Given reduction potentials:

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^). A higher (more positive) E^ indicates a stronger oxidizing agent. Given reduction potentials: Cr_2O_7^2-(aq) + 14H^+(aq) + 6e^- 2Cr^3+(aq) + 7H_2O(l) E^ = +1.33 V I_2(aq) + 2e^- 2I^-(aq) E^ = +0.54 V Fe^2+(aq) + 2e^- Fe(s) E^ = -0.44 V Ca^2+(aq) + 2e^- Ca(s) E^ = -2.87 V Arranging the species (Cr_2O_7^2-, I_2, Fe^2+, Ca^2+) in decreasing order of oxidizing strength: Cr_2O_7^2- > I_2 > Fe^2+ > Ca^2+ ii) When granules of calcium are added to a solution of potassium dichromate, a redox reaction will occur. Calcium metal (Ca) is a strong reducing agent (has a very negative reduction potential, meaning it is easily oxidized), while dichromate ions (Cr_2O_7^2-) are strong oxidizing agents (have a very positive reduction potential, meaning they are easily reduced). The relevant half-reactions are: Oxidation: Ca(s) Ca^2+(aq) + 2e^- (E^_oxidation = +2.87 V) Reduction: Cr_2O_7^2-(aq) + 14H^+(aq) + 6e^- 2Cr^3+(aq) + 7H_2O(l) (E^_reduction = +1.33 V) Since the overall cell potential would be positive (2.87 V + 1.33 V = 4.20 V), the reaction is spontaneous. Observations: The calcium granules will disappear/dissolve as they are oxidized to Ca^2+ ions. The orange solution of potassium dichromate will turn green as the Cr_2O_7^2- ions (orange) are reduced to Cr^3+ ions (green). iii) When the calcium and iron half-cells are coupled: Calcium half-cell: Ca^2+(aq) + 2e^- Ca(s), E^ = -2.87 V Iron half-cell: Fe^2+(aq) + 2e^- Fe(s), E^ = -0.44 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 V < -0.44 V: Calcium will be oxidized (anode): Ca(s) Ca^2+(aq) + 2e^- Iron(II) ions will be reduced (cathode): Fe^2+(aq) + 2e^- Fe(s) Cell diagram: The cell diagram represents the anode on the left and the cathode on the right, separated by a salt bridge (||). Ca(s) | Ca^2+(aq) || Fe^2+(aq) | Fe(s) Calculation of the EMF of the cell (E^_cell): E^_cell = E^_cathode - E^_anode E^_cell = E^(Fe^2+/Fe) - E^(Ca^2+/Ca) E^_cell = (-0.44 V) - (-2.87 V) E^_cell = -0.44 V + 2.87 V E^_cell = +2.43 V