Copper can be deposited from acidified copper sulphate and alkaline cuprous cyanide. If the same cur — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Oxidation States of Copper in Both Electrolytes To compare the mass of copper deposited from both solutions, we must first determine the oxidation state of copper ($\ce{Cu}$) in each solution: 1. **Acidic Copper Sulphate ($\ce{CuSO4}$):** In this solution, copper exists as divalent cupric cations ($\ce{Cu^{2+}}$). The cathodic reduction half-reaction is: $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ Thus, the valency factor ($n$-factor, $n_1$) of copper in acidic copper sulphate is: $$n_1 = 2$$ 2. **Alkaline Cuprous Cyanide ($\ce{Na3[Cu(CN)4]}$ or similar cuprous complex):** In this solution, copper exists as monovalent cuprous cations ($\ce{Cu^+}$). The cathodic reduction half-reaction is: $$\ce{Cu^+(aq) + e^- -> Cu(s)}$$ Thus, the valency factor ($n$-factor, $n_2$) of copper in alkaline cuprous cyanide is: $$n_2 = 1$$ Step 2 - Calculate the Chemical Equivalent Weights The chemical equivalent weight ($E$) of a metal is calculated using the formula: $$E = \frac{\text{Atomic Mass of metal } (M)}{\text{Valency factor } (n)}$$ Using the standard atomic mass of copper ($M_{\ce{Cu}} \approx 63.5\text{ g/mol}$): * **Equivalent weight in acidic copper sulphate ($E_1$):** $$E_1 = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ * **Equivalent weight in alkaline cuprous cyanide ($E_2$):** $$E_2 = \frac{63.5\text{ g/mol}}{1} = 63.5\text{ g/eq}$$ Step 3 - Apply Faraday's First Law of Electrolysis Faraday's First Law states that the mass ($W$) of a substance deposited at an electrode is directly proportional to the quantity of electricity ($Q = I \cdot t$) passed through the cell: $$W = \frac{E \cdot I \cdot t}{F}$$ Where: * $I$ is the electric current. * $t$ is the time of electrolysis. * $F$ is Faraday's constant ($\approx 96500\text{ C/eq}$). Since the **same current** ($I$) is passed for a **definite (equal) time** ($t$), the total electric charge ($Q$) passed through both solutions is identical. Consequently, the mass of copper deposited is directly proportional to its chemical equivalent weight: $$W \propto E$$ Step 4 - Calculate the Ratio of Deposited Masses Using the direct proportionality of mass to equivalent weight: $$\frac{W_{\text{alkaline}}}{W_{\text{acidic}}} = \frac{E_2}{E_1}$$ Substitute the values calculated in Step 2: $$\frac{W_{\text{alkaline}}}{W_{\text{acidic}}} = \frac{63.5\text{ g/eq}}{31.75\text{ g/eq}} = 2$$ $$W_{\text{alkaline}} = 2 \cdot W_{\text{acidic}}$$ Thus, the mass of copper deposited from the alkaline cuprous cyanide solution is exactly twice the mass of copper deposited from the acidic copper sulphate solution under identical electrical conditions. Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** The amount of copper deposited from acidic copper sulphate is lower because of its higher valency factor ($2$), which halves its equivalent weight. * **Option (B) is correct:** As mathematically demonstrated, the lower valency factor ($1$) of copper in alkaline cuprous cyanide doubles its equivalent weight, yielding a higher amount of deposited copper. * **Option (C) is incorrect:** Different amounts of copper are deposited because the copper ions have different oxidation states in the two solutions. * **Option (D) is incorrect:** Copper deposition occurs successfully at the cathode in both setups upon the passage of electric current. $$\text{Correct Option: } \boxed{\text{B}}$$