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Silver chloride is far too insoluble for its solubility to be weighed easily, but a cell measures…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Silver chloride is far too insoluble for its solubility to be weighed easily, but a cell measures it well. The figure shows the cell at $298\ \text{K}$ :

\displaystyle \text{Ag}(s)\,|\,\text{AgCl}(s)\,|\,\text{KCl}(aq,\ 0.100\ \text{mol L}^{-1})\,\|\,\text{AgNO}_3(aq,\ 0.0100\ \text{mol L}^{-1})\,|\,\text{Ag}(s)

On the left, a silver wire coated with solid silver chloride dips into potassium chloride solution, which is saturated with silver chloride. On the right, a clean silver wire dips into silver nitrate solution. The salt bridge holds potassium nitrate. The emf is $0.398\ \text{V}$ , and the right electrode is the cathode. Both electrodes are silver electrodes, so the emf comes only from the difference in $[\text{Ag}^+]$ . Take $E^\circ(\text{Ag}^+/\text{Ag}) = 0.80\ \text{V}$ , $\dfrac{2.303RT}{F} = 0.059\ \text{V}$ and $F = 96487\ \text{C mol}^{-1}$ . Treat the solutions as ideal. The voltmeter draws almost no current, so the concentrations do not change while it is read. The chloride from the dissolved silver chloride itself is far too little to count next to the potassium chloride. Use $\Delta G^\circ = -nFE^\circ$ for (e).

A cell of two beakers. On the left, a silver wire coated with solid silver chloride dips into potassium chloride solution, 0.100 mol per litre. On the right, a clean silver wire dips into silver nitrate solution, 0.0100 mol per litre. A salt bridge of potassium nitrate joins them, and a voltmeter between the wires reads 0.398 V.

What is $[\text{Ag}^+]$ in the left compartment?
What is the solubility product of silver chloride?
What is the standard potential of the electrode $\text{AgCl}(s) + e^- \rightarrow \text{Ag}(s) + \text{Cl}^-(aq)$ ?
What is the solubility of silver chloride in pure water?
What is $\Delta_r G^\circ$ for $\text{AgCl}(s) \rightarrow \text{Ag}^+(aq) + \text{Cl}^-(aq)$ ?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The cell is a silver concentration cell. Its emf gives $[\text{Ag}^+]$ on the left, and the chloride there turns that into $K_{sp}$ .

(a) Silver ions on the left. $\log\dfrac{0.0100}{[\text{Ag}^+]_{\text{left}}} = \dfrac{0.398}{0.059} = 6.7458$ , so $[\text{Ag}^+]_{\text{left}} = 0.0100 \times 10^{-6.7458}\ \text{mol L}^{-1} = 1.80 \times 10^{-9}\ \text{mol L}^{-1}$ .

(b) Solubility product. $[\text{Cl}^-] = 0.100\ \text{mol L}^{-1}$ (the AgCl adds only $1.8 \times 10^{-9}\ \text{mol L}^{-1}$ ). So $K_{sp} = 1.796 \times 10^{-9} \times 0.100 = 1.80 \times 10^{-10}\ \text{mol}^2\ \text{L}^{-2}$ , and $\log K_{sp} = -9.7458$ .

(c) The AgCl electrode. With $[\text{Cl}^-] = 1\ \text{mol L}^{-1}$ , $[\text{Ag}^+] = K_{sp}$ , so $E^\circ(\text{AgCl}/\text{Ag}) = 0.80 + 0.059\log K_{sp} = 0.80 + (-0.5750)\ \text{V} = 0.225\ \text{V}$ . NCERT's table lists $0.22\ \text{V}$ for this electrode.

(d) Solubility in water. $s = \sqrt{K_{sp}} = 1.34 \times 10^{-5}\ \text{mol L}^{-1}$ . In the $0.100\ \text{mol L}^{-1}$ KCl it is only $1.80 \times 10^{-9}\ \text{mol L}^{-1}$ : the common ion effect.

(e) Gibbs energy. For $\text{AgCl}(s) \rightarrow \text{Ag}^+ + \text{Cl}^-$ , $E^\circ = E^\circ(\text{AgCl}/\text{Ag}) - E^\circ(\text{Ag}^+/\text{Ag}) = -0.5750\ \text{V}$ with $n = 1$ . So $\Delta_r G^\circ = -96487 \times (-0.5750)\ \text{J mol}^{-1} = 55.5\ \text{kJ mol}^{-1}$ . The route $\Delta G^\circ = -2.303RT\log K_{sp}$ gives $55.6\ \text{kJ mol}^{-1}$ ; the small gap comes from rounding $2.303RT/F$ to $0.059\ \text{V}$ .

Watch out. A potassium chloride salt bridge would put chloride into the silver nitrate and precipitate silver chloride there. That is why the bridge holds potassium nitrate.

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