Understanding the Half-Cell and Its Oxidation Potential Variation

The given half-cell is: Hg, Hg₂Cl₂ | Cl^–(aq). This is a calomel electrode. We are to plot the electrode oxidation potential (E_ox) against log[Cl^–]. The standard reduction potential for Hg₂²⁺/Hg is 0.79 V, and the solubility product (K_sp) for Hg₂Cl₂ is 10^–18.

Step 1: Write the Half-Cell Reduction Reaction

The reduction reaction for this electrode is: ${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)+{2e}^{-<!--\; –\; -->}\to <!--\; \to \; -->\mathrm{2Hg}\left(l\right)+{\mathrm{2Cl}}^{-<!--\; –\; -->}\left(aq\right)$

Step 2: Relate the Reaction to the Hg₂²⁺/Hg Couple

The calomel electrode's potential is governed by the mercurous ion (Hg₂²⁺) concentration, which is controlled by the chloride ion concentration through the solubility product.

The solubility product is defined as: $K_{sp}=\left[{\mathrm{Hg}}_2^{2+}\right]{\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2={10}^{-<!--\; –\; -->18}$ Therefore, we can solve for [Hg₂²⁺]: $\left[{\mathrm{Hg}}_2^{2+}\right]=\frac{K_{sp}}{{\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2}=\frac{{10}^{-<!--\; –\; -->18}}{{\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2}$

Step 3: Apply the Nernst Equation for the Hg₂²⁺/Hg Couple

The standard reduction potential is given for Hg₂²⁺/Hg: ${\mathrm{Hg}}_2^{2+}+{2e}^{-<!--\; –\; -->}\to <!--\; \to \; -->\mathrm{2Hg}\left(l\right);E^{\circ <!--\; º\; -->}=0.79\text{ V}$ The Nernst equation for the reduction potential (E_red) is: $E_{red}=E^{\circ <!--\; º\; -->}-<!--\; –\; -->\frac{0.059}{2}\log \frac{1}{\left[{\mathrm{Hg}}_2^{2+}\right]}\text{ at 298 K}$ Substitute the expression for [Hg₂²⁺] from the K_sp: $E_{red}=0.79-<!--\; –\; -->\frac{0.059}{2}\log \frac{{\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2}{{10}^{-<!--\; –\; -->18}}$ Simplify the logarithmic term: $E_{red}=0.79-<!--\; –\; -->\frac{0.059}{2}\log \left({10}^{18}{\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2\right)=0.79-<!--\; –\; -->\frac{0.059}{2}[\log {10}^{18}+\log {\left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]}^2]$ $E_{red}=0.79-<!--\; –\; -->\frac{0.059}{2}[18+2\log [{\mathrm{Cl}}^{-<!--\; –\; -->}\left]\right]=0.79-<!--\; –\; -->\frac{0.059}{2}\times <!--\; \times \; -->18-<!--\; –\; -->\frac{0.059}{2}\times <!--\; \times \; -->2\log \left[{\mathrm{Cl}}^{-<!--\; –\; -->}\right]$ <math xmlns="