Are these official VCAA Chemistry questions?

No. Every question is original and written by us to the public VCE Chemistry study design. They are VCAA-style for practice only, and ATARMAxxing is not affiliated with or endorsed by the VCAA. Always check rules, constants and formulae against your VCAA data book and the current study design.

Which topics do these VCE Chemistry practice questions cover?

They span Units 3 and 4: energy and redox (galvanic cells, electrolysis and fuel cells), rates and equilibrium, organic chemistry reactions and naming, and analytical chemistry including titration and chromatography. These are the same areas of study assessed on the end-of-year exam.

Do I need a calculator and data book to answer these?

Yes, for the calculation questions. Treat them like the real exam: use your VCAA data book for constants and standard potentials, show full working, carry correct units throughout, and round only at the final step to an appropriate number of significant figures.

How should I use these practice questions effectively?

Do each one under timed conditions without your notes, then mark honestly against the scheme and write down why you lost each mark. Revisit weak topics within a few days. Spacing your practice this way, rather than cramming, is what builds reliable exam performance over a term.

Where can I find more Chemistry practice?

You can find more free material on our free resources page, and see the full range of subjects and plans on the pricing page. Pair the questions with past VCAA exams and the official assessment reports, which set out exactly how marks are awarded for each item.

10 Free VCE Chemistry Practice Questions + Answers

These ten VCE Chemistry practice questions are written to the Units 3 and 4 study design and cover energy and redox, rates and equilibrium, organic chemistry and analytical chemistry. They are VCAA-style and original, not real VCAA exam questions, and ATARMAxxing is not affiliated with or endorsed by the VCAA. Always confirm any rule, constant or formula against your data book and the current study design.

Each question lists its marks and a worked mark scheme.
Calculation questions need a calculator and the VCAA data book.
Carry units throughout and round only at the final step.
Topics span both Units 3 and 4, as assessed on the end-of-year exam.

Sit each VCE Chemistry practice question under timed conditions, then mark yourself against the scheme below. For more on technique, see whether practice exams are worth it and how to build a VCE study schedule. You can also grab free material on the free resources page.

Question 1 (4 marks)

A galvanic cell is constructed from a zinc half-cell and a copper half-cell connected by a salt bridge. Using standard half-cell reduction potentials (Zn2+/Zn = -0.76 V; Cu2+/Cu = +0.34 V), write the overall cell equation, identify the anode, and calculate the cell's maximum EMF.

Show the mark scheme

1 mark: overall equation Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s). 1 mark: zinc electrode is the anode (oxidation, more negative reduction potential). 1 mark: correct method, EMF = E(cathode) - E(anode) = +0.34 - (-0.76). 1 mark: EMF = +1.10 V (must include the positive sign and unit V). Deduct the final mark if the value is correct but unsigned or unitless.

Question 2 (4 marks)

An electrolytic cell is used to electroplate a spoon with silver from a silver nitrate solution, passing a current of 2.50 A for 30.0 minutes. Calculate the mass of silver deposited. (F = 96500 C/mol; M(Ag) = 107.9 g/mol; Ag+ + e- -> Ag.)

Show the mark scheme

1 mark: Q = I x t = 2.50 x (30.0 x 60) = 4500 C. 1 mark: n(e-) = Q/F = 4500/96500 = 0.04663 mol. 1 mark: n(Ag) = n(e-) = 0.04663 mol (1:1 from the half-equation). 1 mark: m(Ag) = n x M = 0.04663 x 107.9 = 5.03 g (accept 5.0 to 5.03 g, must include unit g).

Question 3 (3 marks)

Explain why a hydrogen-oxygen fuel cell is described as more energy efficient than burning hydrogen in a combustion engine, and write the half-equation occurring at the negative electrode in an acidic hydrogen-oxygen fuel cell.

Show the mark scheme

1 mark: a fuel cell converts chemical energy directly into electrical energy, whereas combustion converts chemical energy to heat then to mechanical or electrical energy, with energy lost as heat at each conversion. 1 mark: the fuel cell therefore loses less energy as heat / has fewer energy-conversion steps. 1 mark: negative electrode (anode, oxidation) half-equation H2(g) -> 2H+(aq) + 2e-.

Question 4 (4 marks)

For the equilibrium 2SO2(g) + O2(g) <-> 2SO3(g), the forward reaction is exothermic. State and justify the effect on the equilibrium yield of SO3 of (i) increasing the temperature and (ii) increasing the total pressure by reducing the volume.

Show the mark scheme

Part (i): 1 mark: yield of SO3 decreases. 1 mark: increasing temperature favours the endothermic (reverse) reaction, so the system shifts left (Le Chatelier). Part (ii): 1 mark: yield of SO3 increases. 1 mark: increased pressure shifts the equilibrium to the side with fewer gas moles (3 mol reactants to 2 mol product), i.e. to the right. No mark for a direction stated without the supporting reason.

Question 5 (4 marks)

The equilibrium constant Kc for N2(g) + 3H2(g) <-> 2NH3(g) is 0.500 at a given temperature. At equilibrium a 2.00 L vessel contains 0.400 mol N2 and 1.20 mol H2. Calculate the equilibrium concentration of NH3.

Show the mark scheme

1 mark: convert to concentrations: [N2] = 0.200 M, [H2] = 0.600 M. 1 mark: correct Kc expression Kc = [NH3]^2 / ([N2][H2]^3). 1 mark: rearrange [NH3]^2 = Kc x [N2] x [H2]^3 = 0.500 x 0.200 x (0.600)^3 = 0.0216. 1 mark: [NH3] = sqrt(0.0216) = 0.147 M (accept 0.147 to 0.15 M, must include unit).

Question 6 (3 marks)

Increasing temperature increases the rate of a reaction. Using collision theory, explain the two distinct reasons why, and state which is the more significant.

Show the mark scheme

1 mark: at higher temperature particles move faster, so collisions are more frequent. 1 mark: a greater proportion of particles have kinetic energy equal to or above the activation energy, so a higher proportion of collisions are successful. 1 mark: the increase in the proportion of particles with energy at or above the activation energy is the more significant factor.

Question 7 (4 marks)

Name the products and the structural feature responsible when 1-propanol is oxidised. Then write a balanced equation, using [O] to represent the oxidant, for the formation of the final organic product.

Show the mark scheme

1 mark: 1-propanol is a primary alcohol (-OH on a terminal carbon), which is why it can be oxidised first to an aldehyde then to a carboxylic acid. 1 mark: naming propanal (aldehyde, partial oxidation) and propanoic acid (carboxylic acid, full oxidation). 1 mark: balanced equation CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. 1 mark: equation correctly balanced for atoms (accept condensed or semi-structural formulae).

Question 8 (4 marks)

Distinguish between an addition reaction of ethene with chlorine and a substitution reaction of ethane with chlorine, writing one balanced equation for each and naming the organic product in each case.

Show the mark scheme

1 mark: ethene (alkene, C=C) undergoes addition: CH2=CH2 + Cl2 -> CH2ClCH2Cl, product 1,2-dichloroethane. 1 mark: correct identification of this as addition because the double bond opens and no atoms are lost. 1 mark: ethane (alkane) undergoes substitution: CH3CH3 + Cl2 -> CH3CH2Cl + HCl (UV light), product chloroethane. 1 mark: correct identification of this as substitution because a H atom is replaced and HCl is produced.

Question 9 (4 marks)

A 25.00 mL sample of vinegar is titrated against 0.150 mol/L NaOH, requiring 22.40 mL to reach the endpoint. Calculate the concentration of acetic acid (CH3COOH) in the vinegar in mol/L. (CH3COOH + NaOH -> CH3COONa + H2O.)

Show the mark scheme

1 mark: n(NaOH) = c x V = 0.150 x 0.02240 = 3.36 x 10^-3 mol. 1 mark: n(CH3COOH) = n(NaOH) = 3.36 x 10^-3 mol (1:1 mole ratio). 1 mark: correct method c = n/V = 3.36 x 10^-3 / 0.02500. 1 mark: c(CH3COOH) = 0.134 mol/L (accept 0.134 to 0.135 mol/L, must include unit).

Question 10 (4 marks)

Explain how high-performance liquid chromatography (HPLC) separates the components of a mixture, and describe how a calibration curve is used to determine the concentration of an unknown component.

Show the mark scheme

1 mark: components separate based on their differing affinity for the stationary phase versus the mobile phase. 1 mark: components with greater attraction to the stationary phase are retained longer and have a longer retention time. 1 mark: a calibration curve is prepared by plotting peak area (or peak height) against concentration for a series of standard solutions of known concentration. 1 mark: the unknown's peak area is measured and read off the line of best fit to determine its concentration (accept interpolation from the calibration line).

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