HSC Exam Practice

Sample response. A galvanic cell is an electrochemical device that converts chemical energy into electrical energy by harnessing a spontaneous redox reaction. It separates the oxidation and reduction half-reactions into two half-cells connected by an external wire (for electron flow) and a salt bridge (for ion flow).

Marking notes. 1 mark for identifying the conversion of chemical energy to electrical energy via a spontaneous redox reaction. 1 mark for describing the separation of half-reactions (two half-cells, external circuit, salt bridge) or equivalent structural description.

Sample response. The salt bridge allows ions to migrate between the two half-cells, maintaining electrical neutrality in each solution. Anions migrate toward the anode half-cell (to balance accumulating metal cations) and cations migrate toward the cathode half-cell (to replace metal cations being removed by reduction). Without the salt bridge, charge imbalance would build up in both half-cells, opposing further electron flow, and the current would drop to zero almost immediately.

Marking notes. 1 mark for identifying that the salt bridge allows ion migration. 1 mark for explaining its role is to maintain electrical neutrality (or prevent charge build-up). 1 mark for correctly describing the consequence of removal (charge imbalance stops current).

Sample response. The anode is where oxidation occurs; the electrode dissolves and decreases in mass as metal atoms lose electrons to form cations. The anode is the negative terminal of a galvanic cell. The cathode is where reduction occurs; metal cations from solution gain electrons and deposit as solid metal, so the electrode increases in mass. The cathode is the positive terminal of a galvanic cell. Mnemonic: “An ox, red cat.”

Marking notes. 1 mark for anode: oxidation, negative polarity. 1 mark for anode mass change (decreases — metal dissolves). 1 mark for cathode: reduction, positive polarity. 1 mark for cathode mass change (increases — metal deposits). All four required for full marks.

Sample response. Cell notation lists components from left to right in the order: anode | anode solution || cathode solution | cathode. A single vertical bar (|) represents a phase boundary between the electrode and its solution. The double vertical bar (||) represents the salt bridge. Example: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

Marking notes. 1 mark for correctly stating the order (anode on left, cathode on right) with single and double bar symbols. 1 mark for a correct worked example of cell notation.

Sample response. When zinc is placed directly in copper sulfate solution, electrons transfer directly from zinc atoms to copper cations at the surface of the zinc. No pathway separates the oxidation and reduction sites, so no directed electron flow occurs and the energy is released as heat. A galvanic cell separates the zinc (anode) and copper sulfate (cathode half-cell) into different containers. Because electrons cannot travel through solution, they are forced to flow through the external wire from the anode to the cathode — this directed electron flow is electrical current. The chemical energy of the spontaneous redox reaction is therefore converted to useful electrical work rather than heat.

Marking notes. 1 mark for explaining that in the direct reaction electrons transfer directly between reactants, not through a wire, so no useful electrical work is done (energy released as heat). 1 mark for explaining that the galvanic cell separates the half-reactions so electrons must travel through the external wire. 1 mark for linking directed electron flow through the wire explicitly to electrical current / electrical energy output.

Sample response. Anode (zinc): zinc metal oxidises (Zn → Zn²⁺ + 2e⁻). Anode mass decreases as zinc dissolves. Concentration of Zn²⁺ ions in the anode solution increases. Cathode (copper): Cu²⁺ ions are reduced and deposit as solid copper (Cu²⁺ + 2e⁻ → Cu). Cathode mass increases. Concentration of Cu²⁺ ions in the cathode solution decreases. Also: anions from the salt bridge migrate into the anode half-cell; cations from the salt bridge migrate into the cathode half-cell. Cell voltage decreases over time as concentrations shift.

Marking notes. 1 mark for anode change (mass decreases + Zn²⁺ concentration increases). 1 mark for cathode change (mass increases + Cu²⁺ concentration decreases). 1 mark for correct half-equation at each electrode (written or described). 1 mark for describing ion migration in the salt bridge OR stating that cell voltage decreases over time with explanation.

Part (a) — 3 marks. Zn/Cu E°cell = +1.10 V; Mg/Ag E°cell = +3.17 V — a difference of 2.07 V [1]. The large difference arises because magnesium has a much more negative standard reduction potential (−2.37 V) than zinc (−0.76 V): magnesium is more readily oxidised, so it drives a larger difference between the anode and cathode potentials [1]. Silver has a more positive E° (+0.80 V) than copper (+0.34 V), further increasing the gap [1].

Part (b) — 4 marks. Highest E°cell: Mg/Ag at +3.17 V [1]. Cell notation: Mg(s) | Mg²⁺(aq) || Ag⁺(aq) | Ag(s) [1]. Anode (oxidation): Mg(s) → Mg²⁺(aq) + 2e⁻. Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s). Electron balance: cathode involves 1e⁻, anode 2e⁻ — multiply cathode × 2: 2Ag⁺(aq) + 2e⁻ → 2Ag(s) [1]. Overall: Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s) [1].

Part (c) — 1 mark. Acceptable responses: (i) Silver is a rare and expensive metal, making large-scale commercial batteries using a silver electrode prohibitively costly. (ii) Magnesium is a highly reactive metal that reacts vigorously with water and many electrolyte solutions, creating engineering challenges in preventing unwanted side reactions. Either valid limitation accepted for 1 mark.

Part (a) — 2 marks. Ag⁺/Ag has the more positive E° (+0.80 V) → more readily reduced → silver is the cathode [1]. Fe²⁺/Fe has the more negative E° (−0.44 V) → more readily oxidised → iron is the anode [1].

Part (b) — 2 marks. Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻ [half-mark]. Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s) [half-mark]. The anode releases 2e⁻ and the cathode consumes 1e⁻ per Ag⁺; multiply cathode half-equation by 2 to balance electrons: 2Ag⁺(aq) + 2e⁻ → 2Ag(s) [1 mark for identifying the step and performing it].

Part (c) — 2 marks. Add balanced half-equations (cancel 2e⁻): Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s) [1]. Charge check: left = 0 + 2(+1) = +2; right = +2 + 0 = +2. ✓ [1].

Part (d) — 1 mark. E°cell = E°cathode − E°anode = +0.80 − (−0.44) = +0.80 + 0.44 = +1.24 V. E°cell > 0 → spontaneous. A positive E°cell indicates that the forward reaction is thermodynamically favoured under standard conditions [1 mark for correct value + spontaneity statement + sign interpretation].

Part (e) — 1 mark. Standard reduction potentials apply under standard conditions: 25°C, 1.00 mol/L for all dissolved species, 1 atm for gases. In real laboratory conditions, concentrations will differ from 1.00 mol/L, temperature may not be exactly 25°C, and junction potentials at the salt bridge may affect the measured voltage. Any one valid assumption for 1 mark.

Sample response. The fundamental electrochemical principles governing a Daniell cell apply fully to a lithium-ion (Li-ion) battery used in an Australian EV: both are galvanic cells in which a spontaneous redox reaction drives electron flow through an external circuit. During discharge of the Li-ion battery, oxidation occurs at the graphite anode (lithium ions are released: LiC₆ → C₆ + Li⁺ + e⁻, simplified) and reduction occurs at the lithium cobalt oxide cathode (Li⁺ + CoO₂ + e⁻ → LiCoO₂), exactly as the Daniell cell has oxidation at the zinc anode (Zn → Zn²⁺ + 2e⁻) and reduction at the copper cathode (Cu²⁺ + 2e⁻ → Cu). Electrons in both cells flow from anode to cathode through the external circuit, constituting the electrical current that charges the EV battery or powers the motor. Ions also migrate internally through the electrolyte in both systems: in the Li-ion cell, Li⁺ ions travel through the liquid organic electrolyte from the anode to the cathode during discharge (the equivalent of ion migration through the salt bridge in the Daniell cell), maintaining electrical neutrality. The higher cell voltage of a Li-ion cell (~3.7 V vs ~1.10 V for Daniell) is explained by the concept of standard reduction potential: E°cell = E°cathode − E°anode. The lithium/graphite anode has a more negative effective E° (stronger tendency to be oxidised) than zinc, while the cobalt oxide cathode has a more positive effective E° than copper, giving a larger E° difference and thus higher voltage. One significant difference is reversibility: the Daniell cell is not practically rechargeable because the electrode materials (zinc dissolving; copper depositing) cannot be reliably restored by applying a reverse voltage. The Li-ion cell is rechargeable because the Li⁺ insertion/extraction process in both electrodes is structurally reversible — applying an external voltage drives Li⁺ back into the graphite anode, restoring the original compositions. This reversibility is what makes Li-ion batteries practical for EV applications where thousands of charge/discharge cycles are required.

Marking notes.

• 1 mark — Identifies both systems as galvanic cells in which a spontaneous redox reaction drives electron flow through an external circuit during discharge.
• 1 mark — Correctly identifies oxidation at the anode and reduction at the cathode in the Li-ion cell (simplified half-equations or description acceptable).
• 1 mark — Correctly identifies the same oxidation/reduction pattern in the Daniell cell and explicitly draws the parallel.
• 1 mark — Correctly describes electron flow from anode to cathode through the external circuit in both cells.
• 1 mark — Correctly describes ion migration through the electrolyte (Li-ion cell) as functionally equivalent to ion migration through the salt bridge (Daniell cell), maintaining electrical neutrality.
• 1 mark — Applies E°cell = E°cathode − E°anode to explain why the Li-ion voltage (~3.7 V) is higher than the Daniell cell (1.10 V), with reference to the larger difference in standard reduction potentials.
• 1 mark — Identifies and explains at least one significant difference between the two systems (e.g. reversibility/rechargeability, cell voltage magnitude, electrode materials, electrolyte type).
• 1 mark — Response is coherent and evaluative: reaches an explicit conclusion about the extent to which Daniell cell principles apply to the Li-ion EV battery (e.g. “the core principles apply fully — both are galvanic cells; the key difference is reversibility and scale of voltage”).