S4 Chemistry · Unit 17

Reduction and
Oxidation Reactions

Oxidation states · OIL RIG · Half-equations · Balancing redox · Electrochemical cells · Standard electrode potentials · Electrolysis · Reactivity series

17.1 Oxidation & Reduction 17.2 Oxidation Numbers 17.3 Half-Equations 17.4 Balancing Redox 17.5 Electrochemical Cells 17.6 Standard Electrode Potentials 17.7 Electrolysis 17.8 Reactivity Series Exercises Quiz Unit Test
17.1

Oxidation and Reduction — Definitions

Multiple Definitions of Oxidation and Reduction

Definition setOxidationReduction
Oxygen (original)Gain of oxygenLoss of oxygen
HydrogenLoss of hydrogenGain of hydrogen
Electrons (most general)Loss of electrons (LEO)Gain of electrons (GER)
Oxidation stateIncrease in oxidation stateDecrease in oxidation state
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OIL RIG — the essential memory aid Oxidation Is Loss (of electrons) — Reduction Is Gain (of electrons). In every redox reaction, one species loses electrons (is oxidised) and another gains electrons (is reduced). You cannot have one without the other.

Oxidising and Reducing Agents

Oxidising Agent A substance that oxidises another species — it causes oxidation by accepting electrons from the other species. The oxidising agent itself is reduced (gains electrons → OS decreases).
Reducing Agent A substance that reduces another species — it causes reduction by donating electrons to the other species. The reducing agent itself is oxidised (loses electrons → OS increases).
Example: Zn + Cu²⁺ → Zn²⁺ + Cu Zn → Zn²⁺ + 2e⁻ (Zn is OXIDISED — loses e⁻ — Zn is the REDUCING AGENT) Cu²⁺ + 2e⁻ → Cu (Cu²⁺ is REDUCED — gains e⁻ — Cu²⁺ is the OXIDISING AGENT)
17.2

Oxidation Numbers (Oxidation States)

Rules for Assigning Oxidation Numbers

  1. Pure elements: OS = 0 (Na, Cl₂, O₂, Fe, etc.)
  2. Simple monoatomic ions: OS = charge of ion (Na⁺ = +1; Cl⁻ = −1; Al³⁺ = +3; S²⁻ = −2)
  3. Fluorine: always −1 in compounds (most electronegative element)
  4. Oxygen: usually −2 (except in F₂O = +2; peroxides like H₂O₂ = −1; superoxides = −½)
  5. Hydrogen: usually +1 (except in metal hydrides like NaH = −1)
  6. Sum of OS in neutral compound = 0
  7. Sum of OS in polyatomic ion = charge of ion
Worked Example 17.1 — Assigning Oxidation Numbers

Find the OS of the underlined element in each:

a

Cr in K₂Cr₂O₇: 2(+1) + 2Cr + 7(−2) = 0 → 2 + 2Cr − 14 = 0 → 2Cr = 12 → Cr = +6

b

Mn in MnO₄⁻: Mn + 4(−2) = −1 → Mn = −1 + 8 = +7

c

S in S₂O₃²⁻: 2S + 3(−2) = −2 → 2S = 4 → S = +2 (average; one S = 0, one S = +4)

d

N in NH₄⁺: N + 4(+1) = +1 → N = −3

e

Fe in Fe₃O₄: 3Fe + 4(−2) = 0 → 3Fe = 8 → Fe = +8/3 ≈ 2.67 (mixed Fe²⁺ and Fe³⁺)

17.3

Half-Equations

Writing Half-Equations

A half-equation shows either the oxidation or the reduction process separately, including electrons explicitly:

Oxidation half-equation: Species → product + ne⁻ Reduction half-equation: Species + ne⁻ → product Common half-equations to know: Zn → Zn²⁺ + 2e⁻ (oxidation) Cu²⁺ + 2e⁻ → Cu (reduction) Fe³⁺ + e⁻ → Fe²⁺ (reduction) Fe²⁺ → Fe³⁺ + e⁻ (oxidation) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (reduction — acidic) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (reduction — acidic) H��O₂ + 2H⁺ + 2e⁻ → 2H₂O (reduction) I₂ + 2e⁻ → 2I⁻ (reduction) 2I⁻ → I₂ + 2e⁻ (oxidation) Cl₂ + 2e⁻ → 2Cl⁻ (reduction) H�� → 2H⁺ + 2e⁻ (oxidation) 2H⁺ + 2e⁻ → H₂ (reduction — standard hydrogen electrode)

Balancing Half-Equations in Acidic Solution — Step-by-step

Method for balancing a half-equation in acidic solution:

  1. Write the unbalanced species on each side
  2. Balance atoms other than O and H first
  3. Balance O by adding H₂O to the side needing O
  4. Balance H by adding H⁺ to the side needing H
  5. Balance charge by adding electrons (e⁻) to the more positive side
  6. Check: atoms and charges balanced
Worked Example — Balance MnO₄⁻ → Mn²⁺
1

MnO₄⁻ → Mn²⁺ (Mn balanced: 1 each side ✓)

2

Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O

3

Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

4

Balance charge: Left = −1 + 8 = +7; Right = +2. Add 5e⁻ to left:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O ✓ (charge: −1+8−5 = +2 = +2 ✓)

17.4

Balancing Full Redox Equations

Combining Half-Equations

To write a balanced redox equation: combine the two half-equations so that the number of electrons lost = number gained (multiply each half-equation if necessary).

Worked Example 17.2 — MnO₄⁻ oxidising Fe²⁺ in acid
1

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

2

Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply × 5 to equalise electrons)

3

Multiply oxidation × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻

4

Add: MnO₄⁻ + 8H⁺ + 5e⁻ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ + 5e⁻

5

Cancel electrons: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Check: Mn: 1=1✓ O: 4=4✓ H: 8=8✓ Fe: 5=5✓ Charge: −1+8+10=+17 = +2+15=+17✓

Worked Example 17.3 — Cr₂O₇²⁻ oxidising I⁻
1

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

2

Oxidation: 2I⁻ → I₂ + 2e⁻ (multiply × 3)

3

6I⁻ → 3I₂ + 6e⁻

4

Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂

17.5

Electrochemical Cells

The Electrochemical Cell — Galvanic/Voltaic Cell

An electrochemical (galvanic) cell converts chemical energy to electrical energy using a spontaneous redox reaction. The two half-reactions are separated and connected by an external circuit and a salt bridge.

ZnSO₄(aq) Zn²⁺ + SO₄²⁻ Zn ANODE (−) Zn→Zn²⁺+2e⁻ CuSO₄(aq) Cu²⁺ + SO₄²⁻ Cu CATHODE (+) Cu²⁺+2e⁻→Cu V e⁻ flow → Salt bridge (KNO₃)
Zn-Cu galvanic cell: Zn anode (oxidation, −), Cu cathode (reduction, +), electrons flow external circuit, ions through salt bridge

Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Convention: anode (oxidation) on the LEFT | salt bridge || cathode (reduction) on the RIGHT.

Salt bridge: Contains an electrolyte (KNO₃, KCl, NH₄NO₃) in agar gel. Completes the circuit by allowing ion flow. Maintains electrical neutrality in each half-cell (K⁺ flows into cathode half-cell; NO₃⁻ flows into anode half-cell).

17.6

Standard Electrode Potentials (E°)

The Standard Hydrogen Electrode (SHE)

All electrode potentials are measured relative to the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V by definition:

SHE: 2H⁺(aq, 1 mol dm⁻³) + 2e⁻ ⇌ H₂(g, 100 kPa) E° = 0.00 V

Standard conditions: 298 K (25°C), all solutions at 1 mol dm⁻³, all gases at 100 kPa (1 bar). The E° for any half-cell is measured against the SHE.

Selected Standard Electrode Potentials

Half-equation (reduction)E° (V)Tendency
F₂ + 2e⁻ → 2F⁻+2.87Strongest oxidising agent
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O+1.51Strong oxidant
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O+1.33Strong oxidant
Cl₂ + 2e⁻ → 2Cl⁻+1.36Strong oxidant
Br₂ + 2e⁻ → 2Br⁻+1.07
Fe³⁺ + e⁻ → Fe²⁺+0.77
I₂ + 2e⁻ → 2I⁻+0.54
Cu²⁺ + 2e⁻ → Cu+0.34
2H⁺ + 2e⁻ → H₂0.00Reference (SHE)
Fe²⁺ + 2e⁻ → Fe−0.44
Zn²⁺ + 2e⁻ → Zn−0.76
Al³⁺ + 3e⁻ → Al−1.66
Mg²⁺ + 2e⁻ → Mg−2.37
Na⁺ + e⁻ → Na−2.71Strongest reducing agent

Using E° Values

Cell EMF (Electromotive Force)

E°cell = E°(cathode/reduction) − E°(anode/oxidation) = E°(more positive) − E°(more negative) Example: Zn-Cu cell: E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = +0.34 − (−0.76) = +1.10 V Positive E°cell → spontaneous reaction → galvanic cell can produce electricity

Predicting Feasibility of Redox Reactions

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Feasibility rule: E°cell > 0 → spontaneous (feasible) A reaction is thermodynamically feasible (spontaneous) if E°cell > 0. The species with the MORE POSITIVE E° will be reduced (acts as oxidising agent); the species with the MORE NEGATIVE E° will be oxidised (acts as reducing agent). This is equivalent to saying: the reaction goes in the direction that decreases free energy (ΔG = −nFE°cell; if E°cell > 0 → ΔG < 0 → spontaneous).
Can Cl₂ oxidise Br⁻? E°(Cl₂/Cl⁻) = +1.36V; E°(Br₂/Br⁻) = +1.07V E°cell = +1.36 − (+1.07) = +0.29V > 0 → YES, spontaneous ✓ Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ Can I₂ oxidise Fe²⁺ to Fe³⁺? E°(I₂/I⁻) = +0.54V; E°(Fe³⁺/Fe²⁺) = +0.77V E°cell = +0.54 − (+0.77) = −0.23V < 0 → NO, not spontaneous ✗ (But Fe³⁺ CAN oxidise I⁻: E°cell = +0.77−0.54 = +0.23V > 0 ✓)
17.7

Electrolysis

Principles of Electrolysis

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction — the reverse of a galvanic cell.

Cathode (negative electrode) — reduction occurs: positive ions (cations) gain electrons here. Metal deposits or gases are produced at the cathode.
Anode (positive electrode) — oxidation occurs: negative ions (anions) lose electrons here. Gases (Cl₂, O₂) or dissolved metal is released at the anode.
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ANODE vs CATHODE in electrolysis vs galvanic cells In BOTH: Anode = oxidation; Cathode = reduction. This never changes. What changes: in a galvanic cell, anode is negative (−) and cathode is positive (+). In electrolysis, anode is positive (+) and cathode is negative (−) — the external power supply forces current in the "wrong" direction.

Electrolysis of Molten and Aqueous Electrolytes

Molten NaCl (Downs Process)

Cathode (−): Na⁺ + e⁻ → Na(l) (sodium metal deposited) Anode (+): 2Cl⁻ → Cl₂(g) + 2e⁻ (chlorine gas produced) Overall: 2NaCl → 2Na + Cl₂ Industrial use: Downs cell (commercial production of Na metal and Cl₂)

Aqueous NaCl (Chlor-Alkali Process)

Cathode: 2H₂O + 2e⁻ → H₂↑ + 2OH⁻ (H₂O preferentially reduced over Na⁺) Anode: 2Cl⁻ → Cl₂↑ + 2e⁻ (Cl⁻ preferentially oxidised over H₂O at inert anode) Overall: 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂ Products: NaOH (industrial alkali), Cl₂ (bleach, PVC), H₂ (fuel cells, Haber process)

Electrolysis of Dilute H₂SO₄

Cathode: 2H⁺ + 2e⁻ → H₂↑ (H⁺ reduced) Anode: 2H₂O → O₂↑ + 4H⁺ + 4e⁻ (water oxidised — inert Pt anode) Ratio: H₂:O₂ = 2:1 (by volume)

Electrolytic Refining of Copper

Anode (+): Cu(impure) → Cu²⁺ + 2e⁻ (impure copper dissolves) Cathode (−): Cu²⁺ + 2e⁻ → Cu(pure) (pure copper deposited) More reactive impurities (Zn, Fe, Ni) → pass into solution but not deposited Less reactive impurities (Ag, Au, Pt) → fall as "anode sludge" and are recovered

Faraday's Laws of Electrolysis

Faraday's First Law: The mass of substance deposited/liberated at an electrode is proportional to the quantity of charge passed.

Faraday's Second Law: For the same charge, the masses of different substances deposited are proportional to their equivalent masses (M/n).

Charge (C) = Current (A) × Time (s): Q = I × t Moles of electrons = Q / F where F = 96,485 C mol⁻¹ (Faraday constant) Moles of substance = (moles of electrons) / n where n = number of electrons per ion Mass deposited = moles × molar mass = (It/nF) × M Example: Deposit Cu²⁺ for 30 min at 2.0 A: Q = 2.0 × 30 × 60 = 3600 C Moles e⁻ = 3600/96485 = 0.0373 mol Moles Cu = 0.0373/2 = 0.0187 mol (Cu²⁺ + 2e⁻ → Cu) Mass Cu = 0.0187 × 63.5 = 1.19 g
17.8

Reactivity Series and Displacement

The Electrochemical Series (Reactivity Series)

Metals listed in order of decreasing reactivity (increasing E° for M^n+/M reduction). The electrochemical series is derived from standard electrode potentials:

Most reactive → K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au → Least reactive Increasing E° value (from K at −2.93V to Au at +1.50V) More reactive metal = stronger reducing agent = more negative E°

Displacement Reactions

A more reactive metal displaces a less reactive metal from its salt solution: Zn + CuSO₄(aq) → ZnSO₄(aq) + Cu↓ (Zn more reactive than Cu — E°=-0.76 vs +0.34V) Fe + CuSO₄(aq) → FeSO₄(aq) + Cu↓ (Fe more reactive than Cu) Cu + ZnSO₄(aq) → NO REACTION (Cu less reactive than Zn) Mg + 2HCl → MgCl₂ + H₂↑ (Mg above H in reactivity series) Cu + HCl → NO REACTION (Cu below H in reactivity series)

Corrosion and Protection

Rusting of iron (electrochemical process): Anode (Fe): Fe → Fe²⁺ + 2e⁻ Cathode (water droplet): O₂ + 2H₂O + 4e⁻ → 4OH⁻ Fe²⁺ + 2OH⁻ → Fe(OH)₂ → Fe(OH)₃ → Fe₂O₃·nH₂O (rust) Protection methods: 1. Painting/coating — barrier prevents O₂/H₂O contact 2. Galvanising (Zn coating) — barrier + sacrificial anode (Zn more reactive than Fe) 3. Sacrificial anode — Mg or Zn blocks attached (more reactive → oxidise preferentially) 4. Tin plating (food cans) — barrier only; if Sn damaged, Fe oxidises faster (Sn less reactive) 5. Cathodic protection — iron made cathode by connecting to more reactive metal or power supply

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Exercises

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Multiple Choice Quiz — 25 Questions

Unit 17: Redox Reactions & Electrochemistry

25 Questions
Q1

In the reaction 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺, the reducing agent is:

Sn²⁺ is oxidised (OS +2 → +4, loses 2 electrons) → Sn²⁺ is the reducing agent. Fe³⁺ is reduced (OS +3 → +2, gains 1 electron) → Fe³⁺ is the oxidising agent. OIL RIG: the reducing agent is oxidised (loses electrons); the oxidising agent is reduced (gains electrons).
Q2

The oxidation state of Cr in K₂Cr₂O₇ is:

K₂Cr₂O₇: 2(+1) + 2Cr + 7(−2) = 0. 2 + 2Cr − 14 = 0. 2Cr = 12. Cr = +6. K₂Cr₂O₇ (potassium dichromate) is an important oxidising agent in titrimetry and in the breathalyser test (Cr goes +6 → +3; the orange dichromate turns green when reduced).
Q3

In the standard hydrogen electrode, E° is defined as:

The SHE (2H⁺ + 2e⁻ ⇌ H₂) is assigned E° = 0.00 V by international convention. All other electrode potentials are measured relative to this reference. Conditions: [H⁺] = 1 mol dm⁻³, p(H₂) = 100 kPa, T = 298 K. It uses a platinum electrode (inert conductor) in contact with H₂ gas and H⁺ solution.
Q4

For the galvanic cell Mg | Mg²⁺ || Cu²⁺ | Cu, E°cell = ?
(E°(Mg²⁺/Mg) = −2.37V; E°(Cu²⁺/Cu) = +0.34V)

E°cell = E°(cathode) − E°(anode) = E°(Cu²⁺/Cu) − E°(Mg²⁺/Mg) = +0.34 − (−2.37) = +2.71 V. Mg is the anode (more negative E°, oxidised: Mg → Mg²⁺ + 2e⁻). Cu²⁺ is the cathode (more positive E°, reduced: Cu²⁺ + 2e⁻ → Cu). Positive E°cell → spontaneous → galvanic cell works.
Q5

The balanced ionic equation for MnO₄⁻ oxidising Fe²⁺ in acid is:

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (5 electrons). Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×5): 5Fe²⁺ → 5Fe³⁺ + 5e⁻. Combined: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Check charge: −1+8+10 = +17; +2+15 = +17 ✓. The H⁺ and H₂O are needed to balance oxygen in acidic solution.
Q6

In electrolysis, which process occurs at the cathode?

Cathode = negative electrode in electrolysis. Cations (positive ions) are attracted to the cathode and gain electrons → REDUCTION occurs. Examples: Cu²⁺ + 2e⁻ → Cu; Na⁺ + e⁻ → Na; 2H⁺ + 2e⁻ → H₂. Anode = positive electrode → oxidation: Cl⁻ → ½Cl₂ + e⁻; H₂O → ½O₂ + 2H⁺ + 2e⁻.
Q7

A current of 3.0 A is passed for 10 minutes through molten AlCl₃. What mass of Al is deposited? (F = 96,485 C/mol; Ar(Al) = 27)

Q = 3.0 × (10×60) = 1800 C. Moles e⁻ = 1800/96485 = 0.01866 mol. Al³⁺ + 3e⁻ → Al: moles Al = 0.01866/3 = 0.00622 mol. Mass = 0.00622 × 27 = 0.168 g ≈ 0.17 g. Note: 3 electrons per Al ion — must divide moles of electrons by 3 to get moles of Al.
Q8

In the electrolysis of aqueous sodium chloride (brine), the gas produced at the cathode is:

Cathode (−) of brine electrolysis: Na⁺ is present but water is preferentially reduced (Na⁺ has a very negative E°; water is more easily reduced at the cathode): 2H₂O + 2e⁻ → H₂ + 2OH⁻. H₂ gas is produced at cathode. The OH⁻ and Na⁺ remaining in solution → NaOH. Anode (+): 2Cl⁻ → Cl₂ + 2e⁻ (Cl⁻ oxidised preferentially despite water also being present). This is the chlor-alkali process: NaCl, NaOH, H₂, Cl₂ all produced.
Q9

The oxidation state of N in HNO₂ (nitrous acid) is:

HNO₂: H(+1) + N + 2O(−2) = 0. 1 + N − 4 = 0. N = +3. Compare: HNO₃ N = +5; NH₃ N = −3; NO N = +2; NO₂ N = +4; N₂O N = +1; N₂ N = 0. The trend: the more oxygen atoms, the higher the oxidation state of N.
Q10

Galvanising protects iron from corrosion even when the coating is scratched because:

E°(Zn²⁺/Zn) = −0.76V < E°(Fe²⁺/Fe) = −0.44V. When both Zn and Fe are exposed: Zn is more easily oxidised (more negative E°) → Zn → Zn²⁺ + 2e⁻. Fe acts as cathode (protected). This is cathodic protection via sacrificial anode. Zinc corrodes, Fe is saved. Contrast with tin plating: Sn less reactive than Fe → when Sn scratched, Fe corrodes faster (becomes the anode).
Q11

Which half-equation correctly represents the reduction of dichromate ions in acid?

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. This is a reduction (Cr goes from +6 to +3). Check: Cr: 2=2 ✓; O: 7=7 ✓; H: 14=14 ✓; Charge: (−2)+(+14)+(−6) = +6; right: +6 ✓. H⁺ and H₂O are needed in acid solution to balance O and H atoms. This half-equation has E° = +1.33V (strong oxidising agent).
Q12

The cell notation Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) indicates:

Cell notation convention: anode (oxidation) on LEFT | salt bridge || cathode (reduction) on RIGHT. Single | = electrode/solution interface. || = salt bridge. So: Zn is anode (Zn → Zn²⁺ + 2e⁻), Cu is cathode (Cu²⁺ + 2e⁻ → Cu). Electrons flow externally from Zn (−) to Cu (+). E°cell = +0.34 − (−0.76) = +1.10V.
Q13

In a galvanic cell, the function of the salt bridge is to:

As oxidation occurs at the anode (e.g. Zn → Zn²⁺), positive ions accumulate → solution would become positive, stopping the reaction. As reduction occurs at cathode (Cu²⁺ → Cu), cations are removed → solution would become negative. Salt bridge (KNO₃ in agar): K⁺ migrates to cathode half-cell; NO₃⁻ migrates to anode half-cell → maintains neutrality → current continues. Electrons flow through external wire (not salt bridge). Salt bridge carries ionic current.
Q14

Which of the following will react spontaneously? (E° values: F₂/F⁻ +2.87; Cl₂/Cl⁻ +1.36; Br₂/Br⁻ +1.07; I₂/I⁻ +0.54 V)

E°cell = E°(cathode/reduction) − E°(anode/oxidation). (C) Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂: Cl₂ reduced (+1.36), Br⁻ oxidised (+1.07 as anode). E°cell = +1.36 − +1.07 = +0.29V > 0 → SPONTANEOUS ✓. (A) E°cell = +0.54 − 1.36 = −0.82V ✗. (B) E°cell = +1.07 − 1.36 = −0.29V ✗. (D) Br₂ + 2I⁻: E°cell = +1.07 − 0.54 = +0.53V > 0 → actually spontaneous (answer D description is wrong). Answer C is the only clearly correctly stated spontaneous reaction.
Q15

Rusting of iron is an electrochemical process. The ANODIC (oxidation) reaction is:

Rusting is an electrochemical cell on the iron surface. Anode (oxidation): Fe → Fe²⁺ + 2e⁻ (iron dissolves into solution). Cathode (reduction): O₂ + 2H₂O + 4e⁻ → 4OH⁻ (oxygen reduced in neutral/slightly acidic water). Fe²⁺ + 2OH⁻ → Fe(OH)₂ → Fe(OH)₃ → Fe₂O₃·nH₂O (rust). The process is accelerated by: salt (increases conductivity), acid rain (provides H⁺), bimetallic contact (galvanic cell).
Q16

What is the oxidation state of Fe in Fe₃O₄?

Fe₃O₄: 3Fe + 4(−2) = 0. 3Fe = 8. Fe = 8/3 = +2.67. This non-integer OS indicates Fe₃O₄ is a mixed oxide (Fe²⁺ · Fe₂O₃): one Fe²⁺ and two Fe³⁺. Fe₃O₄ = magnetite, magnetic, found in lodestone. Similarly, Pb₃O₄ (red lead) contains Pb²⁺ and Pb⁴⁺ (average OS = +8/3).
Q17

Which change represents REDUCTION?

Reduction = decrease in OS (gain of electrons). (A) Zn: 0→+2 = oxidation. (B) Fe: +2→+3 = oxidation. (C) MnO₄⁻: Mn = +7 → Mn²⁺: Mn = +2 → OS decreased → REDUCTION ✓. (D) 2I⁻: −1→0 = oxidation. Remember: OIL RIG — Reduction Is Gain of electrons → OS decreases.
Q18

In electrolytic copper refining, the impure copper is:

Anode (+) = impure copper (dissolves: Cu → Cu²⁺ + 2e⁻). Cathode (−) = thin pure copper sheet (Cu²⁺ + 2e⁻ → Cu pure). [Cu²⁺] stays nearly constant (Cu dissolves at anode, deposits at cathode at the same rate). Reactive impurities (Zn, Fe, Ni) dissolve at anode but don't deposit at cathode (too negative E°). Precious metals (Ag, Au, Pt) don't dissolve — fall as "anode sludge" with economic value. The product is 99.99%+ pure copper for electrical uses.
Q19

The Faraday constant (F = 96,485 C mol⁻¹) represents:

F = N_A × e = 6.022×10²³ × 1.602×10⁻¹⁹ = 96,485 C mol⁻¹. F is the charge carried by one mole of electrons. It connects macroscopic (moles) to electrical (Coulombs) quantities. Q = n(e⁻) × F; n(e⁻) = Q/F; n(substance) = n(e⁻)/z where z = electrons per formula unit. F ≈ 96,500 C mol⁻¹ is commonly used as an approximation.
Q20

The breathalyser test uses K₂Cr₂O₇ (orange) → Cr₂(SO₄)₃ (green). In this reaction, ethanol is:

Ethanol (CH₃CH₂OH): C of CH₂OH has OS = −1. Ethanoic acid (CH₃COOH): C of COOH has OS = +3. OS increases → oxidation. Cr₂O₇²⁻ (Cr = +6) → Cr³⁺ (OS decreases) → reduction. Cr₂O₇²⁻ is the oxidising agent (is reduced, turns from orange to green). Ethanol is the reducing agent (is oxidised). Reaction: 3C₂H₅OH + 2K₂Cr₂O₇ + 8H₂SO₄ → 3CH₃COOH + 2Cr₂(SO₄)₃ + 2K₂SO₄ + 11H₂O. Now replaced by IR spectroscopy in modern breathalysers.
Q21

Which statement about E°cell is correct?

E°cell > 0 → ΔG° = −nFE°cell < 0 → reaction is spontaneous/feasible under standard conditions. E°cell < 0 → not spontaneous (reverse reaction is spontaneous). E°cell = 0 → at equilibrium (as in the SHE). Temperature DOES affect E°cell (the Nernst equation relates E to T and concentrations). This is purely a thermodynamic prediction — a kinetic barrier may still prevent the reaction from occurring in practice.
Q22

In the oxidation number method: what change occurs in a reaction where ClO₃⁻ is converted to Cl⁻?

ClO₃⁻: Cl + 3(−2) = −1 → Cl = +5. Cl⁻: OS = −1. Change: +5 → −1 = decrease of 6 → reduction (gain of 6 electrons per Cl). Half-equation: ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O. This reduction is done by strong reducing agents (e.g. SO₂ or I⁻ in acidic conditions). ClO₃⁻ (chlorate) is the +5 state of Cl; Cl⁻ is the −1 state.
Q23

The displacement reaction Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag occurs because:

E°(Ag⁺/Ag) = +0.80V; E°(Cu²⁺/Cu) = +0.34V. Since E°(Ag⁺/Ag) > E°(Cu²⁺/Cu): Ag⁺ is a stronger oxidising agent than Cu²⁺ → Ag⁺ oxidises Cu → Cu → Cu²⁺ + 2e⁻; Ag⁺ + e⁻ → Ag. E°cell = +0.80 − +0.34 = +0.46V > 0 → spontaneous. Beautiful silver crystals form on the copper. Note: Ag is LESS reactive than Cu in the reactivity series (more reactive = more negative E° = stronger reducing agent).
Q24

Hydrogen peroxide (H₂O₂) can act as both an oxidising agent and a reducing agent. In which reaction does H₂O₂ act as a REDUCING agent?

In reaction B: MnO₄⁻ (+7) → Mn²⁺ (+2) — Mn is reduced → MnO₄⁻ is the oxidising agent. H₂O₂ (O = −1) → O₂ (O = 0) — O is oxidised → H₂O₂ is the reducing agent (is oxidised from −1 to 0 in O₂). H₂O₂ is both oxidising (when it gains electrons, e.g. against I⁻) AND reducing (when it loses electrons, e.g. against MnO₄⁻). OS of O in H₂O₂ = −1, which is intermediate between 0 (O₂) and −2 (H₂O), so H₂O₂ can go either way.
Q25

Standard electrode potentials are measured under standard conditions. Which of these is NOT a standard condition?

Standard conditions for E°: T = 298 K, concentration of all aqueous species = 1 mol dm⁻³, pressure of all gases = 100 kPa (1 bar). pH is NOT separately specified at 7 — for acid/base-involving half-equations (like MnO₄⁻/Mn²⁺), [H⁺] = 1 mol dm⁻³ → pH = 0 (not 7). Standard conditions refer to concentrations (1 mol dm⁻³), not pH = 7. So "pH = 7" is NOT a standard condition.
📝

Unit Test — 50 Marks

Section A — Short Answer

30 marks
Q1 [5 marks]

Find the oxidation state of the bold element in each species: (a) Mn in KMnO₄ (b) S in SO₄²⁻ (c) Cr in Cr₂O₇²⁻ (d) N in N₂H₄ (e) Fe in [Fe(CN)₆]³⁻ (f) Cl in NaClO₄ (g) P in H₃PO₃ (h) C in C₆H₁₂O₆ (i) I in IO₃⁻ (j) Xe in XeO₄. [5 — ½ each]

(a) KMnO₄: +1+Mn+4(−2)=0 → Mn=+7. (b) SO₄²⁻: S+4(−2)=−2 → S=+6. (c) Cr₂O₇²⁻: 2Cr+7(−2)=−2 → Cr=+6. (d) N₂H₄: 2N+4(+1)=0 → N=−2. (e) [Fe(CN)₆]³⁻: CN⁻ ligands = −1 each; Fe+6(−1)=−3 → Fe=+3. (f) NaClO₄: +1+Cl+4(−2)=0 → Cl=+7. (g) H₃PO₃: 3(+1)+P+3(−2)=0 → P=+3. (Note: one P−H bond, two P−OH bonds, one P=O; effective +3 for the formula). (h) C₆H₁₂O₆: 6C+12(+1)+6(−2)=0 → 6C=0 → C=0. (i) IO₃⁻: I+3(−2)=−1 → I=+5. (j) XeO₄: Xe+4(−2)=0 → Xe=+8.
Q2 [5 marks]

Balance the following redox equation using the half-equation method: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ (acidic solution). Show all half-equations and the combined equation. [5]

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O [Mn: +7→+2, 5e⁻ gained]. [1.5] Oxidation: C₂O₄²⁻ → 2CO₂ + 2e⁻ [C: +3→+4, 2e⁻ lost; balance O: 4O each side ✓; charge: −2 = 0−2 = −2 ✓]. [1.5] LCM of electrons: 5×2e⁻ = 10e⁻ = 2×5e⁻. Multiply reduction ×2; oxidation ×5: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O; 5C₂O₄²⁻ → 10CO₂ + 10e⁻. Combined: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂. [1.5] Check: Mn 2=2 ✓; C 10=10 ✓; O 8+20=28; 8+20=28 ✓; H 16=16 ✓; charge: −2+16−10=+4; +4+0=+4 ✓. [0.5]
Q3 [5 marks]

Describe the construction and operation of a Zn-Cu galvanic cell. Include: the half-reactions at each electrode, the direction of electron flow, the role of the salt bridge, the cell notation, and calculate E°cell. (E°(Zn²⁺/Zn) = −0.76V; E°(Cu²⁺/Cu) = +0.34V) [5]

Construction: two half-cells connected by external wire and salt bridge. Left (anode): Zn rod in ZnSO₄ solution. Right (cathode): Cu rod in CuSO₄ solution. Salt bridge (KNO₃ in agar) connects the two solutions. [1] Anode (−, oxidation): Zn → Zn²⁺ + 2e⁻. Zn dissolves; blue solution forms if ZnSO₄ is absent. [1] Cathode (+, reduction): Cu²⁺ + 2e⁻ → Cu. Copper deposits on Cu electrode; blue CuSO₄ solution fades. [1] Electrons flow through external circuit from Zn (−) to Cu (+). K⁺ ions migrate from salt bridge into cathode solution; NO₃⁻ ions into anode solution → maintains electrical neutrality. [1] Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). E°cell = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V. [1]
Q4 [5 marks]

Describe the electrolysis of aqueous sodium chloride (brine). State what is produced at each electrode, write the half-equations, and name three important products of this industrial process with one use of each. [5]

Cathode (−): water is preferentially reduced (not Na⁺, too negative). 2H₂O + 2e⁻ → H₂↑ + 2OH⁻. Hydrogen gas produced; OH⁻ accumulates. [1.5] Anode (+): Cl⁻ is preferentially oxidised (over H₂O at high [Cl⁻]). 2Cl⁻ → Cl₂↑ + 2e⁻. Chlorine gas produced. [1.5] Overall: 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂. [0.5] Products and uses: (1) NaOH — used in soap/detergent manufacture, paper pulp, drain cleaner. (2) Cl₂ — used in making bleach (NaOCl), PVC, disinfecting water. (3) H₂ — used as fuel in fuel cells, in Haber process to make NH₃, in hydrogenation of oils. [1.5]
Q5 [5 marks]

A current of 2.5 A is passed through silver nitrate solution for 45 minutes using platinum electrodes. (a) Which electrode does Ag deposit on? (b) Write the half-equation for this deposition. (c) Calculate the mass of Ag deposited. (F = 96,485 C/mol; Ar(Ag) = 108) [5]

(a) Ag deposits on the cathode (negative electrode) — reduction occurs there: cations (Ag⁺) gain electrons. [1] (b) Ag⁺ + e⁻ → Ag(s). (1 electron per Ag⁺.) [1] (c) Q = I × t = 2.5 × (45 × 60) = 2.5 × 2700 = 6750 C. [1] Moles of electrons = 6750/96485 = 0.06996 mol. [1] Moles of Ag = 0.06996/1 = 0.06996 mol (1 electron per Ag⁺). Mass = 0.06996 × 108 = 7.56 g ≈ 7.6 g. [1]
Q6 [5 marks]

Use the following E° values to: (a) predict which reactions are spontaneous (b) write the ionic equation for the spontaneous reaction (c) calculate E°cell for each. E°(Fe³⁺/Fe²⁺) = +0.77V; E°(I₂/I⁻) = +0.54V; E°(Zn²⁺/Zn) = −0.76V; E°(Cu²⁺/Cu) = +0.34V. Reactions to test: (i) I₂ + 2Fe²⁺ vs 2I⁻ + 2Fe³⁺ (ii) Zn + Cu²⁺ vs Zn²⁺ + Cu. [5]

(i) Forward: I₂ + 2Fe²⁺ → 2I⁻ + 2Fe³⁺. I₂ reduced (cathode, +0.54); Fe²⁺ oxidised (anode, +0.77). E°cell = +0.54 − +0.77 = −0.23V < 0 → NOT spontaneous. Reverse: 2Fe³⁺ + 2I⁻ → I₂ + 2Fe²⁺. E°cell = +0.77 − +0.54 = +0.23V > 0 → SPONTANEOUS. Ionic equation: 2Fe³⁺ + 2I⁻ → I₂ + 2Fe²⁺. (Fe³⁺ is a stronger oxidant than I₂.) [2.5] (ii) Forward: Zn + Cu²⁺ → Zn²⁺ + Cu. Zn oxidised (anode, −0.76); Cu²⁺ reduced (cathode, +0.34). E°cell = +0.34 − (−0.76) = +1.10V > 0 → SPONTANEOUS ✓. Ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Blue CuSO₄ solution fades; copper deposits on zinc. [2.5]

Section B — Extended Answer

20 marks
Q7 [10 marks]

(a) Define the terms: oxidation, reduction, oxidising agent, reducing agent. Illustrate each with the reaction of Fe with Cl₂. [3]
(b) Explain the principles of electrochemical cells. How does the standard electrode potential E° relate to the feasibility of a reaction? Use the relationship ΔG° = −nFE°cell. [4]
(c) Explain the corrosion of iron as an electrochemical process. Why does iron rust faster in salt water than in pure water? Describe two methods of corrosion protection with chemical explanations. [3]

(a) Oxidation: loss of electrons / increase in OS. Reduction: gain of electrons / decrease in OS. Oxidising agent: species that accepts electrons (is itself reduced). Reducing agent: species that donates electrons (is itself oxidised). Fe + Cl₂ reaction: 2Fe + 3Cl₂ → 2FeCl₃. Fe: 0→+3 (oxidised, loses 3e⁻, is the reducing agent). Cl₂: 0→−1 (reduced, gains e⁻, is the oxidising agent). [3] (b) An electrochemical cell converts chemical energy to electrical energy via a spontaneous redox reaction in two separated half-cells. E° values (tabulated as reduction potentials) allow prediction: the half-cell with higher E° undergoes reduction (cathode); the half-cell with lower E° undergoes oxidation (anode). E°cell = E°(cathode) − E°(anode). When E°cell > 0: ΔG° = −nFE°cell < 0 (n = moles of electrons; F = 96,485 C/mol) → spontaneous. E°cell < 0 → ΔG° > 0 → non-spontaneous. E°cell = 0 → ΔG° = 0 → equilibrium. Higher E°cell → more negative ΔG° → more energy available → stronger driving force. [4] (c) Iron rusting is a galvanic cell on the iron surface. Different areas of iron act as anode/cathode (due to stress, impurities). Anode: Fe → Fe²⁺ + 2e⁻ (Fe dissolves). Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (O₂ reduced). Fe²⁺ + 2OH⁻ → Fe(OH)₂ → oxidised further → Fe₂O₃·nH₂O (rust). Salt water: NaCl dissolves → Na⁺ and Cl⁻ greatly increase conductivity of the solution → ion flow easier → galvanic cell current flows faster → Fe corrodes much faster. Protection: (1) Galvanising (Zn coat): Zn more reactive (E° −0.76 vs −0.44V) → sacrificial anode (Zn oxidised, Fe protected). (2) Cathodic protection: Fe made cathode by connecting Mg blocks (Mg E° = −2.37V → oxidised preferentially) or by impressing an external negative current onto the structure (pipelines, ships). [3]
Q8 [10 marks]

(a) Potassium manganate(VII) (KMnO₄) is a versatile oxidising agent. Describe the colour changes observed when KMnO₄ reacts in: (i) acidic solution (ii) neutral/alkaline solution. Give the manganese species produced and their colours in each case. Write the half-equation for the reduction in acid. [4]
(b) A 25.00 cm³ sample of iron(II) sulfate solution is titrated against 0.020 mol dm⁻³ KMnO₄. 22.50 cm³ of KMnO₄ is needed. (i) Write the balanced ionic equation. (ii) Calculate the concentration of Fe²⁺. [3]
(c) Explain the industrial importance of electrolysis with reference to THREE different industrial processes. Include: the electrolyte used, the electrodes, what is produced at each electrode, and the economic/industrial significance. [3]

(a)(i) Acid solution: MnO₄⁻ (purple/violet) → Mn²⁺ (almost colourless/very pale pink). Colour change: purple → colourless. KMnO₄ is self-indicating — the endpoint is when the last drop of KMnO₄ added does not decolour. Mn goes from +7 to +2. Half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. [2] (ii) Neutral/alkaline: MnO₄⁻ (purple) → MnO₂ (brown precipitate). Mn goes from +7 to +4. Half-equation: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂↓ + 4OH⁻. Strong alkali (excess): MnO₄⁻ + e⁻ → MnO₄²⁻ (manganate, green). [2] (b)(i) In acid: 5Fe²⁺ → 5Fe³⁺ + 5e⁻ (×1); MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Combined: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. [1] (ii) Moles KMnO₄ = 0.020 × 22.50/1000 = 4.5×10⁻⁴ mol. 1 MnO₄⁻ : 5 Fe²⁺ → moles Fe²⁺ = 5 × 4.5×10⁻⁴ = 2.25×10⁻³ mol. [Fe²⁺] = 2.25×10⁻³ / 0.02500 = 0.090 mol dm⁻³. [2] (c) Three industrial electrolysis processes (any 3): (1) Chlor-alkali (brine): Electrolyte = conc. NaCl(aq). Cathode = steel mesh: H₂ produced (fuel, Haber process). Anode = titanium with DSA: Cl₂ produced (bleach, PVC, solvents). Left: NaOH solution (soaps, detergents, paper). Major global industry. (2) Aluminium extraction (Hall-Héroult): Electrolyte = molten Al₂O₃ dissolved in cryolite (Na₃AlF₆) at 950°C. Cathode = graphite-lined steel vessel: Al(l) deposited. Anode = graphite (consumed): CO₂/O₂ produced. 15 kWh per kg Al — energy-intensive; hence Al recycling is important. (3) Copper refining: Electrolyte = CuSO₄(aq). Anode = impure Cu. Cathode = thin pure Cu sheet. Pure Cu (99.99%) deposited at cathode. Anode sludge contains Ag, Au, Pt — recovered economically. Essential for electrical wire grade copper. [3]
← Unit 16: Acids & Bases S4 Course Home Unit 18: Energy Changes →

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