Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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R3.2.1 Oxidation
Oxidation
Loss of electrons by a species, shown by an increase in the oxidation state of an element. In suitable cases oxidation can also be recognized as gain of oxygen or loss of hydrogen. Example: Fe²⁺ → Fe³⁺ + e⁻ (iron's oxidation state rises from +2 to +3).
Reduction
Gain of electrons by a species, shown by a decrease in the oxidation state of an element. In suitable cases reduction can also be recognized as loss of oxygen or gain of hydrogen. Oxidation and reduction always occur together in a redox reaction.
Oxidation state
A number assigned to an atom in a species as if all its bonds were ionic, written with the sign first (+3, −2). Elements are 0; the oxidation states in a compound sum to 0 and in an ion sum to the ion's charge; O is usually −2 (−1 in peroxides), H is usually +1 (−1 in metal hydrides), F is always −1. Example: in Cr₂O₇²⁻, 2x + 7(−2) = −2, so x = +6.
Oxidizing agent and reducing agent
An oxidizing agent is the reactant that causes another species to be oxidized; it accepts electrons and is itself reduced. A reducing agent causes another species to be reduced; it donates electrons and is itself oxidized. In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, Cl₂ is the oxidizing agent and Fe²⁺ the reducing agent.
Variable oxidation states
Transition elements form ions in more than one oxidation state (iron +2 and +3; manganese +2, +4, +6 and +7; chromium +3 and +6), and most main-group non-metals show a range of states (nitrogen from −3 in NH₃ to +5 in NO₃⁻; sulfur −2 in H₂S, +4 in SO₂, +6 in SO₄²⁻; chlorine −1 in Cl⁻ to +7 in ClO₄⁻).
Oxidation numbers in names
A Roman numeral in a name gives the oxidation state of the element it follows, not the number of atoms: iron(II) chloride is FeCl₂, iron(III) chloride is FeCl₃, potassium manganate(VII) is KMnO₄ and potassium dichromate(VI) is K₂Cr₂O₇.
Students often think The oxidation state of the central atom of a polyatomic ion is the charge on the ion. In fact −3. The four H atoms contribute 4(+1) = +4 and the states must sum to the ion's charge, +1, so N is +1 − 4 = −3.
Students often think The oxidation states in any species add up to zero. In fact The charge on the ion (+1 for NH₄⁺, −2 for MnO₄²⁻), not zero.
R3.2.2 Half-equation
Half-equation
An equation showing either the oxidation or the reduction part of a redox reaction, with electrons written explicitly: as products in an oxidation (Zn → Zn²⁺ + 2e⁻) and as reactants in a reduction (Cl₂ + 2e⁻ → 2Cl⁻). Atoms and charge must both balance.
Balancing half-equations in acidic or neutral solution
Balance the element changing oxidation state; balance O by adding H₂O; balance H by adding H⁺; balance charge by adding e⁻. Example: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. To combine two half-equations, multiply each so that the electrons lost equal the electrons gained, then add and cancel the electrons.
Students often think The number of electrons in a half-equation equals the charge on the metal ion formed. In fact No. Five electrons are needed: manganese goes from +7 to +2, and the charges balance only with 5e⁻ (−1 + 8 − 5 = +2).
Students often think Reduction is a loss (of electrons), so electrons appear as products in a reduction half-equation. In fact As reactants: reduction is gain of electrons, e.g. Cl₂ + 2e⁻ → 2Cl⁻.
R3.2.3 Relative ease of oxidation of metals
Relative ease of oxidation of metals
Metals are oxidized more readily down a group and towards the left of a period, because the outer electrons are further from the nucleus or are attracted by a smaller effective nuclear charge. Potassium is oxidized more readily than sodium, and sodium more readily than magnesium.
Relative ease of reduction of halogens
Halogens are reduced (gain electrons to form halide ions) more readily going up group 17, because the smaller atom attracts an added electron more strongly. Order: F₂ > Cl₂ > Br₂ > I₂. A halogen oxidizes the halide ions of halogens below it: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
Metal/metal-ion displacement
A metal that is oxidized more readily reduces the aqueous ions of a metal that is oxidized less readily, displacing that metal: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The observations (a coating of metal, a change in solution colour) allow metals to be ranked; the ions of the least readily oxidized metal are the strongest oxidizing agents among them.
Students often think Reactivity changes in the same direction down every group, so the trend learned for one group (alkali metals or halogens) applies to the other. In fact No. Metals in group 1 are oxidized more readily going DOWN the group; halogens are reduced more readily going UP the group.
Students often think A metal with more valence electrons has more electrons available to lose, so it is oxidized more readily. In fact No. Aluminium (three valence electrons) is oxidized less readily than sodium (one); more electrons to remove means more energy is needed.
R3.2.4 Reaction of reactive metals with dilute acids
Reaction of reactive metals with dilute acids
A reactive metal reduces the H⁺(aq) of a dilute acid to hydrogen gas and is itself oxidized to its aqueous ion, forming a salt: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g); Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g). Ionic equation: M(s) + 2H⁺(aq) → M²⁺(aq) + H₂(g) for a metal forming M²⁺.
Students often think Equations for metal–acid reactions can be written by copying the pattern of a familiar example (Mg + 2HCl → MgCl₂ + H₂), so every metal gives one H₂ per metal atom. In fact No. The amount of H₂ depends on the charge of the metal ion: Mg → Mg²⁺ gives one H₂ per Mg, but Al → Al³⁺ gives 3/2 H₂ per Al (2Al + 6HCl → 2AlCl₃ + 3H₂).
Students often think Non-metal elements formed in a reaction are written as single atoms (H, I), one for each ion discharged. In fact No. Hydrogen is released as diatomic H₂ molecules: 2H⁺(aq) + 2e⁻ → H₂(g). Likewise iodide is oxidized to I₂, not I.
R3.2.5 Anode and cathode
Anode and cathode
The anode is the electrode at which oxidation occurs; the cathode is the electrode at which reduction occurs. This holds in both voltaic and electrolytic cells; the signs differ. Voltaic cell: anode negative, cathode positive. Electrolytic cell: anode positive (connected to the positive terminal of the supply), cathode negative.
Students often think The anode is always the positive electrode (and the cathode always negative). In fact No. In an electrolytic cell the anode is positive, but in a voltaic cell the anode is negative. The anode is defined by oxidation, not by its sign.
Students often think Because 'cation' and 'cathode' sound alike, the electrode where cations are produced is the cathode. In fact Not necessarily. The cathode is where reduction occurs, and in a voltaic cell cations are formed by oxidation at the anode (Zn → Zn²⁺ + 2e⁻). Cations are REDUCED at the cathode.
R3.2.6 Primary (voltaic) cell
Primary (voltaic) cell
An electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. It is built from two half-cells (each a metal in a solution of its ions), an external circuit (wire) joining the electrodes, and a salt bridge joining the solutions. Electrons flow in the external circuit from the anode (negative) to the cathode (positive).
Salt bridge
A tube or strip containing an unreactive electrolyte such as KNO₃(aq) that connects the two half-cells. It completes the circuit by allowing ions (not electrons) to move: cations move towards the cathode half-cell and anions towards the anode half-cell, keeping each solution electrically neutral.
Students often think Electrons flow through the salt bridge (or the electrolyte) to complete the circuit. In fact No. Electrons travel only through the external wires and electrodes. In the salt bridge and electrolyte, charge is carried by moving ions.
Students often think Ions in the salt bridge move towards the electrode of opposite charge, so anions move to the positive cathode. In fact No. Anions move towards the ANODE half-cell, where cations are being produced and positive charge is building up in solution.
R3.2.7 Secondary (rechargeable) cell
Secondary (rechargeable) cell
A cell in which the discharge redox reactions can be reversed by an external electrical supply. During charging each electrode half-equation is the reverse of its discharge half-equation, so the electrode that was the anode on discharge becomes the cathode on charging. Example (lead–acid, negative electrode): discharge Pb + SO₄²⁻ → PbSO₄ + 2e⁻; charging PbSO₄ + 2e⁻ → Pb + SO₄²⁻.
Fuel cell
A voltaic cell in which the reactants (a fuel such as hydrogen, and oxygen) are supplied continuously and the products removed, so it produces electricity for as long as fuel is supplied. Advantages: continuous operation, high efficiency, with water as the only product of a hydrogen–oxygen cell. Disadvantages: expensive electrode catalysts and the need to produce, store and transport the fuel. Primary cells are cheap and portable but single-use; secondary cells can be recharged many times but cost more initially.
Students often think Charging runs the same reactions as discharge, simply putting more energy into the cell. In fact No. Charging reverses each discharge half-equation, restoring the original reactants.
Students often think A half-equation is reversed by moving only the electrons to the other side, leaving the other species where they are. In fact No. The whole equation is reversed: all reactants become products, including the electrons.
R3.2.8 Electrolytic cell
Electrolytic cell
An electrochemical cell in which electrical energy from a DC power source drives a non-spontaneous redox reaction. It consists of a DC supply connected to two electrodes (anode to the positive terminal, cathode to the negative terminal) dipping into an electrolyte. Current is carried by electrons in the external wires and electrodes and by the movement of ions in the electrolyte.
Electrolysis of a molten salt
In a molten binary salt the only ions are the metal cation and the non-metal anion. The cation is reduced to the metal at the cathode and the anion is oxidized to the non-metal at the anode. Molten NaCl: cathode 2Na⁺ + 2e⁻ → 2Na; anode 2Cl⁻ → Cl₂ + 2e⁻.
Students often think A molten ionic compound conducts because melting releases delocalized electrons, as in a metal. In fact No. A molten ionic compound conducts because its ions are free to move; it contains no delocalized electrons.
Students often think Passing a current splits the compound into ions, and these ions then carry the current. In fact No. The ions are already present in the molten or dissolved ionic compound; the current makes them move and brings about their discharge at the electrodes.
R3.2.9 Oxidation of alcohols
Oxidation of alcohols
Primary alcohols are oxidized in two steps: first to an aldehyde, then to a carboxylic acid (CH₃CH₂OH + [O] → CH₃CHO + H₂O; CH₃CHO + [O] → CH₃COOH). Secondary alcohols are oxidized to ketones. Tertiary alcohols are not oxidized under similar conditions, because the carbon bearing the OH group has no hydrogen atom.
Distillation and reflux
Distillation: the mixture is heated and vapour passes into a condenser and is collected elsewhere, so a volatile aldehyde is removed as soon as it forms, before it can be oxidized further. Reflux: a vertical condenser returns vapour to the flask, so the aldehyde stays in contact with the oxidizing agent and is oxidized to the carboxylic acid.
Primary, secondary and tertiary alcohols
Classified by the number of carbon atoms bonded to the carbon that carries the OH group: one (primary, e.g. CH₃CH₂OH), two (secondary, e.g. CH₃CH(OH)CH₃) or three (tertiary, e.g. (CH₃)₃COH).
Students often think Reflux removes volatile products from the flask, and distillation keeps them in it. In fact No. Reflux returns the condensed vapour to the flask, keeping the aldehyde in contact with the oxidizing agent. Distillation removes it.
Students often think A primary alcohol and its carboxylic acid are converted into one another directly, in a single step. In fact No. The conversion goes through the aldehyde in both directions: RCH₂OH → RCHO → RCOOH on oxidation, and RCOOH → RCHO → RCH₂OH on reduction.
R3.2.10 Reduction of carbonyl compounds and carboxylic acids
Reduction of carbonyl compounds and carboxylic acids
Carboxylic acids are reduced to primary alcohols via the aldehyde (RCOOH → RCHO → RCH₂OH); aldehydes are reduced to primary alcohols and ketones to secondary alcohols (CH₃COCH₃ → CH₃CH(OH)CH₃). Each step is a gain of hydrogen.
Hydride ion
The ion H⁻: a hydrogen atom carrying an extra electron. In the reduction of carbonyl compounds and carboxylic acids the reducing agent acts as a source of hydride ions, which add hydrogen to the carbon of the C=O group, so the organic compound gains hydrogen and is reduced.
Students often think Aldehydes and ketones are not distinguished, so the carbonyl compound formed from, or lying between, an alcohol and a carboxylic acid can be written as either an aldehyde or a ketone. In fact No. The C=O stays on the carbon that carried the functional group: a primary alcohol and a carboxylic acid are linked via the aldehyde (C=O on the end carbon); a secondary alcohol is linked to a ketone (C=O within the chain).
Students often think Reduction of any carbonyl compound gives a primary alcohol with the OH on the end carbon. In fact No. Aldehydes (and carboxylic acids) give primary alcohols, but ketones give secondary alcohols.
R3.2.11 Hydrogenation and degree of unsaturation
Hydrogenation and degree of unsaturation
Addition of hydrogen across C=C or C≡C bonds, using a catalyst such as nickel. One mole of H₂ saturates one C=C bond; a C≡C bond takes one mole of H₂ to give the alkene and a second to give the alkane. Each addition lowers the degree of unsaturation and is a reduction (gain of hydrogen). Example: CH₃C≡CCH₃ + 2H₂ → CH₃CH₂CH₂CH₃.
Students often think The amount of H₂ (in mol) equals the amount of hydrogen atoms added or released. In fact n/2: each H₂ molecule supplies two hydrogen atoms.
Students often think Each multiple bond, whether double or triple, reacts with one H₂ molecule. In fact No. A C=C bond takes one mole of H₂, but a C≡C bond takes two moles to become a single bond.
R3.2.12 Standard hydrogen electrode HL
Standard hydrogen electrode
A half-cell of hydrogen gas at 100 kPa bubbled over a platinum electrode in a solution with [H⁺] = 1 mol dm⁻³ at 298 K. By convention its standard electrode potential is 0 V (exactly), and the E⦵ of any other half-cell is measured as the potential difference between it and this reference.
Standard electrode potential, E⦵
The potential difference, in volts (V), between a half-cell under standard conditions (298 K, 100 kPa, 1 mol dm⁻³ solutions) and the standard hydrogen electrode, quoted for the reduction half-equation. A more positive E⦵ means the oxidized form is more easily reduced (stronger oxidizing agent); a more negative E⦵ means the reduced form is more easily oxidized (stronger reducing agent).
Students often think A standard electrode potential of zero means that no oxidation or reduction occurs at that electrode. In fact No. H⁺(aq) + e⁻ ⇌ ½H₂(g) does occur at the platinum surface; the value 0 V is assigned by convention.
Students often think Each half-cell has a potential that can be measured on its own, and the hydrogen half-cell happens to measure zero. In fact No. Only a potential difference between two electrodes can be measured, so a reference half-cell is needed and assigned 0 V.
R3.2.13 Standard cell potential, E⦵cell HL
Standard cell potential, E⦵cell
E⦵cell = E⦵(cathode) − E⦵(anode), in V, where the cathode is the half-cell at which reduction occurs. E⦵ values are not multiplied by stoichiometric coefficients. A positive E⦵cell means the reaction is spontaneous as written; a negative value means the reverse reaction is spontaneous.
Students often think E⦵ values must be multiplied by the same factor as the half-equation, like ΔH values. In fact No. E⦵ is an intensive property (energy per unit charge) and is not multiplied by stoichiometric coefficients.
Students often think E⦵cell is calculated as E⦵(anode) − E⦵(cathode), i.e. the two electrode roles are swapped in the subtraction. In fact No. E⦵cell = E⦵(cathode) − E⦵(anode), where the cathode is where reduction occurs.
R3.2.14 ΔG⦵ from E⦵cell HL
ΔG⦵ from E⦵cell
ΔG⦵ = −nFE⦵cell, where n is the amount of electrons transferred per mole of reaction as written, F = 96 500 C mol⁻¹ (the Faraday constant) and E⦵cell is in V; the product is in J mol⁻¹, divided by 1000 for kJ mol⁻¹. A positive E⦵cell gives a negative ΔG⦵ (spontaneous).
Students often think ΔG⦵ = −FE⦵cell, i.e. n is always 1 or can be ignored. In fact No. n is the amount of electrons transferred per mole of reaction as written (2 for Mg + 2Ag⁺ → Mg²⁺ + 2Ag).
Students often think n is the number of electrons written in one half-equation (3 for Al → Al³⁺ + 3e⁻, or 2 for Cu²⁺ + 2e⁻ → Cu), not the number transferred in the balanced overall equation. In fact n = 6. Two Al atoms each lose three electrons (6e⁻ in total) and three Cu²⁺ ions each gain two (6e⁻ in total); n is the number of moles of electrons transferred in the equation as balanced.
R3.2.15 Competing reactions in aqueous electrolysis HL
Competing reactions in aqueous electrolysis
In aqueous solutions water can be reduced at the cathode (2H₂O + 2e⁻ → H₂ + 2OH⁻) and oxidized at the anode (2H₂O → O₂ + 4H⁺ + 4e⁻), competing with the ions of the solute. E⦵ values predict the favoured product, but concentration and electrode material can change it: concentrated NaCl(aq) gives chlorine rather than oxygen at the anode, and a copper anode in CuSO₄(aq) is itself oxidized to Cu²⁺.
Students often think Electrolysis always produces hydrogen and oxygen from water, whatever the electrolyte. In fact No. Water competes with the ions of the solute and does not always react: Cl⁻ is oxidized in concentrated NaCl(aq), and Cu²⁺ is reduced in CuSO₄(aq).
Students often think The product of electrolysis is always the one predicted by E⦵ values alone; concentration has no effect. In fact No. E⦵ values alone favour oxygen (O₂/H₂O +1.23 V vs Cl₂/Cl⁻ +1.36 V), but in concentrated chloride solution chlorine is the main product.
R3.2.16 Electroplating HL
Electroplating
Electrolytic coating of an object with a thin layer of metal. The object is the cathode, the anode is usually the plating metal, and the electrolyte contains ions of that metal. Copper plating: cathode Cu²⁺(aq) + 2e⁻ → Cu(s); anode Cu(s) → Cu²⁺(aq) + 2e⁻, so the concentration of Cu²⁺ stays constant.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 What is the oxidation state of nitrogen in the ammonium ion, NH₄⁺?
Answer and reasoning
+1 — A student who takes the charge on the ion as the oxidation state of the central atom picks this. The +1 is the SUM of the oxidation states of all five atoms: x + 4(+1) = +1 gives N = −3.
−4 — A student who makes the oxidation states add up to zero gets x + 4 = 0, x = −4. NH₄⁺ is an ion, so the states must add up to its charge, +1, giving −3.
−3 — Each H is +1, contributing +4 in total. The oxidation states must add up to the ion's charge, +1: x + 4 = +1, so x = −3.
+4 — A student who equates oxidation state with the number of bonds formed counts four N–H bonds. Oxidation state assigns the bonding electrons to the more electronegative atom (N), so N is −3, not +4.
Working Let the oxidation state of N be x. H is +1. Sum = charge on ion: x + 4(+1) = +1, so x = −3.
2 Which half-equation correctly shows the conversion of MnO₄⁻(aq) into Mn²⁺(aq) in acidic solution?
Answer and reasoning
MnO₄⁻ + 8H⁺ + 2e⁻ → Mn²⁺ + 4H₂O — A student who sets the number of electrons equal to the charge on Mn²⁺ writes 2e⁻. The charges do not balance (left +5, right +2). Mn changes from +7 to +2, so 5e⁻ are needed.
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5e⁻ — A student who thinks reduction means losing electrons writes them as products. Mn falls from +7 to +2, which is a gain of electrons, and the charges balance only with 5e⁻ on the left (−1 + 8 − 5 = +2).
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 3e⁻ — A student who balances charge between MnO₄⁻ (−1) and Mn²⁺ (+2) only, ignoring the 8H⁺, adds 3e⁻ to the right to bring it to −1. Including the H⁺, the left side is −1 + 8 = +7, so 5e⁻ must be added to the LEFT to reach +2.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O — O is balanced with 4H₂O, H with 8H⁺, and charge with electrons: left −1 + 8 − 5 = +2, right +2. Mn goes from +7 to +2, a gain of five electrons (reduction), so electrons are reactants.
Working O: 4 O on the left → 4H₂O on the right. H: 8 H in 4H₂O → 8H⁺ on the left. Charge: left −1 + 8 = +7, right +2, so add 5e⁻ to the left. Check: Mn +7 → +2 is a gain of 5 electrons.
3 Which equation correctly represents the reaction of aluminium with dilute hydrochloric acid?
Answer and reasoning
Al(s) + 3HCl(aq) → AlCl₃(aq) + H₂(g) — A student who copies the pattern of Mg + 2HCl → MgCl₂ + H₂ releases one H₂ per metal atom. Each Al gives three electrons, reducing 3H⁺ to 1½H₂; this equation is not balanced for H, and doubling gives 3H₂ per 2Al.
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) — Aluminium forms Al³⁺, so the salt is AlCl₃, and the H⁺ ions are reduced to diatomic hydrogen. Balancing: 2 Al, 6 H and 6 Cl on each side.
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 6H(g) — A student who gives off one H atom for each H⁺ reduced writes 6H(g). Hydrogen is diatomic: 6H⁺ gain 6 electrons to form 3H₂ molecules.
Al(s) + 3HCl(aq) → AlCl₃(aq) + H₂O(l) — A student who applies 'acid + base → salt + water' to a metal writes water as the product. A metal contains no oxygen; the H⁺ ions are reduced to hydrogen gas: 2Al + 6HCl → 2AlCl₃ + 3H₂.
4 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?
Answer and reasoning
It is positive, as oxidation takes place at a positive anode. — A student who thinks the anode is always positive picks this. That is true only in electrolytic cells. In a voltaic cell the anode supplies electrons to the circuit and is negative.
It is the cathode, as positive zinc ions are formed there. — A student who links 'cation' with 'cathode' picks this. Forming Zn²⁺ is oxidation, which defines the anode. Cations are REDUCED at the cathode.
It is negative, as zinc is oxidized there and releases electrons. — Oxidation (Zn → Zn²⁺ + 2e⁻) makes zinc the anode, and the electrons it releases into the wire make it the negative electrode of the voltaic cell.
It is the cathode, as the oxidation of zinc atoms takes place there. — A student who pairs oxidation with the cathode picks this. Zinc atoms are indeed oxidized at this electrode, but oxidation defines the ANODE; in a voltaic cell the anode is negative.
5 Which statement about fuel cells, primary cells and secondary cells is correct?
Answer and reasoning
A fuel cell supplies electricity for as long as its fuel and oxygen are supplied. — In a fuel cell the reactants are fed in continuously and the products removed, so it does not run down like a primary or secondary cell, which contains a fixed amount of reactants.
A secondary cell stores electricity, which it releases unchanged on discharge. — A student who pictures a battery as a container of electricity picks this. A secondary cell stores chemical energy; electricity is generated by a redox reaction during discharge.
A hydrogen fuel cell produces only water, so using one causes no pollution of any kind. — A student who considers only the cell reaction picks this. Producing, compressing and storing hydrogen uses energy and, at present, mostly fossil fuels: a recognized disadvantage of hydrogen fuel cells.
A fuel cell burns its fuel and converts the heat released to electrical energy. — A student who links 'fuel' with burning picks this. In a fuel cell hydrogen is oxidized at the anode and oxygen reduced at the cathode; chemical energy is converted directly into electrical energy, without combustion.
6 Propan-1-ol is converted into propanoic acid by heating it under reflux with an acidified oxidizing agent. Why is reflux used?
Answer and reasoning
Reflux removes propanal from the flask as soon as it has formed. — A student who swaps the roles of reflux and distillation picks this. Removing propanal as it forms is what distillation does, giving the aldehyde, not the acid.
Reflux converts propan-1-ol straight to the acid, with no propanal formed. — A student who thinks the acid forms in one step picks this. Oxidation goes via the aldehyde, propanal; reflux keeps that intermediate in the flask so it is oxidized too.
Reflux keeps air in contact with the mixture so its oxygen can react. — A student who thinks organic oxidation needs oxygen gas picks this. The oxidizing agent in the flask supplies the oxidation; reflux simply stops volatile compounds escaping.
Reflux returns condensed vapour, so propanal stays in the flask to be oxidized. — The vertical condenser returns volatile propanal (formed in the first step) to the flask, where it is oxidized further to propanoic acid.
7 Carbonyl compounds are reduced by reducing agents that act as a source of hydride ions, H⁻. What is the role of the hydride ion?
Answer and reasoning
It adds hydrogen to the carbon of the C=O group, so the compound is reduced. — The hydride ion supplies hydrogen, with an electron pair, to the carbonyl carbon. The organic compound gains hydrogen, which is reduction.
It removes hydrogen from the carbonyl compound, which reduces the compound. — A student who has the hydrogen definitions reversed picks this. Loss of hydrogen is oxidation; the hydride ion ADDS hydrogen, reducing the compound.
It acts as an acid, donating a proton, H⁺, to the oxygen of the C=O group. — A student who treats every hydrogen ion as a proton picks this. H⁻ carries two electrons and a negative charge; it is not an acid. Its role is to add hydrogen to the C=O carbon.
It is a catalyst, written above the arrow as it is not used up in the reaction. — A student who assumes anything over the arrow is a catalyst picks this. The hydride source is consumed: it is the reducing agent, and it adds hydrogen to the compound.
8 Why is the standard electrode potential of the hydrogen half-cell, H⁺(aq) + e⁻ ⇌ ½H₂(g), given as 0 V? HL
Answer and reasoning
It is assigned by convention as the reference for other half-cells. — Only potential differences can be measured, so one half-cell must be chosen as the reference and given a value. By convention it is the standard hydrogen electrode, set at 0 V.
No oxidation or reduction occurs at the hydrogen electrode. — A student who reads zero as 'nothing happens' picks this. H⁺/H₂ electron transfer does occur at the platinum surface; the electrode acts as anode or cathode depending on the other half-cell.
Its absolute potential has been measured directly and found to be zero. — A student who thinks single electrode potentials can be measured picks this. A voltmeter always measures a difference between two electrodes; 0 V is a defined, not a measured, value.
The platinum electrode is inert, so no potential develops. — A student who links inert platinum with zero potential picks this. Other couples measured on platinum, such as Fe³⁺/Fe²⁺, have non-zero E⦵; the 0 V comes from convention.
9 Standard electrode potentials:
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E⦵ = −0.76 V
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E⦵ = −0.45 V
Sn²⁺(aq) + 2e⁻ ⇌ Sn(s) E⦵ = −0.14 V
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E⦵ = +0.34 V
Which reaction is spontaneous under standard conditions? HL
Answer and reasoning
Fe(s) + Zn²⁺(aq) → Fe²⁺(aq) + Zn(s) — A student who reads −0.76 V as the 'largest' E⦵ and so takes Zn²⁺ to be the strongest oxidizing agent picks this. The sign matters: the most negative E⦵ makes Zn²⁺ the weakest oxidizing agent of the four. E⦵cell = −0.76 − (−0.45) = −0.31 V, so the reaction is not spontaneous; the reverse (Zn + Fe²⁺) is.
Cu(s) + Sn²⁺(aq) → Cu²⁺(aq) + Sn(s) — A student who adds the two E⦵ values gets +0.34 + (−0.14) = +0.20 V, the only positive sum among the options, and concludes this reaction is spontaneous. E⦵cell is a difference: −0.14 − (+0.34) = −0.48 V. Cu²⁺/Cu has the more positive E⦵, so Cu²⁺ is reduced and Sn oxidized, not the reverse.
Cu(s) + Fe²⁺(aq) → Cu²⁺(aq) + Fe(s) — A student who takes a negative E⦵cell to mean 'spontaneous', as for ΔG⦵, picks the most negative value: −0.45 − (+0.34) = −0.79 V. The sign convention for E⦵cell is the opposite of that for ΔG⦵: because ΔG⦵ = −nFE⦵cell, a POSITIVE E⦵cell gives a negative ΔG⦵ and a spontaneous reaction. This reaction is the least favourable of the four.
Fe(s) + Sn²⁺(aq) → Fe²⁺(aq) + Sn(s) — Sn²⁺ is reduced (E⦵ = −0.14 V) and Fe is oxidized (E⦵ = −0.45 V): E⦵cell = −0.14 − (−0.45) = +0.31 V. A positive E⦵cell means the reaction is spontaneous under standard conditions; the couple with the more positive E⦵ is the one reduced.
Working E⦵cell = E⦵(cathode, species reduced) − E⦵(anode, species oxidized).
Fe + Zn²⁺ → Fe²⁺ + Zn: Zn²⁺ reduced, Fe oxidized: −0.76 − (−0.45) = −0.31 V (not spontaneous).
Fe + Sn²⁺ → Fe²⁺ + Sn: Sn²⁺ reduced, Fe oxidized: −0.14 − (−0.45) = +0.31 V (spontaneous).
Cu + Sn²⁺ → Cu²⁺ + Sn: Sn²⁺ reduced, Cu oxidized: −0.14 − (+0.34) = −0.48 V (not spontaneous).
Cu + Fe²⁺ → Cu²⁺ + Fe: Fe²⁺ reduced, Cu oxidized: −0.45 − (+0.34) = −0.79 V (not spontaneous).
Only Fe(s) + Sn²⁺(aq) has E⦵cell > 0.
10 When water is electrolysed, a little sulfuric acid is usually added. Hydrogen forms at the cathode and oxygen at the anode. What is the role of the sulfuric acid? HL
Answer and reasoning
It is the source of the hydrogen and oxygen gases produced. — A student who expects the added substance to be what reacts picks this. The gases come from water; the amount of sulfuric acid stays the same.
It supplies ions so that the water conducts a useful current. — Pure water contains very few ions and conducts poorly. The added ions carry the current; water is decomposed at the electrodes, 2H₂O → 2H₂ + O₂, and the acid is not used up.
It lets electrons travel through the water between the electrodes. — A student who thinks electrons flow through the electrolyte picks this. Electrons travel only in the wires and electrodes; ions carry the current through the solution.
It lets the current split the water into ions that carry charge. — A student who thinks the current creates ions picks this. The ions from the acid are present before any current flows; the current moves them.
Read the ones marked not yet in Learn, then Verify.
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32 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Chlorine reacts with aqueous iron(II) ions:
2Fe²⁺(aq) + Cl₂(aq) → 2Fe³⁺(aq) + 2Cl⁻(aq)
Which species is the oxidizing agent in this reaction?
Answer and reasoning
Fe²⁺, since it loses electrons and so is itself oxidized — A student who thinks the oxidizing agent is the species oxidized picks Fe²⁺. Fe²⁺ is oxidized (+2 → +3), which makes it the REDUCING agent; the oxidizing agent is Cl₂, which takes its electrons.
Cl₂, since it gains electrons and so is itself reduced — Chlorine's oxidation state falls from 0 to −1: each Cl₂ gains two electrons from Fe²⁺. The species that accepts electrons, and so is reduced, is the oxidizing agent.
Cl⁻, since it is the species holding the gained electrons — A student who looks for the species that ends up with the electrons picks Cl⁻. Agents are reactants: Cl⁻ is the product of reduction, and the reactant that accepted the electrons, Cl₂, is the oxidizing agent.
None, since no species here gains or loses any oxygen atoms — A student who defines redox only by oxygen transfer decides this is not redox. Oxidation states change (Fe +2 → +3, Cl 0 → −1), so electrons are transferred and the reaction is redox, with Cl₂ as the oxidizing agent.
K₂Cr₂O₇, potassium dichromate(VI) — 2(+1) + 2x + 7(−2) = 0 gives 2x = +12, so each Cr is +6: potassium dichromate(VI). The Roman numeral is the oxidation state of chromium, per atom.
K₂MnO₄, potassium manganate(IV) — A student who takes the Roman numeral as the number of oxygen atoms writes (IV). 2(+1) + x + 4(−2) = 0 gives Mn = +6: potassium manganate(VI).
Fe₂(SO₄)₃, iron(VI) sulfate — A student who forgets to divide by the two Fe atoms gets +6. 2x + 3(−2) = 0 gives 2x = +6, so each Fe is +3: iron(III) sulfate.
MnCl₂, manganese(VII) chloride — A student who fixes manganese at +7, as in its best-known compound KMnO₄, writes (VII). In MnCl₂, x + 2(−1) = 0 gives Mn = +2: manganese(II) chloride. Manganese, a transition element, has variable oxidation states.
3 In acidic solution, MnO₄⁻(aq) is reduced to Mn²⁺(aq) and Fe²⁺(aq) is oxidized to Fe³⁺(aq). In the balanced overall redox equation, what amount of Fe²⁺ reacts with 1 mol of MnO₄⁻?
Answer and reasoning
1 mol — A student who writes both half-equations but adds them as written, without matching the electrons, gets 1 mol. Electrons lost must equal electrons gained: 5 Fe²⁺ are needed to supply the 5 electrons one MnO₄⁻ accepts.
5 mol — Manganese goes from +7 to +2, so each MnO₄⁻ gains 5 electrons (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O); each Fe²⁺ loses 1. The iron half-equation is multiplied by 5: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
2 mol — A student who takes the number of electrons to be the charge on the ion formed, Mn²⁺, writes the manganese half-equation with 2e⁻ and uses 2 Fe²⁺. Manganese goes from +7 to +2, a gain of 5e⁻, so 5 Fe²⁺ are needed.
3 mol — A student who balances charge using only MnO₄⁻ (−1) and Mn²⁺ (+2), ignoring the H⁺ ions, writes the manganese half-equation with 3e⁻ and uses 3 Fe²⁺. With 8H⁺ included, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O balances, so 5 Fe²⁺ react.
Working Mn: +7 → +2, a gain of 5e⁻ per MnO₄⁻ (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O). Fe: +2 → +3, a loss of 1e⁻ per Fe²⁺ (Fe²⁺ → Fe³⁺ + e⁻). Electrons lost = electrons gained, so the iron half-equation is multiplied by 5: 5 mol Fe²⁺ react with 1 mol MnO₄⁻.
4 Using positions in the periodic table, which of these metals would be expected to be oxidized most readily?
Answer and reasoning
lithium — A student who applies the halogen trend (most reactive at the top) to metals picks lithium. For metals, ease of oxidation increases DOWN a group because the outer electron is further from the nucleus.
aluminium — A student who thinks more valence electrons make a metal easier to oxidize picks aluminium. Aluminium must lose three electrons from a small atom with a larger nuclear charge; it is oxidized much less readily than rubidium.
rubidium — Rubidium is in group 1, period 5. Its single outer electron is far from the nucleus and well shielded, so it is lost most easily; all the other options are higher in a group or further to the right.
sodium — A student who judges reactivity from the most dramatic reaction seen in class (sodium in water) picks sodium. Sodium is above rubidium in group 1, so its outer electron is closer to the nucleus and is lost less readily; rubidium is simply too hazardous to demonstrate.
5 Which statement about the relative ease of reduction of the halogens is correct?
Answer and reasoning
Iodine is reduced more readily than bromine, as it is further down group 17. — A student who applies the alkali-metal trend (more reactive down the group) to halogens picks this. Halogens GAIN electrons, which is easier for smaller atoms, so reactivity decreases down group 17.
Bromine is reduced more readily than chlorine, as its nucleus has more protons. — A student who thinks more protons always means stronger attraction picks this. Bromine's extra protons are outweighed by an extra shell and more shielding, so an added electron is held less strongly than in chlorine.
Iodide ions are reduced more readily than iodine, as they are charged. — A student who confuses halide ions with halogens picks this. I⁻ already has a complete outer shell and cannot gain another electron; it is the halogen molecule, I₂, that is reduced to I⁻.
Chlorine is reduced more readily than bromine, as its atoms are smaller. — An electron added to the smaller chlorine atom is closer to the nucleus and less shielded, so it is attracted more strongly. Ease of reduction of the halogens increases up group 17.
6 A student places three metals, X, Y and Z, in solutions of each other's nitrates. X forms a coating in solutions of Y²⁺ and of Z²⁺; Y forms a coating in the solution of Z²⁺ only; Z shows no change in either solution.
Which of these is the strongest oxidizing agent?
Answer and reasoning
the ion Z²⁺ — Z is oxidized least readily (it displaces neither of the others), so its ion, Z²⁺, is the most readily reduced: it accepts electrons from both X and Y. That makes Z²⁺ the strongest oxidizing agent.
the ion X²⁺ — A student who assumes the most reactive metal forms the most reactive ion picks X²⁺. X is oxidized most readily, so X²⁺ has the least tendency to regain electrons: it is the weakest oxidizing agent of the three ions.
the metal X — A student who thinks the oxidizing agent is the species most readily oxidized picks X. X is the strongest REDUCING agent: it gives electrons to Y²⁺ and Z²⁺.
the metal Z — A student who reads 'unreactive' as 'holds electrons strongly' treats metal Z as an electron acceptor. Metal atoms here can only lose electrons; it is Z's ion, Z²⁺, that is readily reduced.
7 Aluminium reacts with excess dilute sulfuric acid. What amount of hydrogen gas is produced when 0.10 mol of aluminium reacts completely?
Answer and reasoning
0.10 mol — A student who copies the pattern of Mg + H₂SO₄ → MgSO₄ + H₂ assumes one H₂ per metal atom. Each Al atom gives three electrons, enough to reduce 3H⁺ to 1.5 H₂, so 0.15 mol H₂ forms.
0.30 mol — A student who counts the hydrogen atoms released (three per Al atom) as moles of H₂ gets 0.30 mol. Hydrogen gas is diatomic: 0.30 mol of H atoms form 0.15 mol of H₂.
0.07 mol — A student who uses the ratio 2 : 3 upside down gets 0.10 × 2/3 = 0.07 mol. From 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂, 2 mol Al give 3 mol H₂: 0.10 × 3/2 = 0.15 mol.
0.15 mol — Aluminium forms Al³⁺, so the equation is 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g). 2 mol Al give 3 mol H₂, so 0.10 mol Al gives 0.10 × 3/2 = 0.15 mol H₂.
Working Al → Al³⁺ + 3e⁻ and 2H⁺ + 2e⁻ → H₂, so 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g). n(H₂) = 0.10 mol × 3/2 = 0.15 mol.
8 Molten sodium chloride is electrolysed. Which statement about the electrode connected to the positive terminal of the DC supply is correct?
Answer and reasoning
It is the anode, where chloride ions lose electrons. — The positive electrode draws electrons from the melt: 2Cl⁻ → Cl₂ + 2e⁻. Oxidation occurs there, so it is the anode; in an electrolytic cell the anode is positive.
It is the cathode, as the cathode is the positive electrode. — A student who fixed 'anode negative, cathode positive' from voltaic cells picks this. In an electrolytic cell the positive electrode is where oxidation occurs, so it is the anode.
It is the cathode, where chloride ions become oxidized. — A student who pairs oxidation with the cathode picks this. Oxidation of Cl⁻ does happen at this electrode, but oxidation defines the ANODE.
It is the anode, where the chloride ions gain electrons. — A student who thinks oxidation is gain of electrons picks this. Chloride ions are oxidized by LOSING electrons: 2Cl⁻ → Cl₂ + 2e⁻.
9 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?
Answer and reasoning
Electrons move through it from the zinc half-cell to the copper half-cell. — A student who pictures the whole circuit as a loop of electrons picks this. Electrons move only in the wire and electrodes; in the salt bridge the charge is carried by ions.
NO₃⁻ ions move to the copper half-cell, attracted by its positive electrode. — A student who transfers the electrolysis picture of ions attracted to electrodes picks this. The copper solution is LOSING cations, so anions move away from it, towards the zinc half-cell.
Zn²⁺ ions pass through it to be deposited on the copper electrode. — A student who looks for a way for metal to move between electrodes picks this. The copper deposited comes from Cu²⁺ already in the copper half-cell; the salt bridge carries its own K⁺ and NO₃⁻ ions.
K⁺ ions move to the copper half-cell and NO₃⁻ ions to the zinc half-cell. — Zn²⁺ is produced in the zinc half-cell and Cu²⁺ removed from the copper half-cell. Nitrate ions move to the zinc side and potassium ions to the copper side to keep both solutions neutral.
10 A voltaic cell is made from a Mg(s)|Mg²⁺(aq) half-cell and an Ag(s)|Ag⁺(aq) half-cell. Magnesium is oxidized far more readily than silver.
Which describes the flow of electrons in this cell?
Answer and reasoning
From Ag to Mg through the wire, since current leaves the positive terminal — A student who treats conventional current as electron flow picks this. Conventional current flows from + (Ag) to − (Mg), but electrons move the opposite way, from Mg to Ag.
From Mg to Ag through the wire, released as Mg atoms are oxidized — Mg → Mg²⁺ + 2e⁻ at the anode releases electrons; they travel through the external wire to the silver cathode, where Ag⁺ + e⁻ → Ag. Inside the cell, ions carry the charge.
From Mg to Ag through the wire, returning to Mg through the salt bridge — A student who thinks electrons must travel round the whole loop picks this. Electrons do not pass through the salt bridge; ions move there to keep the solutions neutral.
From Ag to Mg through the wire, released as Ag⁺ ions are discharged — A student who thinks a positive ion becomes an atom by losing electrons picks this. Ag⁺ becomes Ag by GAINING an electron, which it receives from the wire at the silver cathode.
11 During discharge of a lead–acid cell, the reaction at the negative electrode is:
Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Which half-equation shows the reaction at this electrode while the cell is being charged?
Answer and reasoning
Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻ — A student who thinks charging repeats the discharge reaction with extra energy picks this. Charging must regenerate the Pb that discharge used up, so the reaction is reversed.
PbSO₄(s) → Pb(s) + SO₄²⁻(aq) + 2e⁻ — A student who thinks reduction means losing electrons writes them as products. Converting PbSO₄ (Pb +2) to Pb (0) is a gain of electrons, and this equation does not balance for charge.
PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq) — Charging reverses the discharge reaction: every species changes side, including the electrons. PbSO₄ is reduced back to Pb, so this electrode is the cathode during charging.
Pb(s) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) — A student who reverses a half-equation by moving only the electrons writes this. It still forms PbSO₄, and the charges do not balance (−4 on the left, 0 on the right).
12 Molten lead(II) bromide is electrolysed using graphite electrodes. How is the electric current conducted through the molten lead(II) bromide itself?
Answer and reasoning
By electrons moving through the melt from the cathode to the anode — A student who pictures the circuit as a single loop of electrons picks this. Free electrons do not travel through an ionic melt; the ions carry the current there.
By Pb²⁺ and Br⁻ ions, already present, moving to the electrodes — Lead(II) bromide consists of ions. Melting frees them to move: Pb²⁺ moves to the cathode and Br⁻ to the anode. Electrons carry the current only in the wires and electrodes.
By delocalized electrons released when the solid melts — A student who applies the metallic model of conduction picks this. An ionic compound has no delocalized electrons, solid or molten; the melt conducts because its ions become mobile.
By ions that the current creates by splitting up the lead(II) bromide — A student who reads 'electrolysis' as 'splitting into ions by electricity' picks this. The ions exist in the solid already; the current moves them and discharges them at the electrodes.
13 Molten potassium iodide, KI(l), is electrolysed with inert electrodes. Which half-equation shows the reaction at the anode?
Answer and reasoning
2I⁻ → I₂ + 2e⁻ — The anode is where oxidation occurs. The only anion in the melt is I⁻, which loses electrons to form iodine.
2K⁺ + 2e⁻ → 2K — A student who thinks the anode is always negative sends K⁺ there. In an electrolytic cell the anode is positive, and K⁺ is reduced at the cathode, not the anode.
2K⁺ → 2K + 2e⁻ — A student who thinks a positive ion becomes an atom by losing electrons treats K⁺ → K as a loss of electrons, i.e. oxidation, and places it at the anode. K⁺ must GAIN an electron to become K (reduction, at the cathode); this equation does not balance for charge. At the anode I⁻ is oxidized: 2I⁻ → I₂ + 2e⁻.
2I⁻ + 2e⁻ → I₂ — A student who thinks oxidation is gain of electrons writes electrons on the left. Iodide ions are oxidized by LOSING electrons, and this equation does not balance for charge.
14 2-methylpropan-2-ol, (CH₃)₃COH, is heated with an acidified oxidizing agent under the conditions that oxidize ethanol. Which statement is correct?
Answer and reasoning
It is oxidized to a ketone, as its OH group is on a central carbon. — A student who uses 'OH in the middle of the chain gives a ketone' picks this. That rule describes secondary alcohols; a tertiary C–OH carbon has no hydrogen to lose, so no ketone forms.
It is oxidized to a carboxylic acid if it is heated for long enough. — A student who thinks every alcohol ends up as an acid picks this. Only primary alcohols give carboxylic acids; tertiary alcohols are not oxidized under these conditions.
It is not oxidized, as its C–OH carbon carries no hydrogen atom. — (CH₃)₃COH is a tertiary alcohol: the carbon bearing the OH is bonded to three carbons and no hydrogen, so it cannot be oxidized under these conditions.
It is not oxidized, as only primary alcohols can be oxidized. — The conclusion is right but the reason is wrong: secondary alcohols are oxidized to ketones. (CH₃)₃COH is not oxidized because its C–OH carbon carries no hydrogen atom.
15 Propan-1-ol is heated under reflux with an excess of acidified oxidizing agent, represented by [O]. Which equation represents the overall change?
Answer and reasoning
CH₃CH₂CH₂OH + [O] → CH₃COCH₃ + H₂O — A student who does not distinguish aldehydes from ketones writes propanone. The OH of propan-1-ol is on the end carbon, so oxidation gives the aldehyde propanal and then propanoic acid; a ketone forms only from a secondary alcohol.
CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O — A student who thinks oxidation of a primary alcohol always stops at the aldehyde picks this. Under reflux with excess oxidizing agent, propanal is oxidized further to propanoic acid.
CH₃CH₂CH₂OH + 9[O] → 3CO₂ + 4H₂O — A student who equates oxidation of an organic compound with combustion picks this. Oxidation of the alcohol functional group keeps the carbon skeleton and gives propanoic acid.
CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O — Under reflux both steps occur: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O, then CH₃CH₂CHO + [O] → CH₃CH₂COOH. Overall, two [O] are used and propanoic acid forms.
16 Butanone, CH₃COCH₂CH₃, is reduced. What is the organic product?
Answer and reasoning
CH₃CH₂CH₂CH₂OH — A student who thinks reduction always gives a primary alcohol picks butan-1-ol. The C=O of butanone is on carbon 2, so the OH ends up on carbon 2: a secondary alcohol.
CH₃CH₂CH₂CH₃ — A student who takes reduction as loss of oxygen to mean loss of ALL the oxygen picks butane. Reduction of a ketone adds hydrogen across the C=O bond; the product, butan-2-ol, keeps its oxygen.
CH₃CH(OH)CH₂CH₃ — Reduction adds hydrogen to the C=O carbon and oxygen of the ketone. The OH stays on carbon 2, so the product is the secondary alcohol butan-2-ol.
CH₃CH₂CH₂COOH — A student who swaps oxidation and reduction thinks reduction adds oxygen and picks butanoic acid. Reduction is gain of hydrogen: butanone gains two H atoms to form butan-2-ol.
17 Which describes the reduction of propanoic acid, CH₃CH₂COOH?
Answer and reasoning
Propan-1-ol forms in a single step, with no aldehyde formed. — A student who thinks acid and alcohol interconvert directly picks this. Reduction proceeds via the aldehyde, propanal.
Propanal forms first and is then reduced to propan-1-ol. — Carboxylic acids are reduced to primary alcohols via the aldehyde: CH₃CH₂COOH → CH₃CH₂CHO → CH₃CH₂CH₂OH.
Propanone forms first and is then reduced to propan-2-ol. — A student who does not distinguish aldehydes from ketones picks this. The COOH carbon is at the end of the chain, so the intermediate is the aldehyde propanal and the product the primary alcohol.
Propane forms, as reduction removes both of its oxygen atoms. — A student who takes 'loss of oxygen' to mean loss of all oxygen picks this. The acid is reduced to the aldehyde and then the primary alcohol, which still contains oxygen.
18 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?
Answer and reasoning
2 mol — A student who adds one H₂ to each multiple bond counts 2 mol. The triple bond needs two H₂ (one per π bond) to become a single bond.
5 mol — A student who adds H₂ to every bond in each multiple bond counts 2 + 3 = 5. Only the π bonds react (one in C=C, two in C≡C); the σ bonds remain.
6 mol — A student who counts hydrogen atoms added (6) as moles of H₂ gets 6 mol. Hexane has 6 more H atoms than C₆H₈, which is 3 H₂ molecules.
3 mol — The C=C bond takes 1 mol of H₂ and the C≡C bond takes 2 mol, giving hexane, CH₃CH₂CH₂CH₂CH₂CH₃. Total 3 mol.
Working π bonds: C=C has 1, C≡C has 2, total 3. One H₂ per π bond: 3 mol H₂ per mol. Check: C₆H₈ + 3H₂ → C₆H₁₄.
19 Standard electrode potentials:
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s) E⦵ = −2.37 V
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E⦵ = −0.76 V
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E⦵ = +0.34 V
Ag⁺(aq) + e⁻ ⇌ Ag(s) E⦵ = +0.80 V
Which species is the strongest oxidizing agent? HL
Answer and reasoning
Mg²⁺(aq), as its E⦵ value is the largest in size — A student who compares sizes and ignores signs picks Mg²⁺. −2.37 V is the most NEGATIVE value: Mg²⁺ is the hardest ion to reduce and the weakest oxidizing agent listed.
Mg(s), as it is the species most readily oxidized — A student who thinks the oxidizing agent is the species oxidized picks Mg. Mg is the strongest REDUCING agent; the strongest oxidizing agent is the most easily reduced species, Ag⁺.
Ag⁺(aq), as its E⦵ value is the most positive — The most positive E⦵ belongs to the oxidized form that is most easily reduced. Ag⁺ (+0.80 V) accepts electrons most readily, so it is the strongest oxidizing agent listed.
Ag(s), as silver has the most positive E⦵ value — A student who attaches E⦵ to the metal rather than to the ion picks Ag. E⦵ describes the reduction of the left-hand species: Ag⁺ is the oxidizing agent; Ag(s) is a weak reducing agent.
20 Standard electrode potentials:
Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E⦵ = +0.77 V
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E⦵ = +0.34 V
Which statement about the reaction Cu(s) + 2Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺(aq) is correct? HL
Answer and reasoning
It is spontaneous as written, since E⦵cell = +1.20 V — A student who doubles the Fe³⁺/Fe²⁺ value because of the coefficient 2 gets 2(0.77) − 0.34 = +1.20 V. E⦵ values are not multiplied by coefficients; E⦵cell = +0.43 V.
It is spontaneous as written, since E⦵cell = +0.43 V — Fe³⁺ is reduced (cathode) and Cu is oxidized (anode): E⦵cell = E⦵(cathode) − E⦵(anode) = +0.77 − (+0.34) = +0.43 V. A positive E⦵cell means the forward reaction is spontaneous.
It is spontaneous in reverse, since E⦵cell = −0.43 V — A student who calculates E⦵(anode) − E⦵(cathode) gets 0.34 − 0.77 = −0.43 V and concludes that the reverse reaction is spontaneous. The species reduced is Fe³⁺, so its value is E⦵(cathode): E⦵cell = +0.43 V and the forward reaction is spontaneous.
It is spontaneous in reverse, since E⦵cell = +0.43 V — A student who carries over 'negative means spontaneous' from ΔG⦵ calculates E⦵cell correctly but reads it the wrong way. A POSITIVE E⦵cell means the reaction is spontaneous as written, since ΔG⦵ = −nFE⦵cell is then negative.
Working Cathode (reduction): Fe³⁺ + e⁻ → Fe²⁺, E⦵ = +0.77 V. Anode (oxidation): Cu → Cu²⁺ + 2e⁻, E⦵ = +0.34 V. E⦵cell = +0.77 − (+0.34) = +0.43 V > 0, so the reaction is spontaneous as written.
21 For the cell reaction Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s):
E⦵(Mg²⁺/Mg) = −2.37 V; E⦵(Ag⁺/Ag) = +0.80 V; F = 96 500 C mol⁻¹
What is ΔG⦵ for the reaction as written? HL
Answer and reasoning
−306 kJ mol⁻¹ — A student who leaves out n (takes n = 1) gets −96 500 × 3.17 = −305 905 J mol⁻¹ ≈ −306 kJ mol⁻¹. Two electrons are transferred per Mg, so n = 2.
−766 kJ mol⁻¹ — A student who doubles the silver E⦵ for the coefficient 2 gets E⦵cell = 2(0.80) + 2.37 = 3.97 V and ΔG⦵ = −2 × 96 500 × 3.97 = −766 kJ mol⁻¹. E⦵ is not multiplied by coefficients.
+303 kJ mol⁻¹ — A student who adds the two E⦵ values gets 0.80 + (−2.37) = −1.57 V, so ΔG⦵ = −2 × 96 500 × (−1.57) = +303 kJ mol⁻¹. E⦵cell = E⦵(cathode) − E⦵(anode) = +3.17 V.
22 Concentrated NaCl(aq) is electrolysed with inert electrodes.
Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) E⦵ = +1.36 V
O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l) E⦵ = +1.23 V
What is the main product at the anode, and why? HL
Answer and reasoning
Oxygen, as E⦵ values decide the product whatever the Cl⁻ concentration — A student who uses E⦵ values alone picks oxygen. E⦵ values refer to 1 mol dm⁻³ solutions; in concentrated NaCl(aq), chlorine is observed to be the main anode product.
Hydrogen, as H⁺ ions from the water are attracted to the anode — A student who thinks the anode is negative sends H⁺ there. In electrolysis the anode is positive and oxidation occurs there; hydrogen forms at the cathode.
Chlorine, as the high Cl⁻ concentration favours oxidation of Cl⁻ over water — E⦵ values refer to 1 mol dm⁻³ solutions and, used alone, predict oxygen. Experimentally, concentrated NaCl(aq) gives mainly chlorine at the anode (2Cl⁻ → Cl₂ + 2e⁻), while dilute NaCl(aq) gives mainly oxygen: the high Cl⁻ concentration favours Cl⁻ oxidation. This is why E⦵ values alone cannot predict the product here.
Oxygen, as water is oxidized in preference to any ion in solution — A student who expects water always to react picks this. Water competes with Cl⁻, but it does not always win: in concentrated NaCl(aq), Cl⁻ is oxidized and chlorine is the main anode product.
23 CuSO₄(aq) is electrolysed first with graphite electrodes and then with copper electrodes. Which statement explains the difference at the anode? HL
Answer and reasoning
With copper electrodes, the anode itself is oxidized to Cu²⁺, so no oxygen forms. — Copper is oxidized far more readily than water (E⦵ +0.34 V vs +1.23 V), so a copper anode dissolves: Cu → Cu²⁺ + 2e⁻. With inert graphite, water is oxidized to O₂.
There is no difference: water is oxidized to oxygen at both types of anode. — A student who treats electrodes as mere connectors picks this. A copper anode takes part in the reaction, dissolving as Cu²⁺ instead of water being oxidized.
With copper electrodes, Cu²⁺ ions are reduced to copper at the anode. — A student who thinks the anode is negative places the reduction of Cu²⁺ there. Reduction occurs at the cathode; at the anode, the copper is oxidized.
With graphite electrodes, sulfate ions are oxidized at the anode, not water. — A student who assumes the salt's anion is always discharged picks this. Sulfate ions are not oxidized in aqueous solution; at a graphite anode water is oxidized to oxygen.
24 A steel key is electroplated with copper using a copper anode and CuSO₄(aq) as the electrolyte. Which half-equation shows the reaction at the key? HL
Answer and reasoning
Cu(s) → Cu²⁺(aq) + 2e⁻ — A student who places oxidation at the cathode picks this. This is the reaction at the copper ANODE; at the key (cathode), reduction occurs.
Cu²⁺(aq) → Cu(s) + 2e⁻ — A student who thinks an ion becomes an atom by losing electrons picks this. Cu²⁺ must GAIN two electrons to become Cu, and this equation does not balance for charge.
Cu²⁺(aq) + 2e⁻ → Cu(s) — The key is the cathode. Electrons supplied by the power source reduce Cu²⁺ ions to copper, which coats the key. At the copper anode, Cu → Cu²⁺ + 2e⁻ replaces the ions.
4H⁺(aq) + 4e⁻ → 2H₂(g) — A student who expects every aqueous electrolysis to give hydrogen at the cathode picks this. Cu²⁺ (E⦵ = +0.34 V) is reduced far more readily than H⁺ (E⦵ = 0.00 V) or water, so copper, not hydrogen, is deposited on the key: Cu²⁺(aq) + 2e⁻ → Cu(s).
25 Na₂SO₄(aq) is electrolysed with inert electrodes.
Na⁺(aq) + e⁻ ⇌ Na(s) E⦵ = −2.71 V
2H₂O(l) + 2e⁻ ⇌ H₂(g) + 2OH⁻(aq) E⦵ = −0.83 V
O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l) E⦵ = +1.23 V
What forms at the cathode, and why? HL
Answer and reasoning
Hydrogen, as water is reduced more readily than Na⁺(aq) ions — Reduction occurs at the cathode. The two species that could be reduced are Na⁺ (E⦵ = −2.71 V) and water (E⦵ = −0.83 V). The less negative E⦵ belongs to water, so water is reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻. Oxygen forms at the anode, so overall this is the electrolysis of water.
Sodium, as Na⁺/Na has the E⦵ value that is largest in size — A student who compares sizes and ignores signs reads −2.71 V as the 'largest' value and so takes Na⁺ as the most easily reduced species. −2.71 V is the most NEGATIVE value: Na⁺ is the hardest species here to reduce, and water (−0.83 V) is reduced instead.
Sodium, as the metal ion of the dissolved salt is reduced there — A student who carries the molten-salt pattern over to solutions expects the salt's metal to be deposited. In aqueous solution water competes; its E⦵ (−0.83 V) is far less negative than that of Na⁺ (−2.71 V), so hydrogen forms, not sodium.
Oxygen, as water molecules are oxidized at the cathode — A student who pairs oxidation with the cathode picks this. Oxidation of water to oxygen does occur, but at the ANODE; the cathode is where reduction occurs, giving hydrogen.
26 Which statement about primary cells and secondary (rechargeable) cells is correct?
Answer and reasoning
A primary cell holds a store of electricity that flows out until it is empty. — A student who pictures a cell as a container of electricity picks this. A cell stores chemical energy in its reactants; electrical energy is produced by the redox reaction as the reactants are used up.
A secondary cell can be recharged endlessly with no loss of capacity. — A student who takes 'rechargeable' to mean 'good for ever' picks this. Side reactions and changes to the electrodes reduce the capacity over many cycles; a limited cycle life is a disadvantage of secondary cells.
A secondary cell is recharged by running its discharge reaction again. — A student who thinks charging repeats the discharge reaction with extra energy picks this. Charging drives the discharge reaction in REVERSE, regenerating the original reactants.
A primary cell costs less to buy but can never be reused once it runs down. — A primary cell's reaction cannot practicably be reversed, so the cell is used once and thrown away: cheap and convenient at purchase, but more waste over time. A secondary cell costs more initially but can be recharged many times.
27 Nitrogen has the oxidation state −3 in ammonia, NH₃. What is the oxidation state of nitrogen in the nitrate ion, NO₃⁻?
Answer and reasoning
−3 — A student who gives nitrogen one fixed oxidation state, the −3 of ammonia, picks this. Nitrogen's oxidation state varies with what it is bonded to: here x + 3(−2) = −1 gives +5.
+5 — O is −2, and the oxidation states must add up to the ion's charge: x + 3(−2) = −1, so x = +5. Like most main-group non-metals, nitrogen has variable oxidation states: −3 in NH₃, 0 in N₂, +4 in NO₂ and +5 in NO₃⁻.
−1 — A student who takes the charge on the ion as the oxidation state of the central atom picks this. The −1 is the SUM of the oxidation states of all four atoms: x + 3(−2) = −1 gives N = +5.
+6 — A student who makes the oxidation states add up to zero gets x − 6 = 0, x = +6. NO₃⁻ is an ion, so the states must add up to its charge, −1, giving +5.
Working Let the oxidation state of N be x. O is −2. Sum = charge on the ion: x + 3(−2) = −1, so x = +5.
28 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?
Answer and reasoning
Copper dipping into a solution containing Cu²⁺ ions — Each half-cell is a metal dipping into a solution of its own ions. At the copper cathode, Cu²⁺ ions from this solution are reduced: Cu²⁺ + 2e⁻ → Cu. (The zinc half-cell is likewise zinc dipping into Zn²⁺(aq).)
Copper dipping into a solution of zinc ions, Zn²⁺ — A student who thinks the Zn²⁺ ions formed at the anode cross to the cathode and are deposited there expects Zn²⁺ to be the ions in the copper half-cell. The ions reduced at the copper cathode are Cu²⁺ from its own solution; the salt bridge carries only its own K⁺ and NO₃⁻ ions.
Copper dipping into a beaker of distilled water — A student who thinks electrons carry the current through the solutions expects no dissolved ions to be needed. Electrons move only in the wire and electrodes; ions carry charge in the solutions, and Cu²⁺ ions are needed as the species reduced at the cathode.
Copper dipping into KNO₃(aq), the salt bridge solution — A student who treats the less reactive metal itself as the electron acceptor thinks any conducting solution will do. Copper atoms cannot gain electrons; the species reduced at the cathode is Cu²⁺, so the copper must dip into a solution of Cu²⁺ ions.
29 Propan-1-ol is heated with an acidified oxidizing agent in apparatus set up for distillation, and the distillate is collected. What is the main organic product collected, and why?
Answer and reasoning
Propanoic acid, as distillation keeps propanal in the flask — A student who swaps the roles of distillation and reflux picks this. Keeping propanal in the flask is what reflux does; distillation removes it, so the aldehyde is collected.
Propanone, as the alcohol is oxidized to a ketone — A student who does not distinguish aldehydes from ketones picks propanone. The OH of propan-1-ol is on the end carbon, so it is a primary alcohol and its first oxidation product is the aldehyde propanal; ketones come from secondary alcohols.
Propanal, as it distils out before it can be oxidized further — The first step gives propanal, CH₃CH₂CHO, which is more volatile than the alcohol and the acid. In distillation the vapour passes into the condenser and is collected elsewhere, so propanal leaves the flask before it can be oxidized to propanoic acid.
Propanoic acid, as it forms directly in one step — A student who thinks the acid forms in a single step picks this. Oxidation goes via the aldehyde; with distillation, that aldehyde is removed as it forms and is the product collected.
30 Oct-4-yne, CH₃CH₂CH₂C≡CCH₂CH₂CH₃, reacts completely with excess hydrogen in the presence of a nickel catalyst. What is the organic product?
Answer and reasoning
CH₃CH₂CH₂CH=CHCH₂CH₂CH₃ — A student who adds one H₂ to the multiple bond and stops picks oct-4-ene. That is the product of only one H₂; the remaining C=C is still unsaturated and, with excess hydrogen, takes a second H₂ to give octane.
CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ — The C≡C bond has two π bonds, and each takes one H₂. With excess hydrogen both are hydrogenated, so the triple bond becomes a single bond and the product is the saturated alkane octane, CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ (C₈H₁₈).
CH₃CH₂CH₂CH₃ and CH₃CH₂CH₂CH₃ — A student who thinks hydrogen breaks every bond in the C≡C group, separating the two carbons, writes two molecules of butane. Only the two π bonds react; the σ bond between C4 and C5 remains, so the carbon chain stays intact and the product is octane.
CH₃CH₂CH₂CHOHCHOHCH₂CH₂CH₃ — A student who confuses hydrogenation with hydration adds H and OH across each of the two π bonds and writes octane-4,5-diol. Hydrogen gas contains no oxygen, so no OH group can form; each π bond takes one H₂, and the product is octane, CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃.
31 For the reaction 2Al(s) + 3Sn²⁺(aq) → 2Al³⁺(aq) + 3Sn(s), ΔG⦵ = −880 kJ mol⁻¹. F = 96 500 C mol⁻¹. What is E⦵cell for this reaction? HL
Answer and reasoning
3.04 V — A student who takes n = 3 from the single half-equation Al → Al³⁺ + 3e⁻ gets 880 000 / (3 × 96 500) = 3.04 V. The equation as written involves two Al atoms, so 6 mol of electrons are transferred: n = 6 and E⦵cell = 1.52 V.
1.52 V — Two Al atoms lose 6e⁻ and three Sn²⁺ ions gain 6e⁻, so n = 6. E⦵cell = −ΔG⦵/(nF) = 880 000 J mol⁻¹ / (6 × 96 500 C mol⁻¹) = 1.52 V.
1.82 V — A student who adds the electrons in the two half-equations (3 + 2 = 5) gets 880 000 / (5 × 96 500) = 1.82 V. The electrons lost by aluminium are the same electrons gained by the tin ions; scaled to cancel, they number 6, so n = 6.
9.12 V — A student who leaves n out (takes n = 1) gets 880 000 / 96 500 = 9.12 V, an impossibly large potential for a single cell. n = 6 for the equation as written, giving 1.52 V.
Working Half-equations: 2Al → 2Al³⁺ + 6e⁻ and 3Sn²⁺ + 6e⁻ → 3Sn, so n = 6 mol of electrons per mole of reaction as written.
ΔG⦵ = −nFE⦵cell, so E⦵cell = −ΔG⦵/(nF).
ΔG⦵ = −880 kJ mol⁻¹ = −880 000 J mol⁻¹.
E⦵cell = 880 000 J mol⁻¹ / (6 × 96 500 C mol⁻¹) = 880 000 / 579 000 = 1.52 V (J C⁻¹ = V).
Consistency check: E⦵(Sn²⁺/Sn) − E⦵(Al³⁺/Al) = −0.14 − (−1.66) = +1.52 V.
32 A steel bracelet is to be electroplated with silver. It is placed in AgNO₃(aq) with a silver rod as the second electrode; the bracelet is connected to the negative terminal of a DC power supply and the silver rod to the positive terminal. Which statement about this cell is correct? HL
Answer and reasoning
The silver rod is the cathode, because Ag⁺(aq) ions are produced at it — A student who names the electrode by the ions formed there ('cations, so cathode') picks this. Electrodes are named by the reaction: the rod is oxidized (Ag → Ag⁺ + e⁻), and oxidation defines the anode. Cations are consumed at the cathode (the bracelet), not produced there.
The bracelet is the anode, because it is connected to the negative terminal — A student who fixes the anode as the negative electrode picks this. In an electrolytic cell the negative electrode is the cathode: it receives electrons from the supply and reduction occurs there, Ag⁺(aq) + e⁻ → Ag(s), which is why silver deposits on the bracelet.
Water is oxidized to O₂(g) at the silver rod, which only conducts the current — A student who treats every electrode as inert expects the anode reaction of water electrolysis. Silver is easier to oxidize than water (E⦵(Ag⁺/Ag) = +0.80 V is less positive than E⦵(O₂/H₂O) = +1.23 V), so the silver rod itself is oxidized, Ag(s) → Ag⁺(aq) + e⁻, and no oxygen forms.
The silver rod is the anode and dissolves, forming Ag⁺(aq) ions — The electrode joined to the positive terminal is the anode, where oxidation occurs: Ag(s) → Ag⁺(aq) + e⁻. A silver anode is oxidized in preference to water, so it dissolves and replaces the Ag⁺ ions that are reduced onto the bracelet (Ag⁺(aq) + e⁻ → Ag(s)), keeping the electrolyte concentration steady.
That was your twenty minutes. Real practice on R3.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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