Consider the following electrochemical cell : Pt | O2(g)(1 bar) | HCl(aq) || M2+(aq, 1.0 M) | M(s) The pH above which, oxygen gas would start to evolve at anode is (nearest integer). [Given : . arrayl E M2+/M = 0.994 V \ E O2/H2O = 1.23 V array \ standard reduction potential and RTF(2.303) = 0.059 V at the given condition]
Answer
4
Solution
Anode (oxidation): 2H2O O2 + 4H+ + 4e-
Cathode (reduction): M2+ + 2e- M
Reduction potential of the oxygen electrode (pO2 = 1 bar):
EO2/H2O = 1.23 - 0.0594 1[H+]4
= 1.23 - 0.059 pH
Ecell = Ecathode - Eanode = 0.994 - (1.23 - 0.059 pH)
= 0.059 pH - 0.236
O₂ is evolved (cell works) only when Ecell > 0
0.059 pH > 0.236
pH > 4
Answer: 4