IChO 1991 · Theory problem 3

Silver/silver chloride and calomel electrodes

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Type II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride (\(\ce{Ag},\,\ce{AgCl}/\ce{Cl^-}\)) and the calomel electrode (\(\ce{Hg},\,\ce{Hg2Cl2}/\ce{Cl^-}\)) are examples of such electrodes. The standard emf of a cell built of those electrodes \((-)\ \ce{Ag},\ce{AgCl}/\ce{Cl^-}\parallel\ce{Hg2Cl2}/\ce{Hg}\ (+)\) is \(E^0=0.0455\ \mathrm{V}\) at \(T=298\ \mathrm{K}\). The temperature coefficient for this cell is \(dE^0/dT=3.38\times10^{-4}\ \mathrm{V\,K^{-1}}\).

3.1 Give the equations of the reactions taking place at both the cell electrodes and the overall cell reaction.

3.2 Calculate the Gibbs free energy change (\(\Delta G^0\)) for the process taking place in the cell at 298 K. What does its sign imply?

3.3 Calculate the enthalpy change for the process taking place at 298 K.

\[ \Delta S=n\,F\,\Delta E/\Delta T. \]

3.4 Knowing the standard potential of \(\ce{Ag}/\ce{Ag+}\) electrode is \(E^0=0.799\ \mathrm{V}\) and the solubility product of \(\ce{AgCl}\) \(K_{sp}=1.73\times10^{-10}\), calculate the standard electrode potential value of the silver/silver chloride electrode. Derive an expression showing the dependence between \(E^0(\ce{Ag}/\ce{Ag+})\) and \(E^0(\ce{Ag},\ce{AgCl}/\ce{Cl^-})\).

3.5 Calculate the solubility product of \(\ce{Hg2Cl2}\) knowing that the standard potential of the calomel electrode is \(E^0=0.798\ \mathrm{V}\).

\[ F=96487\ \mathrm{C\,mol^{-1}},\qquad R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}},\qquad T=298\ \mathrm{K} \]

Official English text of the problem.

Answers

  • 3.1 Negative electrode, oxidation: \[ \ce{2Ag(s) + 2Cl^-(aq) -> 2AgCl(s) + 2e^-}. \] Positive electrode, reduction: \[ \ce{Hg2Cl2(s) + 2e^- -> 2Hg(l) + 2Cl^-(aq)}. \] Overall: \[ \boxed{\ce{2Ag(s) + Hg2Cl2(s) -> 2AgCl(s) + 2Hg(l)}}. \]

  • 3.2 \(\boxed{\Delta G^\circ=-8.78\ \mathrm{kJ\,mol^{-1}}}\). The forward reaction is thermodynamically spontaneous under standard conditions at \(298\ \mathrm{K}\).

  • 3.3 \(\boxed{\Delta H^\circ=+10.7\ \mathrm{kJ\,mol^{-1}}}\). The reaction is endothermic.

  • 3.4 \[ \boxed{ E^\circ_{\mathrm{AgCl}} =E^\circ_{\ce{Ag+}/\ce{Ag}} +\frac{RT}{F}\ln K_{sp}(\ce{AgCl}) =0.222\ \mathrm{V} }. \]

  • 3.5 The printed potential is inconsistent with the preceding data. If \(0.798\ \mathrm{V}\) denotes \(E^\circ_{\ce{Hg2^2+}/\ce{Hg}}\), the result is \[ \boxed{K_{sp}(\ce{Hg2Cl2})=1.12\times10^{-18}}. \]

Solution

All electrode potentials below use the reduction convention. Thermodynamic changes are per mole of the overall reaction written in 3.1. Solubility products are expressed using dimensionless activities.

The species and their SMILES representations are:

Species SMILES
Silver, \(\ce{Ag}\) [Ag]
Mercury, \(\ce{Hg}\) [Hg]
Chloride ion, \(\ce{Cl^-}\) [Cl-]
Silver(I) ion, \(\ce{Ag+}\) [Ag+]
Mercury(I) dimer ion, \(\ce{Hg2^2+}\) [Hg+][Hg+]
Silver(I) chloride, \(\ce{AgCl}\) [Ag+].[Cl-]
Mercury(I) chloride (calomel), \(\ce{Hg2Cl2}\) Cl[Hg][Hg]Cl

These SMILES represent atoms, ions, or formula-unit connectivity, not bulk metallic or crystal structures. The calomel representation does not encode the linear geometry of its \(\ce{Cl-Hg-Hg-Cl}\) unit.

3.1 Electrode reactions and overall reaction

The negative electrode of this galvanic cell is the anode, where oxidation occurs. Silver changes from oxidation state \(0\) to \(+1\), forming insoluble silver(I) chloride:

\[ \ce{Ag(s) + Cl^-(aq) -> AgCl(s) + e^-}. \]

At the positive electrode, the cathode, mercury(I) in calomel is reduced to elemental mercury. Each formula unit contains two mercury atoms, each requiring one electron:

\[ \ce{Hg2Cl2(s) + 2e^- -> 2Hg(l) + 2Cl^-(aq)}. \]

Multiplying the silver half-reaction by two and adding cancels both electrons and chloride ions:

\[ \boxed{\ce{2Ag(s) + Hg2Cl2(s) -> 2AgCl(s) + 2Hg(l)}}. \]

Thus the electron-transfer number for this reaction is \(n=2\).

3.2 Gibbs free energy change

For a reversible electrochemical cell,

\[ \Delta G^\circ=-nFE^\circ_{\mathrm{cell}}. \]

Therefore,

\[ \begin{aligned} \Delta G^\circ &=-2(96487\ \mathrm{C\,mol^{-1}})(0.0455\ \mathrm{V})\\ &=-8780.317\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-8.78\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The negative sign means that the reaction as written proceeds spontaneously in the thermodynamic sense under standard conditions at \(298\ \mathrm{K}\), and can supply electrical work.

3.3 Enthalpy change

Using the given temperature coefficient,

\[ \Delta S^\circ =nF\frac{dE^\circ_{\mathrm{cell}}}{dT}. \]

Hence,

\[ \begin{aligned} \Delta S^\circ &=2(96487)(3.38\times10^{-4})\\ &=65.2252\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Since \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\),

\[ \begin{aligned} \Delta H^\circ &=\Delta G^\circ+T\Delta S^\circ\\ &=-8780.317+(298)(65.2252)\\ &=10656.8\ \mathrm{J\,mol^{-1}}\\ &=\boxed{+10.7\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The reaction is endothermic. Its positive entropy change makes \(T\Delta S^\circ\) sufficiently large to give a negative \(\Delta G^\circ\).

3.4 Standard silver/silver chloride electrode potential

The dissolution equilibrium is

\[ \ce{AgCl(s) <=> Ag+(aq) + Cl^-(aq)}, \]

with

\[ K_{sp}(\ce{AgCl})=a_{\ce{Ag+}}a_{\ce{Cl^-}}. \]

For the reduction

\[ \ce{Ag+(aq) + e^- -> Ag(s)}, \]

the Nernst equation is

\[ E=E^\circ_{\ce{Ag+}/\ce{Ag}} +\frac{RT}{F}\ln a_{\ce{Ag+}}. \]

Substituting \(a_{\ce{Ag+}}=K_{sp}/a_{\ce{Cl^-}}\) gives

\[ E= E^\circ_{\ce{Ag+}/\ce{Ag}} +\frac{RT}{F}\ln K_{sp} -\frac{RT}{F}\ln a_{\ce{Cl^-}}. \]

The silver/silver chloride reduction is

\[ \ce{AgCl(s) + e^- -> Ag(s) + Cl^-(aq)}. \]

Its standard potential is obtained at \(a_{\ce{Cl^-}}=1\). Thus, in the notation of the question,

\[ \boxed{ E^0(\ce{Ag},\ce{AgCl}/\ce{Cl^-}) = E^0(\ce{Ag}/\ce{Ag+}) +\frac{RT}{F}\ln K_{sp}(\ce{AgCl}) }. \]

Numerically,

\[ \begin{aligned} E^\circ_{\mathrm{AgCl}} &=0.799+ \frac{(8.314)(298)}{96487} \ln(1.73\times10^{-10})\\ &=0.2218218\ \mathrm{V}\\ &=\boxed{0.222\ \mathrm{V}}. \end{aligned} \]

3.5 Solubility product of calomel

First, the given cell emf fixes the calomel electrode potential:

\[ E^\circ_{\mathrm{cell}} =E^\circ_{\mathrm{calomel}}-E^\circ_{\mathrm{AgCl}}. \]

Therefore,

\[ E^\circ_{\mathrm{calomel}} =0.0455+0.2218218 =0.2673218\ \mathrm{V}. \]

This conflicts with the \(0.798\ \mathrm{V}\) explicitly called the calomel potential in part 3.5. A consistent calculation is possible if that value instead refers to

\[ \ce{Hg2^2+(aq) + 2e^- -> 2Hg(l)}, \qquad E^\circ_{\ce{Hg2^2+}/\ce{Hg}}=0.798\ \mathrm{V}. \]

Under this interpretation, calomel dissolves according to

\[ \ce{Hg2Cl2(s) <=> Hg2^2+(aq) + 2Cl^-(aq)}, \]

so

\[ K_{sp}(\ce{Hg2Cl2}) =a_{\ce{Hg2^2+}}a_{\ce{Cl^-}}^2. \]

Adding this dissolution reaction to the mercury(I) reduction gives the calomel electrode reaction. The corresponding standard Gibbs energies therefore satisfy

\[ -2FE^\circ_{\mathrm{calomel}} =-RT\ln K_{sp} -2FE^\circ_{\ce{Hg2^2+}/\ce{Hg}}. \]

Rearranging,

\[ \boxed{ K_{sp} = \exp\!\left[ \frac{2F}{RT} \left( E^\circ_{\mathrm{calomel}} -E^\circ_{\ce{Hg2^2+}/\ce{Hg}} \right) \right] }. \]

Using the unrounded intermediate potential,

\[ \begin{aligned} K_{sp} &=\exp\!\left[ \frac{2(96487)}{(8.314)(298)} (0.2673218-0.798) \right]\\ &=\boxed{1.12\times10^{-18}}. \end{aligned} \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.