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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

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Answer

To find the total energy required for the 13-week period:

extTotalenergyneeded=13extweeksimes800extMJ/weekimes106extkJ/MJ=1.04imes109extkJ ext{Total energy needed} = 13 ext{ weeks} imes 800 ext{ MJ/week} imes 10^6 ext{ kJ/MJ} = 1.04 imes 10^9 ext{ kJ}

Using the given reaction, we can see that burning 1 mole of ethanol produces 1370 kJ of energy. We can calculate the moles of ethanol required:

ext{Moles of ethanol needed} = rac{1.04 imes 10^9 ext{ kJ}}{1370 ext{ kJ/mol}} \ = 759,170 ext{ mol}

Next, we find the molar mass of ethanol (C₂H₆O):

  • C: 12.01 g/mol (x2) = 24.02 g/mol
  • H: 1.008 g/mol (x6) = 6.048 g/mol
  • O: 16.00 g/mol (x1) = 16.00 g/mol
  • Total = 24.02 + 6.048 + 16.00 = 46.068 g/mol

Now calculate the total mass of ethanol:

extMassofethanol=extMolesimesextMolarmass =759,170extmolimes46.068extg/mol =34,299,140.36extg =34,299.14extkg ext{Mass of ethanol} = ext{Moles} imes ext{Molar mass} \ = 759,170 ext{ mol} imes 46.068 ext{ g/mol} \ = 34,299,140.36 ext{ g} \ = 34,299.14 ext{ kg}

Step 2

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

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The half reaction occurring at the anode during the oxidation of ethanol in the fuel cell can be written as:

extC2extH5extOH+12extH++12e−→2extCO2+6extH2extO ext{C}_2 ext{H}_5 ext{OH} + 12 ext{H}^+ + 12e^- → 2 ext{CO}_2 + 6 ext{H}_2 ext{O}

Step 3

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

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The electrical energy produced can be calculated using the formula:

extElectricalenergy=extVoltageimesextCharge ext{Electrical energy} = ext{Voltage} imes ext{Charge}

From the problem, the voltage across the fuel cell is 1.15 V. We need to calculate the total charge transferred in the reaction:

Since the reaction at the anode involves the transfer of 12 electrons (from the oxidation half-reaction), and knowing that 1 mole of electrons corresponds to 96,485 coulombs:

extChargepermole=12imes96,485extC=1,157,820extC/mol ext{Charge per mole} = 12 imes 96,485 ext{ C} = 1,157,820 ext{ C/mol}

Thus, the electrical energy provided per mole of ethanol is:

extElectricalenergy=1.15extVimes1,157,820extC =1,333,130extJ/mol =1.33extMJ/mol ext{Electrical energy} = 1.15 ext{ V} imes 1,157,820 ext{ C} \ = 1,333,130 ext{ J/mol} \ = 1.33 ext{ MJ/mol}

Step 4

Suggest one important reason why the fuel cell would be better than the generator for this purpose

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Answer

One important reason is that fuel cells are generally more efficient than internal combustion engines. They convert chemical energy directly into electrical energy with less energy loss in the form of heat, leading to a cleaner and more efficient energy production method.

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