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.
ScanSolve AI Answer
Hey bazuunelson — ready when you are. Here are the answers for 24 a): a) (i) Step 1: Identify the definition of a weakest reducing agent. A reducing agent is a species that gets oxidized. The weakest reducing agent is the one that is most difficult to oxidize, meaning its standard reduction potential (E^) is the most positive. Step 2: Find the most positive E^ value from the table. The most positive E^ value is +2.87 V for the half-reaction Q^3+(aq) + 3e^- Q(s). The weakest reducing agent is Q(s). The question asks for the E^ value associated with it. The E^ value of the weakest reducing agent is +2.87 V. a) (ii) Step 1: Recall the standard electrode potential for hydrogen. The standard electrode potential for hydrogen (2H^+(aq) + 2e^- H_2(g)) is defined as 0.00 V. Step 2: Compare this value to the given E^ values in the table. The element with an E^ value of 0.00 V is T. Element T is likely to be hydrogen. Reason: The standard electrode potential of hydrogen is defined as 0.00 V, which matches the E^ value for element T. a) (iii) Step 1: Determine which element will be oxidized (anode) and which will be reduced (cathode). For elements S and V: S^2+(aq) + 2e^- S(s) E^ = +1.23 V V^2+(aq) + 2e^- V(s) E^ = -0.76 V The species with the more negative (or less positive) E^ will be oxidized (anode), and the species with the more positive E^ will be reduced (cathode). Since E^_V (-0.76 V) is more negative than E^_S (+1.23 V), V will be oxidized at the anode, and S^2+ will be reduced at the cathode. Step 2: Draw the cell diagram. Anode (oxidation): V(s) → V^2+(aq) + 2e^- Cathode (reduction): S^2+(aq) + 2e^- → S(s) The diagram should show: • A V electrode immersed in a V^2+ solution (anode). • An S electrode immersed in an S^2+ solution (cathode). • A salt bridge connecting the two solutions. • An external wire connecting the two electrodes, with an arrow indicating electron flow from V to S. • A voltmeter in the external circuit. ` V(s) | V²⁺(aq) || S²⁺(aq) | S(s) (Anode) (Cathode) ` (Due to text-based format, a visual diagram cannot be drawn, but the description above outlines the key components.) a) (iv) Step 1: Use the formula for cell e.m.f. The e.m.f. of the cell (E^_cell) is calculated as: E^_cell = E^_cathode - E^_anode Step 2: Substitute the E^ values for the cathode and anode. From part (iii), S is the cathode and V is the anode. E^_cathode = E^_S = +1.23 V E^_anode = E^_V = -0.76 V Step 3: Calculate the e.m.f. E^_cell = (+1.23 V) - (-0.76 V) E^_cell = 1.23 V + 0.76 V E^_cell = 1.99 V The e.m.f. of the electrochemical cell is 1.99 V. Send me the next one 📸