At the end of the electrolysis, how many grams of the gaseous product appear at the anode? — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Determine the total charge and electrolysis duration.** From the passage: current $= 15.0\ \text{A}$, electrolysis time $= 1\ \text{h} = 3600\ \text{s}$. $$Q_{\text{total}} = I \times t = 15.0 \times 3600 = 54{,}000\ \text{C}$$ **Step 2: Identify the anode reaction.** At the anode (inert electrode in NiSO₄ solution), water is oxidised: $$\ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}$$ All electrons that flow through the circuit must originate at the anode. The 60% current efficiency applies only to the cathode. At the anode, **all** charge produces $\text{O}_2$ (no competing anode reaction with inert electrodes in sulphate solution). **Step 3: Calculate moles of O₂.** $$n_{e^-} = \frac{Q}{F} = \frac{54{,}000}{96{,}500} = 0.5596\ \text{mol}$$ Each mole of $\text{O}_2$ requires 4 moles of electrons: $$n_{\ce{O2}} = \frac{0.5596}{4} = 0.1399\ \text{mol}$$ **Step 4: Calculate mass of O₂.** $$m_{\ce{O2}} = 0.1399 \times 32 = 4.48\ \text{g}$$ $$\boxed{\text{Answer: A — } 4.48\ \text{g of O}_2\text{ forms at the anode}}$$