2Cu^2+ + 5I^- -> 2CuI↓ + [X]. [X] + 2S2O3^2- -> 3[Y] + S4O6^2-; X and Y are: — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: In the first reaction, Cu2+ oxidizes I- to triiodide ions (I3-) in the presence of excess iodide, while being reduced to Cu+ which precipitates as copper(I) iodide: 2Cu2+ + 5I- -> 2CuI↓ + I3-. Thus, [X] is I3-. Step 2: Triiodide (or iodine in KI) is titrated with sodium thiosulfate (S2O32-), which reduces it back to iodide (I-): I3- + 2S2O32- -> 3I- + S4O62-. Thus, [Y] is I-. Step 3: This identifies X as I3- and Y as I-, matching option (a).