Question

Consider the following electrochemical cell:
\[ \text{Pt } | \ \mathrm{O_2(g,1\,bar)} \ | \ \mathrm{HCl(aq)} \ || \ \mathrm{M^{2+}(aq,1.0\,M)} \ | \ \mathrm{M(s)} \] The pH above which oxygen gas would start to evolve at the anode is ____________ (nearest integer). Given:
\[ E^\circ_{\mathrm{M^{2+}/M}} = 0.994\,\text{V} \] \[ E^\circ_{\mathrm{O_2/H_2O}} = 1.23\,\text{V} \] \[ \frac{RT}{F}(2.303)=0.059\,\text{V} \]

Hint:

Oxygen evolution at an anode depends strongly on pH due to the involvement of protons in the half-reaction.

Answer 1

Correct Answer: 2

Step 1: Write Nernst equation for oxygen electrode.
\[ E_{\mathrm{O_2/H_2O}} = 1.23 - \frac{0.059}{4}\log\left(\frac{1}{[H^+]^4}\right) \] \[ E_{\mathrm{O_2/H_2O}} = 1.23 - 0.059\,\text{pH} \]
Step 2: Condition for oxygen evolution.
Oxygen will start evolving when
\[ E_{\mathrm{O_2/H_2O}} = E_{\mathrm{M^{2+}/M}} \] \[ 1.23 - 0.059\,\text{pH} = 0.994 \]
Step 3: Solve for pH.
\[ 0.059\,\text{pH} = 0.236 \] \[ \text{pH} = 4 \] Oxygen evolution starts above this pH, hence nearest integer value is
\[ \boxed{2} \]
Final Answer:
\[ \boxed{2} \]