The copper anode of a cell containing silver nitrate solution weighs 60.0 g. After passing current f — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Electrode Reactions and Equivalent Weights During the electrolysis of a silver nitrate ($\ce{AgNO3}$) solution using a copper anode, the following half-cell reactions occur: 1. **At the Platinum Cathode (Reduction):** Silver ions ($\ce{Ag^+}$) gain electrons and deposit as solid silver metal: $$\ce{Ag^+(aq) + e^- -> Ag(s)}$$ The number of electrons transferred per silver atom is $n_{\ce{Ag}} = 1$. The equivalent weight ($E_{\ce{Ag}}$) of silver is: $$E_{\ce{Ag}} = \frac{\text{Atomic Mass of Ag}}{n_{\ce{Ag}}}$$ $$E_{\ce{Ag}} = \frac{108\text{ g/mol}}{1\text{ eq/mol}} = 108\text{ g/eq}$$ 2. **At the Copper Anode (Oxidation):** Since copper is an active anode, it undergoes oxidation, dissolving into the solution as divalent copper cations ($\ce{Cu^{2+}}$): $$\ce{Cu(s) -> Cu^{2+}(aq) + 2e^-}$$ The number of electrons transferred per copper atom is $n_{\ce{Cu}} = 2$. The equivalent weight ($E_{\ce{Cu}}$) of copper is: $$E_{\ce{Cu}} = \frac{\text{Atomic Mass of Cu}}{n_{\ce{Cu}}}$$ $$E_{\ce{Cu}} = \frac{64\text{ g/mol}}{2\text{ eq/mol}} = 32\text{ g/eq}$$ Step 2 - Apply Faraday's Second Law of Electrolysis According to Faraday's Second Law of Electrolysis, when the same quantity of electricity is passed through different electrodes, the number of equivalents of substances reacting or depositing at the electrodes is equal: $$\text{Equivalents of Cu oxidized at anode} = \text{Equivalents of Ag deposited at cathode}$$ Using the relation $\text{Equivalents} = \frac{\text{Mass}}{\text{Equivalent Weight}}$: $$\frac{W_{\ce{Cu}}}{E_{\ce{Cu}}} = \frac{W_{\ce{Ag}}}{E_{\ce{Ag}}}$$ Step 3 - Substitute the Values and Calculate the Mass of Dissolved Copper We are given: * Mass of silver deposited ($W_{\ce{Ag}}$) = $3.24\text{ g}$ * Equivalent weight of silver ($E_{\ce{Ag}}$) = $108\text{ g/eq}$ * Equivalent weight of copper ($E_{\ce{Cu}}$) = $32\text{ g/eq}$ Substitute these values into our relation: $$\frac{W_{\ce{Cu}}}{32\text{ g/eq}} = \frac{3.24\text{ g}}{108\text{ g/eq}}$$ $$\frac{3.24}{108} = 0.03\text{ equivalents of Ag deposited}$$ Therefore, the mass of copper dissolved ($W_{\ce{Cu}}$) is: $$W_{\ce{Cu}} = 32\text{ g/eq} \times 0.03\text{ eq}$$ $$W_{\ce{Cu}} = 0.96\text{ g}$$ Step 4 - Calculate the Final Weight of the Copper Anode Since the copper anode undergoes oxidation and dissolves, its mass decreases over time: $$\text{Final weight of anode} = \text{Initial weight of anode} - W_{\ce{Cu}}$$ $$\text{Final weight of anode} = 60.0\text{ g} - 0.96\text{ g}$$ $$\text{Final weight of anode} = \boxed{59.04\text{ g}}$$ Step 5 - Explanation of Options * **Option (A) is incorrect** because $0.96\text{ g}$ is the mass of copper that dissolved from the anode during electrolysis, not its remaining weight. * **Option (B) is incorrect** because $60\text{ g}$ is the initial mass of the copper anode before any reaction took place. * **Option (C) is correct** because subtracting the dissolved copper mass ($0.96\text{ g}$) from the initial mass ($60.0\text{ g}$) gives the correct final weight of $59.04\text{ g}$. * **Option (D) is incorrect** because it adds the dissolved copper mass to the initial mass, whereas an active anode must always lose mass during oxidation. $$\text{Correct Option: } \boxed{\text{C}}$$