Here are the solutions to the electrochemistry problems: e) i) To arrange the species in order of decreasing oxidizing strength, we compare their standard reduction potentials ($E^\circ$). A higher (more positive) $E^\circ$ indicates a stronger oxidizing agent. Given reduction potentials: $\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \to 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)$ $E^\circ = +1.33 \text{ V}$ $\text{I}_2(aq) + 2e^- \to 2\text{I}^-(aq)$ $E^\circ = +0.54 \text{ V}$ $\text{Fe}^{2+}(aq) + 2e^- \to \text{Fe}(s)$ $E^\circ = -0.44 \text{ V}$ $\text{Ca}^{2+}(aq) + 2e^- \to \text{Ca}(s)$ $E^\circ = -2.87 \text{ V}$ Arranging the species ($\text{Cr}_2\text{O}_7^{2-}$, $\text{I}_2$, $\text{Fe}^{2+}$, $\text{Ca}^{2+}$) in decreasing order of oxidizing strength: $\boxed{\text{Cr}_2\text{O}_7^{2-} > \text{I}_2 > \text{Fe}^{2+} > \text{Ca}^{2+}}$ ii) When granules of calcium are added to a solution of potassium dichromate, a redox reaction will occur. Calcium metal ($\text{Ca}$) is a strong reducing agent (has a very negative reduction potential, meaning it is easily oxidized), while dichromate ions ($\text{Cr}_2\text{O}_7^{2-}$) are strong oxidizing agents (have a very positive reduction potential, meaning they are easily reduced). The relevant half-reactions are: Oxidation: $\text{Ca}(s) \to \text{Ca}^{2+}(aq) + 2e^-$ ($E^\circ_{\text{oxidation}} = +2.87 \text{ V}$) Reduction: $\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \to 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)$ ($E^\circ_{\text{reduction}} = +1.33 \text{ V}$) Since the overall cell potential would be positive ($2.87 \text{ V} + 1.33 \text{ V} = 4.20 \text{ V}$), the reaction is spontaneous. Observations: The calcium granules will disappear/dissolve as they are oxidized to $\text{Ca}^{2+}$ ions. The orange solution of potassium dichromate will turn green as the $\text{Cr}_2\text{O}_7^{2-}$ ions (orange) are reduced to $\text{Cr}^{3+}$ ions (green). iii) When the calcium and iron half-cells are coupled: Calcium half-cell: $\text{Ca}^{2+}(aq) + 2e^- \to \text{Ca}(s)$, $E^\circ = -2.87 \text{ V}$ Iron half-cell: $\text{Fe}^{2+}(aq) + 2e^- \to \text{Fe}(s)$, $E^\circ = -0.44 \text{ V}$ The half-cell with the more negative reduction potential will undergo oxidation (anode), and the half-cell with the more positive reduction potential will undergo reduction (cathode). Since $-2.87 \text{ V} < -0.44 \text{ V}$: Calcium will be oxidized (anode): $\text{Ca}(s) \to \text{Ca}^{2+}(aq) + 2e^-$ Iron(II) ions will be reduced (cathode): $\text{Fe}^{2+}(aq) + 2e^- \to \text{Fe}(s)$ Cell diagram: The cell diagram represents the anode on the left and the cathode on the right, separated by a salt bridge ($||$). $$\boxed{\text{Ca}(s) | \text{Ca}^{2+}(aq) || \text{Fe}^{2+}(aq) | \text{Fe}(s)}$$ Calculation of the EMF of the cell ($E^\circ_{\text{cell}}$): $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ(\text{Fe}^{2+}/\text{Fe}) - E^\circ(\text{Ca}^{2+}/\text{Ca})$$ $$E^\circ_{\text{cell}} = (-0.44 \text{ V}) - (-2.87 \text{ V})$$ $$E^\circ_{\text{cell}} = -0.44 \text{ V} + 2.87 \text{ V}$$ $$E^\circ_{\text{cell}} = \boxed{+2.43 \text{ V}}$$