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

To find the pH above which oxygen gas starts evolving at the anode, consider the oxygen evolution half-reaction:

\[ \mathrm{O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)} \]

The standard electrode potential for this reaction is:

\[ E^\circ_{\mathrm{O_2/H_2O}} = 1.23~\text{V} \]

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 Nernst Equation

Under non-standard conditions, the Nernst equation for this half-reaction is:

\[ E = E^\circ_{\mathrm{O_2/H_2O}} - \frac{0.059}{4}\log\!\left(\frac{1}{(a_{\mathrm{H^+}})^4}\right) \]

Since \( a_{\mathrm{H^+}} = 10^{-\text{pH}} \), the equation simplifies to:

\[ E = 1.23 + 0.059\,\text{pH} \]

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Condition for Oxygen Evolution

The standard reduction potential for the metal electrode is:

\[ E^\circ_{\mathrm{M^{2+}/M}} = 0.994~\text{V} \]

For oxygen evolution to occur at the anode:

\[ E(\mathrm{O_2}) > E_{\mathrm{M^{2+}/M}} \]

Substituting values:

\[ 1.23 + 0.059\,\text{pH} > 0.994 \]

Solving:

\[ 0.059\,\text{pH} > -0.236 \]

\[ \text{pH} > -4 \]

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Physical Interpretation

Since pH values are non-negative in practical systems, oxygen evolution becomes feasible only when the pH is sufficiently high under realistic electrochemical conditions.

Thus, the minimum practical pH above which oxygen evolution begins is:

\[ \boxed{2} \]