Question

Consider the electrochemical reaction between $Ag(s)$ and $Cl_2(g)$ electrodes in 1 litre of $0.1\,M\,KCl$ aqueous solution. Solubility product of $AgCl$ is $1.8\times10^{-10}$ and $F=96500\,\text{C mol}^{-1}$. At $1\,\mu A$ current, calculate the time required to start observing the $AgCl$ precipitation in the galvanic cell

• A. \(173 \, s \)
• B. \(346 \, s \)
• C. \(1.25 \times 10^6 \, s \)
• D. \(1.25 \times 10^5 \, s \)
• E. \(101 \, s \)

Hint:

Precipitation problems in electrochemistry link the Solubility Product (\(K_{sp}\)) to Faraday's Laws. Always find the threshold concentration first.

Answer 1

The Correct Option is A

Concept: Precipitation of \(AgCl\) begins when the ionic product \([Ag^+][Cl^-]\) exceeds the solubility product \(K_{sp}\). In this galvanic cell, silver is oxidized to \(Ag^+\) ions. \[ Ag(s) \rightarrow Ag^+(aq) + e^- \]

Step 1: Determine the concentration of \(Ag^+\) required for precipitation.
Given \([Cl^-] = 0.1 \, M\) (from \(KCl\)) and \(K_{sp} = 1.8 \times 10^{-10}\): \[ [Ag^+][Cl^-] = K_{sp} \implies [Ag^+](0.1) = 1.8 \times 10^{-10} \] \[ [Ag^+] = 1.8 \times 10^{-9} \, mol/L \]

Step 2: Calculate total moles of \(Ag^+\) needed.
Since the volume is \(1 \, L\): \[ \text{Moles of } Ag^+ = 1.8 \times 10^{-9} \, mol \]

Step 3: Use Faraday's Law to find time \(t\).
The charge \(Q = n \cdot F\). Also \(Q = I \cdot t\). \[ I \cdot t = n \cdot F \implies (1 \times 10^{-6} \, A) \cdot t = (1.8 \times 10^{-9} \, mol) \cdot (96500 \, C/mol) \] \[ t = \frac{1.8 \times 10^{-9} \times 96500}{1 \times 10^{-6}} = 1.8 \times 10^{-3} \times 96500 \] \[ t \approx 173.7 \, s \]