HSC Exam Practice

Sample response. The standard reduction potential (E°) is the voltage of a half-cell measured under standard conditions (25°C, 1 mol L⁻¹ all aqueous species, 1 atm for gases) relative to the standard hydrogen electrode (E° = 0.00 V). A positive E° indicates the species is readily reduced (has a high tendency to gain electrons — strong oxidising agent). A negative E° indicates the species is more readily oxidised than H₂ (has a lower tendency to gain electrons; the species in the oxidised form in the half-reaction is a stronger reducing agent than H⁺).

Marking notes. 1 mark: defines E° as a voltage measured under standard conditions vs the SHE. 1 mark: positive E° = readily reduced / strong oxidant. 1 mark: negative E° = more easily oxidised than H₂ / stronger reducing agent.

Sample response. An active electrode participates directly in the half-reaction — its mass changes as the cell operates (dissolves at the anode or has metal deposited at the cathode). Example: Zn(s) → Zn²⁺(aq) + 2e⁻ requires a zinc electrode because solid zinc is consumed. An inert electrode does not participate in the half-reaction — it only provides a conducting surface; its mass does not change. Example: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ requires a platinum (inert) electrode because both species are aqueous — no solid electrode material is produced or consumed.

Marking notes. 1 mark: active electrode definition (participates, mass changes). 1 mark: correct example with justification. 1 mark: inert electrode definition (does not participate, mass unchanged). 1 mark: correct example with justification (half-reaction must have both species as aqueous/gas).

Sample response. For dissolution in dilute HCl: E°cell = E°(H⁺/H₂) − E°(Mⁿ⁺/M) = 0.00 − E°(M). (a) Copper: E°cell = 0.00 − (+0.34) = −0.34 V < 0 → non-spontaneous; Cu does NOT dissolve. (b) Zinc: E°cell = 0.00 − (−0.76) = +0.76 V > 0 → spontaneous; Zn dissolves. (c) Iron: E°cell = 0.00 − (−0.44) = +0.44 V > 0 → spontaneous; Fe dissolves.

Marking notes. 1 mark per metal: correct E°cell calculation AND correct conclusion. Calculation must be shown for full marks.

Sample response. E°(Zn²⁺/Zn) = −0.76 V is more negative than E°(Fe²⁺/Fe) = −0.44 V. When zinc is electrically connected to steel in seawater (the electrolyte), a galvanic cell forms. Zinc, with the lower E°, is preferentially oxidised and acts as the anode: Zn(s) → Zn²⁺(aq) + 2e⁻. Iron is forced to be the cathode (where reduction occurs); iron cannot be oxidised while it functions as the cathode. The iron structure is protected from corrosion because the oxidation reaction always occurs at the zinc anode instead. The zinc sacrifices itself — it corrodes away — while the iron remains intact. E°cell = −0.44 − (−0.76) = +0.32 V > 0, confirming the galvanic cell is spontaneous.

Marking notes. 1 mark: compare E° values and identify zinc as having more negative E°. 1 mark: zinc is the anode and is oxidised (correct half-equation). 1 mark: iron is the cathode and therefore cannot be oxidised/corrode. 1 mark: calculate E°cell = +0.32 V and confirm spontaneity.

Sample response. Cell notation convention: anode on the left, cathode on the right; | = phase boundary; || = salt bridge. Cell notation: Pt | Fe²⁺(aq), Fe³⁺(aq) || MnO₄⁻(aq), Mn²⁺(aq), H⁺(aq) | Pt. E°cell = E°cathode − E°anode = +1.51 − (+0.77) = +0.74 V.

Marking notes. 1 mark: correct cell notation with Pt on both sides and anode left/cathode right. 1 mark: state the convention (anode left, | for phase, || for salt bridge). 1 mark: E°cell = +0.74 V with correct calculation.

Sample response. Despite the thermodynamic prediction (E°cell = +1.66 V > 0), aluminium reacts rapidly with oxygen in air to form a thin, dense, strongly adhering aluminium oxide (Al₂O₃) passivation layer on its surface. This layer is impermeable to further oxidation and prevents the underlying aluminium from reacting with dilute acid. The E° table predicts thermodynamic feasibility but cannot predict whether kinetic barriers such as passivation layers will prevent the reaction.

Marking notes. 1 mark: formation of a protective Al₂O₃ passivation layer on the surface. 1 mark: this kinetic barrier prevents further reaction even though E° predicts thermodynamic feasibility.

Sample response (a). The cell voltage decreases approximately linearly as temperature increases from 10°C to 60°C, dropping from 1.09 V to 1.02 V (a decrease of 0.07 V over 50°C). At 25°C, interpolating from the graph, the measured cell voltage is approximately 1.07–1.08 V.

Marking notes (a). 1 mark: describes the downward linear trend with reference to values or range. 1 mark: estimated value at 25°C within 0.02 V of the interpolated value (~1.07–1.08 V).

Sample response (b). The standard concentration of 1 mol L⁻¹ may not be exactly achieved in the classroom (ion concentrations drift as the cell operates, or may not be precisely 1 mol L⁻¹ to begin with). Reduced ion concentration changes the Nernst-equivalent voltage, reducing the measured cell voltage below E°cell. Internal resistance of the salt bridge and solution also causes a voltage drop under load. Accept also: temperature not at standard 25°C; electrode surface impurities.

Marking notes (b). 1 mark: identifies one valid non-standard condition (concentration, temperature, internal resistance). 1 mark: correctly explains how that condition reduces measured voltage below E°cell.

Sample response (c). Both electrodes in the Daniell cell are active. The zinc anode dissolves (Zn(s) → Zn²⁺(aq) + 2e⁻), so its mass decreases. The copper cathode has copper metal deposited onto it (Cu²⁺(aq) + 2e⁻ → Cu(s)), so its mass increases. Neither electrode is inert — both participate directly in the half-reactions.

Marking notes (c). 1 mark: both electrodes correctly identified as active with correct mass changes (zinc loses mass, copper gains mass). 1 mark: explains that neither is inert because both solid metals participate in the half-reactions.

Sample response (a). Fe³⁺/Fe²⁺: E° = +0.77 V. Cu²⁺/Cu: E° = +0.34 V. Cu²⁺/Cu has the more negative E° (+0.34 V < +0.77 V), so the Cu/Cu²⁺ half-reaction is preferentially oxidised — Half-cell B (copper) is the anode. Wait: +0.34 V is less positive than +0.77 V, so Cu²⁺/Cu has the lower E° → Cu is the anode (oxidised). Fe³⁺/Fe²⁺ has the higher E° (+0.77 V) → Half-cell A is the cathode (Fe³⁺ reduced to Fe²⁺). Anode (Half-cell B): Cu(s) → Cu²⁺(aq) + 2e⁻; active electrode (solid copper dissolves). Cathode (Half-cell A): Fe³⁺(aq) + e⁻ → Fe²⁺(aq); inert platinum electrode (both Fe³⁺ and Fe²⁺ are aqueous).

Marking notes (a). 1 mark: correctly identifies Half-cell A (Fe³⁺/Fe²⁺) as cathode and Half-cell B (Cu²⁺/Cu) as anode with E° justification. 1 mark: correct half-reactions written for each. 1 mark: correctly identifies active (Cu) and inert (Pt) electrodes with justification.

Sample response (b). E°cell = E°cathode − E°anode = +0.77 − (+0.34) = +0.43 V. Electron balance: anode produces 2e⁻; cathode consumes 1e⁻, so multiply cathode by 2. Overall equation: Cu(s) + 2Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺(aq).

Marking notes (b). 1 mark: E°cell = +0.43 V correctly calculated. 1 mark: balanced overall equation with electrons correctly cancelled (Cu + 2Fe³⁺ → Cu²⁺ + 2Fe²⁺).

Sample response (c). The platinum electrode does not change mass because the half-reaction at the cathode is Fe³⁺(aq) + e⁻ → Fe²⁺(aq). Both reactant (Fe³⁺) and product (Fe²⁺) remain in aqueous solution — no solid iron is deposited on the platinum surface and no platinum dissolves. The platinum simply conducts electrons from the external circuit into the solution, where Fe³⁺ ions accept them. Platinum is inert in this context: it does not participate chemically in the half-reaction.

Marking notes (c). 1 mark: identifies that both Fe species are aqueous so no solid is deposited or dissolved at the electrode. 1 mark: links this to the definition of an inert electrode — platinum only conducts electrons and does not participate in the half-reaction.

Sample response. Standard reduction potentials (E° values) are a powerful quantitative tool for predicting which half-reaction will be the cathode (more positive E°) and which will be the anode (more negative E°) in any galvanic cell, and for calculating whether a cell will be spontaneous (E°cell = E°cathode − E°anode; spontaneous if positive). They also determine whether an active or inert electrode is needed by identifying which species are involved in each half-reaction.

For a cell with an active electrode — for example, the Daniell cell (Zn|Zn²⁺||Cu²⁺|Cu) — E° values correctly predict that zinc (E° = −0.76 V) is the anode and copper (E° = +0.34 V) is the cathode, giving E°cell = +1.10 V. This is a spontaneous cell. The zinc electrode dissolves and the copper electrode gains mass, as the half-reactions involve solid metals.

For a cell with an inert electrode — for example, the Fe²⁺/Fe³⁺–MnO₄⁻/Mn²⁺ cell — E° values correctly predict that MnO₄⁻/Mn²⁺ (E° = +1.51 V) is the cathode and Fe²⁺/Fe³⁺ (E° = +0.77 V) is the anode, giving E°cell = +0.74 V. Both half-reactions involve only aqueous ions, so platinum inert electrodes are required — E° values alone do not reveal this requirement; it must be deduced by examining whether solid electrode material is produced or consumed.

However, E° values have limitations. They predict thermodynamic feasibility under standard conditions (25°C, 1 mol L⁻¹, 1 atm), not actual cell behaviour under real conditions. A notable example is aluminium: E°(Al³⁺/Al) = −1.66 V predicts aluminium should dissolve readily in dilute acid (E°cell = +1.66 V), yet in practice aluminium is protected by a dense Al₂O₃ passivation layer that kinetically inhibits the reaction. The E° table predicts thermodynamic spontaneity but not the rate, and cannot account for kinetic barriers such as passivation, electrode surface fouling, or activation energy barriers.

In summary, E° values are essential for predicting the direction of redox reactions, assigning anode and cathode roles, and calculating cell voltage. They reliably guide electrode material selection (active vs inert) by clarifying what species are involved in the half-reaction. Their limitation is that they describe equilibrium thermodynamics, not kinetic feasibility — a cell with a large positive E°cell may still react slowly or not at all if kinetic barriers dominate.

Marking criteria.

• 1 mark — Correctly states the prediction rule: E°cell = E°cathode − E°anode; spontaneous when positive; more positive E° = cathode.
• 1 mark — Analyses a galvanic cell with an active electrode (e.g. Daniell cell) and correctly assigns anode and cathode, using E° values, and describes electrode mass changes.
• 1 mark — Analyses a galvanic cell with an inert electrode, identifies the requirement for platinum (or graphite), and links the need for inert electrodes to the absence of solid metal species in the half-reaction.
• 1 mark — Describes a situation where E° predictions do not align with observed behaviour (e.g. aluminium passivation, or non-standard concentrations reducing measured voltage below E°cell).
• 1 mark — Explains that E° values predict thermodynamic spontaneity but not reaction rate or kinetic feasibility.
• 1 mark — Provides an overall analytical conclusion that integrates both the utility and the limitation of E° values as a predictive tool.
• 1 mark — Response uses precise lesson terminology throughout (standard reduction potential, anode, cathode, active/inert electrode, spontaneous, E°cell, passivation/kinetic barrier) and is logically structured.