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e) (i) Select the species which is the strongest reducing agent? (ii) Write a cell diagram for the cell obtained by coupling A and B.

Chemistry

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Here's the solution for question e):

e) (i) Select the species which is the strongest reducing agent?

  1. List the given standard reduction potentials (E∘E^\circ) for each half-reaction.

    • A: Fe3+(aq)+e−⇌Fe2+(aq)\text{Fe}^{3+} (\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+} (\text{aq}) E∘=+0.77 VE^\circ = +0.77 \text{ V}
    • B: Cr2O72−(aq)+14H+(aq)+6e−⇌2Cr3+(aq)+7H2O\text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 14\text{H}^+ (\text{aq}) + 6\text{e}^- \rightleftharpoons 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O} E∘=+1.33 VE^\circ = +1.33 \text{ V}
    • C: Cl2(g)+2e−⇌2Cl−(aq)\text{Cl}_2 (\text{g}) + 2\text{e}^- \rightleftharpoons 2\text{Cl}^- (\text{aq}) E∘=+1.6 VE^\circ = +1.6 \text{ V}
  2. Identify the strongest reducing agent.

    A reducing agent is a species that gets oxidized (loses electrons). In a table of standard reduction potentials, the strongest reducing agents are the products of the half-reactions with the lowest (most negative) standard reduction potentials.
    Comparing the E∘E^\circ values: +0.77 V+0.77 \text{ V} (A), +1.33 V+1.33 \text{ V} (B), and +1.6 V+1.6 \text{ V} (C), the lowest potential is +0.77 V+0.77 \text{ V} for half-reaction A.
    The reducing agent in half-reaction A is Fe2+(aq)\text{Fe}^{2+} (\text{aq}).

    The strongest reducing agent is Fe2+(aq)\boxed{\text{Fe}^{2+} (\text{aq})}.

    e) (ii) Write a cell diagram for the cell obtained by coupling A and B.

  3. Determine the anode and cathode.

    When coupling two half-cells, the one with the lower standard reduction potential will undergo oxidation (anode), and the one with the higher standard reduction potential will undergo reduction (cathode).

    • Half-cell A: E∘=+0.77 VE^\circ = +0.77 \text{ V}
    • Half-cell B: E∘=+1.33 VE^\circ = +1.33 \text{ V} Since E∘(A)<E∘(B)E^\circ (\text{A}) < E^\circ (\text{B}), half-cell A will be the anode (oxidation) and half-cell B will be the cathode (reduction).
  4. Write the half-reactions for the anode and cathode.

    • Anode (oxidation): Fe2+(aq)→Fe3+(aq)+e−\text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}^{3+} (\text{aq}) + \text{e}^-
    • Cathode (reduction): Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O\text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 14\text{H}^+ (\text{aq}) + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}
  5. Construct the cell diagram.

    The general convention for a cell diagram is: Anode∣Anode solution∣∣Cathode solution∣Cathode\text{Anode} | \text{Anode solution} || \text{Cathode solution} | \text{Cathode}.
    For half-cells involving only ions in solution, an inert electrode (like Platinum

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