Electrochemical cell oxygen evolution pH threshold JEE Main
Q. 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 :
E°M2+/M = 0.994 V
E°O2/H2O = 1.23 V (standard)
Correct Answer: 4

Explanation

At anode, two possible oxidation processes compete:

1. Oxidation of metal M to M2+
2. Oxidation of water to oxygen gas

Oxygen evolution reaction at anode is:

O2 + 4H+ + 4e− ⇌ 2H2O

Using Nernst equation:

E = E° − (0.059 / 4) log(1 / [H+]4)

E = 1.23 − 0.059 × pH

For oxygen evolution to start, oxidation potential of water must become equal to oxidation potential of metal:

1.23 − 0.059 × pH = 0.994

0.059 × pH = 1.23 − 0.994 = 0.236

pH = 0.236 / 0.059 ≈ 4

Hence, oxygen gas starts evolving at anode when pH is greater than approximately 4.

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