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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Another one Mr. — let's solve it. Here's the solution for question e): e) (i) Select the species which is the strongest reducing agent? Step 1: List the given standard reduction potentials (E^) for each half-reaction. A: Fe^3+ (aq) + e^- Fe^2+ (aq) E^ = +0.77 V B: Cr_2O_7^2- (aq) + 14H^+ (aq) + 6e^- 2Cr^3+ (aq) + 7H_2O E^ = +1.33 V C: Cl_2 (g) + 2e^- 2Cl^- (aq) E^ = +1.6 V Step 2: Identify the strongest reducing agent. A reducing agent is a species that gets oxidized (loses electrons). In a table of standard reduction potentials, the strongest reducing agents are the products of the half-reactions with the lowest (most negative) standard reduction potentials. Comparing the E^ values: +0.77 V (A), +1.33 V (B), and +1.6 V (C), the lowest potential is +0.77 V for half-reaction A. The reducing agent in half-reaction A is Fe^2+ (aq). The strongest reducing agent is Fe^2+ (aq). e) (ii) Write a cell diagram for the cell obtained by coupling A and B. Step 1: Determine the anode and cathode. When coupling two half-cells, the one with the lower standard reduction potential will undergo oxidation (anode), and the one with the higher standard reduction potential will undergo reduction (cathode). Half-cell A: E^ = +0.77 V Half-cell B: E^ = +1.33 V Since E^ (A) < E^ (B), half-cell A will be the anode (oxidation) and half-cell B will be the cathode (reduction). Step 2: Write the half-reactions for the anode and cathode. Anode (oxidation): Fe^2+ (aq) → Fe^3+ (aq) + e^- Cathode (reduction): Cr_2O_7^2- (aq) + 14H^+ (aq) + 6e^- → 2Cr^3+ (aq) + 7H_2O Step 3: Construct the cell diagram. The general convention for a cell diagram is: Anode | Anode solution || Cathode solution | Cathode. For half-cells involving only ions in solution, an inert electrode (like Platinum