VCAA Chemistry What are the current and future options for supplying energy?

15 sample questions with marking guides and sample answers

Q6
2025
VCAA
1 mark
Q6
1 mark

The molar enthalpy for combustion of glucose, C6H12O6C_6H_{12}O_6, is −2840 kJ mol−1-2840\ \mathrm{kJ\ mol^{-1}}.

Which one of the following describes what occurs to one mole of carbon dioxide, CO2CO_2, during photosynthesis?

A

473.3 kJ of energy is absorbed.

B

2840 kJ of energy is absorbed.

C

473.3 kJ of energy is released.

D

2840 kJ of energy is released.

Reveal Answer
A

473.3 kJ of energy is absorbed.

Correct Answer

Photosynthesis is the reverse of combustion, so it absorbs 2840 kJ2840\ \mathrm{kJ} per mole of glucose produced. Since 6 moles of CO2CO_2 are required to produce one mole of glucose, 2840/6=473.3 kJ2840 / 6 = 473.3\ \mathrm{kJ} of energy is absorbed per mole of CO2CO_2.

B

2840 kJ of energy is absorbed.

This is the total energy absorbed to produce one full mole of glucose. Producing one mole of glucose requires 6 moles of CO2CO_2, not just one.

C

473.3 kJ of energy is released.

Photosynthesis is an endothermic process that absorbs energy from sunlight. This option incorrectly states that energy is released.

D

2840 kJ of energy is released.

This describes the energy released during the combustion of one mole of glucose, rather than the energy absorbed per mole of CO2CO_2 during photosynthesis.

Q5
2025
VCAA
8 marks
Q5

A constant current of 1.80 A at a potential difference of 4.85 V was passed through a heating coil to calibrate a calorimeter. The temperature of the water increased by 2.1 °C. The calibration factor was calculated as 374 J °C−1^{-1}.

Q5a (i)
2 marks

Calculate the time that the current was passed through the heating coil during the calibration of the calorimeter.

Reveal Answer
Marking Criteria
DescriptorMarks

Calculates the correct energy (E=785.4E = 785.4 J).

1

Calculates the correct time based on the value of energy determined, including correct units (t=90t = 90 s or 1.51.5 min).

1
Q5a (ii)
3 marks

A student correctly states, ‘the calorimeter contained less than 100 g of water’.
Justify this statement with reference to the results of the experiment.

Reveal Answer
Marking Criteria
DescriptorMarks

Identifies that the heat capacity of water was required.

1

Demonstrates an understanding that at least 418418 J of energy would have been needed to heat 100100 g of water by 11 °C.

1

Demonstrates an understanding that since only 374374 J of energy was needed to heat the calorimeter by 11 °C, there must have been less than 100100 g of water present.

1
Q5b
3 marks

The reaction below is performed in the calibrated calorimeter.

2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}

80.0 mL of 0.40 M H2_2SO4_4(aq) and 0.060 mol of NaOH(aq) were reacted in the calorimeter.

Calculate the amount, in mol, of H2_2SO4_4(aq) in excess.

Reveal Answer
Marking Criteria
DescriptorMarks

Calculates the correct n(H2SO4)n(\text{H}_2\text{SO}_4) required (0.0300.030 mol).

1

Calculates the correct amount of n(H2SO4)n(\text{H}_2\text{SO}_4) originally available (0.0320.032 mol).

1

Calculates the correct n(H2SO4)n(\text{H}_2\text{SO}_4) in excess (0.0020.002 mol).

1
Q19
2022
QCAA
Paper 1
1 mark
Q19
1 mark

Three voltaic cells are constructed with metal Q as one electrode and metals R, S or T as the other electrode. The potential differences for the cells are shown in the table.

Voltaic cellHalf-cellHalf-cellPotential difference (V)
1Q(s)/Q2+(aq)\text{Q(s)} / \text{Q}^{2+}\text{(aq)}R+(aq)/R(s)\text{R}^+\text{(aq)} / \text{R(s)}1.18
2Q(s)/Q2+(aq)\text{Q(s)} / \text{Q}^{2+}\text{(aq)}S2+(aq)/S(s)\text{S}^{2+}\text{(aq)} / \text{S(s)}0.72
3T(s)/T3+(aq)\text{T(s)} / \text{T}^{3+}\text{(aq)}Q2+(aq)/Q(s)\text{Q}^{2+}\text{(aq)} / \text{Q(s)}0.95

The relative strength of the reducing agents from strongest to weakest is

A

T > Q > S > R

B

S > Q > T > R

C

R > Q > S > T

D

Q > R > T > S

Reveal Answer
A

T > Q > S > R

Correct Answer

In Cell 3, T is oxidized while Q is reduced, so T is a stronger reducing agent than Q (T>QT > Q). In Cells 1 and 2, Q is oxidized while R and S are reduced, so Q is stronger than both (Q>S,RQ > S, R). Since the potential with R (1.18 V1.18\text{ V}) is higher than with S (0.72 V0.72\text{ V}), R has a higher reduction potential, making S the stronger reducing agent (S>RS > R).

B

S > Q > T > R

This option incorrectly ranks S as stronger than Q. In Cell 2, Q acts as the anode (oxidized) and S as the cathode (reduced), which demonstrates that Q is a stronger reducing agent than S.

C

R > Q > S > T

This option incorrectly identifies R as the strongest reducing agent. R produces the largest voltage when reduced by Q, indicating it has the highest reduction potential and is therefore the weakest reducing agent.

D

Q > R > T > S

This option incorrectly ranks Q as stronger than T. In Cell 3, T acts as the anode (oxidized) and Q as the cathode (reduced), which demonstrates that T is a stronger reducing agent than Q.

Q4
2023
VCAA
1 mark
Q4
1 mark

Bioethanol is a fuel produced by

A

fermentation.

B

anaerobic digestion.

C

transesterification of fats.

D

a substitution reaction of ethene.

Reveal Answer
A

fermentation.

Correct Answer

Bioethanol is produced through the fermentation of sugars (such as glucose from crops) by microorganisms like yeast.

B

anaerobic digestion.

Anaerobic digestion of organic matter produces biogas, which is primarily a mixture of methane and carbon dioxide, not bioethanol.

C

transesterification of fats.

Transesterification of fats and oils is the chemical process used to produce biodiesel, not bioethanol.

D

a substitution reaction of ethene.

While ethanol can be produced from ethene, it is through an addition reaction (hydration), and this process yields synthetic ethanol from petrochemicals rather than bioethanol.

Q9
2024
VCAA
1 mark
Q9
1 mark

When 5.0 g of ethanol, C2_2H5_5OH(l), undergoes complete combustion in air, the mass of carbon dioxide, CO2_2, gas produced is

A

2.4 g

B

4.8 g

C

9.6 g

D

19 g

Reveal Answer
A

2.4 g

Incorrect. This mass is obtained if the moles of ethanol are incorrectly divided by 2 instead of multiplied by 2 based on the stoichiometry of the reaction.

B

4.8 g

Incorrect. This mass is obtained by assuming a 1:1 molar ratio between ethanol and CO2_2, forgetting that each mole of ethanol (C2_2H5_5OH) produces two moles of CO2_2.

C

9.6 g

Correct Answer

Correct. The balanced equation shows a 1:2 ratio of ethanol to CO2_2. Converting 5.0 g of ethanol to moles (0.1085 mol) and multiplying by 2 gives 0.217 mol of CO2_2, which equals 9.6 g.

D

19 g

Incorrect. This mass would be produced if the molar ratio of ethanol to CO2_2 was incorrectly assumed to be 1:4 instead of 1:2.

Q17
2022
QCAA
Paper 1
1 mark
Q17
1 mark

Identify the redox reaction.

A

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}

B

CaO(s)+H2O(l)→Ca(OH)2(s)\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(s)}

C

Cl2(g)+H2O(l)→HCl(aq)+HClO(aq)\text{Cl}_2\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{HCl(aq)} + \text{HClO(aq)}

D

NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

Reveal Answer
A

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}

This is a thermal decomposition reaction where the oxidation states of Calcium (+2), Carbon (+4), and Oxygen (-2) remain unchanged.

B

CaO(s)+H2O(l)→Ca(OH)2(s)\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(s)}

This is a combination reaction where no elements change their oxidation states (Ca remains +2, O remains -2, H remains +1).

C

Cl2(g)+H2O(l)→HCl(aq)+HClO(aq)\text{Cl}_2\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{HCl(aq)} + \text{HClO(aq)}

Correct Answer

This is a disproportionation redox reaction where chlorine is simultaneously reduced from 0 to -1 in HCl\text{HCl} and oxidized from 0 to +1 in HClO\text{HClO}.

D

NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

This is an acid-base neutralization reaction where ions exchange partners without any change in oxidation numbers.

Q2
2021
VCAA
1 mark
Q2
1 mark

Biodiesel and petrodiesel

A

have different viscosities.

B

have the same environmental impact.

C

contain molecules with no polar groups.

D

will flow easily through fuel lines in very cold climate conditions.

Reveal Answer
A

have different viscosities.

Correct Answer

This is correct because biodiesel contains polar ester groups that create stronger intermolecular forces, resulting in a higher viscosity compared to non-polar petrodiesel.

B

have the same environmental impact.

This is incorrect because biodiesel is derived from renewable biomass and generally produces fewer net carbon emissions and particulates compared to fossil-fuel-derived petrodiesel.

C

contain molecules with no polar groups.

This is incorrect because while petrodiesel consists of non-polar hydrocarbons, biodiesel consists of fatty acid methyl esters which contain polar ester groups.

D

will flow easily through fuel lines in very cold climate conditions.

This is incorrect because biodiesel has a higher cloud point than petrodiesel, meaning it is more prone to gelling and restricting flow in very cold climates.

Q17
2023
VCAA
1 mark
Q17
1 mark

Lignite is a type of brown coal. When lignite is completely combusted in a power station, 19.0 MJ/tonne of energy is released. The efficiency of the power station is 39%.

What mass of lignite is required to produce 42.0 MJ of usable energy in the power station?

A

0.862 tonnes

B

1.16 tonnes

C

2.21 tonnes

D

5.67 tonnes

Reveal Answer
A

0.862 tonnes

This incorrect answer is obtained by calculating 42.0/(19.0/0.39)42.0 / (19.0 / 0.39), which incorrectly divides the energy density by the efficiency instead of multiplying them to find the usable energy per tonne.

B

1.16 tonnes

This incorrect answer is obtained by calculating 19.0/(42.0×0.39)19.0 / (42.0 \times 0.39), which incorrectly divides the energy density by the required energy instead of dividing the required energy by the usable energy per tonne.

C

2.21 tonnes

This incorrect answer is obtained by calculating 42.0/19.042.0 / 19.0, which represents the mass needed if the power station were 100% efficient, completely ignoring the 39% efficiency factor.

D

5.67 tonnes

Correct Answer

The usable energy per tonne is 19.0×0.39=7.41 MJ/tonne19.0 \times 0.39 = 7.41 \text{ MJ/tonne}. To produce 42.0 MJ of usable energy, the mass required is 42.0/7.41=5.67 tonnes42.0 / 7.41 = 5.67 \text{ tonnes}.

Q12
2025
VCAA
1 mark
Q12
1 mark

A fuel cell utilises the reaction between hydrogen, H2H_2, and oxygen, O2O_2, to produce water, as shown in the reaction below.

2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

When a fuel cell produces 36000 C of charge, the mass of O2O_2 consumed is closest to

A

11.9 g

B

2.98 g

C

1.49 g

D

0.373 g

Reveal Answer
A

11.9 g

This incorrect answer assumes that only 1 mole of electrons is transferred per mole of O2O_2, rather than the actual 4 moles required by the half-equation.

B

2.98 g

Correct Answer

First, calculate the moles of electrons: n(e−)=3600096500=0.373n(e^-) = \frac{36000}{96500} = 0.373 mol. Since the reduction of O2O_2 requires 4 electrons (O2+4H++4e−→2H2OO_2 + 4H^+ + 4e^- \rightarrow 2H_2O), n(O2)=0.3734=0.0933n(O_2) = \frac{0.373}{4} = 0.0933 mol. The mass is 0.0933×32.0=2.980.0933 \times 32.0 = 2.98 g.

C

1.49 g

This incorrect answer uses the molar mass of atomic oxygen (16.016.0 g/mol) instead of molecular oxygen (O2O_2, 32.032.0 g/mol) to calculate the final mass.

D

0.373 g

This value represents the number of moles of electrons produced (n=3600096500=0.373n = \frac{36000}{96500} = 0.373 mol), not the mass of O2O_2 consumed in grams.

Q6
2020
VCAA
8 marks
Q6

Methane gas, CH4\text{CH}_4, can be captured from the breakdown of waste in landfills. CH4\text{CH}_4 is also a primary component of natural gas. CH4\text{CH}_4 can be used to produce energy through combustion.

Q6a
1 mark

Write the equation for the incomplete combustion of CH4\text{CH}_4 to produce carbon monoxide, CO.

Reveal Answer

2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) or
CH4(g) + 1.5O2(g) → CO(g) + 2H2O(l)

Marking Criteria
DescriptorMarks

Correct response.

1
Q6b
2 marks

If 20.0 g of CH4\text{CH}_4 is kept in a 5.0 L sealed container at 25∘C25 ^\circ\text{C}, what would be the pressure in the container?

Reveal Answer

n(CH4)=20.0/16.0=1.25 mol∗n(\text{CH}_4) = 20.0 / 16.0 = 1.25 \text{ mol} *
Pressure(CH4)=nRT/V=1.25×8.31×298/5.0=6.2×102 kPa∗\text{Pressure}(\text{CH}_4) = nRT / V = 1.25 \times 8.31 \times 298 / 5.0 = 6.2 \times 10^2 \text{ kPa} *

Marking Criteria
DescriptorMarks

Calculates correct amount of CH4 (1.25 mol).

1

Calculates correct pressure of CH4 (6.2 x 10^2 kPa).

1
Q6c
3 marks

A Bunsen burner is used to heat a beaker containing 350.0 g of water. Complete combustion of 0.485 g of CH4\text{CH}_4 raises the temperature of the water from 20∘C20 ^\circ\text{C} to 32.3∘C32.3 ^\circ\text{C}.

Calculate the percentage of the Bunsen burner's energy that is lost to the environment.

Reveal Answer

Energy from CH4=0.485 g×55.6 kJ g−1=27.0 kJ\text{Energy from CH}_4 = 0.485 \text{ g} \times 55.6 \text{ kJ g}^{-1} = 27.0 \text{ kJ}

Energy absorbed by water=4.18 J g−1 K−1×350.0×(32.3−20.0)\text{Energy absorbed by water} = 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 350.0 \times (32.3 - 20.0)
=1.80×104 J= 1.80 \times 10^4 \text{ J}
=18.0 kJ= 18.0 \text{ kJ}

Energy lost to environment=27.0−18.0=9.0 kJ\text{Energy lost to environment} = 27.0 - 18.0 = 9.0 \text{ kJ}

% energy lost=(9.0/27.0)×100=33.3%\% \text{ energy lost} = (9.0 / 27.0) \times 100 = 33.3 \%

Marking Criteria
DescriptorMarks

Calculating the energy from CH4.

1

Calculating the energy absorbed by water.

1

Working out the percentage of energy loss.

1
Q6d
2 marks

Compare the environmental impact of CH4\text{CH}_4 obtained from landfill to the environmental impact of CH4\text{CH}_4 obtained from natural gas.

Reveal Answer

Similarity – methane from both sources

  • Both produce atmospheric carbon dioxide through combustion.
  • Methane from both sources contains small amounts of nitrogen and sulfur; combustion of natural gas leads to the formation of acidic oxides such as SOx and NOx.

Difference – landfill versus natural gas

  • Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon.
  • Methane from landfill is more carbon neutral, methane from natural gas increases atmospheric CO2 levels.
  • Obtaining methane from natural gas via fracking causes additional significant environmental damage, whereas when obtaining methane from a landfill the damage has already been done in the formation of the landfill.
  • Landfill gases contain less methane and release more CO2 (for the same amount of energy generated), natural gas contains more methane and releases comparatively less CO2.
  • Methane captured from landfill and used as a source on energy may have a positive impact as it is a more potent greenhouse gas than CO2.
  • CH4 from landfill is more easily collected compared to fracking/sourcing methane from fossil fuels.
Marking Criteria
DescriptorMarks

1 mark for each valid comparison point (any 2 of): Both produce atmospheric carbon dioxide through combustion; Methane from both sources contains small amounts of nitrogen and sulfur; Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon; Methane from landfill is more carbon neutral; Obtaining methane from natural gas via fracking causes additional significant environmental damage; Landfill gases contain less methane and release more CO2; Methane captured from landfill may have a positive impact as it is a more potent greenhouse gas than CO2; CH4 from landfill is more easily collected.

2
Q22
2025
SCSA
1 mark
Q22
1 mark

The main reason a vehicle powered by a hydrogen fuel cell has lower polluting emissions than a vehicle powered by an internal combustion engine is because

A

hydrogen fuel cells release carbon dioxide into the atmosphere; however, it is not produced by the burning of fossil fuels.

B

the only by-products of the hydrogen fuel cell are water and heat.

C

fuel cells convert chemical energy directly into electrical energy.

D

fuel cells will not work unless the reactants are constantly supplied.

Reveal Answer
A

hydrogen fuel cells release carbon dioxide into the atmosphere; however, it is not produced by the burning of fossil fuels.

Hydrogen fuel cells do not produce carbon dioxide (CO2CO_2) during operation; they generate electricity through an electrochemical reaction rather than combustion.

B

the only by-products of the hydrogen fuel cell are water and heat.

Correct Answer

The chemical reaction in a hydrogen fuel cell combines hydrogen (H2H_2) and oxygen (O2O_2) to produce only water (H2OH_2O) and heat, eliminating the harmful tailpipe emissions associated with burning fossil fuels.

C

fuel cells convert chemical energy directly into electrical energy.

While this describes the efficient energy conversion mechanism of a fuel cell, it does not explain the composition of the emissions or why they are less polluting.

D

fuel cells will not work unless the reactants are constantly supplied.

This statement describes the operational requirement of a fuel cell (continuous fuel supply) but is unrelated to the environmental impact or chemical composition of its emissions.

Q16
2022
QCAA
Paper 1
1 mark
Q16
1 mark

Determine the oxidation state of manganese in MnO4−\text{MnO}_4^-.

A

+1

B

+2

C

+7

D

+8

Reveal Answer
A

+1

This is incorrect. If manganese had an oxidation state of +1, the total charge of the ion would be +1+4(−2)=−7+1 + 4(-2) = -7, which does not match the actual charge of -1.

B

+2

This is incorrect. While +2 is a very common oxidation state for manganese (e.g., in Mn2+\text{Mn}^{2+} salts), it is not the oxidation state found in the permanganate ion.

C

+7

Correct Answer

This is correct. Oxygen typically has an oxidation state of -2. Using the formula x+4(−2)=−1x + 4(-2) = -1 (where -1 is the overall charge), we solve for xx to find that manganese is +7.

D

+8

This is incorrect. Manganese is in Group 7 and has only 7 valence electrons, so its maximum possible oxidation state is +7. It cannot reach +8.

Q22
2020
VCAA
1 mark
Q22
1 mark

The combustion of which fuel provides the most energy per 100 g?

A

pentane (M=72 g mol−1\text{M} = 72 \text{ g mol}^{-1}), which releases 49 097 MJ tonne−149\,097 \text{ MJ tonne}^{-1}

B

nitromethane (M=61 g mol−1\text{M} = 61 \text{ g mol}^{-1}), which releases 11.63 kJ g−111.63 \text{ kJ g}^{-1}

C

butanol (M=74 g mol−1\text{M} = 74 \text{ g mol}^{-1}), which releases 2670 kJ mol−12670 \text{ kJ mol}^{-1}

D

ethyne (M=26 g mol−1\text{M} = 26 \text{ g mol}^{-1}), which releases 1300 kJ mol−11300 \text{ kJ mol}^{-1}

Reveal Answer
A

pentane (M=72 g mol−1\text{M} = 72 \text{ g mol}^{-1}), which releases 49 097 MJ tonne−149\,097 \text{ MJ tonne}^{-1}

Converting 49 097 MJ tonne−149\,097 \text{ MJ tonne}^{-1} to kJ g−1\text{kJ g}^{-1} gives 49.1 kJ g−149.1 \text{ kJ g}^{-1}, which yields 4910 kJ4910 \text{ kJ} per 100 g100 \text{ g}. This is slightly less than the energy provided by ethyne.

B

nitromethane (M=61 g mol−1\text{M} = 61 \text{ g mol}^{-1}), which releases 11.63 kJ g−111.63 \text{ kJ g}^{-1}

An energy release of 11.63 kJ g−111.63 \text{ kJ g}^{-1} yields 1163 kJ1163 \text{ kJ} per 100 g100 \text{ g}, which is the lowest energy output among the given choices.

C

butanol (M=74 g mol−1\text{M} = 74 \text{ g mol}^{-1}), which releases 2670 kJ mol−12670 \text{ kJ mol}^{-1}

Dividing 2670 kJ mol−12670 \text{ kJ mol}^{-1} by the molar mass of 74 g mol−174 \text{ g mol}^{-1} gives 36.1 kJ g−136.1 \text{ kJ g}^{-1}, resulting in 3610 kJ3610 \text{ kJ} per 100 g100 \text{ g}.

D

ethyne (M=26 g mol−1\text{M} = 26 \text{ g mol}^{-1}), which releases 1300 kJ mol−11300 \text{ kJ mol}^{-1}

Correct Answer

Dividing 1300 kJ mol−11300 \text{ kJ mol}^{-1} by the molar mass of 26 g mol−126 \text{ g mol}^{-1} gives 50 kJ g−150 \text{ kJ g}^{-1}. This yields 5000 kJ5000 \text{ kJ} per 100 g100 \text{ g}, which is the highest energy output of all the options.

Q22
2021
VCAA
1 mark
Q22
1 mark

1 L of octane has a mass of 703 g at SLC. The efficiency of the reaction when octane undergoes combustion in the petrol engine of a car is 25.0%.

What volume of octane stored in a petrol tank at SLC is required to produce 528 MJ of usable energy in a combustion engine?

A

3.92 L

B

11.8 L

C

15.7 L

D

62.7 L

Reveal Answer
A

3.92 L

This result comes from incorrectly multiplying the theoretical volume by the efficiency (15.7 L×0.2515.7 \text{ L} \times 0.25), rather than dividing by it to account for the extra fuel needed due to energy loss.

B

11.8 L

This value is obtained by incorrectly multiplying the theoretical volume by 0.750.75 (or 100%−25%100\% - 25\%), which is an incorrect application of the efficiency percentage.

C

15.7 L

This is the volume of octane required if the engine were 100%100\% efficient (528 MJ/33.67 MJ/L528 \text{ MJ} / 33.67 \text{ MJ/L}). It fails to account for the 25.0%25.0\% efficiency of the engine, which requires more fuel to be burned.

D

62.7 L

Correct Answer

The total energy needed is 528 MJ/0.25=2112 MJ528 \text{ MJ} / 0.25 = 2112 \text{ MJ}. With an energy density of 33.67 MJ/L33.67 \text{ MJ/L} (calculated from octane's heat of combustion of 5460 kJ/mol5460 \text{ kJ/mol}, molar mass of 114.0 g/mol114.0 \text{ g/mol}, and density of 703 g/L703 \text{ g/L}), the required volume is 2112 MJ/33.67 MJ/L=62.7 L2112 \text{ MJ} / 33.67 \text{ MJ/L} = 62.7 \text{ L}.

Q3
2022
QCAA
Paper 1
1 mark
Q3
1 mark

Which option is true for the redox equation?

Fe(s)+CuCl2(aq)→FeCl2(aq)+Cu(s)\text{Fe(s)} + \text{CuCl}_2\text{(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{Cu(s)}

A

Fe is oxidised and Cu is the oxidising agent

B

Fe is oxidised and Cu2+\text{Cu}^{2+} is the oxidising agent

C

Fe2+\text{Fe}^{2+} is oxidised and Cu is the oxidising agent

D

Fe2+\text{Fe}^{2+} is oxidised and Cu2+\text{Cu}^{2+} is the oxidising agent

Reveal Answer
A

Fe is oxidised and Cu is the oxidising agent

While Fe is oxidised, the oxidising agent is the specific species that accepts electrons. In this reaction, the copper(II) ion (Cu2+\text{Cu}^{2+}) accepts electrons, not neutral copper (Cu).

B

Fe is oxidised and Cu2+\text{Cu}^{2+} is the oxidising agent

Correct Answer

Fe loses electrons (oxidation state changes from 0 to +2), so it is oxidised. Cu2+\text{Cu}^{2+} gains electrons (oxidation state changes from +2 to 0), so it acts as the oxidising agent.

C

Fe2+\text{Fe}^{2+} is oxidised and Cu is the oxidising agent

Oxidation involves the reactant losing electrons. Here, solid Fe is oxidised, not the product Fe2+\text{Fe}^{2+}. Additionally, the oxidising agent is the ion Cu2+\text{Cu}^{2+}, not neutral Cu.

D

Fe2+\text{Fe}^{2+} is oxidised and Cu2+\text{Cu}^{2+} is the oxidising agent

The species being oxidised is the reactant Fe, which loses electrons to become Fe2+\text{Fe}^{2+}. The ion Fe2+\text{Fe}^{2+} is the product of oxidation, not the substance being oxidised.

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How many VCAA Chemistry questions cover What are the current and future options for supplying energy??
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