Here's the solution for question 35: 35) The standard reduction potentials are: $\text{P}^{3+}(aq) + \text{e}^- \rightleftharpoons \text{P}^{2+}(aq)$ ; $E^\circ = +0.61 \text{ V}$ $\text{Q}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{Q}(s)$ ; $E^\circ = +0.77 \text{ V}$ $\text{R}_2(g) + 2\text{e}^- \rightleftharpoons 2\text{R}^-(aq)$ ; $E^\circ = +0.54 \text{ V}$ $\text{S}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{S}(s)$ ; $E^\circ = -0.44 \text{ V}$ $\text{T}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{T}(s)$ ; $E^\circ = -0.74 \text{ V}$ a) The strongest oxidizing agent is the species with the most positive standard reduction potential. Comparing the $E^\circ$ values, $+0.77 \text{ V}$ is the highest. The strongest oxidizing agent is $\boxed{\text{Q}^{2+}(aq)}$. b) To oxidize $\text{R}^-$ ion to $\text{R}_2$ (atom R), we need a substance that has a standard reduction potential greater than $+0.54 \text{ V}$ (the reduction potential of $\text{R}_2$). Both $\text{P}^{3+}$ ($+0.61 \text{ V}$) and $\text{Q}^{2+}$ ($+0.77 \text{ V}$) can oxidize $\text{R}^-$. The substance that would be used is $\boxed{\text{Q}^{2+}(aq)}$ (as it is the stronger oxidizing agent). c) The cell is represented as $\text{T}(s) / \text{T}^{2+}(aq) // \text{S}^{2+}(aq) / \text{S}(s)$. i) The electrodes are: Anode: T(s) Cathode: S(s) ii) Equations for the half-cell reactions: Anode (oxidation): $$\text{T}(s) \to \text{T}^{2+}(aq) + 2\text{e}^-$$ Cathode (reduction): $$\text{S}^{2+}(aq) + 2\text{e}^- \to \text{S}(s)$$ iii) Cell equation: $$\boxed{\text{T}(s) + \text{S}^{2+}(aq) \to \text{T}^{2+}(aq) + \text{S}(s)}$$ Electromotive force (e.m.f) of the cell: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{S}^{2+}/\text{S}} - E^\circ_{\text{T}^{2+}/\text{T}}$$ $$E^\circ_{\text{cell}} = (-0.44 \text{ V}) - (-0.74 \text{ V})$$ $$E^\circ_{\text{cell}} = -0.44 \text{ V} + 0.74 \text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{+0.30 \text{ V}}$$ 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. $$t = 15 \text{ min} \times 60 \text{ s/min} = 900 \text{ s}$$ Step 2: Calculate the quantity of electricity ($Q$). $$Q = I \times t$$ $$Q = 2.5 \text{ A} \times 900 \text{ s}$$ $$Q = \boxed{2250 \text{ C}}$$ ii) Step 1: Determine the moles of electrons ($n_e$) passed. $$n_e = \frac{Q}{F}$$ $$n_e = \frac{2250 \text{ C}}{96500 \text{ C/mol e⁻}}$$ $$n_e \approx 0.023316 \text{ mol e⁻}$$ Step 2: Use the stoichiometry for the deposition of a divalent ion ($\text{M}^{2+}$). $$\text{M}^{2+}(aq) + 2\text{e}^- \to \text{M}(s)$$ From the equation, $2$ moles of electrons are required to deposit $1$ mole of M. $$n_M = \frac{n_e}{2}$$ $$n_M = \frac{0.023316 \text{ mol e⁻}}{2}$$ $$n_M \approx 0.011658 \text{ mol M}$$ Step 3: Calculate the relative atomic mass of M ($M_M$). $$M_M = \frac{\text{mass of M}}{\text{moles of M}}$$ $$M_M = \frac{0.74 \text{ g}}{0.011658 \text{ mol}}$$ $$M_M \approx 63.47 \text{ g/mol}$$ The relative atomic mass of M is $\boxed{63.5}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.