11. Three Faradays of electricity are passed through three electrolytic cells in series containing , — Electrochemistry Chemistry Question

💡 Solution & Explanation

Moles deposited = $\frac{\text{Faradays passed}}{\text{n-factor}}$. For $Ag^+$ (n=1) = 3/1 = 3. For $Ca^{2+}$ (n=2) = 3/2 = 1.5. For $Al^{3+}$ (n=3) = 3/3 = 1. The molar ratio is 3 : 1.5 : 1, which scales to 6 : 3 : 2.