A certain current liberated 0.50 g of hydrogen in 2 h. How many grams of copper can be liberated by — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Cathodic Reactions and Determine the Valency Factors ($n$-factors) During electrolysis, reduction reactions take place at the cathode of both cells: 1. **Hydrogen liberation at the cathode:** $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ From this balanced half-reaction, the number of moles of electrons required to liberate $1\text{ mole}$ of hydrogen gas ($\ce{H2}$) is: $$n_{\ce{H2}} = 2$$ 2. **Copper deposition at the cathode:** In a copper sulphate ($\ce{CuSO4}$) solution, copper exists as divalent cupric cations ($\ce{Cu^{2+}}$). These ions undergo reduction to form solid copper metal: $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ From this balanced half-reaction, the number of moles of electrons required to deposit $1\text{ mole}$ of copper metal ($\ce{Cu}$) is: $$n_{\ce{Cu}} = 2$$ Step 2 - Calculate the Chemical Equivalent Weights ($E$) The equivalent weight ($E$) of a substance is defined as its molar mass ($M$) divided by its valency factor ($n$-factor): $$E = \frac{M}{n}$$ * **Equivalent weight of hydrogen gas ($\ce{H2}$):** Using the molar mass of $\ce{H2}$ ($2\text{ g/mol}$): $$E_{\ce{H2}} = \frac{2\text{ g/mol}}{2} = 1\text{ g/eq}$$ * **Equivalent weight of copper metal ($\ce{Cu}$):** Using the atomic mass of copper ($M_{\ce{Cu}} = 63.5\text{ g/mol}$): $$E_{\ce{Cu}} = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ Step 3 - Apply Faraday's Second Law of Electrolysis Faraday's Second Law of Electrolysis states that when the same quantity of electricity ($Q$) is passed through different electrolyte solutions for the same duration of time, the masses ($W$) of the substances liberated or deposited at the electrodes are directly proportional to their chemical equivalent weights ($E$): $$W \propto E$$ Therefore, the ratio of the mass of copper deposited ($W_{\ce{Cu}}$) to the mass of hydrogen gas liberated ($W_{\ce{H2}}$) is given by: $$\frac{W_{\ce{Cu}}}{W_{\ce{H2}}} = \frac{E_{\ce{Cu}}}{E_{\ce{H2}}}$$ We are given: * Mass of hydrogen liberated ($W_{\ce{H2}}$) = $0.50\text{ g}$ Substitute the known values into the equation to calculate $W_{\ce{Cu}}$: $$\frac{W_{\ce{Cu}}}{0.50\text{ g}} = \frac{31.75\text{ g/eq}}{1\text{ g/eq}}$$ $$W_{\ce{Cu}} = 31.75 \times 0.50\text{ g}$$ $$W_{\ce{Cu}} = \mathbf{15.875\text{ g}} \approx \mathbf{15.88\text{ g}}$$ Thus, the mass of copper liberated under the same conditions is approximately $15.88\text{ g}$. Step 4 - Evaluate the Options * **Option (A) is incorrect:** $12.7\text{ g}$ is mathematically incorrect and does not correspond to the proportional equivalent weight relationship. * **Option (B) is correct:** As shown by our calculation using Faraday's second law, the mass of copper liberated is exactly $15.875\text{ g}$, which rounds to $15.88\text{ g}$. * **Option (C) is incorrect:** $31.75\text{ g}$ is the equivalent weight of copper itself. This would be the mass deposited only if exactly $1.0\text{ g}$ (which is $1\text{ equivalent}$) of hydrogen were liberated. * **Option (D) is incorrect:** $63.5\text{ g}$ is the atomic mass of copper, which would require twice the amount of electricity passed in this experiment to be deposited. $$\text{Correct Option: } \boxed{\text{B}}$$