Electro Chemistry Part 3

Concept:Oxygen gas evolves at the anode when the reduction potential of the oxygen electrode becomes lower than that of the metal electrode.Explanation:For the oxygen electrode, the reduction reaction is:$\mathrm{O_2(g)} + 4\mathrm{H^+(aq)} + 4e^- \rightarrow 2\mathrm{H_2O(l)}$The Nernst equation gives:$E_{\mathrm{O_2/H_2O}} = E^\circ_{\mathrm{O_2/H_2O}} - \frac{0.059}{4} \log\left( \frac{1}{p_{\mathrm{O_2}}[\mathrm{H^+}]^4} \right)$Given $p_{\mathrm{O_2}} = 1$ bar, so $\log p_{\mathrm{O_2}} = 0$.Thus, $E_{\mathrm{O_2/H_2O}} = 1.23 + 0.059 \log[\mathrm{H^+}]$Since $\log[\mathrm{H^+}] = -\mathrm{pH}$, we get:$E_{\mathrm{O_2/H_2O}} = 1.23 - 0.059 \, \mathrm{pH}$For the metal electrode, $[\mathrm{M^{2+}}] = 1.0$ M, so its reduction potential equals the standard value:$E_{\mathrm{M^{2+}/M}} = E^\circ_{\mathrm{M^{2+}/M}} = 0.994$ VOxygen starts evolving at the anode when the oxygen electrode becomes the anode, i.e., when its reduction potential is less than that of the metal. At the threshold condition, the two potentials are equal:$1.23 - 0.059 \, \mathrm{pH} = 0.994$Solve for pH:$0.059 \, \mathrm{pH} = 1.23 - 0.994 = 0.236$$\mathrm{pH} = \frac{0.236}{0.059} = 4$Therefore, for $\mathrm{pH} > 4$, oxygen gas begins to evolve at the anode.Answer:4