This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Welcome back chris — been a while! Let's pick up where you left off. i) Step 1: Understand what a reducing agent is. A reducing agent is a species that causes another species to be reduced, and in doing so, it itself gets oxidized (loses electrons). Step 2: Relate reducing strength to standard reduction potentials. The standard electrode potentials (E^) provided are for reduction reactions. A stronger reducing agent is the reduced form of a redox couple that has a lower (more negative or less positive) standard reduction potential. This means it has a greater tendency to be oxidized. Step 3: Compare the given standard reduction potentials. • For A: Fe^3+(aq) + e^- Fe^2+(aq), E^ = +0.77 V • For B: Cr_2O_7^2-(aq) + 14H^+(aq) + 6e^- 2Cr^3+(aq) + 7H_2O, E^ = +1.33 V • For C: Cl_2(g) + 2e^- 2Cl^-(aq), E^ = +1.6 V Step 4: Identify the strongest reducing agent. The lowest standard reduction potential among the given options is +0.77 V for the Fe^3+/Fe^2+ couple. The reduced species in this couple is Fe^2+. Therefore, Fe^2+ is the strongest reducing agent. The strongest reducing agent is Fe^2+(aq). ii) Step 1: Determine the anode and cathode. When coupling two half-cells, the half-reaction with the higher standard reduction potential will act as the cathode (reduction), and the half-reaction with the lower standard reduction potential will act as the anode (oxidation). Comparing A (E^ = +0.77 V) and B (E^ = +1.33 V): Since E^(B) > E^(A), half-reaction B will be the cathode, and half-reaction A will be the anode. Step 2: Write the half-reactions for the anode and cathode. Anode (oxidation, reverse of A): Fe^2+(aq) → Fe^3+(aq) + e^- Cathode (reduction, B): Cr_2O_7^2-(aq) + 14H^+(aq) + 6e^- → 2Cr^3+(aq) + 7H_2O Step 3: Construct the cell diagram. The cell diagram notation is: Anode | Anode solution || Cathode solution | Cathode. Since all species involved are aqueous ions, inert platinum (Pt) electrodes are used. Anode compartment: Pt(s) | Fe^2+(aq), Fe^3+(aq) Cathode compartment: Cr_2O_7^2-(aq), H^+(aq), Cr^3+(aq) | Pt(s) The cell diagram is: Pt(s) | Fe^2+(aq), Fe^3+(aq) || Cr_2O_7^2-(aq), H^+(aq), Cr^3+(aq) | Pt(s) Send me the next one 📸