The standard reduction potentials are:

Here's the solution for question 35: 35) The standard reduction potentials are: P^3+(aq) + e^- P^2+(aq) ; E^ = +0.61 V Q^2+(aq) + 2e^- Q(s) ; E^ = +0.77 V R_2(g) + 2e^- 2R^-(aq) ; E^ = +0.54 V S^2+(aq) + 2e^- S(s) ; E^ = -0.44 V T^2+(aq) + 2e^- T(s) ; E^ = -0.74 V a) The strongest oxidizing agent is the species with the most positive standard reduction potential. Comparing the E^ values, +0.77 V is the highest. The strongest oxidizing agent is Q^2+(aq). b) To oxidize R^- ion to R_2 (atom R), we need a substance that has a standard reduction potential greater than +0.54 V (the reduction potential of R_2). Both P^3+ (+0.61 V) and Q^2+ (+0.77 V) can oxidize R^-. The substance that would be used is Q^2+(aq) (as it is the stronger oxidizing agent). c) The cell is represented as T(s) / T^2+(aq) // S^2+(aq) / S(s). i) The electrodes are: Anode: T(s) Cathode: S(s) ii) Equations for the half-cell reactions: Anode (oxidation): T(s) T^2+(aq) + 2e^- Cathode (reduction): S^2+(aq) + 2e^- S(s) iii) Cell equation: T(s) + S^2+(aq) T^2+(aq) + S(s) Electromotive force (e.m.f) of the cell: E^_cell = E^_cathode - E^_anode E^_cell = E^_S^2+/S - E^_T^2+/T E^_cell = (-0.44 V) - (-0.74 V) E^_cell = -0.44 V + 0.74 V E^_cell = +0.30 V iv) Electrons flow from the anode to the cathode in the external circuit. Therefore, electrons flow from the T electrode to the S electrode. d) i) Step 1: Convert time to seconds. t = 15 min × 60 s/min = 900 s Step 2: Calculate the quantity of electricity (Q). Q = I × t Q = 2.5 A × 900 s Q = 2250 C ii) Step 1: Determine the moles of electrons (n_e) passed. n_e = (Q)/(F) n_e = 2250 C96500 C/mol e⁻ n_e ≈ 0.023316 mol e⁻ Step 2: Use the stoichiometry for the deposition of a divalent ion (M^2+). M^2+(aq) + 2e^- M(s) From the equation, 2 moles of electrons are required to deposit 1 mole of M. n_M = (n_e)/(2) n_M = 0.023316 mol e⁻2 n_M ≈ 0.011658 mol M Step 3: Calculate the relative atomic mass of M (M_M). M_M = mass of Mmoles of M M_M = 0.74 g0.011658 mol M_M ≈ 63.47 g/mol The relative atomic mass of M is 63.5. Last free one today — make it count tomorrow, or type /upgrade for unlimited.