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: Al 26.98, Ni 58.69, Cu 63.55, Ag 107.87 g mol⁻¹. You are not asked to label electrodes as positive or negative.
Question 1 (multiple choice · foundation)
What is the main job of the salt bridge in a galvanic cell?
- (A) It carries electrons from the anode half-cell to the cathode half-cell.
- (B) It lets ions move between the half-cells so that each half-cell stays electrically neutral.
- (C) It supplies the ions that are reduced at the cathode.
- (D) It stores charge so that the voltage stays constant as the cell runs.
Answer and explanation
Answer: (B). As the cell runs, the anode half-cell gains cations and the cathode half-cell loses them. Ions from the salt bridge move in to balance the charge, which completes the circuit and keeps the current flowing.
- (A) Electrons never travel through the salt bridge or the solutions; they move only through the wire and electrodes.
- (C) The salt-bridge ions are chosen not to react. The ions reduced at the cathode come from the cathode solution.
- (D) The salt bridge does not store charge; its job is to stop charge building up.
Question 2 (multiple choice · core)
A galvanic cell is made from a cadmium strip in Cd(NO₃)₂(aq) and a copper strip in Cu(NO₃)₂(aq), joined by a wire and by a KNO₃ salt bridge. As the cell runs, the cadmium strip loses mass. Which statement is correct?
- (A) Electrons flow from Cu to Cd in the wire, and K⁺ ions move towards the cadmium half-cell.
- (B) Electrons flow from Cd to Cu in the wire, and NO₃⁻ ions move towards the cadmium half-cell.
- (C) Electrons flow from Cd to Cu in the wire, and NO₃⁻ ions move towards the copper half-cell.
- (D) Electrons flow from Cd to Cu through the salt bridge, and K⁺ ions move towards the copper half-cell.
Answer and explanation
Answer: (B). Cadmium loses mass, so Cd → Cd²⁺ + 2e⁻ (oxidation): cadmium is the anode. Electrons leave the anode through the wire towards the copper cathode. The cadmium half-cell gains Cd²⁺ ions, so anions (NO₃⁻) move in from the salt bridge to balance the charge.
- (A) has the electron direction backwards and sends cations towards the anode.
- (C) has the right electron direction, but anions move towards the anode, not the cathode.
- (D) is right about K⁺, but electrons do not pass through the salt bridge.
Question 3 (multiple choice · core)
Molten potassium bromide, KBr(l), is electrolysed using two graphite electrodes. Which products form?
- (A) Br₂ at the anode and K at the cathode
- (B) K at the anode and Br₂ at the cathode
- (C) Br₂ at the anode and H₂ at the cathode
- (D) O₂ at the anode and K at the cathode
Answer and explanation
Answer: (A). The anode is where oxidation happens: 2Br⁻ → Br₂ + 2e⁻. The cathode is where reduction happens: K⁺ + e⁻ → K.
- (B) swaps the electrodes. Forming K from K⁺ is a reduction, so it cannot happen at the anode.
- (C) and (D) involve water. A molten salt contains no water, so neither H₂ nor O₂ can form.
Question 4 (calculation · core)
A galvanic cell is made from an aluminium strip in Al(NO₃)₃(aq) and a nickel strip in Ni(NO₃)₂(aq). After some time, the nickel strip has gained 0.880 g.
(a) Write the two half-reactions and the balanced overall equation. (b) Calculate the change in mass of the aluminium strip.
Worked solution
(a) Nickel gains mass, so Ni²⁺ is reduced at the nickel cathode; aluminium is the anode.
- Anode: Al(s) → Al³⁺(aq) + 3e⁻
- Cathode: Ni²⁺(aq) + 2e⁻ → Ni(s)
- Balance electrons (6 each): 2Al(s) + 3Ni²⁺(aq) → 2Al³⁺(aq) + 3Ni(s)
(b)
- n(Ni) = 0.880 g ÷ 58.69 g mol⁻¹ = 0.01499 mol.
- Electrons: 2 × 0.01499 = 0.02999 mol.
- n(Al) = 0.02999 ÷ 3 = 0.009996 mol (or 0.01499 × 2/3).
- Mass of Al: 0.009996 mol × 26.98 g mol⁻¹ = 0.270 g lost from the aluminium strip.
Suggested mark points (3): 1 for correct half-reactions with electrons balanced in the overall equation; 1 for using the 2 : 3 mole ratio of Al to Ni (through electrons); 1 for 0.270 g stated as a loss.
Common error: a 1 : 1 ratio gives 0.405 g.
Question 5 (constructed response · core)
A student builds a galvanic cell from a nickel strip in Ni(NO₃)₂(aq) and a silver strip in AgNO₃(aq), with a KNO₃ salt bridge and a voltmeter. The silver strip gains mass. Predict and explain the effect of each change on the cell.
(a) The KNO₃ salt bridge is replaced by one filled with KCl(aq). (b) The salt bridge is replaced by a solid glass rod dipping into both solutions. (c) The nickel strip is replaced by a much larger nickel strip, with all else the same.
Worked solution and suggested Marlbridge rubric
(a) Where the bridge dips into the silver half-cell, Cl⁻ ions from the bridge meet Ag⁺ ions and form a solid precipitate, AgCl. This removes Ag⁺, the ion reduced at the cathode, and the solid can block the bridge, so the cell works less well (the current and voltage fall).
(b) Solid glass contains no ions that can move. No ions can pass between the half-cells to balance the charge, so charge builds up in each beaker and the current through the wire stops almost at once. The rod looks like a bridge, but it does not complete the circuit.
(c) The voltage reading is unchanged, because the voltage depends on which reaction takes place (and on the concentrations), not on the size of the electrode. The cell can run for longer, because more nickel is available to be oxidised.
| Point | What earns it |
|---|---|
| 1 | (a) AgCl precipitate forms, removing Ag⁺ (or blocking the bridge) |
| 1 | (a) Consequence: cell performance (current or voltage) falls |
| 1 | (b) No moving ions in the glass, so charge builds up and the current stops |
| 1 | (c) Voltage unchanged and cell can run longer, with a reason |
Question 6 (constructed response · stretch)
A copper strip in Cu(NO₃)₂(aq) and a silver strip in AgNO₃(aq) are joined with a wire, a voltmeter and a KNO₃ salt bridge. The copper strip loses mass.
(a) Identify the anode, write the overall reaction and state whether ΔG° is positive or negative. (b) The voltmeter is now replaced by a power supply, which forces the reaction to run the other way. Identify the anode in this new set-up and write its half-reaction. (c) In the set-up from (b), the silver strip loses 0.540 g. Calculate the change in mass of the copper strip. (d) Explain, in terms of thermodynamic favourability, why the set-up in (b) needs a power supply but the set-up in (a) does not.
Worked solution and suggested Marlbridge rubric
(a) Copper loses mass, so it is oxidised: copper is the anode. Overall: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s). The cell runs by itself, so ΔG° is negative.
(b) Now silver is oxidised, so the silver strip is the anode: Ag(s) → Ag⁺(aq) + e⁻. (Copper is now the cathode: Cu²⁺ + 2e⁻ → Cu.)
(c) n(Ag) = 0.540 ÷ 107.87 = 0.005006 mol = moles of electrons. n(Cu) = 0.005006 ÷ 2 = 0.002503 mol. Mass = 0.002503 × 63.55 = 0.159 g gained by the copper strip.
(d) The reaction in (a) is thermodynamically favourable (ΔG° < 0), so it proceeds by itself and releases electrical energy: a galvanic cell. The reaction in (b) is its reverse, so ΔG° > 0; it is unfavourable and happens only while the power supply provides electrical energy: an electrolytic cell.
| Point | What earns it |
|---|---|
| 1 | (a) Copper anode, correct overall equation and ΔG° negative |
| 1 | (b) Silver identified as the new anode with Ag → Ag⁺ + e⁻ |
| 1 | (c) Uses the 2 Ag : 1 Cu ratio via electrons |
| 1 | (c) 0.159 g, stated as a gain |
| 1 | (d) Links (a) to ΔG° < 0 (galvanic) and (b) to the reverse reaction with ΔG° > 0 driven by electrical energy (electrolytic) |
Question 7 (explanation · stretch)
A galvanic cell is made from a magnesium strip in Mg(NO₃)₂(aq) and a lead strip in Pb(NO₃)₂(aq), with an NaNO₃ salt bridge. The magnesium strip loses mass. A student writes four statements:
- “Electrons flow through the wire from the lead strip to the magnesium strip.”
- “Electrons travel through the salt bridge to complete the circuit.”
- “Na⁺ ions in the salt bridge move towards the magnesium half-cell.”
- “Magnesium is the cathode, because that is where the reaction starts.”
(a) Correct each statement, giving a reason. (b) Describe what happens to particles at the surface of the lead strip as the cell runs, and what you would observe in the lab.
Model answer and suggested Marlbridge rubric
(a)
- Electrons flow from magnesium to lead. Magnesium is oxidised (Mg → Mg²⁺ + 2e⁻), so electrons leave from it.
- Electrons never pass through the salt bridge; ions move through it, and electrons move only through the wire.
- Na⁺ ions move towards the lead half-cell (the cathode), where Pb²⁺ is being removed; NO₃⁻ moves towards the magnesium half-cell.
- Magnesium is the anode, because oxidation happens there. “Where the reaction starts” is not a definition.
(b) Pb²⁺ ions in the solution reach the lead surface, each gains two electrons that have come through the wire, and becomes a Pb atom attached to the strip. Observation: the lead strip gains mass (a deposit of lead forms), while the magnesium strip gets thinner.
| Point | What earns it |
|---|---|
| 1 | Statements 1 and 4 corrected: electrons Mg → Pb, and Mg is the anode because it is oxidised |
| 1 | Statement 2 corrected: ions, not electrons, move in the salt bridge |
| 1 | Statement 3 corrected: Na⁺ moves towards the lead (cathode) half-cell, with a charge-balance reason |
| 1 | (b) Particle-level description: Pb²⁺ gains 2e⁻ at the surface to form Pb atoms |
| 1 | (b) Lab-scale observation: lead strip gains mass |
How did you do?
- Q1 or Q7(a) wrong: re-read “What each part of a galvanic cell does” in the study guide.
- Q2 or Q3 wrong: practise “anode = oxidation” first, then trace electrons and ions (Figures 1 and 2).
- Q4 or Q6(c) wrong: balance the electrons before comparing moles (Worked example 1).
- Q5 wrong: review the table “Changing the set-up: what happens?”.
Then tick off the topic checklist.