Year 11 Chemistry Module 3 Checkpoint 2

Checkpoint 2

Covering Lessons 07–10: metal activity series, displacement reactions, redox reactions, oxidation states, half-equations, galvanic cells, standard reduction potentials, inert electrodes, and cathodic protection.

⏱ ~20 min 10 MC · 4 Short Answer Lessons 07–10

What’s Covered

L07
Metal Activity Series
  • Reactivity and atomic properties
  • NESA activity series order
  • Displacement reactions
  • Predicting reactions
  • Sacrificial protection
L08
Redox & Oxidation States
  • OIL RIG definition
  • Oxidation numbers (7 rules)
  • Half-equations (6-step method)
  • Combining half-equations
  • Bleach chemistry
L09
Galvanic Cells
  • Cell components & function
  • Standard reduction potentials
  • E°cell = E°cathode − E°anode
  • Predicting spontaneity
  • Li-ion batteries
L10
Inert Electrodes & Cathodic Protection
  • Active vs inert electrodes
  • Predicting from E° table
  • Sacrificial anode protection
  • ICCP systems
  • Metals in acid

Multiple Choice — 10 marks

L07 — Activity Series

1. According to the NESA activity series, which metal is the most reactive?

A Magnesium (Mg)
B Calcium (Ca)
C Potassium (K)
D Sodium (Na)
L07 — Displacement Reactions

2. An iron nail is placed in copper(II) sulfate solution. Which observation would be expected?

A No reaction occurs; iron is less reactive than copper
B A red-brown solid forms on the nail and the blue colour of the solution fades
C A blue precipitate forms in the solution
D Hydrogen gas bubbles are produced at the nail surface
L08 — OIL RIG / Redox

3. In the reaction 2Mg(s) + O₂(g) → 2MgO(s), which statement correctly identifies what happens to magnesium?

A Magnesium is reduced because it gains oxygen
B Magnesium is oxidised because it loses electrons (oxidation number increases from 0 to +2)
C Magnesium is neither oxidised nor reduced — only oxygen changes
D Magnesium is the oxidising agent in this reaction
L08 — Oxidation Numbers

4. What is the oxidation number of nitrogen in the nitrate ion (NO₃⁻)?

A +3
B +5
C −3
D +1
L08 — Half-Equations

5. In the half-equation Fe²⁻ → Fe³⁻ + e⁻, the iron is:

A Being oxidised at the cathode and acts as the oxidising agent
B Being reduced at the anode and acts as the reducing agent
C Being oxidised (loses an electron) and acts as the reducing agent
D Being reduced (gains an electron) and acts as the oxidising agent
L09 — Galvanic Cell Electron Flow

6. In a galvanic cell with a zinc anode and a copper cathode, in which direction do electrons flow through the external circuit?

A From the copper electrode to the zinc electrode
B From the zinc electrode to the copper electrode
C Through the salt bridge from zinc to copper
D Through the electrolyte from copper to zinc
L09 — E°cell Calculation

7. Given E°(Zn²⁻/Zn) = −0.76 V and E°(Cu²⁻/Cu) = +0.34 V, what is E°cell for a galvanic cell with a zinc anode and copper cathode?

A −1.10 V
B −0.42 V
C +0.42 V
D +1.10 V
L10 — Inert Electrodes

8. An inert electrode (such as platinum) is required in a galvanic cell when:

A The cell voltage is greater than 1.0 V
B The electrolyte is a solid rather than a solution
C Only the cathode half-reaction involves a metal
D The electrode material would participate in the reaction or dissolve into the electrolyte if an active metal were used
L10 — Spontaneity

9. Which of the following correctly identifies the condition for a galvanic cell reaction to be spontaneous?

A E°cell is negative and current flows from cathode to anode through the wire
B E°cell is positive, with electrons flowing from anode to cathode through the external circuit
C The cathode dissolves and the anode gains mass during operation
D No current flows and E°cell = 0 V (equilibrium)
L07–L10 — Synthesis

10. A silver wire is placed in a solution of copper(II) nitrate. Given E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁻/Cu) = +0.34 V, which prediction is correct?

A No reaction occurs; E°cell = +0.34 − 0.80 = −0.46 V (negative → non-spontaneous)
B Silver is oxidised and Cu²⁻ is reduced; E°cell = +0.46 V
C Copper metal forms on the silver wire; E°cell = +1.14 V
D Silver dissolves and a blue solution forms; E°cell = −1.14 V

Short Answer — 12 marks

L07 — Activity Series & Displacement

SA1. A student separately adds strips of zinc and copper to silver nitrate solution. (a) Using the NESA activity series, predict whether each metal will react with AgNO₃(aq). (b) Write the balanced equation for each reaction that occurs, or state ‘no reaction’ if not. (3 marks)

1 mark for correct prediction for both metals (Zn above Ag → reacts; Cu above Ag → reacts); 1 mark each for balanced equations: Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag; Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

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L08 — Oxidation States & Half-Equations

SA2. For the reaction between permanganate ions and iron(II) ions in acidic solution: MnO₄⁻(aq) + Fe²⁻(aq) → Mn²⁻(aq) + Fe³⁻(aq). (a) Determine the oxidation number of Mn in MnO₄⁻ and in Mn²⁻. (b) Identify which species is oxidised and which is reduced. (c) Write the half-equation for the oxidation of Fe²⁻. (3 marks)

1 mark: Mn in MnO₄⁻ = +7 (O₄=−8, overall −1, so Mn=+7); Mn in Mn²⁻ = +2; 1 mark: Mn reduced (+7→+2); Fe oxidised (+2→+3); 1 mark: Fe²⁻ → Fe³⁻ + e⁻

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L09 — Galvanic Cell Calculations

SA3. A galvanic cell is constructed using a zinc electrode in ZnSO₄(aq) and a silver electrode in AgNO₃(aq) connected by a salt bridge. E°(Zn²⁻/Zn) = −0.76 V; E°(Ag⁺/Ag) = +0.80 V. (a) Identify the anode and cathode. (b) Calculate E°cell and determine if the cell is spontaneous. (c) Write the balanced overall cell reaction. (3 marks)

1 mark: Zn = anode (lower E°, oxidised); Ag = cathode (higher E°, reduced); 1 mark: E°cell = +0.80 − (−0.76) = +1.56 V > 0 → spontaneous; 1 mark: Zn + 2Ag⁺ → Zn²⁻ + 2Ag (balance electrons: Zn → Zn²⁻ + 2e⁻; 2×(Ag⁺ + e⁻ → Ag))

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L10 — Cathodic Protection

SA4. A steel (iron) pipeline is buried underground in moist soil. An engineer attaches blocks of magnesium metal to the pipeline to prevent corrosion. E°(Fe²⁻/Fe) = −0.44 V; E°(Mg²⁻/Mg) = −2.37 V. (a) Name this type of cathodic protection. (b) Explain why magnesium is effective for this purpose using E° values. (c) Predict what will happen to the magnesium blocks over time. (3 marks)

1 mark: sacrificial anode protection; 1 mark: Mg has lower (more negative) E° than Fe → Mg is more readily oxidised → Mg acts as anode (is preferentially oxidised), Fe is protected as cathode; E°cell = −0.44 − (−2.37) = +1.93 V > 0 confirms spontaneous; 1 mark: Mg blocks gradually dissolve/corrode (Mg → Mg²⁻ + 2e⁻) and must be periodically replaced

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  • Q1 — C: The NESA activity series from most to least reactive: K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au. Potassium (K) is the most reactive metal on this list.
  • Q2 — B: Iron is above copper in the activity series → Fe displaces Cu from its salt solution. Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). Observations: red-brown copper metal deposits on the nail; the blue colour of the CuSO₄ solution fades as Cu²⁻ is removed from solution.
  • Q3 — B: Magnesium goes from oxidation number 0 (in Mg metal) to +2 (in MgO). Increase in oxidation number = oxidation = loss of electrons (OIL). Magnesium is the reducing agent (it is oxidised and causes O₂ to be reduced). Option A incorrectly applies the definition — gaining oxygen IS oxidation, not reduction.
  • Q4 — B: In NO₃⁻: charge = −1. O = −2 each × 3 = −6. N + (−6) = −1 → N = +5.
  • Q5 — C: Fe²⁻ loses one electron to become Fe³⁻ — this is oxidation (OIL). A species that is oxidised acts as the reducing agent (it reduces something else by donating electrons). Oxidation occurs at the anode in a galvanic cell, not at the cathode.
  • Q6 — B: At the anode (Zn), oxidation occurs: Zn → Zn²⁻ + 2e⁻. Electrons released at the zinc anode travel through the external wire to the copper cathode, where reduction occurs: Cu²⁻ + 2e⁻ → Cu. Electrons always flow from anode to cathode through the external circuit.
  • Q7 — D: E°cell = E°cathode − E°anode = +0.34 − (−0.76) = +0.34 + 0.76 = +1.10 V. Never reverse the sign of the anode reduction potential before subtracting.
  • Q8 — D: Inert electrodes (Pt, graphite) are used when the electrode material itself would otherwise react — for example, when the half-reaction involves only aqueous ions (no solid metal electrode species) or when an active metal electrode would dissolve into the electrolyte and interfere with the cell reaction.
  • Q9 — B: A positive E°cell indicates the reaction is spontaneous under standard conditions. In a spontaneous galvanic cell, electrons flow spontaneously from the anode (oxidation) to the cathode (reduction) through the external circuit.
  • Q10 — A: For Ag to be oxidised and Cu²⁻ to be reduced, E°cell = E°(Cu²⁻/Cu) − E°(Ag⁺/Ag) = +0.34 − 0.80 = −0.46 V. A negative E°cell means the reaction is non-spontaneous — no reaction occurs. Silver is BELOW copper in the activity series (less reactive), so it cannot displace copper from solution.

SA1: (a) Zinc is above silver in the NESA activity series → Zn will react with AgNO₃. Copper is also above silver → Cu will also react with AgNO₃. (b) Zn(s) + 2AgNO₃(aq) → Zn(NO₃)₂(aq) + 2Ag(s); Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s). In both cases, a silver-grey solid deposits on the metal strip and the metal ions in solution change: Zn²⁻ replaces Ag⁺ (colourless solution remains); Cu²⁻ replaces Ag⁺ (solution turns blue as Cu²⁻ forms).

SA2: (a) In MnO₄⁻: O = −2 × 4 = −8; overall charge = −1; Mn = −1 − (−8) = +7. In Mn²⁻: Mn = +2. (b) Mn decreases from +7 to +2 → reduced. Fe increases from +2 to +3 → oxidised. (c) Fe²⁻(aq) → Fe³⁻(aq) + e⁻ [Note: the permanganate half-reaction requires H⁺ and H₂O to balance in acidic solution; that full balancing is beyond this checkpoint scope, but the Fe²⁻ half-equation is straightforward].

SA3: (a) Anode = Zn (lower reduction potential, E° = −0.76 V; Zn is oxidised). Cathode = Ag (higher reduction potential, E° = +0.80 V; Ag⁺ is reduced). (b) E°cell = E°cathode − E°anode = +0.80 − (−0.76) = +1.56 V. E°cell > 0 → spontaneous. (c) Half-equations: Zn → Zn²⁻ + 2e⁻; 2 × (Ag⁺ + e⁻ → Ag). Overall: Zn(s) + 2Ag⁺(aq) → Zn²⁻(aq) + 2Ag(s).

SA4: (a) Sacrificial anode protection (a form of cathodic protection). (b) Mg has a more negative standard reduction potential (E° = −2.37 V) than Fe (E° = −0.44 V). This means Mg is more readily oxidised than Fe. When both metals are in electrical contact in the moist soil (acting as an electrolyte), Mg acts as the anode (is preferentially oxidised: Mg → Mg²⁻ + 2e⁻) and Fe is protected as the cathode — electrons flow from Mg to Fe rather than Fe being oxidised. E°cell = −0.44 − (−2.37) = +1.93 V > 0 confirms the process is spontaneous. (c) The magnesium blocks gradually corrode and dissolve as Mg is continuously oxidised. Over time they shrink and must be periodically inspected and replaced to maintain protection.

Track Your Score

Multiple Choice (Q1–10) / 10
Short Answer (SA1–4) / 12
Total — / 22

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