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AS 6

Electrochemistry

AS Level · Physical chemistry · Papers 1, 2 and 3 · extended in A2 24

At AS Level, electrochemistry means redox: keeping track of electrons through oxidation numbers. It is a short topic, but it is used everywhere — in Group 17, in nitrogen and sulfur chemistry, in titrations, and in all of A Level topic 24, where electrode potentials and electrolysis are added.

🎯What you need to be able to do

  • Work out the oxidation number of any element in a compound or ion.
  • Use changes in oxidation number to balance equations.
  • Explain redox, oxidation, reduction and disproportionation in terms of electrons and of oxidation numbers.
  • Identify oxidising agents and reducing agents, and explain what each does.
  • Use a Roman numeral to show the oxidation number of an element in a name, such as iron(III) or chlorate(V).

📚The chemistry

Oxidation numbers

An oxidation number (oxidation state) is the charge an atom would have if all its bonds were fully ionic, with the electrons of each bond given to the more electronegative atom. For simple ions it is just the charge; in covalent compounds it is bookkeeping, but consistent bookkeeping is what makes redox work. Apply these rules in order:

  1. An uncombined element is 0: Fe, O2, Cl2, S8.
  2. The oxidation numbers in a compound add up to 0; in an ion they add up to the charge of the ion.
  3. Fluorine is always −1. Group 1 metals are +1, Group 2 metals +2, aluminium +3.
  4. Hydrogen is +1 — except in metal hydrides such as NaH, where it is −1.
  5. Oxygen is −2 — except in peroxides such as H2O2 (−1) and in compounds with fluorine.
  6. Chlorine is −1 in chlorides, but takes positive values when combined with oxygen or fluorine.

Then find the unknown by subtraction. In the dichromate ion, Cr2O72−: 7 oxygens contribute −14; the ion’s charge is −2; so the two chromiums contribute +12, and each is +6. Always write the sign: “+6”, not “6”.

The Roman numeral in a name is the oxidation number of the element before it. Iron(II) sulfate contains Fe at +2; iron(III) chloride, Fe at +3. For oxoanions it goes on the central atom: in sodium chlorate(I), NaClO, chlorine is +1; in sodium chlorate(V), NaClO3, it is +5. Manganate(VII) is MnO4−, with Mn at +7.

Redox

Oxidation and reduction can be defined two ways, and you need both:

Oxidation — loss of electrons; increase in oxidation number
Reduction — gain of electrons; decrease in oxidation number

(“OIL RIG”: oxidation is loss, reduction is gain.) Electrons lost by one species must be gained by another, so the two always happen together: a redox reaction. In

\[ \mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)} \]

zinc goes from 0 to +2 (oxidised, losing two electrons) and copper from +2 to 0 (reduced, gaining them). The oxidation number definition is the more useful one when no ions are obvious, as in the reaction of hydrogen with oxygen: H goes from 0 to +1, O from 0 to −2.

Oxidising and reducing agents

An oxidising agent oxidises something else, and is itself reduced — it is an electron acceptor. A reducing agent reduces something else and is itself oxidised — an electron donor. In the zinc reaction, Cu2+ is the oxidising agent and Zn the reducing agent. The naming feels backwards at first; the test is always “what does it do to the other species?”

Common oxidising agents: O2, Cl2, acidified MnO4− (purple → colourless Mn2+), acidified Cr2O72− (orange → green Cr3+), concentrated H2SO4. Common reducing agents: reactive metals, H2, C, CO, I−, Fe2+.

Disproportionation

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. The classic examples are chlorine with sodium hydroxide, which you will meet again in topic 11 (Group 17):

\[ \mathrm{Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O} \quad \text{(cold, dilute)} \] \[ \mathrm{3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O} \quad \text{(hot, concentrated)} \]

In the cold reaction chlorine goes from 0 to −1 (in NaCl) and to +1 (in NaClO). In the hot one, from 0 to −1 and to +5. To show disproportionation in an answer, give the oxidation numbers of the element before and after, and say which change is oxidation and which is reduction.

Balancing equations with oxidation numbers

For a redox equation that will not balance by inspection, balance the electrons first:

  1. Find the elements whose oxidation numbers change, and by how much each changes.
  2. Choose multipliers so the total increase equals the total decrease — electrons lost equal electrons gained.
  3. Balance any other atoms except O and H.
  4. In acidic solution, balance O by adding H2O, then H by adding H+.
  5. Check that the charges balance on both sides.

Step 5 is not optional: if the charges do not balance, the equation is wrong, however good the atom count looks.

✏️Worked example

Acidified potassium manganate(VII) oxidises iron(II) ions to iron(III) ions, and is itself reduced to manganese(II) ions. (a) Give the oxidation number of manganese in MnO4− and in the product. (b) Construct the balanced ionic equation. (c) Identify the oxidising agent and the reducing agent. (d) In a titration, 0.00150 mol of MnO4− reacts. How many moles of Fe2+ were present?

(a) In MnO4−: 4 × (−2) = −8 from oxygen; the charge is −1; so Mn is +7. The product Mn2+ is +2. (That is what the (VII) in the name says.)

(b) Mn goes from +7 to +2: a decrease of 5 (gains 5 electrons). Fe goes from +2 to +3: an increase of 1 (loses 1 electron). For the changes to balance, one MnO4− must react with five Fe2+:

\[ \mathrm{MnO_4^{-} + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+}} \]

Four O atoms on the left: add 4H2O on the right. That puts 8 H on the right: add 8H+ on the left.

\[ \mathrm{MnO_4^{-}(aq) + 8H^{+}(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)} \]

Charge check: left −1 + 8 + 10 = +17; right +2 + 15 = +17. Balanced.

(c) MnO4− is reduced, so it is the oxidising agent. Fe2+ is oxidised, so it is the reducing agent.

(d) The equation ratio is 1 : 5, so n(Fe2+) = 5 × 0.00150 = 0.00750 mol.

Check it. Count electrons directly: five Fe2+ each lose one, so five electrons are released, and one Mn(+7) needs exactly five to reach +2. The electron count and the charge count agree, which is what a correct redox equation always shows. The 1 : 5 ratio is also worth remembering — it is the ratio in every manganate(VII)–iron(II) titration you will ever do.
Treating the oxidation number as a charge on the atom, and balancing only atoms. Manganese in MnO4− is not an Mn7+ ion — the bonds are covalent — so do not write it as one. And an equation such as MnO4− + 8H+ + Fe2+ → Mn2+ + Fe3+ + 4H2O balances in atoms but not in charge (+9 against +5). Using a 1 : 1 ratio here makes every later titration answer five times too small.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Give the oxidation number of the named element in: (a) N in NH4+; (b) S in S2O32−; (c) H in CaH2; (d) O in H2O2; (e) Cl in ClO4−.
(a) −3: 4 H at +1 = +4; the ion is +1, so N = +1 − 4. (b) +2: 3 O = −6; ion −2; two S together = +4. (c) −1: a metal hydride; Ca is +2, so each H is −1. (d) −1: a peroxide; 2 H = +2, total 0, so the two O together = −2. (e) +7: 4 O = −8; ion −1; Cl = +7. The ion is chlorate(VII).
2. Use oxidation numbers to balance: Cu + HNO3 → Cu(NO3)2 + NO + H2O.
Cu: 0 → +2 (increase of 2). N: +5 in HNO3 → +2 in NO (decrease of 3). Electrons balance with 3Cu (increase 6) and 2NO (decrease 6). 3Cu give 3Cu(NO3)2, which contains 6 more nitrate N whose oxidation number does not change, so HNO3 needed = 6 + 2 = 8. Hydrogen: 8 H → 4H2O. 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O. Check O: left 24; right 18 + 2 + 4 = 24. The key point is that only some of the nitric acid is reduced; the rest just supplies nitrate ions.
3. Construct the ionic equation for acidified dichromate(VI) ions, Cr2O72−, oxidising iodide ions to iodine; chromium ends as Cr3+.
Cr: +6 → +3, a decrease of 3 per Cr, so 6 per Cr2O72−. I: −1 → 0, an increase of 1 per I, so 6 I− are needed, giving 3I2. Seven O on the left → 7H2O on the right → 14H+ on the left. Cr2O72− + 14H+ + 6I− → 2Cr3+ + 3I2 + 7H2O. Charge: left −2 + 14 − 6 = +6; right 2 × (+3) = +6.
4. Explain, using oxidation numbers, why the reaction of chlorine with cold dilute sodium hydroxide is described as disproportionation.
Cl2 + 2NaOH → NaCl + NaClO + H2O. Chlorine starts at 0 in Cl2. In NaCl it is −1, a decrease, so that chlorine has been reduced. In NaClO (sodium chlorate(I)) it is +1, an increase, so that chlorine has been oxidised. The same element in the same species is both oxidised and reduced in one reaction, which is the definition of disproportionation. Na, O and H do not change oxidation number.
5. In the reaction 2Fe3+ + 2I− → 2Fe2+ + I2, identify the oxidising agent and the reducing agent, explaining in terms of electrons.
Each Fe3+ gains one electron (+3 → +2): it is reduced, so it takes electrons from the iodide — Fe3+ is the oxidising agent (electron acceptor). Each I− loses one electron (−1 → 0): it is oxidised, so it gives electrons to the iron — I− is the reducing agent (electron donor). Two electrons are transferred for each I2 formed, which is why two Fe3+ are needed.
6. Name, using Roman numerals: (a) FeSO4; (b) KClO3; (c) Cu2O; (d) NaNO2.
(a) Iron(II) sulfate — sulfate is 2−, so Fe is +2. (b) Potassium chlorate(V) — K +1, 3 O −6, so Cl +5. (c) Copper(I) oxide — O −2 shared between two Cu, so each is +1. (d) Sodium nitrate(III) — Na +1, 2 O −4, so N +3; the syllabus also uses its traditional name, sodium nitrite.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the redox pages, including a longer treatment of building ionic equations from half-equations
  • Royal Society of Chemistry — practical guides to manganate(VII) and iodine–thiosulfate titrations, the two redox titrations Paper 3 uses most