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An isolated research station is to be staffed by a small group of scientists for 13 weeks - VCE - SSCE Chemistry - Question 6 - 2002 - Paper 1

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An isolated research station is to be staffed by a small group of scientists for 13 weeks. Part of the exercise is to test the effectiveness of liquid ethanol (CH₃CH... show full transcript

Worked Solution & Example Answer:An isolated research station is to be staffed by a small group of scientists for 13 weeks - VCE - SSCE Chemistry - Question 6 - 2002 - Paper 1

Step 1

Calculate the total mass of ethanol needed for heating and cooking

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Answer

  1. Calculate Total Energy Requirements: The total energy required for heating and cooking over 13 weeks is:

    800extMJ/week×13 weeks=10400 MJ=1.04×1010 kJ800 ext{ MJ/week} \times 13 \text{ weeks} = 10400 \text{ MJ} = 1.04 \times 10^{10} \text{ kJ}

  2. Determine Moles of Ethanol Required: From the reaction given:

    ΔH=−1370 kJ/mol\Delta H = -1370 \text{ kJ/mol}

    The number of moles of ethanol needed is:

    moles of ethanol=1.04×1010 kJ1370 kJ/mol≈7591.9 mol\text{moles of ethanol} = \frac{1.04 \times 10^{10} \text{ kJ}}{1370 \text{ kJ/mol}} \approx 7591.9 \text{ mol}

  3. Calculate Mass of Ethanol: The molar mass of ethanol (C₂H₅OH) is approximately 46 g/mol. Thus, the total mass of ethanol required is:

    mass=7591.9 mol×46 g/mol≈349,200 g=349.2 kg \text{mass} = 7591.9 \text{ mol} \times 46 \text{ g/mol} \approx 349,200 \text{ g} = 349.2 \text{ kg}

Step 2

Give the half reaction occurring at the anode where the ethanol is oxidised in the fuel cell

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Answer

The half reaction at the anode for the oxidation of ethanol in the fuel cell is:

C2H5OH(l)+6H++6e−→2CO2(g)+3H2O(l)\text{C}_2\text{H}_5\text{OH}(l) + 6\text{H}^+ + 6e^- \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)

Step 3

Calculate the electrical energy provided per mole of ethanol consumed in the fuel cell

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Answer

  1. Calculate the total charge transfer: The number of electrons transferred in the oxidation of ethanol is 6, therefore:

    Energy=Voltage×Charge\text{Energy} = \text{Voltage} \times \text{Charge}

    The charge (Q) is given by:

    Q=n×FQ = n \times F where:

    • n = number of moles of electrons (6 moles per mole of ethanol)
    • F = Faraday's constant (approximately 96500 C/mol)

    Therefore, for 1 mole of ethanol:

    Q=6×96500=579000 CQ = 6 \times 96500 = 579000 \text{ C}

  2. Substituting into the energy formula: The electrical energy per mole of ethanol can be calculated as:

    Energy=1.15 V×579000 C≈666850 J=666.85 kJ\text{Energy} = 1.15 \text{ V} \times 579000 \text{ C} \approx 666850 \text{ J} = 666.85 \text{ kJ}

Step 4

Suggest one reason why the fuel cell would be better than the generator

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Answer

One important reason why the fuel cell would be better than the generator is efficiency. Fuel cells are generally more efficient than internal combustion engines, converting a higher percentage of the energy in the fuel directly into electrical energy, while generators typically waste energy as heat.

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