/H+): H2(g) 2H+(aq) + 2e-, n = 2
[H+] = 0.02 M = 2 10-2 M, pH2 = 2 atm
E = E - 0.0592 [H+]2pH2
⇒ E = 0 - 0.0295 (2 10-2)22
⇒ E = -0.0295 (2 10-4)
⇒ E = -0.0295 (0.3010 - 4)
⇒ E = -0.0295 (-3.699) = +0.109 V
Answer: (2)