Consider the following electrochemical cell : $$Pt| O_{2}(g)(1 bar) | HCl(aq)||M^{2+}(aq$$, $$1.0M)|M(s)$$
The pH above which , oxygen gas would start to evolve at anode is ____ (nearest integer).
[Given: $$E_{M^{2+}/m}^\circ = 0.994V$$ , $$E_{O^{2}/H_{2}O}^\circ = 1.23V$$ are standard reduction potential and $$\frac{RT}{F}(2.303)=0.059V$$ at the given condition]

We have the electrochemical cell: $$Pt \mid O_2(g)(1 \text{ bar}) \mid HCl(aq) \| M^{2+}(aq, 1.0M) \mid M(s)$$

Given: $$E^\circ_{M^{2+}/M} = 0.994$$ V, $$E^\circ_{O_2/H_2O} = 1.23$$ V, and $$\frac{RT}{F}(2.303) = 0.059$$ V.

At the anode (oxidation): $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$

At the cathode (reduction): $$M^{2+} + 2e^- \rightarrow M$$

For the cell to function (and O$$_2$$ to evolve), the cell EMF must be positive: $$E_{cell} \geq 0$$.

$$E_{cell} = E_{cathode} - E_{anode}$$

The anode half-cell is: $$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$$, with $$E^\circ = 1.23$$ V.

Using the Nernst equation:

$$E_{anode} = 1.23 - \frac{0.059}{4} \log\left(\frac{1}{[H^+]^4 \cdot P_{O_2}}\right) = 1.23 - 0.059 \times pH$$

(since $$P_{O_2} = 1$$ bar and $$\log(1/[H^+]^4) = 4 \times pH$$)

$$E_{cathode} = E^\circ_{M^{2+}/M} = 0.994$$ V (since $$[M^{2+}] = 1.0$$ M)

$$E_{cathode} \geq E_{anode}$$

$$0.994 \geq 1.23 - 0.059 \times pH$$

$$0.059 \times pH \geq 1.23 - 0.994 = 0.236$$

$$pH \geq \frac{0.236}{0.059} = 4.0$$

The pH above which oxygen gas would start to evolve at the anode is 4.

The correct answer is 4.