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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I) Step 1: Understand what an oxidizing agent is. An oxidizing agent is a species that gains electrons (gets reduced) in a redox reaction. The standard electrode potential (E^) measures the tendency of a species to be reduced. Step 2: Relate E^ to oxidizing strength. A more positive E^ indicates a stronger tendency for reduction, meaning the species is a stronger oxidizing agent. Conversely, a more negative E^ indicates a weaker tendency for reduction, meaning the species is a weaker oxidizing agent. Step 3: Identify the species with the most negative E^. From the table, the standard electrode potentials for the oxidizing agents (species on the left side of the half-equations) are: 1. R^2+: -2.38 V 2. (1)/(2)S^+: +2.87 V 3. T^2+: +0.34 V 4. U^2+: -2.90 V 5. (1)/(2)V^+: +0.80 V The most negative E^ value is -2.90 V, which corresponds to U^2+. Step 4: State the weakest oxidizing agent and explain. The weakest oxidizing agent is U^2+. Explanation: U^2+ has the most negative standard electrode potential (E^ = -2.90 V), indicating it has the least tendency to gain electrons and be reduced. Therefore, it is the weakest oxidizing agent among the given species. II) Step 1: Understand how to maximize e.m.f. The electromotive force (e.m.f) of a cell is calculated as E^_cell = E^_cathode - E^_anode. To produce the highest e.m.f, we need to select the half-cell with the most positive E^ to act as the cathode (where reduction occurs) and the half-cell with the most negative E^ to act as the anode (where oxidation occurs). Step 2: Identify the half-cell with the most positive E^. The most positive E^ value in the table is +2.87 V, corresponding to the half-cell: (1)/(2)S^+(aq) + e^- S(s) Step 3: Identify the half-cell with the most negative E^. The most negative E^ value in the table is -2.90 V, corresponding to the half-cell: U^2+(aq) + 2e^- U(s) Step 4: State the two half-cells. The two half-cells that will produce the highest e.m.f are: (1)/(2)S^+(aq) + e^- S(s) U^2+(aq) + 2e^- U(s) The highest e.m.f produced would be: E^_cell = (+2.87 V) - (-2.90 V) = +5.77 V 3 done, 2 left today. You're making progress.