The potential of electrode in a saturated solution of AgI at 298 K is — Electrochemistry Chemistry Question

💡 Solution & Explanation

In a saturated solution of AgI, the concentration of $Ag^+$ ions is $[Ag^+] = \sqrt{K_{sp}} = \sqrt{6.4 \times 10^{-17}} = \sqrt{64 \times 10^{-18}} = 8 \times 10^{-9}$ M. Using the Nernst equation for the half-cell: $E = E^\circ_{Ag^+\|Ag} - \frac{0.06}{1} \log \frac{1}{[Ag^+]} = 0.80 + 0.06 \log(8 \times 10^{-9}) = 0.80 + 0.06(\log 8 - 9) = 0.80 + 0.06(0.9 - 9) = 0.80 + 0.06(-8.1) = 0.80 - 0.486 = 0.314$ V.