[Single-digit Integer] A test for complete removal of Cu^2+ ions from a solution of Cu^2+ is to add — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Initial Moles of Copper Ions and the Total Charge Transferred We first determine the initial moles of copper(II) ions ($\ce{Cu^2+}$) in the solution: $$n_{\text{initial}}(\ce{Cu^2+}) = \text{Molarity} \times \text{Volume}$$ $$n_{\text{initial}}(\ce{Cu^2+}) = 0.1\text{ mol L}^{-1} \times 0.250\text{ L} = 0.025\text{ mol}$$ Next, we calculate the total electrical charge ($Q$) passed through the electrolytic cell using the current ($I$) and time ($t$): $$Q = I \times t$$ $$Q = 5\text{ A} \times 1351\text{ s} = 6755\text{ C}$$ Using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$), we calculate the total moles of electrons ($n_{e^-}$) transferred: $$n_{e^-} = \frac{Q}{F}$$ $$n_{e^-} = \frac{6755\text{ C}}{96,500\text{ C mol}^{-1}} = 0.07\text{ mol}$$ Step 2 - Analyze the Cathode and Anode Half-Reactions during Electrolysis At the cathode, the reduction of copper(II) ions to metallic copper takes place: $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$$ According to the stoichiometry of this reaction, the moles of electrons required to completely reduce all $0.025\text{ mol}$ of $\ce{Cu^2+}$ ions is: $$n_{e^-, \text{required}} = 2 \times n_{\text{initial}}(\ce{Cu^2+})$$ $$n_{e^-, \text{required}} = 2 \times 0.025\text{ mol} = 0.05\text{ mol}$$ Since the total moles of electrons passed ($0.07\text{ mol}$) is strictly greater than the moles required for complete reduction ($0.05\text{ mol}$): * All bulk $\ce{Cu^2+}$ ions are completely reduced and deposited as metallic copper ($\ce{Cu(s)}$) on the cathode. * The excess $0.02\text{ mol}$ of electrons ($0.07\text{ mol} - 0.05\text{ mol}$) goes toward the reduction of hydronium ions (or water) to release hydrogen gas ($\ce{H2}$). * At the anode, the oxidation of water occurs, releasing oxygen gas ($\ce{O2}$) which remains saturated in the aqueous solution: $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ Step 3 - Evaluate the Spontaneous Dissolution of Copper in Ammonia When a sufficient quantity of ammonia ($\ce{NH3}$) is added to the electrolyzed solution, the acid is neutralized, and a free ammonia concentration of $0.10\text{ M}$ is maintained. Although the bulk $\ce{Cu^2+}$ ions were reduced to metallic copper, the copper metal on the cathode remains in contact with the solution. In the presence of dissolved oxygen gas ($\ce{O2}$) and ammonia ($\ce{NH3}$), metallic copper undergoes spontaneous oxidation and dissolution to form the highly stable tetraamminecopper(II) complex, $\ce{[Cu(NH3)4]^2+}$: $$\ce{2Cu(s) + 8NH3(aq) + O2(g) + 2H2O(l) <=> 2[Cu(NH3)4]^2+(aq) + 4OH^-(aq)}$$ We can verify the thermodynamic feasibility of this dissolution by calculating the standard cell potential ($E^\circ_{\text{dissolution}}$) for this process: * The reduction half-potential of the complex is: $$E^\circ_{\ce{[Cu(NH3)4]^2+/Cu}} = E^\circ_{\ce{Cu^2+/Cu}} - \frac{0.0591}{2}\log_{10}(K_f)$$ $$E^\circ_{\ce{[Cu(NH3)4]^2+/Cu}} = 0.34\text{ V} - 0.03\log_{10}(1.1 \times 10^{13}) \approx -0.05\text{ V}$$ * The standard reduction potential of oxygen in a basic medium is: $$\ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)} \quad E^\circ = +0.40\text{ V}$$ * The overall potential for the dissolution process is: $$E^\circ_{\text{cell}} = 0.40\text{ V} - (-0.05\text{ V}) = +0.45\text{ V} > 0\text{ V}$$ Because $E^\circ_{\text{cell}}$ is positive, the dissolution is highly spontaneous. Consequently, a trace amount of copper metal will spontaneously dissolve back into the solution as $\ce{[Cu(NH3)4]^2+}$ complex ions. Step 4 - Compare with the Detection Limit and Determine the Final Code The concentration of the formed complex, $\ce{[Cu(NH3)4]^2+}$, easily reaches and exceeds the minimum detectable concentration of $1 \times 10^{-5}\text{ M}$. This results in a visible blue color in the electrolyzed solution upon the addition of ammonia. Therefore, the answer to whether a blue color would be shown is **yes**, which is represented by marking **1**. $$\text{Final Numerical Code: } \boxed{1}$$