Redox II
Notes for every section, eight new exam-style questions with full worked solutions, and online recall cards for your review days. Nothing here is copied from the book.
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14.1 Oxidation and reduction, two ways
Oxidation is loss of electrons; the oxidation number increases.
Reduction is gain of electrons; the oxidation number decreases.
OIL RIG: Oxidation Is Loss, Reduction Is Gain.
A redox reaction has both at once. The oxidising agent accepts electrons, so it is itself reduced. The reducing agent donates electrons, so it is itself oxidised.
Example: Zn + Cu2+ → Zn2+ + Cu. Zinc goes 0 → +2 (oxidised; it is the reducing agent). Copper goes +2 → 0 (reduced; Cu2+ is the oxidising agent).
Always name the species and give both oxidation numbers when asked to identify what is oxidised or reduced.
Check yourself: In Cl2 + 2I− → 2Cl− + I2, which species is the reducing agent?
I− (it is oxidised from −1 to 0)
14.2 Writing and combining half-equations
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14.3 Standard electrode potentials and the hydrogen electrode
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14.4 Three kinds of half-cell
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14.5 Cell emf and cell diagrams
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14.6 The electrochemical series
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14.7 Predicting whether a reaction is feasible
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14.8 E⦵cell, total entropy change and K
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14.9 Conditions and the limits of predictions
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14.10 Disproportionation
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14.11 Storage cells
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14.12 Fuel cells
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14.13 Redox titrations: the methods
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14.14 Redox titration calculations
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14.15 Core Practical 10: investigating some electrochemical cells
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14.16 Core Practical 11: a redox titration
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Extra worked questions
Cover the solution, answer on paper, then mark yourself against each step. Each step shows the mark it earns.
All 8 worked solutions are free for members. The first one is open to everyone.
The zinc–copper cell
Calculate the emf of a cell made from Zn2+/Zn (E⦵ = −0.76 V) and Cu2+/Cu (E⦵ = +0.34 V), and state which electrode is negative.
Try it first, then show the worked solution
| emf | +0.34 − (−0.76) = +1.10 V | 1 |
|---|---|---|
| Negative electrode | Zinc (more negative E⦵; it is oxidised) | 1 |
Answer: 1.10 V; zinc is the negative electrode.
Examiner tip: Subtracting a negative value is where marks go. Write the brackets in.
A magnesium–silver cell
Write the cell diagram and calculate the emf for a cell made from Mg2+/Mg (−2.37 V) and Ag+/Ag (+0.80 V).
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The standard hydrogen electrode
Describe the standard hydrogen electrode, giving all the standard conditions, and explain why platinum is used.
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A reaction that will not go
Explain, using E⦵ values, why bromine does not oxidise chloride ions. E⦵: Br2/Br− +1.07 V; Cl2/Cl− +1.36 V.
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The alkaline fuel cell
E⦵: O2 + 2H2O + 4e− ⇌ 4OH−, +0.40 V; 2H2O + 2e− ⇌ H2 + 2OH−, −0.83 V. Give the reaction at the negative electrode, the overall equation and the emf. Give one reason a hydrogen fuel cell is not automatically carbon-neutral.
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Cell emf, entropy and K
Reaction A has E⦵cell = +1.10 V; reaction B has E⦵cell = +0.03 V. (a) Which reaction has the larger equilibrium constant? (b) Describe the equilibrium mixture for reaction B. (c) State the sign of ΔStotal for both.
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A dichromate titration
25.0 cm3 of an Fe2+ solution needs 18.75 cm3 of 0.0200 mol dm−3 K2Cr2O7 in acid. Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+. Calculate [Fe2+].
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Iron in a tablet
A 0.950 g iron tablet is dissolved in dilute sulfuric acid and made up to 250 cm3. 25.0 cm3 portions need a mean of 11.20 cm3 of 0.00500 mol dm−3 KMnO4. Calculate the percentage by mass of iron in the tablet. Ar(Fe) = 55.8.
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Video walkthroughs
Narrated walkthroughs of exam-style questions from this topic of the book, showing where each mark comes from. Try the question first, then watch. Captions are available in the player.
A nickel and silver cell
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Manganate(VII) titration of iron(II)
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Iodine-thiosulfate titration of bleach
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Hydrogen and methanol fuel cells
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Measuring cell emfs in the lab
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Water of crystallisation by manganate(VII) titration
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Recall cards
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