By strictly applying Faraday's laws to the electrolysis of aqueous using completely inert electrodes — Electrochemistry Chemistry Question

💡 Solution & Explanation

In aqueous $MgSO_4$, water is reduced at the cathode and oxidized at the anode. Cathode reaction: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (n-factor for $H_2 = 2$). Anode reaction: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ (n-factor for $O_2 = 4$). Faraday's law dictates equivalents are equal: Equiv($H_2$) = Equiv($O_2$). Therefore, Moles($H_2$) $\times 2$ = Moles($O_2$) $\times 4$. The ratio Moles($H_2$) / Moles($O_2$) = $4 / 2 = 2$.