Secondary 3 Chemistry Redox Electrochemistry Quiz

Secondary 3 Chemistry Quiz - Redox Electrochemistry: Answer Key

Total Marks: 40 marks

Section A: Multiple Choice and Short Response

Question 1 [1 mark]

Answer: B (+6)

Explanation: To find the oxidation state of sulfur in H₂SO₄:

• Hydrogen has oxidation state +1 (rule: +1 in compounds, except with metals)
• Oxygen has oxidation state −2 (rule: −2 in compounds, except peroxides)
• The compound is neutral, so sum of oxidation states = 0

Let oxidation state of S = x: $2(+1) + x + 4(-2) = 0$ $2 + x - 8 = 0$ $x - 6 = 0$ $x = +6$

Marking note: Deduct mark if working not shown or arithmetic error made.

Question 2 [1 mark]

Answer: B (Zinc is oxidised from 0 to +2)

Explanation: In the reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

• Zinc starts as elemental metal: oxidation state = 0
• In zinc sulfate, Zn²⁺ ion has oxidation state = +2
• Increase in oxidation state = oxidation (loss of electrons)
• Zinc loses 2 electrons: Zn → Zn²⁺ + 2e⁻

This is a displacement reaction where more reactive zinc displaces less reactive copper. The oxidation state increases because electrons are lost to copper ions.

Common mistake: Students confuse oxidation and reduction. Use OILRIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

Question 3 [1 mark]

Answer: Oxidation is the loss of electrons.

Explanation: Oxidation and reduction can be defined in three ways:

1. Electron transfer: Oxidation is loss of electrons (OIL)
2. Oxygen gain: Oxidation is gain of oxygen
3. Hydrogen loss: Oxidation is loss of hydrogen

For ionic/electrochemical contexts, the electron transfer definition is most precise and universally applicable. When a species loses electrons, it becomes more positively charged (or less negatively charged), which we describe as an increase in oxidation state.

Question 4 [3 marks]

(a) [1 mark] Answer: −1

Explanation: In H₂O₂, using normal rules would give oxygen as −2, but this is a peroxide (contains O−O single bond). For peroxides, the special rule applies: oxygen has oxidation state −1.

Calculation: 2(+1) + 2(x) = 0, so x = −1.

(b) [2 marks] Answer: In H₂O₂, oxygen has an intermediate oxidation state of −1, between the values in O₂ (0) and typical oxides (−2). Therefore oxygen in H₂O₂ can be:

• Reduced to −2 (acting as oxidising agent, gaining electrons)
• Oxidised to 0 (acting as reducing agent, losing electrons)

Explanation for oxidation half-equation: H₂O₂ → O₂ + 2H⁺ + 2e⁻ (oxygen oxidised from −1 to 0)

Explanation for reduction half-equation: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (oxygen reduced from −1 to −2)

Marking breakdown:

• [1 mark] for explaining that −1 is an intermediate oxidation state
• [1 mark] for identifying both directions of change with correct species

Question 5 [2 marks]

Answer: $\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$

Working:

• Given reduction: Cl₂ + 2e⁻ → 2Cl⁻ (chlorine gains electrons, is reduced)
• Oxidation of bromide: 2Br⁻ → Br₂ + 2e⁻ (bromide loses electrons, is oxidised)
• Electrons cancel: add the half-equations directly

Explanation: This is a displacement reaction where more reactive chlorine (stronger oxidising agent) displaces less reactive bromine from its compound. The E° value for Cl₂/Cl⁻ (+1.36 V) is more positive than Br₂/Br⁻ (+1.07 V), so chlorine has greater tendency to be reduced.

Marking breakdown:

• [1 mark] for correct formula of products and reactants
• [1 mark] for correctly balanced equation (species and charge)

Section B: Structured Questions

Question 6 [3 marks]

(a) [2 marks] Answer: Magnesium is oxidised. It loses two electrons to form Mg²⁺ ions: Mg → Mg²⁺ + 2e⁻. The oxidation state increases from 0 to +2.

Explanation:

• In elemental Mg, oxidation state = 0
• In MgCl₂, Mg has oxidation state = +2
• Increase in oxidation state indicates oxidation
• Magnesium is more reactive than hydrogen, so it displaces H⁺ from the acid

Marking breakdown:

• [1 mark] for identifying Mg as oxidised
• [1 mark] for correct explanation with electron loss/oxidation state change

(b) [1 mark] Answer: 2H⁺ + 2e⁻ → H₂

Explanation: Hydrogen ions from the acid gain electrons (reduction). This is the reduction half-equation. Two protons combine with two electrons to form one molecule of hydrogen gas.

Question 7 [3 marks]

(a) [1 mark] Answer: The salt bridge completes the circuit by allowing ions to flow between the two half-cells, maintaining electrical neutrality.

Explanation: Without the salt bridge, positive charge would build up in the anode compartment (as Zn²⁺ enters solution) and negative charge would build up in the cathode compartment (as Cu²⁺ is removed). The salt bridge allows NO₃⁻ ions to flow toward the anode and K⁺ ions to flow toward the cathode, balancing charge without mixing solutions.

(b) [1 mark] Answer: Electrons flow from zinc electrode (negative) to copper electrode (positive), or left to right as drawn.

Explanation: Zinc is more reactive than copper, so zinc metal is oxidised (loses electrons). Electrons travel through the external wire from the zinc half-cell to the copper half-cell, where Cu²⁺ ions are reduced to Cu metal.

(c) [1 mark] Answer: Cu²⁺ + 2e⁻ → Cu

Explanation: At the copper electrode (cathode), copper ions gain electrons and are reduced to copper metal. This is where reduction occurs because copper has a more positive reduction potential than zinc.

Question 8 [4 marks]

(a) [2 marks] Answer: A standard hydrogen electrode (SHE) consists of hydrogen gas at 1 atm pressure bubbled over a platinum electrode coated with platinum black, immersed in a solution containing H⁺ ions at 1 mol/dm³ concentration, at 298 K (25°C). By convention, its electrode potential is defined as 0.00 V.

Marking breakdown:

• [1 mark] for correct physical description (H₂ gas, Pt electrode, H⁺ at 1 mol/dm³)
• [1 mark] for standard conditions (298 K, 1 atm, 1 mol/dm³) and E° = 0.00 V

(b) [2 marks] Answer: Yes, Fe³⁺ will oxidise I⁻ to I₂. E°cell = E°(cathode) − E°(anode) = +0.77 − (+0.54) = +0.23 V. Since E°cell is positive, the reaction is feasible.

Explanation:

• Fe³⁺/Fe²⁺ has more positive E° (+0.77 V) than I₂/I⁻ (+0.54 V)
• Therefore Fe³⁺ is the stronger oxidising agent (more tendency to be reduced)
• Fe³⁺ will accept electrons from I⁻, oxidising I⁻ to I₂ while being reduced to Fe²⁺

Marking breakdown:

• [1 mark] for correct prediction with reasoning about E° values
• [1 mark] for calculating or stating that E°cell is positive

Question 9 [3 marks]

(a) [1 mark] Answer: From +4 to +2, a decrease of 2.

Working:

• In MnO₂: x + 2(−2) = 0, so x = +4
• In MnCl₂: x + 2(−1) = 0, so x = +2
• Change: +4 → +2

(b) [2 marks] Answer:

• Manganese is reduced: oxidation state decreases from +4 to +2, gaining electrons
• Chlorine is oxidised: in HCl, Cl has oxidation state −1; in Cl₂, Cl has oxidation state 0. Some chloride ions lose electrons to form chlorine gas

Explanation: Not all HCl is oxidised—only the chloride ions that form Cl₂. The chloride in MnCl₂ remains at −1. This is a disproportionation-related observation where HCl acts as both acid source and reducing agent.

Marking breakdown:

• [1 mark] for identifying manganese reduction with oxidation state evidence
• [1 mark] for identifying chlorine oxidation with correct oxidation states

Question 10 [3 marks]

(a) [1 mark] Answer: Graphite electrodes are inert because they do not react with the molten electrolyte or products; they merely provide a surface for electron transfer without being consumed or altered.

(b) (i) [1 mark] Answer: Pb²⁺ + 2e⁻ → Pb

Explanation: Lead ions are attracted to the negative cathode, where they gain electrons and are reduced to molten lead metal.

(ii) [1 mark] Answer: 2Br⁻ → Br₂ + 2e⁻

Explanation: Bromide ions are attracted to the positive anode, where they lose electrons and are oxidised to bromine vapour.

Question 11 [3 marks]

(a) [1 mark] Answer: This is electroplating, because a thin layer of one metal (copper) is deposited onto another metal object (the rod) using electricity.

(b) [2 marks] Answer: Iron is more reactive than copper, so if the iron surface is not perfectly clean and oxide-free, or if the electrolyte composition is not carefully controlled, the copper deposit may be powdery, non-uniform, or poorly adherent. Iron may also form a thin oxide layer that prevents direct metal-to-metal bonding.

Marking breakdown:

• [1 mark] for noting iron's greater reactivity or oxide layer issues
• [1 mark] for explaining consequence (poor adhesion, powdery deposit)

Question 12 [2 marks]

(a) [1 mark] Answer: Mg > Zn > Fe > Cu (most reactive to least reactive)

Explanation from results:

• Mg displaces Cu (Mg most reactive so far)
• Cu cannot displace Zn (Zn more reactive than Cu)
• Zn displaces Fe (Zn more reactive than Fe, but we need to place Fe relative to Mg)

Order: Mg > Zn > Fe > Cu

(b) [1 mark] Answer: Copper is less reactive than zinc, so copper cannot displace zinc from zinc nitrate; zinc ions have greater tendency to remain as ions than copper ions have to displace them.

Question 13 [3 marks]

(a) [1 mark] Answer: 2:1 (hydrogen:oxygen)

(b) [2 marks] Answer: At cathode: 4H⁺ + 4e⁻ → 2H₂ (two molecules of H₂ from 4 electrons) At anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (one molecule of O₂ from 4 electrons)

Since the same quantity of electricity (4 moles of electrons) produces 2 volumes of H₂ and 1 volume of O₂, the ratio is 2:1.

Alternative using water as source: Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻

To match electrons: multiply cathode equation by 2: 4H₂O + 4e⁻ → 2H₂ + 4OH⁻

Now 4 electrons produce 2H₂ and 1O₂, giving 2:1 ratio.

Marking breakdown:

• [1 mark] for correct relevant half-equations
• [1 mark] for electron matching leading to 2:1 ratio

Question 14 [3 marks]

(a) [1 mark] Answer: The bicycle handlebar (steel) is made the cathode (negative electrode).

(b) [2 marks] Answer: A low current for longer time allows controlled, gradual deposition of chromium ions, permitting them to arrange regularly in the metallic lattice and settle smoothly on the surface. A high current causes rapid deposition, producing a rough, brittle, nodular coating with internal stress and poor adhesion.

Marking breakdown:

• [1 mark] for explaining that low current allows ordered/gradual deposition
• [1 mark] for explaining that high current causes rapid, uneven, rough deposition

Question 15 [4 marks]

(a) (i) [1 mark] Answer: −1

Working: Let oxidation state of N = x. In NH₂OH: x + 2(+1) + (−2) + (+1) = 0, so x + 2 − 2 + 1 = 0, thus x = −1.

(Alternatively: x + 2 + (−2) + 1 = 0 for H₂N-O-H structure)

(ii) [1 mark] Answer: +1

Working: In N₂O: 2x + (−2) = 0, so 2x = +2, x = +1.

(b) [2 marks] Answer: Nitrogen's oxidation state increases from −1 in NH₂OH to +1 in N₂O (increase of 2 per N atom, or 4 overall for N₂O formation). An increase in oxidation state indicates loss of electrons, which is oxidation. The species that is oxidised causes reduction in another species, so hydroxylamine acts as a reducing agent.

Marking breakdown:

• [1 mark] for correct oxidation state change (−1 to +1)
• [1 mark] for linking oxidation to reducing agent function

Section C: Extended Response

Question 16 [3 marks]

Answer structure and marking guidance:

Electrochemical corrosion process [2 marks]:

• Anodic region (iron surface impurities/stressed areas): Fe → Fe²⁺ + 2e⁻ (oxidation, iron loses electrons) [1 mark]
• Cathodic region (oxygen-rich areas): O₂ + 2H₂O + 4e⁻ → 4OH⁻ (reduction) [1 mark]
• Fe²⁺ and OH⁻ combine, further oxidise to Fe(OH)₃, which dehydrates to rust (Fe₂O₃·nH₂O)

Sacrificial protection with zinc [1 mark]:

• Zinc is more reactive than iron (higher in reactivity series), so zinc is oxidised preferentially: Zn → Zn²⁺ + 2e⁻
• Zinc sacrifices itself, protecting iron from oxidation; electrons flow from zinc to iron, keeping iron as the cathode in local cells
• Zinc blocks must be replaced periodically as they corrode away

Common mistakes to flag:

• Students describe zinc as "coating" rather than understanding electrochemical protection
• Confusing anode and cathode in rusting mechanism
• Not mentioning that zinc is oxidised preferentially due to higher reactivity

Question 17 [2 marks]

(a) [1 mark] Answer: The silver anode dissolves as Ag → Ag⁺ + e⁻, releasing silver ions into the solution to replace those deposited on the cathode, maintaining constant Ag⁺ concentration in the electrolyte.

(b) [1 mark] Answer: $n = \frac{m}{M_r} = \frac{0.67}{108} = 6.2 \times 10^{-3} \text{ mol} = 0.0062 \text{ mol}$

Or approximately 6.204 × 10⁻³ mol

Calculation: $\frac{0.67}{108} = 0.0062037... \approx 6.20 \times 10^{-3} \text{ mol}$

Question 18 [3 marks]

(a) [1 mark] Answer: In SO₂: +4; In SO₄²⁻: +6

Working for SO₂: x + 2(−2) = 0, x = +4 Working for SO₄²⁻: x + 4(−2) = −2, x − 8 = −2, x = +6

(b) [2 marks] Answer: The oxidising agent is Cr₂O₇²⁻ (potassium dichromate(VI) / dichromate ion).

Explanation: Chromium's oxidation state decreases from +6 in Cr₂O₇²⁻ to +3 in Cr³⁺. A decrease in oxidation state indicates gain of electrons (reduction). The species that is reduced is the oxidising agent because it causes oxidation in another species (SO₂ to SO₄²⁻, where sulfur increases from +4 to +6).

Marking breakdown:

• [1 mark] for correct identification with oxidation state evidence for Cr
• [1 mark] for explaining the oxidising agent is the species being reduced, linking to oxidation of sulfur

Question 19 [3 marks]

(a) [2 marks] Answer: $2\text{MnO}_4^- + 16\text{H}^+ + 5\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$

Working:

• MnO₄⁻ equation needs 5e⁻, H₂C₂O₄ equation produces 2e⁻
• LCM of 5 and 2 = 10 electrons
• Multiply MnO₄⁻ equation by 2 (10 electrons, 2 MnO₄⁻, 16 H⁺, 2 Mn²⁺, 8 H₂O)
• Multiply H₂C₂O₄ equation by 5 (10 electrons, 5 H₂C₂O₄, 10 CO₂, 10 H⁺)
• Add and simplify: 16H⁺ − 10H⁺ (from products side when reversing) = 6H⁺ remaining, but check carefully:

Actually: 2 × [MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] gives 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O

5 × [H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻] gives 5H₂C₂O₄ → 10CO₂ + 10H⁺ + 10e⁻

Adding: 2MnO₄⁻ + 16H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂ + 10H⁺

Simplify: 2MnO₄⁻ + 6H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂

Check atoms and charges balance on both sides.

Marking breakdown:

• [1 mark] for correct multiplication to balance electrons
• [1 mark] for correctly simplified final equation with all species balanced

(b) [1 mark] Answer: Purple/pink to colourless (very pale pink) / or decolourised

Explanation: MnO₄⁻ ions are intensely purple. As they are reduced to nearly colourless Mn²⁺ ions (very pale pink, appears colourless in dilute solution), the purple colour disappears. This colour change makes the reaction self-indicating.

Question 20 [3 marks]

Answer structure with marking guidance:

Marking breakdown [3 marks]:

• [1 mark] for correct identification and comparison of oxidising agents
• [1 mark] for clear contrast of reaction conditions (temperature, pH, time, catalyst/enzyme)
• [1 mark] for environmental comparison: chemical method produces toxic waste vs biochemical method being green/sustainable with oxygen/water as by-products

Teaching note: The biochemical method is the basis of traditional vinegar production. Acetobacter bacteria oxidise ethanol to ethanoic acid aerobically. Industrially, this is done in aerated tanks (acetifiers) at ~30°C, with the bacteria living as a biofilm on wood shavings or synthetic packing material.

END OF ANSWER KEY