Consider the following electrochemical cell : Pt|O_ 2 ( g)(1 bar)|...

In this electrochemical cell, at the anode oxygen evolution occurs from water oxidation. Using Nernst equation for the O₂/H₂O couple: $E_{anode} = E°_{O_2/H_2O} - \frac{0.059}{4} \log \frac{1}{[H^+]^4} = 1.23 - 0.059 \times pH$

At the cathode: $E_{cathode} = 0.994$ V (standard potential for M²⁺/M reduction)

For oxygen to start evolving, the cell potential becomes zero: $E_{cell} = E_{cathode} - E_{anode} = 0$

$0.994 = 1.23 - 0.059 \times pH$

$0.059 \times pH = 0.236$

$pH = 4.0$