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IB Diploma Chemistry HL · Reactivity 3.2 — electron transfer reactions
Mini-Lesson

Electron transfer — redox

5 missions. Guess, learn, then check. Your place is saved.

  1. Mission 1When iron rusts, does it gain or lose electrons?
  2. Mission 2Why does zinc fizz in acid when a gold ring doesn't?
  3. Mission 3Why can you recharge a phone battery but not a standard AA battery?
  4. Mission 4Why does a bulb light only once solid lead(II) bromide has melted?
  5. Mission 5How does a breath tester detect alcohol?
Mission 1

Mission 1: When iron rusts, does it gain or lose electrons?

Make a guess: When iron rusts, does it gain or lose electrons?

Mission 1

Four ways to spot oxidation

OxidationReduction
Electronslostgained
Oxidation staterisesfalls
Oxygengainedlost
Hydrogenlostgained
  • The two always happen together: a redox reaction.
  • The electron and oxidation-state definitions work for every redox reaction.
Mission 1

Oxidation states: the rules

  • An uncombined element is 0: Fe, O₂, Cl₂.
  • A monatomic ion's state is its charge: Fe³⁺ is +3.
  • Usually O is −2 and H is +1. F is −1 in all its compounds.
  • In compounds, group 1 metals are +1; group 2 metals, +2.
  • The states add up to the charge on the species: 0 for a compound.
  • Write the sign first: +3, not 3+.
Mission 1

Worked examples

  • SO₄²⁻: x + 4(−2) = −2, so S is +6.
  • H₂S: 2(+1) + x = 0, so S is −2.
  • P₂O₇⁴⁻: 2x + 7(−2) = −4, so 2x = +10 and each P is +5.
  • Sulfur runs from −2 to +6: most non-metals have several states.
Quick check
?What is the oxidation state of nitrogen in the ammonium ion, NH₄⁺?
Quick check
?In which species does chlorine have an oxidation state of +7?
Mission 1

Oxidation states in names

  • A Roman numeral in a name gives the oxidation state of the element before it.
  • FeCl₂ is iron(II) chloride; FeCl₃ is iron(III) chloride.
  • Transition elements have variable states: Cu₂O is copper(I) oxide; CuO is copper(II) oxide.
  • Oxyanions too: ClO₃⁻ is chlorate(V); SO₃²⁻ is sulfate(IV); SO₄²⁻ is sulfate(VI).
Quick check
?What is the formula of vanadium(V) oxide?
Mission 1

Oxidizing and reducing agents

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transfer from a zinc atom to a copper(II) ionoxidation state shown above each species0+2ZnCu2+e−e−oxidized:reducing agentreduced:oxidizing agent+20Zn2+Cu+Zn + Cu2+ → Zn2+ + Cu
  • The oxidizing agent takes electrons, so it is itself reduced.
  • The reducing agent gives electrons, so it is itself oxidized.
  • Agents are reactants, never products.
  • Zn + Cu²⁺ → Zn²⁺ + Cu. Zn: 0 → +2, oxidized: the reducing agent.
  • Cu²⁺: +2 → 0, reduced: the oxidizing agent.
Quick check
?Chlorine reacts with aqueous iron(II) ions: 2Fe²⁺(aq) + Cl₂(aq) → 2Fe³⁺(aq) + 2Cl⁻(aq) Which species is the oxidizing agent in this reaction?
Mission 1

Half-equations

  • A half-equation shows the oxidation or the reduction alone, with the electrons written in.
  • Oxidation: electrons are products. Zn → Zn²⁺ + 2e⁻.
  • Reduction: electrons are reactants. Cl₂ + 2e⁻ → 2Cl⁻.
  • Both atoms and charge must balance.
Mission 1

Balancing in acid: worked example

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Cr2O7 2- to Cr3+ in acid: four steps1 Balance CrCr2O72− → 2Cr3+2 Add 7H2O (balances O)Cr2O72− → 2Cr3+ + 7H2O3 Add 14H+ (balances H)Cr2O72− + 14H+ → 2Cr3+ + 7H2O4 Add 6e− (balances charge)Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2Ocharge before e−: left +12, right +6after adding 6e−: left +6, right +6

Cr₂O₇²⁻ → Cr³⁺ in acidic solution:

  • Balance Cr: Cr₂O₇²⁻ → 2Cr³⁺.
  • Balance O with H₂O: add 7H₂O on the right.
  • Balance H with H⁺: add 14H⁺ on the left.
  • Balance charge with e⁻: left +12, right +6, so add 6e⁻ on the left.
  • Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Check: 2 Cr each fall by 3, so 6e⁻.
Quick check
?Which half-equation correctly shows the conversion of MnO₄⁻(aq) into Mn²⁺(aq) in acidic solution?
Mission 1

Combining half-equations

  • Scale so the electrons lost equal the electrons gained, then add and cancel them.
  • Oxidation: 2I⁻ → I₂ + 2e⁻. Multiply by 3 to give 6e⁻.
  • Add to the dichromate half-equation; cancel 6e⁻:
  • Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
  • Check charge: left −2 + 14 − 6 = +6; right 2(+3) = +6.
Quick check
?Which half-equation shows the reduction of chlorate(V) ions, ClO₃⁻, to chloride ions in acidic solution?
Quick check
?Combine ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O with 2I⁻ → I₂ + 2e⁻. Which is the balanced overall equation?
Mission check

?A simplified equation for rusting is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). Which statement is correct?
Mission 2

Mission 2: Why does zinc fizz in acid when a gold ring doesn't?

Make a guess: Why does zinc fizz in acid when a gold ring doesn't?

Mission 2

Metals: the trend in ease of oxidation

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table outline with the trends in ease of oxidation and reductiongroup 1group 17period 3metals: oxidized morereadily down group 1and to the left of a periodhalogens: reduced morereadily up group 17
  • Metals react by losing electrons: they are oxidized, so they act as reducing agents.
  • Down a group, oxidation gets easier: outer electrons are further out and more shielded.
  • Towards the left of a period it gets easier too: Na > Mg > Al.
  • So K is oxidized more readily than Na, and Na more readily than Mg.
Quick check
?Using positions in the periodic table, which of these metals would be expected to be oxidized most readily?
Mission 2

Halogens: the opposite trend

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table outline with the trends in ease of oxidation and reductiongroup 1group 17period 3metals: oxidized morereadily down group 1and to the left of a periodhalogens: reduced morereadily up group 17
  • Halogens react by gaining electrons: X₂ + 2e⁻ → 2X⁻. They are oxidizing agents.
  • Reduction gets easier up group 17: F₂ > Cl₂ > Br₂ > I₂.
  • A smaller atom holds an added electron closer to the nucleus, with less shielding.
  • A halogen oxidizes the halide ions below it: Cl₂ + 2I⁻ → 2Cl⁻ + I₂.
  • Br₂ with Cl⁻: no reaction.
Quick check
?Which statement about the relative ease of reduction of the halogens is correct?
Quick check
?Which mixture reacts, as shown by its equation?
Mission 2

Metal displacement reactions

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strip in copper(II) solution: before and after, with a zoom on electron transferBeforeAfterzinc strip inblue Cu2+(aq)copper coating,solution palerzoom: at the zinc surfaceZnCu2+e−e−Zn gives 2e− to Cu2+Zn + Cu2+ → Zn2+ + Cu
  • A metal oxidized more readily displaces a less readily oxidized metal from its ions.
  • Zn + Cu²⁺ → Zn²⁺ + Cu: copper coats the zinc; the blue colour fades.
  • Cu + Zn²⁺: no reaction.
  • Exams give you the data. Rank the metals by which displaces which.
  • The ion of the least readily oxidized metal is the strongest oxidizing agent in the set.
Quick check
?A strip of nickel is placed in Cu²⁺(aq), and a coating of copper forms. Another strip of nickel is placed in Zn²⁺(aq), and nothing happens. Which statement is correct?
Mission 2

Reactive metals with dilute acids

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fizzes in dilute hydrochloric acid; a gold ring does notdilute HCl in each tubeH2 gaszincfizzes: H2 formsgold ringno bubbles
  • A reactive metal reduces H⁺ to hydrogen gas and is itself oxidized.
  • Metal + acid → salt + hydrogen. This is redox, not neutralization.
  • Mg + 2HCl → MgCl₂ + H₂. Ionic: Mg + 2H⁺ → Mg²⁺ + H₂.
  • Zn + H₂SO₄ → ZnSO₄ + H₂.
  • Each H₂ needs 2e⁻: count the electrons the metal loses.
  • Metals oxidized less readily than hydrogen, such as Cu, Ag and Au, do not react.
Quick check
?Which equation correctly represents the reaction of aluminium with dilute hydrochloric acid?
Calculate
11.31 g of zinc reacts completely with excess dilute hydrochloric acid. Calculate the amount, in mol, of hydrogen gas produced. (Ar of Zn = 65.38, data booklet.)
Mission check

?A strip of zinc fizzes in dilute hydrochloric acid, but a gold ring does not. Which explains this?
Mission 3

Mission 3: Why can you recharge a phone battery but not a standard AA battery?

Make a guess: Why can you recharge a phone battery but not a standard AA battery?

Mission 3

Anode and cathode

  • Anode: the electrode where oxidation happens.
  • Cathode: the electrode where reduction happens.
  • This holds in every electrochemical cell. Their signs depend on the type of cell.
Mission 3

Building a primary (voltaic) cell

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cell: zinc and copper half-cells joined by a salt bridge and an external circuitVZn(s) | Zn2+(aq)anode (−)Zn → Zn2+ + 2e−Cu(s) | Cu2+(aq)cathode (+)Cu2+ + 2e− → Cuexternal circuitvoltmetersalt bridge, KNO3(aq)
  • A voltaic cell turns the chemical energy of a spontaneous redox reaction into electrical energy.
  • Two half-cells: zinc in Zn²⁺(aq); copper in Cu²⁺(aq).
  • Joined by an external circuit (wire) and a salt bridge, e.g. KNO₃(aq).
  • Zinc is oxidized more readily. Anode: Zn → Zn²⁺ + 2e⁻.
  • Cathode: Cu²⁺ + 2e⁻ → Cu.
Mission 3

Which way do electrons and ions go?

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cell: electrons flow in the wire; ions move through the salt bridgeVZn(s) | Zn2+(aq)anode (−)Zn → Zn2+ + 2e−Cu(s) | Cu2+(aq)cathode (+)Cu2+ + 2e− → Cuexternal circuite−e−salt bridgeNO3−K+
  • Electrons flow through the wire, from the anode (Zn) to the cathode (Cu).
  • So in a voltaic cell the anode is negative; the cathode is positive.
  • Electrons never pass through the solutions or the salt bridge.
  • Salt bridge: anions move to the anode half-cell, where Zn²⁺ is building up.
  • Cations move to the cathode half-cell, where Cu²⁺ is being used up.
  • This keeps both solutions neutral, so the current keeps flowing.
Quick check
?A voltaic cell is made from a Zn(s)|Zn²⁺(aq) half-cell and a Cu(s)|Cu²⁺(aq) half-cell. Zinc is oxidized more readily than copper. Which statement about the zinc electrode is correct?
Quick check
?A voltaic cell is to be built from a zinc electrode and a copper electrode, joined by a wire and by a salt bridge containing KNO₃(aq). Zinc is to be the anode. What should the half-cell containing the copper electrode consist of?
Quick check
?A voltaic cell is made from Zn(s)|Zn²⁺(aq) and Cu(s)|Cu²⁺(aq) half-cells joined by a salt bridge containing KNO₃(aq). Zinc is the anode. Which describes how charge moves through the salt bridge?
Mission 3

Primary, secondary and fuel cells

  • Primary cell: single use. Its reaction cannot practically be reversed.
  • Secondary cell: rechargeable. A charger's current drives the reaction backwards.
  • Charging reverses each whole half-equation: products become reactants, electrons included.
  • Fuel cell: fuel (e.g. H₂) and O₂ are fed in continuously; it never needs recharging.
  • A cell stores chemical energy, not electricity.
Mission 3

Charging, worked: the lithium-ion cell

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cell on discharge and on chargeDischarge: the cell supplies currentloade−e−porous separatorsoaked in electrolyte++++++(−) graphite(+) cobalt oxideLiC6 → C6CoO2 → LiCoO2Charge: a charger reverses itchargere−e−porous separatorsoaked in electrolyte++++++(−) graphite(+) cobalt oxideC6 → LiC6LiCoO2 → CoO2
  • Discharge, negative electrode (oxidation): LiC₆ → C₆ + Li⁺ + e⁻.
  • Discharge, positive electrode (reduction, Co +4 → +3): CoO₂ + Li⁺ + e⁻ → LiCoO₂.
  • Charging reverses both: C₆ + Li⁺ + e⁻ → LiC₆ and LiCoO₂ → CoO₂ + Li⁺ + e⁻.
  • During charging, Li⁺ ions move back into the graphite.
Quick check
?A nickel–cadmium cell is rechargeable. While it is being charged, the reaction at its positive electrode is: Ni(OH)₂ + OH⁻ → NiO(OH) + H₂O + e⁻. Which half-equation shows the reaction at this electrode during discharge?
Mission 3

Trade-offs

PrimarySecondaryFuel cell
Forcheap; light and portable; long shelf liferecharged hundreds of times; cheaper per use; less wasteruns while fuel is supplied; efficient; a H₂ cell's only product is water
Againstsingle use; waste to dispose of or recycledearer to buy; capacity falls with use; slow to rechargecostly catalysts (platinum); H₂ hard to store; most H₂ is made from fossil fuels
Quick check
?Which statement about fuel cells, primary cells and secondary cells is correct?
Mission check

?Why can a phone's lithium-ion cell be recharged, while a standard alkaline AA cell cannot?
Mission 4

Mission 4: Why does a bulb light only once solid lead(II) bromide has melted?

Make a guess: Why does a bulb light only once solid lead(II) bromide has melted?

Mission 4

The electrolytic cell

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of molten lead(II) bromide with graphite electrodesDC supply−+bulbmolten lead, Pb(l)cathode(−)anode(+)Pb2+Br−molten PbBr2Br2(g)brownvapourheatcathode: Pb2+ + 2e− → Pb(l)anode: 2Br− → Br2(g) + 2e−done in a fume cupboard
  • An electrolytic cell uses electrical energy to drive a non-spontaneous redox reaction.
  • A DC power supply is connected to two electrodes in an electrolyte.
  • Anode: joined to the positive terminal. Cathode: joined to the negative terminal.
  • Oxidation is still at the anode; reduction at the cathode.
Mission 4

Voltaic and electrolytic cells side by side

Voltaic cellElectrolytic cell
Energy changechemical → electricalelectrical → chemical
Reactionspontaneousnon-spontaneous
Anode (oxidation)negativepositive
Cathode (reduction)positivenegative
Quick check
?Which statement about an electrolytic cell is correct?
Mission 4

How the current is carried

  • In the wires and electrodes: electrons.
  • In the electrolyte: ions moving. Cations go to the cathode; anions go to the anode.
  • In a solid ionic compound the ions are held in the lattice: no conduction.
  • Molten (or dissolved), the ions are free to move. No electrons pass through the melt.
Mission 4

Products from a molten salt: worked example

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of molten lead(II) bromide with graphite electrodesDC supply−+bulbmolten lead, Pb(l)cathode(−)anode(+)Pb2+Br−molten PbBr2Br2(g)brownvapourheatcathode: Pb2+ + 2e− → Pb(l)anode: 2Br− → Br2(g) + 2e−done in a fume cupboard

Molten lead(II) bromide, PbBr₂(l), with inert electrodes:

  • Cathode (−): Pb²⁺ + 2e⁻ → Pb(l). Molten lead collects.
  • Anode (+): 2Br⁻ → Br₂(g) + 2e⁻. Brown vapour.
  • Overall: PbBr₂(l) → Pb(l) + Br₂(g).
  • A molten binary salt gives the metal at the cathode and the non-metal at the anode.
Quick check
?Molten zinc chloride is electrolysed with inert electrodes. What forms at each electrode?
Mission check

?Molten lead(II) bromide is electrolysed using graphite electrodes. How is the electric current conducted through the molten lead(II) bromide itself?
Mission 5

Mission 5: How does a breath tester detect alcohol?

Make a guess: How does a breath tester detect alcohol?

Mission 5

Primary, secondary, tertiary alcohols

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secondary and tertiary alcohols as displayed formulasethanolprimaryHHHHHC1COHpropan-2-olsecondaryHHHHHHOHHC1C2C2-methylbutan-2-oltertiaryOHHHHHHHHHHHHC1C2C3CCringed C: the carbon bonded to –OHblue numbers: carbons bonded to itorange H: hydrogen on that carbon
  • Find the carbon bonded to the –OH. Count the carbons attached to it.
  • Primary: one (methanol: none). Ethanol, CH₃CH₂OH.
  • Secondary: two. Propan-2-ol, CH₃CH(OH)CH₃.
  • Tertiary: three. 2-methylbutan-2-ol, (CH₃)₂C(OH)CH₂CH₃.
  • Primary and secondary alcohols have an H on that carbon; tertiary alcohols have none.
Mission 5

What oxidation gives

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map for alcohols with [O]oxidation: [O] is the oxidizing agentprimaryalcoholaldehydecarboxylicacid[O][O]CH3CH2OH → CH3CHO → CH3COOHsecondaryalcoholketone[O]stopsCH3CH(OH)CH3 → CH3COCH3tertiaryalcoholnot oxidizedno H on the C–OH carbon
  • [O] stands for the oxidizing agent; you need not name it.
  • Primary → aldehyde → carboxylic acid, in two steps:
  • CH₃CH₂OH + [O] → CH₃CHO + H₂O, then CH₃CHO + [O] → CH₃COOH.
  • Secondary → ketone: CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O.
  • Tertiary: not oxidized under the same conditions; there is no H on the C–OH carbon.
Quick check
?Which of these is a secondary alcohol?
Mission 5

Distillation or reflux?

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for heating under reflux and for distillationRefluxfor the acidwater inwater outverticalcondenseropen at the topvapour condensesand runs backanti-bumpinggranulesheater orwater bathDistillationfor the aldehydethermometerstillheadreceiverwater inwater outcondenserelectric heater or water bathno naked flame: organic liquids are flammable
  • The aldehyde boils lowest: it has no O–H, so no hydrogen bonding between its molecules.
  • Distillation: the aldehyde boils off as it forms, so it is not oxidized further.
  • Reflux: a vertical condenser returns vapour to the flask, so the aldehyde stays and becomes the acid.
  • For the aldehyde: distil, with the alcohol in excess. For the acid: reflux, with excess oxidizing agent.
Quick check
?Butan-1-ol is to be oxidized to butanoic acid with excess acidified oxidizing agent. Which set-up should be used?
Quick check
?Pentan-3-ol is heated under reflux with excess acidified oxidizing agent. Which equation is correct?
Mission 5

Reduction: adding hydrogen to C=O

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map: carbonyl compounds and acids with [H]reduction: [H] is the reducing agentcarboxylicacidaldehydeprimaryalcohol[H][H]CH3COOH + 2[H] → CH3CHO + H2OCH3CHO + 2[H] → CH3CH2OHCH3COOH + 4[H] → CH3CH2OH + H2O(overall)ketonesecondaryalcohol[H]CH3COCH3 + 2[H] → CH3CH(OH)CH3H− adds to the C=O carbonOCRR′H−
  • Reducing a carbonyl compound or acid is a gain of hydrogen. [H] stands for the reducing agent.
  • The reducing agent supplies hydride ions, H⁻: hydrogen with two electrons, which adds to the C=O carbon.
  • Acid → aldehyde: CH₃COOH + 2[H] → CH₃CHO + H₂O.
  • Aldehyde → primary alcohol: CH₃CHO + 2[H] → CH₃CH₂OH.
  • Overall: CH₃COOH + 4[H] → CH₃CH₂OH + H₂O.
  • Ketone → secondary alcohol: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃.
Quick check
?Carbonyl compounds are reduced by reducing agents that act as a source of hydride ions, H⁻. What is the role of the hydride ion?
Quick check
?Which describes the reduction of propanoic acid, CH₃CH₂COOH?
Quick check
?Which equation shows the complete reduction of butanoic acid, CH₃CH₂CH₂COOH, to an alcohol?
Quick check
?Which compound needs the most [H] per molecule to be reduced completely to an alcohol?
Mission 5

Hydrogenation

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of but-2-yne to but-2-ene to butane, showing sigma and pi bondsbut-2-yneCH3CCCH3C≡C: one σ + two π+ H2Ni catalystbut-2-eneCH3CHCHCH3C=C: one σ + one π+ H2Ni catalystbutaneCH3CH2CH2CH3C–C: σ onlyσ bond: keptπ bond: breaks
  • Hydrogenation: H₂ adds across C=C or C≡C, over a heated nickel catalyst.
  • It lowers the degree of unsaturation: a reduction (gain of hydrogen).
  • One H₂ per π bond: C=C takes 1 H₂; C≡C takes 2 H₂. The σ bond stays.
  • CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃.
  • CH₃C≡CCH₃ + 2H₂ → CH₃CH₂CH₂CH₃, via but-2-ene.
Quick check
?Hex-1-en-5-yne, CH₂=CHCH₂CH₂C≡CH, reacts completely with excess hydrogen in the presence of a nickel catalyst. What amount of H₂ reacts with 1 mol of hex-1-en-5-yne?
Quick check
?Buta-1,3-diene, CH₂=CHCH=CH₂, reacts completely with excess hydrogen over a nickel catalyst. Which statement is correct?
Mission check

?Some breath testers contain a fuel cell. Ethanol in the breath reacts at one electrode, producing a current. Which statement is correct?
Recap

The big ideas

  • Oxidation is loss of electrons, a rise in oxidation state. The oxidizing agent is itself reduced.
  • In acid, balance O with H₂O, H with H⁺, then charge with e⁻.
  • Metals are oxidized more readily down a group; halogens are reduced more readily up group 17.
  • Reactive metal + dilute acid → salt + hydrogen.
  • Anode = oxidation. Voltaic: anode negative. Electrolytic: anode positive. Electrons travel only in wires.
  • Molten salt: ions carry the current; metal at the cathode, non-metal at the anode.
  • Primary alcohol → aldehyde → acid; secondary → ketone; tertiary is not oxidized. Reduction and hydrogenation add hydrogen.
🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Electron transfer — redox for IB Diploma Chemistry HL. 🎉

  1. Mission 1When iron rusts, does it gain or lose electrons?
  2. Mission 2Why does zinc fizz in acid when a gold ring doesn't?
  3. Mission 3Why can you recharge a phone battery but not a standard AA battery?
  4. Mission 4Why does a bulb light only once solid lead(II) bromide has melted?
  5. Mission 5How does a breath tester detect alcohol?

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