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I) The student was surprised that no calcium was produced because in an aqueous solution of calcium chloride, there are Ca²+ ions and water molecules (H₂O) present. At t

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b(iii)

5.30 g\text{5.30 g}

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Here's the solution for part (b):

Part b(ii)

The equation for the reaction that occurred at the anode if the solution was concentrated is

i) The student was surprised that no calcium was produced because in an aqueous solution of calcium chloride, there are Ca2+\text{Ca}^{2+} ions and water molecules (H2O\text{H}_2\text{O}) present. At the cathode (negative electrode), water is preferentially reduced over calcium ions because water has a less negative (more positive) standard reduction potential (E∘(H2O)=−0.83 VE^\circ(\text{H}_2\text{O}) = -0.83 \text{ V}) compared to calcium ions (E∘(Ca2+)=−2.87 VE^\circ(\text{Ca}^{2+}) = -2.87 \text{ V}). Therefore, hydrogen gas is produced instead of calcium metal.

2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^-(aq) \to \text{Cl}_2(g) + 2\text{e}^-

In a concentrated chloride solution, chloride ions are preferentially oxidized over water due to their high concentration and overpotential effects, producing chlorine gas.

Part b(iii)

5.30 g\text{5.30 g}

  1. Convert time to seconds.

    t=1 hour×60 min/hour×60 s/min=3600 st = 1 \text{ hour} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 3600 \text{ s}

  2. Calculate the total charge (QQ) passed.

    Q=I×tQ = I \times t Q=4 A×3600 sQ = 4 \text{ A} \times 3600 \text{ s} Q=14400 CQ = 14400 \text{ C}

  3. Determine the number of moles of electrons (nen_e) transferred.

    ne=QFn_e = \frac{Q}{F} ne=14400 C96500 C/mol e⁻n_e = \frac{14400 \text{ C}}{96500 \text{ C/mol e⁻}} ne≈0.14922 mol e⁻n_e \approx 0.14922 \text{ mol e⁻}

  4. Use the stoichiometry from the anode reaction (2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^-(aq) \to \text{Cl}_2(g) + 2\text{e}^-) to find moles of Cl2\text{Cl}_2.

    From the reaction, 2 moles of electrons produce 1 mole of Cl2\text{Cl}_2.

    nCl2=ne2n_{\text{Cl}_2} = \frac{n_e}{2} nCl2=0.14922 mol e⁻2 mol e⁻/mol Cl2n_{\text{Cl}_2} = \frac{0.14922 \text{ mol e⁻}}{2 \text{ mol e⁻/mol Cl}_2} nCl2≈0.07461 mol Cl2n_{\text{Cl}_2} \approx 0.07461 \text{ mol Cl}_2

  5. Calculate the mass of Cl2\text{Cl}_2 produced.

    The molar mass of Cl2=2×35.5 g/mol=71.0 g/mol\text{Cl}_2 = 2 \times 35.5 \text{ g/mol} = 71.0 \text{ g/mol}.

    mCl2=nCl2×MCl2m_{\text{Cl}_2} = n_{\text{Cl}_2} \times M_{\text{Cl}_2} mCl2=0.07461 mol×71.0 g/molm_{\text{Cl}_2} = 0.07461 \text{ mol} \times 71.0 \text{ g/mol} mCl2≈5.297 gm_{\text{Cl}_2} \approx 5.297 \text{ g}

    The mass of the product at the anode is 5.30 g\boxed{\text{5.30 g}}.

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