The current efficiency of an electrodeposition of copper metal in which 9.8 g of copper is deposited — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Cathodic Reduction Half-Reaction and Determine the Equivalent Weight In an aqueous solution of copper sulphate ($\ce{CuSO4}$), copper exists as divalent cupric cations ($\ce{Cu^{2+}}$). During electrolysis, these ions migrate to the cathode, where they undergo reduction to form solid copper metal: $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ From this balanced half-reaction, the valency factor ($n$-factor) of copper is: $$n = 2$$ Using the standard atomic mass of copper ($M_{\ce{Cu}} \approx 63.5\text{ g/mol}$), we calculate the equivalent weight ($E$) of copper: $$E = \frac{\text{Atomic Mass of Cu}}{n}$$ $$E = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ Step 2 - Calculate the Theoretical Mass of Copper ($W_{\text{th}}$) using Faraday's First Law Faraday's First Law of Electrolysis states that the mass of a substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte: $$W_{\text{th}} = \frac{E \cdot I \cdot t}{F}$$ Where: * $E$ is the equivalent weight of copper = $31.75\text{ g/eq}$ * $I$ is the electric current = $3\text{ A}$ * $t$ is the time of electrolysis = $10000\text{ s}$ * $F$ is Faraday's constant = $96500\text{ C/eq}$ Substituting these values into the formula: $$W_{\text{th}} = \frac{31.75\text{ g/eq} \times 3\text{ A} \times 10000\text{ s}}{96500\text{ C/eq}}$$ $$W_{\text{th}} = \frac{952500}{96500}\text{ g} \approx 9.87\text{ g}$$ Step 3 - Calculate the Current Efficiency ($\eta$) In practical electrolytic systems, side reactions (such as the reduction of water or impurities) can consume a fraction of the passed current, meaning the actual mass deposited ($W_{\text{act}}$) is often less than the theoretical mass. The current efficiency ($\eta$) is defined as: $$\eta = \left( \frac{W_{\text{act}}}{W_{\text{th}}} \right) \times 100\%$$ Given: * Actual mass of copper deposited ($W_{\text{act}}$) = $9.8\text{ g}$ Substitute the actual and theoretical masses into the equation: $$\eta = \left( \frac{9.8\text{ g}}{9.87\text{ g}} \right) \times 100\%$$ $$\eta \approx 99.29\% \approx \mathbf{99\%}$$ Thus, the current efficiency of the electrodeposition process is approximately $99\%$. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** $60\%$ is far below the calculated efficiency, representing a highly inefficient system with massive side reactions. * **Option (B) is correct:** As mathematically calculated, the current efficiency is approximately $99\%$. * **Option (C) is incorrect:** $92\%$ is a mathematical underestimate. * **Option (D) is incorrect:** $75\%$ would correspond to a much lower deposited mass of copper (around $7.4\text{ g}$). $$\text{Correct Option: } \boxed{\text{B}}$$