These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: F = 96,485 C mol⁻¹; I = q / t; R = 0.08206 L atm mol⁻¹ K⁻¹; Cr 52.00, Mg 24.31, Cu 63.55, Ag 107.87 g mol⁻¹. All the situations and measurements are invented for practice.
Question 1 (multiple choice · foundation)
A steady current of 2.00 A flows through an electrolytic cell for 16.1 minutes. How many moles of electrons pass through the circuit?
- (A) 3.34 × 10⁻⁴ mol
- (B) 0.0100 mol
- (C) 0.0200 mol
- (D) 1.93 × 10³ mol
Answer and explanation
Answer: (C). Time: 16.1 × 60 = 966 s. Charge: q = I × t = 2.00 A × 966 s = 1932 C. Moles of electrons: 1932 C ÷ 96,485 C mol⁻¹ = 0.0200 mol.
- (A) leaves the time in minutes (2.00 × 16.1 = 32.2), which is 60 times too small.
- (B) halves the correct answer, as if each electron counted twice. The question asks for electrons, not for the product of a 2-electron half-reaction.
- (D) is the charge in coulombs, not moles. It forgets to divide by F.
Question 2 (multiple choice · core)
Three electrolytic cells are joined in series. Cell 1 contains molten NaCl, cell 2 contains molten CaCl₂, and cell 3 contains a molten mixture with Al³⁺ ions. The current runs until all three cathodes have collected metal. What is the ratio of moles of Na : Ca : Al produced?
- (A) 1 : 1 : 1
- (B) 1 : 2 : 3
- (C) 6 : 3 : 2
- (D) 3 : 2 : 1
Answer and explanation
Answer: (C). In series, the same moles of electrons pass through every cell. Suppose 6 mol of electrons pass. Na⁺ + e⁻ → Na gives 6 mol Na; Ca²⁺ + 2e⁻ → Ca gives 3 mol Ca; Al³⁺ + 3e⁻ → Al gives 2 mol Al. Ratio 6 : 3 : 2.
- (A) assumes one mole of electrons gives one mole of any metal, ignoring the ion charges.
- (B) is the ratio of electrons needed per atom, not the moles of metal formed. It is upside down.
- (D) lowers the amount by one step for each extra unit of ion charge. The electrons must be divided by the charge (6 ÷ 1, 6 ÷ 2, 6 ÷ 3), not counted down.
Question 3 (multiple choice · core)
A student copper-plates a coin from CuSO₄(aq) at a steady current. Which single change would double the mass of copper deposited?
- (A) Doubling the concentration of Cu²⁺(aq), with the same current and time
- (B) Using a coin with twice the surface area, with the same current and time
- (C) Running the cell for twice as long at the same current
- (D) Doubling the current and halving the time
Answer and explanation
Answer: (C). The mass deposited depends on the charge passed, q = I × t. Twice the time at the same current means twice the charge, so twice the moles of electrons and twice the copper.
- (A) Concentration does not appear in the calculation. As long as Cu²⁺ does not run out, the same charge deposits the same copper.
- (B) A larger surface spreads the copper more thinly; the total mass is still set by the charge.
- (D) 2I × ½t = I × t, so the charge, and the mass, are unchanged.
Question 4 (calculation · core)
A spoon is silver-plated in a cell that uses AgNO₃(aq) and a bar of pure silver as the anode. The spoon is the cathode. The plater needs to deposit 1.25 g of silver and uses a steady current of 0.750 A.
(a) Calculate the time needed, in seconds and in minutes. (b) State and explain the change in mass of the silver anode.
Worked solution
(a)
- n(Ag) = 1.25 g ÷ 107.87 g mol⁻¹ = 0.011588 mol.
- Ag⁺ + e⁻ → Ag, so n(e⁻) = 0.011588 mol.
- q = 0.011588 mol × 96,485 C mol⁻¹ = 1118 C.
- t = q / I = 1118 C ÷ 0.750 A = 1.49 × 10³ s, which is 24.8 minutes.
(b) At the anode, silver is oxidised: Ag → Ag⁺ + e⁻. The same 0.011588 mol of electrons leaves the anode, so 0.011588 mol of silver dissolves. The anode loses 1.25 g, the same mass that the spoon gains, because both half-reactions involve the same metal with one electron per atom.
Suggested mark points (3): 1 for moles of electrons equal to moles of Ag; 1 for the time with the correct unit; 1 for a 1.25 g loss at the anode with the same-electrons reason.
Common error: using 2 electrons per silver atom doubles the time to 49.7 minutes.
Question 5 (calculation · core)
A solution of an unknown chromium salt is electrolysed with a current of 0.650 A for 45.0 minutes. The cathode gains 0.315 g of chromium. Determine the charge on the chromium ions in the solution, and write the cathode half-reaction.
Worked solution
- q = I × t = 0.650 A × (45.0 × 60 s) = 1755 C.
- n(e⁻) = 1755 C ÷ 96,485 C mol⁻¹ = 0.018189 mol.
- n(Cr) = 0.315 g ÷ 52.00 g mol⁻¹ = 0.0060577 mol.
- Electrons per Cr atom = 0.018189 ÷ 0.0060577 = 3.00.
The ions are Cr³⁺, and the half-reaction is Cr³⁺(aq) + 3e⁻ → Cr(s).
Suggested mark points (3): 1 for moles of electrons from the charge; 1 for moles of chromium; 1 for the ratio of 3 leading to Cr³⁺ and a balanced half-reaction.
The ratio comes out as a whole number (3.00) because the data are consistent. With real lab data, round to the nearest whole number and say so.
Question 6 (constructed response · core)
A galvanic cell is made from a magnesium strip in Mg(NO₃)₂(aq) and a copper strip in Cu(NO₃)₂(aq), joined by a KNO₃ salt bridge. The cell runs a small motor. Magnesium is the anode. An ammeter shows a steady current of 0.0400 A for 6.00 hours.
(a) Calculate the charge that passes and the moles of electrons. (b) Calculate the change in mass of each electrode, stating whether it is a gain or a loss. (c) A student says: “The same electrons pass through both electrodes, so the magnesium strip must lose the same mass that the copper strip gains.” Evaluate this claim. (d) When the copper strip is weighed, it has gained only 0.270 g. Suggest one reason for the difference from your answer to (b).
Worked solution and suggested Marlbridge rubric
(a) t = 6.00 × 3600 = 21,600 s. q = 0.0400 A × 21,600 s = 864 C. n(e⁻) = 864 ÷ 96,485 = 0.008955 mol.
(b) Anode: Mg → Mg²⁺ + 2e⁻. Cathode: Cu²⁺ + 2e⁻ → Cu. Both need 2 electrons per atom, so n(Mg) = n(Cu) = 0.008955 ÷ 2 = 0.004477 mol.
- Magnesium strip: 0.004477 × 24.31 = 0.109 g lost.
- Copper strip: 0.004477 × 63.55 = 0.285 g gained.
(c) The claim is incorrect. The same moles of electrons pass through both electrodes, and because both ions are 2+, the moles of Mg lost and Cu gained are equal. But the molar masses differ (24.31 and 63.55 g mol⁻¹), so equal moles have different masses: the copper gain is about 2.6 times the magnesium loss.
(d) Any one sensible reason, for example: some of the copper deposit is loose and was lost when the strip was rinsed or dried; the current was not perfectly steady, so the true average current was a little lower than 0.0400 A. (The measured gain is about 95% of the calculated value, so the calculation gives the maximum.)
| Point | What earns it |
|---|---|
| 1 | (a) 864 C and 0.00895 mol of electrons, with time in seconds |
| 1 | (b) Uses 2 electrons per atom to get 0.00448 mol of each metal |
| 1 | (b) 0.109 g lost from Mg and 0.285 g gained by Cu |
| 1 | (c) Says the claim is wrong: equal moles, but different molar masses give different masses |
| 1 | (d) A valid reason that would make the measured gain smaller |
Accept answers carried forward from (a). Do not award (d) for “human error” without saying what went wrong.
Question 7 (constructed response · stretch)
Water containing a little sulfuric acid is electrolysed between two platinum electrodes. Hydrogen gas forms at one electrode and oxygen gas at the other. A current of 1.20 A flows for 30.0 minutes.
(a) Write the half-reaction at each electrode and identify which electrode is the anode. (b) Calculate the moles of H₂ and of O₂ formed. (c) Calculate the volume of each gas at 298 K and 1.00 atm. (d) Explain, in terms of electrons, why the volume of hydrogen is twice the volume of oxygen.
Worked solution and suggested Marlbridge rubric
(a) Cathode (reduction): 2H⁺(aq) + 2e⁻ → H₂(g). Anode (oxidation): 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. Oxygen forms at the anode, because oxidation happens there.
(b) q = 1.20 A × 1800 s = 2160 C. n(e⁻) = 2160 ÷ 96,485 = 0.022387 mol.
- n(H₂) = 0.022387 ÷ 2 = 0.0112 mol.
- n(O₂) = 0.022387 ÷ 4 = 0.00560 mol.
(c) V = nRT / P.
- H₂: 0.011193 × 0.08206 × 298 ÷ 1.00 = 0.274 L.
- O₂: 0.0055967 × 0.08206 × 298 ÷ 1.00 = 0.137 L.
(d) The same moles of electrons pass through both electrodes. Each H₂ molecule needs 2 electrons, but each O₂ molecule releases 4. So the electrons that make one O₂ make two H₂, giving twice as many moles of hydrogen. At the same temperature and pressure, equal moles of gas take up equal volumes, so the hydrogen volume is twice the oxygen volume.
| Point | What earns it |
|---|---|
| 1 | (a) Both half-reactions correct, with oxygen at the anode |
| 1 | (b) 0.0224 mol of electrons from the charge |
| 1 | (b) Divides by 2 for H₂ and by 4 for O₂ |
| 1 | (c) Both volumes, from the ideal gas law |
| 1 | (d) Links 2 e⁻ per H₂ and 4 e⁻ per O₂ to a 2 : 1 mole ratio, and moles to volume at the same T and P |
Alternative for (a): a cathode written as 2H₂O + 2e⁻ → H₂ + 2OH⁻ is also a correct reduction producing hydrogen and earns the point, although H⁺ is the better choice in acid.
How did you do?
- Q1 or Q4 wrong: re-read “Counting electrons with an ammeter and a clock” and Figure 1 in the study guide.
- Q2, Q5 or Q7(b) wrong: practise “The half-reaction sets the electron count” and Worked example 3.
- Q3 or Q6(c) wrong: review “What changes the amount of product, and what does not”.
Then tick off the topic checklist.