Electrochemistry (A Level)

• Time in seconds; Q/F is moles of electrons, not of product; E⦵ is never multiplied. More positive E⦵ = better oxidising agent = reduced; more negative = better reducing agent = oxidised.
• Seconds, electrons, never multiply. More positive E⦵ is reduced; the negative electrode is oxidised and sends the electrons out. Positive E⦵cell, negative ΔG⦵, feasible, but not necessarily fast.

• 24.1.1 I can predict — predict the identities of substances liberated during electrolysis from the state of electrolyte (molten or aqueous), position in the redox series (electrode potential) and concentration
• 24.1.2 I can state — state and apply the relationship F = Le between the Faraday constant, F, the Avogadro constant, L, and the charge on the electron, e
• 24.1.3 I can calculate — calculate: (a) the quantity of charge passed during electrolysis, using Q = It (b) the mass and/or volume of substance liberated during electrolysis
• 24.1.4 I can describe — describe the determination of a value of the Avogadro constant by an electrolytic method
• 24.2.1 I can define — define the terms: (a) standard electrode (reduction) potential (b) standard cell potential
• 24.2.2 I can describe — describe the standard hydrogen electrode
• 24.2.3 I can describe — describe methods used to measure the standard electrode potentials of: (a) metals or non-metals in contact with their ions in aqueous solution (b) ions of the same element in different oxidation states
• 24.2.4 I can calculate — calculate a standard cell potential by combining two standard electrode potentials
• 24.2.5 I can use — use standard cell potentials to: (a) deduce the polarity of each electrode and hence explain/deduce the direction of electron flow in the external circuit of a simple cell (b) predict the feasibility of a reaction
• 24.2.6 I can deduce — deduce from E⦵ values the relative reactivity of elements, compounds and ions as oxidising agents or as reducing agents
• 24.2.7 I can construct — construct redox equations using the relevant half-equations
• 24.2.8 I can predict — predict qualitatively how the value of an electrode potential, E, varies with the concentrations of the aqueous ions
• 24.2.9 I can use — use the Nernst equation, e.g. E = E⦵ + (0.059/z) log([oxidised species]/[reduced species]), to predict quantitatively how the value of an electrode potential varies with the concentrations of the aqueous ions; examples include Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq)
• 24.2.10 I can understand — understand and use the equation ΔG⦵ = −nE⦵cell F

Tie the names to the chemistry, not to the sign. Oxidation is always at the anode and reduction always at the cathode. Only the sign of the electrode changes between the two cell types. If you remember “oxidation at the anode” and which way the electrons are being pushed, the signs follow.

• “0.500 A for 30 minutes: Q = 0.500 × 30 = 15 C.” Correct In Q = It the time is in seconds. 30.0 min = 1800 s, so Q = 900 C. The ampere is a coulomb per second, so any other time unit gives the wrong answer.
• “n(Cu) = Q/F.” Correct Q/F is the number of moles of electrons. The half-equation converts it to moles of product: Cu²⁺ + 2e⁻ → Cu halves it; 2H₂O → O₂ + 4H⁺ + 4e⁻ quarters it.
• “Ag⁺ + e⁻ → Ag is +0.80 V, so 2Ag⁺ + 2e⁻ → 2Ag is +1.60 V.” Correct E⦵ is never multiplied. It measures how strongly a half-cell pulls electrons, not how much reacts, so doubling the half-equation leaves it at +0.80 V. Only ΔG⦵ scales with the number of electrons.
• “Brine gives chlorine because Cl₂/Cl⁻ has a less positive E⦵ than O₂/H₂O.” Correct It does not: +1.36 V against +1.23 V, so dilute chloride gives mainly oxygen. Concentrated chloride gives chlorine because the high [Cl⁻] makes E(Cl₂/Cl⁻) less positive than +1.36 V, narrowing the gap to +1.23 V (the background note in lesson 24.1.1 explains the rest). This is a concentration effect, not a reversal of the standard values.
• “Electrons flow from the positive electrode to the negative electrode.” Correct Electrons leave the negative electrode, which has the more negative E⦵ and is where oxidation happens. They flow through the external wire to the positive electrode. Inside the cell, charge is carried by ions in the salt bridge, never by electrons.
• “The standard hydrogen electrode is hydrogen gas bubbled into acid.” Correct All four parts are needed: a platinum electrode, H⁺(aq) at 1.00 mol dm⁻³, H₂ at 101 kPa, and 298 K. Leave one out and the electrode is not standard.
• “E⦵cell is positive, so the reaction happens.” Correct A positive E⦵cell means the reaction is feasible under standard conditions. It can still be too slow to observe (high activation energy), and a small E⦵cell can change sign when the concentrations are not 1.00 mol dm⁻³.
• “Sodium is deposited at the cathode when NaCl(aq) is electrolysed.” Repair Na⁺/Na is −2.71 V, far more negative than H⁺/H₂ at 0.00 V, so hydrogen is given off. Sodium forms only from the molten salt.
• “Concentrated brine gives chlorine because Cl₂/Cl⁻ has a less positive E⦵ than O₂/H₂O.” Repair It is the other way round: +1.36 V against +1.23 V. The high [Cl⁻] makes E(Cl₂/Cl⁻) less positive (+1.32 V at 5.0 mol dm⁻³), narrowing the gap to +1.23 V. It is a concentration effect, not a property of the standard values.
• “Q = It = 0.500 × 30 = 15 C.” Repair t must be in seconds: 30.0 min = 1800 s, so Q = 900 C.
• “n(Cu) = Q/F.” Repair Q/F is moles of electrons. Cu²⁺ + 2e⁻ → Cu halves it: n(Cu) = Q/2F.
• “n(O₂) = n(e⁻)/2, like hydrogen.” Repair 2H₂O → O₂ + 4H⁺ + 4e⁻: four electrons per O₂, so n(O₂) = n(e⁻)/4, and oxygen is half the volume of hydrogen.
• “F = Le, so F = 6.022 × 10²³ × (−1.60 × 10⁻¹⁹) = −9.64 × 10⁴ C mol⁻¹.” Repair The minus sign on e only says the electron is negative. F is the magnitude of the charge on a mole of electrons, +9.65 × 10⁴ C mol⁻¹.
• “Copper flaking off the cathode makes the calculated value of L too small.” Repair Less copper means fewer moles of electrons for the same Q, so F = Q/n(e⁻) is too large, and so is L.
• “The standard hydrogen electrode is hydrogen gas in acid.” Repair All four parts: a platinum electrode, H⁺(aq) at 1.00 mol dm⁻³, H₂ at 101 kPa, and 298 K. Its potential is 0.00 V by definition.
• “Any voltmeter will do.” Repair A high-resistance voltmeter, so that no current flows and the e.m.f. is read with the concentrations unchanged.
• “The salt bridge lets electrons cross between the two beakers.” Repair Electrons travel only through the external wire. The salt bridge carries ions, which keep each solution electrically neutral without the solutions mixing.
• “For Cu + 2Ag⁺, E⦵cell = 2(0.80) − 0.34 = +1.26 V.” Repair E⦵ is never multiplied: E⦵cell = 0.80 − 0.34 = +0.46 V. Only ΔG⦵ scales with n.
• “Electrons flow from the positive electrode to the negative.” Repair From the negative electrode (more negative E⦵, where oxidation occurs) through the external circuit to the positive.
• “E⦵cell is positive, so the reaction happens.” Repair It is feasible. Whether it is observed depends on the rate (activation energy) and on whether the conditions are standard.
• “Using the Nernst equation with z = 1 for Cu²⁺/Cu.” Repair z is the number of electrons in the half-equation: 2 for Cu²⁺/Cu, 1 for Fe³⁺/Fe²⁺.
• “ΔG⦵ = −nE⦵cellF = −1 × 1.10 × 96 500 = −106 kJ mol⁻¹ for Zn + Cu²⁺.” Repair n is the moles of electrons in the balanced equation, 2 here: −2 × 1.10 × 96 500 = −212 300 J mol⁻¹ = −212 kJ mol⁻¹. V × C mol⁻¹ is J mol⁻¹, so divide by 1000 for kJ.

• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.
• Interleave with the chapters that use this one. Topic 28 (transition elements) uses E⦵ values for every redox reaction of the d-block ions: when you reach it, re-answer cell drill items 4 and 5 and builder drill item 1. Topic 23 shares ΔG⦵: when revising it, redo worked example 6 in reverse. Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Chemistry 9701 has five components. Topic 24 is A Level content, so it is examined in Papers 4 and 5. A Level content: examined in Paper 4 (A Level structured, which also requires the AS content) and, as practical context, Paper 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E.
• There is no multiple-choice paper on A Level content; topic 24 is examined through Paper 4's structured questions. The outcomes ask you to define E⦵ and E⦵cell with their conditions, describe the standard hydrogen electrode and how E⦵ is measured, predict electrolysis products, deduce polarity and feasibility, and construct redox equations from half-equations.
• Q = It and the mass or gas volume liberated; F = Le; E⦵cell from two E⦵ values; E from the Nernst equation; ΔG⦵ from E⦵cell. You are given F, L, e, Vₘ and every E⦵ in the Data section, so the skill is choosing the right values and using them, not recalling them.
• The Avogadro determination supplies a complete planning context: charge as the independent variable, mass gained as the dependent variable, a straight-line graph whose gradient gives F, and a precision argument about the ammeter and the balance. Measuring E⦵cell supplies an evaluation context: non-standard concentrations and the wrong voltmeter.
• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.