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Revision notes on how to construct and write full redox equations for Advanced Level Theoretical Chemistry:

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Doc Brown's Advanced Physical–Theoretical Inorganic Chemistry

REDOX REACTIONS Revision Notes Part 2

Sections 5 to 7 Analysing, constructing and using redox equations

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GCSE level notes on introducing REDOX reactions

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This page explains, with suitable examples, how to analyse redox reactions i.e. to identify the oxidation (rise in oxidation state) and the reduction (oxidation state decrease) in inorganic chemistry.

Then the method of how to construct a fully balanced ionic redox symbol equation from half–cell reactions (often given as the reduction half–reaction) is explained for many examples.

REDOX section–index: 1. Basic redox definitions * 2. Introducing oxidation state (with sub–index) * 3. Oxidation state rules–guidelines & inorganic examples * 4. Naming inorganic compounds * 5. Using oxidation states to describe redox changes in a given inorganic reaction equation (with sub–index) * 6. Constructing full inorganic redox equations from half–equations (with sub–index) * 7. Redox titrations * 8. Organic synthesis reductions (with summary table) * 9. Organic synthesis oxidations (with summary table) * 10. Other Organic Redox Reactions (with sub–index) * 11. Carbon's ox. state in selected organic compounds and functional group level * You are advised to study sections 1. to 6. in strict order and covers the requirements of A level * See also Equilibria Part 7 Redox Reactions for Half cell equilibria, electrode potential, standard hydrogen electrode, Simple cells and notation, Electrochemical Series, EØcell for reaction feasibility, 'batteries' and fuel cell systems etc.

Advanced Inorganic Chemistry Notes

Redox reactions in Inorganic Chemistry sections 5–7

The concepts of oxidation, reduction, oxidation state/number and redox reactions were introduced and explained using a variety of examples in Part 1 sections 1–4. Here in Part 2 sections 5–7, how to redox analyse more complex reactions and combine half–cell equations to write a full ionic–redox equation is explained and how redox chemistry is used in preparative and titration chemistry.

Application of redox concepts to cells are described in Equilibria Part 7 "Redox Equilibria"

You should know that oxidation is the process of electron loss and oxidising agents are electron acceptors. Reduction is the process of electron gain and reducing agents are electron donors.

You need to be able to apply the rules for assigning oxidation states.

You should be able to work out the oxidation state of an element in a compound or ion from the formula, write half-equations identifying the oxidation and reduction processes in redox reactions AND combine half-equations to give an overall redox equation

5. Using oxidation states to describe redox changes in a given reaction equation

Ex 5.1 The reaction between aluminium and a copper(II) salt solution

Ex 5.2 The reaction between sulphur dioxide/sulphite and halogens

  • If neutral molecules or more complex ions are involved, a bit more care must be taken e.g. when the sulphur dioxide is oxidised to sulphate by bromine (or the reduction of bromine to bromide).

  • SO2(aq) + Br2(aq) + 2H2O(l) ==> SO42–(aq) + 2Br–(aq) + 4H+(aq)

  • (i) the oxidation half reaction is: SO2(aq) + 2H2O(l) ==> SO42–(aq) + 4H+(aq) + 2e–

    • The sulphur changes from ox. state +4 to +6 (SO2 to SO42–).

  • (ii) the reduction half–reaction is: Br2(aq) + 2e– ==> 2Br–(aq)

    • Two bromine atoms (as molecule) change from ox. state 0 to –1.

  • The hydrogen (+1) and oxygen (–2) do not change oxidation state.

    • (i) + (ii) equals the balanced equation, 2 electrons gained and lost or an ox. state rise and fall of 2 units.

    • Bromine is the oxidising agent (gain/accept e–s, lowered ox. state),

    • and sulphur dioxide is the reducing agent (loses e–s, inc. ox. state of S).

  • Sulphur dioxide does ionise to a small extent in water to give the sulphite ion, and adding a strong non–oxidising acid like dilute hydrochloric acid to sodium metabisulphite produces the ion, which means another equation can also adequately describe the redox change in terms of sulphur and bromine.

    • e.g. if the sulphite ion acts as the reducing agent the reaction with chlorine would be written as:

    • SO32–(aq) + Cl2(aq) + H2O(l) ==> SO42–(aq) + 2Cl–(aq) + 2H+(aq)

  • Section 5. Index of examples of redox equation analysis

    TOP OF PAGE and sub-index

Ex 5.3 The oxidation of ammonia with molecular oxygen

  • The concept of oxidation state can now be fully applied to reactions which do not involve ions e.g.

  • The oxidation of ammonia via a Pt catalyst at high temperature which is part of the chemistry of nitric acid manufacture.

  • 4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g)  

  • The oxidation number analysis is:

    • 4N at (–3) each in NH3 and 10O all at (0) in O2 change to ...

    • 4N at (+2) each in NH3, 4O at (–2) each and 6 O at (–2) each in H2O.

    • H is +1 throughout i.e. does not undergo an ox. state change.

    • Oxygen is reduced from ox. state (0) to (–2).

    • Nitrogen is oxidised from ox. state (–3) to (+2).

    • The total increase in ox. state change of 4 x (–3 to +2) for nitrogen is balanced by the total decrease in ox. state change of 10 x (0 to –2) for oxygen i.e. 20 e– or ox. state units change in each case.

    • Oxygen is the oxidising agent (gain/accept e–s, lowered ox. state) and ammonia is the reducing agent (loses e–s, inc. ox. state of N).

  • Section 5. Index of examples of redox equation analysis

    TOP OF PAGE and sub-index

Ex 5.4 The reaction between iron and steam at >400oC

Ex 5.5 The formation of titanium(IV) chloride from titanium(IV) oxide in the extraction of titanium metal

  • TiO2(s) + C(s) + 2Cl2(g) ==> TiCl4(l) + CO2(g)

  • Oxidation number changes:

    • The carbon, C is oxidised (0 to +4 in CO2) and the chlorine is reduced (0) in Cl2 to (–1) in TiCl4.

    • The oxidation of 1 x C from (0) to (+4) is balanced by the reduction of 4 x Cl from (0) to (–1).

    • Titanium (+4) and oxygen (–2) do not change.

  • Chlorine is the oxidising agent (gains e–s, lowered ox. state) and carbon is the reducing agent (loses e–s, inc. ox. state of C), but not in sense carbon reduces an iron oxide to iron in a blast furnace because titanium does not change oxidation state and another step is required to obtain the metal.

  • The titanium itself is extracted via another redox reaction by displacement with a more reactive metal.

  • TiCl4(l) + 4Na(s) ==> Ti(s) + 4NaCl(s)

    • Ox. state changes :

    • The titanium is reduced from (+4) in TiO2 to (0) as Ti, and the displacing metal is oxidised (from 0 to >+1).

    • The reduction of 1 x Ti (+4) to (0) is balanced by the oxidation of 4 x Na (0) to (+1)

    • or TiCl4(l) + 2Mg(s) ==> Ti(s) + 2MgCl2(s)

      • Ox. state changes : The 1 x Ti (+4) to (0) reduction is balanced by the 2 x Mg (0) to (+2) oxidation.

    • In both cases titanium(IV) chloride is the oxidising agent (gain/accept e–s, lowered ox. state of Ti) and sodium/magnesium are the reducing agent (lose/donate e–s, inc. their ox. state).

  • Section 5. Index of examples of redox equation analysis

  • For more on titanium chemistry see Periodic Table Advanced Inorganic Chemistry Notes Part 10a "3d block Transition Metals Series Introduction and Elements Sc to Mn on Period 4 – detailed revision notes

    TOP OF PAGE and sub-index

Ex 5.6 The combustion of liquid hydrazine and oxygen

  • This has been used as a liquid rocket fuel mixture.

  • N2H4(l) + O2(l) ==> N2(g) + 2H2O(g)

    • Oxidation state changes:

    • The nitrogen in hydrazine oxidised (N from –2 to 0) and the oxygen molecules are reduced (O from 0 to –2).

    • The oxidation of 2 x N from (–2) to (0) is balanced by the reduction of 2 x O from (0) to (–2).

    • H at (+1) does not change in oxidation state and nitrogen is initially (–2) here, NOT (–3) as in ammonia, NH3.

    • Oxygen is the oxidising agent (gains/accepts e–s, lowered ox. state) and hydrazine is the reducing agent (loses/donates e–s, inc. ox. state of N).

  • Section 5. Index of examples of redox equation analysis

    TOP OF PAGE and sub-index

Ex 5.7 Converting manganese(IV) oxide into potassium manganate(VI)/manganate(VII)

  • 5.7.1: If a mixture of manganese(IV) oxide, potassium hydroxide and potassium chlorate(V) is heated strongly and fuse in a crucible, the following redox reaction takes place:

    • 3MnO2 + 6OH– + ClO3– ==> 3MnO42– + 3H2O + Cl–

    • The manganate(VI) ion is formed and the ox. number changes are ...

    • the oxidation of 3 x Mn from (+4) to (+6), MnO2 ==> MnO4–, total 6 e's lost

    • Is balanced by the reduction of 1 x Cl from (+5) to (–1), ClO3– ==> Cl–, total 6 e's gained

    • and hydrogen (+1) and oxygen (–2) do not change oxidation state.

    • The chlorate(V) ion is the oxidising agent (gains/accepts e–s, lowered ox. state of Cl) and manganese(IV) oxide is the reducing agent (loses/donates e–s, inc. ox. state of Mn).

  • 5.7.2: When the fused mixture is dissolved in water, the initially green solution of the manganate(VI) ion, slowly changes to the purple colour of the manganate(VII) ion and a black precipitate of manganese(IV) oxide.

    • 3MnO42–(aq) + 2H2O(l) ==> 2MnO4–(aq) + MnO2(s) + 4OH–(aq)

    • Oxidation number changes:

    • Initially there are three Mn at (+6).

    • Two Mn at (+6) are oxidised to two Mn(+7), an ox. state total increase of 2 units or 2e– lost,

    • and one Mn at (+6) is reduced to Mn(+4), an ox. state total decrease of 2 units or 2e– gained.

    • This is an example of disproportionation  where an element in one oxidation state simultaneously changes into a higher and lower oxidation state species.

    • It also means that the manganate(VI) ions simultaneously acts as a reducing agent and oxidising agent!

    • As in 5.7.1, hydrogen (+1) and oxygen (–2) do not change in oxidation state.

  • Section 5. Index of examples of redox equation analysis

  • For more on manganese chemistry see Periodic Table Advanced Inorganic Chemistry Notes Part 10a "3d block Transition Metals Series Introduction and Elements Sc to Mn on Period 4 – detailed revision notes

    TOP OF PAGE and sub-index

Ex 5.8 The disproportionation of copper(I) oxide in acid

  • When brown copper(I) oxide is dissolved in dilute sulphuric acid, a deposit of copper is formed and a blue solution of copper(II) sulphate.

    • Cu2O(s) + H2SO4(aq) ==> Cu(s) + CuSO4(aq) + H2O(l)

    • Cu2O(s) + 2H+(aq) ==> Cu(s) + Cu2+(aq) + H2O(l)

    • The hydrated copper(I) ion, Cu+(aq) is unstable and changes into two oxidation states simultaneously,

    • elemental Cu(0) and the blue copper(II) ion (+2), and is another example disproportionation (see also 5.7.2).

    • i.e. 2 x Cu (+1) changes to 1 Cu at (0) by reduction and 1 Cu at (+2) by oxidation.

    • Oxygen in its (–2) ox. state and hydrogen in its (+1) ox. state do not change.

    • It also means that the copper(I) ions simultaneously acts as a reducing agent and oxidising agent and are simultaneously oxidised and reduced.

    • You also observe the same disproportionation if copper(I) sulphate is dissolved in water

    • Cu2SO4(s) + aq ==> Cu(s) + Cu2+(aq) + SO42–(aq)

  • Section 5. Index of examples of redox equation analysis

  • For more on copper chemistry see Part 10b 3d–block Transition Metals Fe to Zn – detailed revision notes

    TOP OF PAGE and sub-index

Ex 5.9 The conversion of cobalt(II) to cobalt(III) via molecular oxygen (air) or hydrogen peroxide

  • When aqueous ammonia is added to cobalt(II) salt solutions the hexaamminecobalt(II) complex is formed which is readily oxidised to the cobalt(II) complex by (i) oxygen dissolving from air or (ii) adding hydrogen peroxide solution.

    • [Co(H2O)6]2+(aq) + 6NH3(aq) ==> [Co(NH3)6]2+(aq) + 6H2O(l) 

      • NOT a redox reaction, just a ligand exchange or ligand substitution reaction.

    • (i) 4[Co(NH3)6]2+(aq) + O2(aq) + 4H+(aq) ==> 4[Co(NH3)6]3+(aq) + 2H2O(l) 

      • Oxidation: In the complexes 4 Co at (+2) change to 4 Co at (+3), total increase of 4 ox. state units.

      • Reduction: 2 O in O2 at (0) change to 2 O at (–2) in 2H2O, total decrease of 4 ox. state units.

      • No redox change involving the NH3 ligand or the H+ ions, N stays at –3 and H at +1.

    • (ii) 2[Co(NH3)6]2+(aq) + H2O2(aq) + 2H+(aq) ==> 2[Co(NH3)6]3+(aq) + 2H2O(l) 

      • Oxidation: 2 Co at (+2) change to 2 Co at (+3), total increase of 2 ox. state units.

      • Reduction: 2 O at (–1) in H2O2 change to 2 O at (–2) in 2H2O, total decrease of 4 ox. state units.

      • No redox change involving the NH3 ligand or the H's of the hydrogen peroxide molecule or hydrogen ions.

    • Oxygen/hydrogen peroxide act as the oxidising agent (gains/accepts e–s, lowered ox. state of O) and the cobalt(II) complex is the reducing agent (loses/donates e–s, inc. ox. state of Co).

  • Section 5. Index of examples of redox equation analysis

  • For more on cobalt chemistry see Part 10b 3d–block Transition Metals Fe to Zn – detailed revision notes

    TOP OF PAGE and sub-index

Ex 5.10 The conversion of chromium(III) to chromium(VI) compounds

Ex 5.11 The oxidation of hydrogen sulphide with iron(III) ions

Ex 5.12 The reaction of copper(II) ions with iodide ion

  • When potassium iodide solution is added to a copper salt solution, a white precipitate of copper(I) iodide is formed which is masked by a brown solution/black ppt. of iodine. If sodium thiosulphate solution is added dropwise carefully, it reacts to remove the iodine giving the colourless iodide ion (see Ex 5.12), hence the CuI precipitate is better scene.

    • 2Cu2+(aq) + 4I–(aq) ==> 2CuI(s) + I2(aq/s)

    • Oxidation: 2 I– at (–1) of the 4I– change to 2 at (0) in I2, total 2e– loss, inc. 2 ox. state units.

    • Reduction: 2 Cu at (+2) change to 2 Cu (+1) in the CuI, total 2e– gain, decrease 2 ox. state units.

    • Two of the I– iodide ions do not change oxidation state, but they do change their physical–chemical situation.

    • The copper(II) ion acts as the oxidising agent (gains/accepts e–s, lowered ox. state of O) and the iodide ion is the reducing agent (loses/donates e–s, inc. ox. state of I).

  • Section 5. Index of examples of redox equation analysis

  • For more on copper chemistry see Part 10b 3d–block Transition Metals Fe to Zn – detailed revision notes

    TOP OF PAGE and sub-index

Ex 5.13 More examples of disproportionation and vice versa (involving O and N ions/compounds)

  • 5.13.1 Hydrogen peroxide decomposition, catalysed by the black solid manganese(IV) oxide, MnO2.

    • 2H2O2(aq) ==> O2(g) + 2H2O(l)

    • Ox. state changes: 4O at (–1) change to 2O at (0) in O2 and 2O at (–2) in H2O

    • and H is unchanged at (+1).

    • A case of disproportionation where an element in a species simultaneously changes into a higher and lower oxidation state i.e. here two oxygen atoms increase their oxidation state and two oxygen atoms decrease their oxidation state.

    • It also means that hydrogen peroxide simultaneously acts as a reducing agent and oxidising agent.

  • 5.13.2 The reaction between ammonium and nitrate(III) (nitrite) ions

    • NH4+(aq) + NO2–(aq) ==> 2H2O(l) + N2(g)

    • Here its the opposite of disproportionation where two species of an element in different oxidation states react to produce one species of a single intermediate oxidation state.

    • Ox. state changes: Nitrogen in a (–3) and a (+3) state both end up in the (0) state.

    • Oxygen at (–2) and hydrogen (+1) remain unchanged in oxidation state.

    • The nitrite ion acts as the oxidising agent and gets reduced (N +3 to 0, 3e's gained, decrease of 3 ox. state units)

    • and the ammonium ion acts as the reducing agent and gets oxidised (N –3 to 0, 3 e's lost, inc. ox. state 3 units).

    • The nitrite ion acts as the oxidising agent (gains/accepts e–s, lowered ox. state of N) and the ammonium ion acts as the reducing agent (loses/donates e–s, inc. ox. state of N).

  • See other disproportionation reactions 5.7.2 MnO42–, 5.8 Cu2O, 5.14 chlorine and an organic example.

  • and the opposite of disproportionation! reactions 6.6 iodate(V) + iodide.

  • Section 5. Index of examples of redox equation analysis

    TOP OF PAGE and sub-index

5.14 Some chlorine, chlorates and chloride redox changes

  • In all the reactions quoted in section 5.14, (i) the oxidation states of hydrogen (+1) and oxygen (–2) remain unchanged and (ii) the process descriptions are over simplified but the main reactions described provide good examples of the redox chemistry of chlorine.

    •  When chlorine reacts with aqueous sodium hydroxide at least two different reaction can occur depending on the temperature and concentration of the strong base–alkali.

  • 5.14.1 With cold dilute sodium hydroxide solution alkali sodium chlorate(I) (NaClO, the bleach sodium hypochlorite) is formed as well as sodium chloride.

    • 2NaOH(aq) + Cl2(aq) ==> NaCl(aq) + NaClO(aq) + H2O(l)

    • 2OH–(aq) + Cl2(aq) ==> Cl–(aq) + ClO–(aq) + H2O(l)

    • The chlorine disproportionates from 2Cl(0) to 1Cl (–1, chloride ion) plus 1Cl(+1, chlorate(I) ion).

    • Overall 1 electron gained, (1 ox. state unit decrease) balanced by 1 electron lost (1 ox. state unit increase).

  • 5.14.2 However, with hot concentrated sodium hydroxide solution, above 75oC, the formation of sodium chlorate(V) predominates as well as sodium chloride.

    • 6NaOH(aq) + 3Cl2(aq) ==> 5NaCl(aq) + NaClO3(aq) + 3H2O(l)

    • 6OH–(aq) + 3Cl2(aq) ==> 5Cl–(aq) + ClO3–(aq) + 3H2O(l)

    • The chlorine disproportionates from 6Cl(0) to 5Cl(–1, chloride ion) plus 1Cl(+5, chlorate(V)ion).

    • Overall 5 electrons gained, (5 ox. state unit decrease) balanced by 5 electrons lost (5 ox. state unit increase).

  • 5.14.3 The change in reaction mode from 5.14.1 to 5.14.2 is due to the instability of the chlorate(I) ion, which at higher temperatures disproportionates into the chloride ion and the chlorate(V) ion.

    • 3NaClO(aq) ==> 2NaCl(aq) + NaClO3(aq)

    • 3ClO–(aq) ==> 2Cl–(aq) + ClO3–(aq)

    • The 'chlorine' in the chlorate ion disproportionates from 3Cl(+1) to 2Cl(–1, chloride ion) plus 1Cl(+5, chlorate(V) ion).

    • Overall 4 electrons gained, (4 ox. state unit decrease) balanced by 4 electrons lost (4 ox. state unit increase).

  • 5.14.4 A concentrated solution of sodium chlorate(I) is a useful source of chlorine in the laboratory because it readily reacts with conc. hydrochloric acid to give off the gas.

    • NaClO(aq) + 2HCl(aq) ==> NaCl(aq) + H2O(l) + Cl2(aq/g)

    • ClO–(aq) + Cl–(aq) + 2H+(aq) ==> H2O(l) + Cl2(aq/g)

    • The 'chlorine' here does the opposite of disproportionation and changes from 1Cl(+1, chlorate(I) ion) plus 1Cl(–1, chloride ion) to give 2Cl(0, chlorine molecule).

    • Overall 1 electron lost, (1 ox. state unit increase) balanced by  electron gained (1 ox. state unit decrease).

  • Section 5. Index of examples of redox equation analysis

    TOP OF PAGE and sub-index

6. Constructing full ionic–redox equations from half–cell equations

REDOX EQUATION CHECKS

  • Use of the correct 'species' of the half–cell equations to be put together, the right way round AND in the correct ratio based on the number of electrons transferred or oxidation state changes.

  • Getting the right ratio of the oxidation/reduction half–cell reactions should ensure the electrons are 'hidden' and you can add them up in a simple algebraic way – see the tabular expression of the way of thinking.

  • If not told, you must decide on the direction of change – which is oxidised or reduced? from EØ data supplied (half–cell potentials) and the most +ve (least –ve) half–cell potential indicates the reduction half–equation. Much more on this in Equilibria Part 7.

  • The total increase in oxidation states of elements = the total decrease in oxidation states of elements,

  • or, total electrons gained by species = total electrons lost by the species involved.

  • Add up the ion charges, the totals should be the same on both sides of the equation. I find this a handy extra check especially with stray H+'s and  H2O's.

  • The 'traditional' atom count – do last because its not completely reliable with redox equations!

Ex 6.1 The reaction between zinc metal and a silver salt solution

All the half–cell equations are presented as a reduction – electron gain, so one must be reversed!

 See also Equilibria Part 7 Redox Reactions for Half cell equilibria, electrode potential, standard hydrogen electrode, Simple cells and notation, Electrochemical Series, EØcell for reaction feasibility, 'batteries' and fuel cell systems etc.

  • Half–cell reaction data:

    • (i) Zn2+(aq) + 2e– ==> Zn(s) (EØ = –0.76V*, Zn will act as reducing agent, EØ less positive)

    • (ii) Ag+(aq) + e– ==> Ag(s) (EØ = +0.80V*, reduction of the oxidising agent with the more positive EØ)

    • * It doesn't matter here if you haven't yet studied EØ, half–cell potentials in detail, but the more +ve half–cell species acts as the oxidising agent and so is the reduction half of the reaction.

  • The more reactive metal Zn, displaces the less reactive metal (Ag) from one of its compounds, which is the reaction feasibility rule at lower academic levels for such a redox reaction (see also halogen displacement reaction 6.2 below).

  • With redox analysis of the reaction we can now say:

    • The zinc is oxidised from 0 to +2 in ox. state,  2e– loss,

    • and the two silver ions are reduced from –1 to 0 ox. state, 2 x 1e– gain.

  • The zinc metal is a stronger reducing agent (more powerful e– donor, less +ve EØ) than silver,

  • or to put it another way,

  • the Ag+ ion is a stronger oxidising agent (more powerful e– acceptor, more +ve EØ) than the Zn2+ ion.

  • So one of the Zn half–cell equations will be balanced by two of the silver half–cell equations giving the complete ionic–redox equation, showing NO electrons.

  • 1 x oxidation half–cell, (i) reversed  Zn(s) ==> Zn2+(aq) + 2e–
    2 x reduction half cell, (ii) 2Ag+(aq) + 2e– ==> 2Ag(s)
    added gives full redox equation Zn(s) + 2Ag+(aq) ==> Zn2+(aq) + 2Ag(s)

  • This sort of displacement reaction can be used to plate more reactive metals with a less reactive metal without the need for electrolysis–electroplating e.g. dipping iron/steel into copper(II) sulphate to give a pink–brown coating of copper.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

  • See also Periodic Table Advanced Inorganic Chemistry Notes Part 10a "3d block Transition Metals Series Introduction and Elements Sc to Mn on Period 4 – detailed revision notes

    TOP OF PAGE and sub-index

Ex 6.2 The reaction between aqueous chlorine and potassium iodide solution

  • Half–cell reaction data:

    • (i) 1/2Cl2(aq) + e– ==> Cl–(aq) (EØ = +1.36, can be Cl2(aq) + 2e– ==> 2Cl–(aq) but EØ still +1.36)

    • (ii) 1/2I2(aq) + e– ==> I–(aq) (EØ = +0.54V, can be I2(aq) + 2e– ==>  2I–(aq) but EØ still +0.54V)

  • Chlorine molecules are reduced from ox. state (0) to (–1) of the chloride ion, 1 electron gain.

  • Iodide ions are oxidised from ox. state (–1) to (0) of the iodine molecule, 1 electron loss.

  • So chlorine molecules are the oxidising agent (more powerful e– acceptor, more +ve EØ) and iodide ions are the reducing agent (e– donor, less +ve EØ).

  • 2 x oxidation half–cell, (ii) reversed  2I–(aq) ==> I2(aq) + 2e–
    2 x reduction half–cell, (i) Cl2(aq) + 2e– ==> 2Cl–(aq)
    added gives full redox equation Cl2(aq) +  2I–(aq) ==> Cl2(aq) + I2(aq)

  • One method of estimating chlorine in water e.g. from bleaches, is to add excess potassium iodide and titrating the liberated iodine with standardised sodium thiosulphate, which itself is another redox reaction (see Ex 6.10)

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

    TOP OF PAGE and sub-index

Ex 6.3 The reaction between hydrogen peroxide and iron(II) ions

  • Half–cell reaction data:

    • (i) H2O2(aq) + 2H+(aq) + 2e– ==> 2H2O(l) (EØ = +1.36, reduction of oxidising agent with the more positive EØ)

    • (ii) Fe3+(aq) +  e– ==> Fe2+(aq) (EØ = +0.77, less positive, so Fe3+ can't oxidise hydrogen peroxide)

  • Both the iron(III) ion and hydrogen peroxide molecule can act as oxidising agents, but hydrogen peroxide is stronger and so oxidises the iron(II) ion to the iron(III) ion.

  • Oxidation: Two iron(II) ions at (+2) lose an electron each to give iron(III) ions at (+3) ox. state.

  • Reduction: 2 O at (–1) in each H2O2 are reduced to the (–2) state in the 2H2O.

  • The hydrogens (+1) do not change oxidation state.

  • 2 x oxidation half–cell, (ii) reversed  2Fe2+(aq) –  2e– ==> 2Fe3+(aq)
    1 x reduction half–cell, (i) H2O2(aq) + 2H+(aq) + 2e– ==> 2H2O(l)
    added gives full redox equation 2Fe2+(aq) + H2O2(aq) + 2H+(aq) ==> 2Fe3+(aq) + 2H2O(l)

  • This reaction is used to convert e.g. iron(II) sulphate, FeSO4, into iron(III) sulphate, Fe2(SO4)3, because dissolving iron in dil. sulphuric acid gives the Fe(II) salt.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

  • For more on iron chemistry see Part 10b 3d–block Transition Metals Fe to Zn – detailed revision notes

    TOP OF PAGE and sub-index

Ex 6.4 The reaction between acidified manganate(VII) ions and iron(II) ions

  • Half–cell reaction data:

    • (i) MnO4–(aq) + 8H+(aq)  + 5e– ==> Mn2+(aq) + 4H2O(l) (EØ = +1.52, reduction of oxidising agent)

    • (ii) Fe3+(aq) +  e– ==> Fe2+(aq) (EØ = +0.77, Fe2+ acts as reducing agent, Fe2+ gets oxidised, EØ less positive)

  • Oxidation: Iron(II) ions, Fe2+, (+2) lose an electron, so oxidised to the iron(III) ion, Fe3+, (+3), Fe +2 to +3 ox. state.

  • Reduction: Manganate(VII) ions, MnO4–, (+7) are reduced to manganese(II) ions, Mn2+, (+2), 5e– gain, so five Fe2+ ions can be oxidised, Mn +7 to +2 ox. state.

  • Hydrogen (+1) and oxygen (–2) do not change oxidation state.

  • 5 x ox. half–cell, (ii) reversed  5Fe2+(aq) ==> 5Fe3+(aq) + 5e–
    1 x reduction half–cell, (i) MnO4–(aq) + 8H+(aq)  + 5e– ==> Mn2+(aq) + 4H2O(l)
    added full redox equation MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) ==> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

  • This reaction is used to quantitatively estimate iron(II) ions and is self–indicating. On the addition of standardised potassium manganate(VII) to the iron solution, decolourisation occurs as the almost colourless Mn(II) ion (a VERY pale pink) is formed from the reduction of the intensely purple manganate(VII) ion, and the end–point is the first permanent pale pink with =< 1 drop excess of the oxidising agent.

  • The presence of dilute sulfuric ('supplier' of the proto, H+ ion), ensures the desired sole reduction of the manganate(VII) ion to the Mn(II) ion, thereby preventing the formation of a manganese(IV) oxide precipitate. Formation of MnO2 which would not give a good end point and cause a duality in the redox reactions occurring, so introducing errors and quantitative complications.

  • There is a 2nd good reason for using dilute sulphuric acid, as opposed to using other common mineral acids. Dil. sulphuric acid does not undergo any redox reactions under the conditions of this titration.

  • Dilute hydrochloric acid cannot be used because the manganate(VII) ion will oxidise the chloride ion (see 6.11) and dil. nitric acid, via the nitrate(V) ion, will oxidise the iron(II) ion, i.e. both acids will lead to false titration results.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

  • For more on manganese chemistry see Periodic Table Advanced Inorganic Chemistry Notes Part 10a "3d block Transition Metals Series Introduction and Elements Sc to Mn on Period 4 – detailed revision notes

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Ex 6.5 The reaction between acidified potassium dichromate(VI) and iron(II) ions

  • Half–cell reaction data:

    • (i) Cr2O72–(aq) + 14H+(aq) + 6e– ==> 2Cr3+(aq) + 7H2O(l) (EØ = +1.33, reduction of the oxidising agent)

    • (ii) Fe3+(aq) +  e– ==> Fe2+(aq) (EØ = +0.77, Fe2+ will act as reducing agent, EØ less positive)

  • Oxidation: Iron(II) ions at (+2) lose an electron each to give an iron(III) ion at (+3), Fe +2 to +3 ox. state.

  • Reduction: Each Cr at (+6) is reduced by gaining 3e– to give Cr at (+3) ox state, Cr +6 to +3 ox. state.

  • the Cr2O72– ion is the oxidising agent and each can oxidise 6 Fe2+ ions.

  • Hydrogen (+1) and oxygen (–2) do not change oxidation state.

  • 6 x ox. half–cell, (ii) rev.  6Fe2+(aq) ==> 6Fe3+(aq) + 6e–
    1 x red'n half–cell, (i) Cr2O72–(aq) + 14H+(aq) + 6e– ==> 2Cr3+(aq) + 7H2O(l)
    added full equation Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) ==> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

  • Like with potassium manganate(VII), standardised potassium dichromate(VI) solution can be used to estimate quantitatively iron(II) ions in solution, though a special redox organic dye* indicator which must be used to detect the end point.

  • The organic dye changes colour when oxidised to another form, but only after the iron is oxidised i.e. it is not as easily oxidised as Fe2+, i.e. the dye's EØ is  more +ve than Fe2+ but lees than for the manganate(VII) ion, hence it is capable of being oxidized by the dichromate(VI) ion to show the end–point.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

  • For more on chromium chemistry see Periodic Table Advanced Inorganic Chemistry Notes Part 10a "3d block Transition Metals Series Introduction and Elements Sc to Mn on Period 4 – detailed revision notes

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Ex 6.6 The reaction between iodate(V) and iodide ions in acidified aqueous solution

  • Half–cell reaction data:

    • (i) 1/2I2(aq) + e– ==> I–(aq) (EØ = +0.54V, iodide gets oxidised, acts as reducing agent, less positive EØ)

    • (ii) IO3–(aq) + 6H+(aq) + 5e– ==> 1/2I2(aq) + 3H2O(l)  (EØ = +1.19V, reduction of oxidising agent, more positive EØ)

  • The iodide ions (I at –1) are oxidised to iodine molecules (I at 0) by electron loss to the iodate(V) ion, I –1 to 0 ox. state.

  • The iodate(V) ions (I at +5) are reduced to iodine molecules (I at 0) by electron gain from the iodide ions (the reducing agent), I +5 to 0 ox. state.

  • Hydrogen (+1) and oxygen (–2) do not change oxidation state.

  • 5 x ox'n half–cell, (i) reversed  5I–(aq) ==> 5/2I2(aq) + 5e–
    1 x reduction half–cell, (ii) IO3–(aq) + 6H+(aq) + 5e– ==> 1/2I2(aq) + 3H2O(l)
    added gives full equation IO3–(aq) + 6H+(aq) + 5I–(aq) ==> 3I2(aq) + 3H2O(l)

  • The reaction can be used to estimate iodate(V) by adding excess potassium iodide and titrating the liberated iodine with standardised sodium thiosulphate or using the liberated iodine from a known quantity of potassium iodate(V) salt with excess KI(aq) salt solution to standardise the sodium thiosulphate.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

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Ex 6.7 The reaction between acidified potassium manganate(VII) and hydrogen peroxide solutions aqueous

Ex 6.8 The titration of ethanedioate with acidified potassium manganate(VII) solution

Ex 6.9 Reduction of the vanadium(IV) oxo–cation by tin(II) salts

Ex 6.10 Titrating iodine with standardised sodium thiosulphate solution

  • Half–cell reaction data:

    • (i) 1/2I2(aq) + e– ==> I–(aq)

      • EØ = +0.54V, the most positive, so iodine (I2) acts as the oxidising agent, which gets reduced in the process.

    • (ii) 1/2S4O62–(aq) + e– ==> S2O32–(aq)

      • EØ = +0.09V, this EØ is less positive, so the thiosulfate ion (S2O32–)  will be oxidised.

  • The iodine is reduced by the thiosulphate ion to form iodide, ox. state of I (0) to (–1).

  • The thiosulphate ion is oxidised to the tetrathionate ion. In doing so the sulphur atom changes ox. state from an average of four at (+2) in the two S2O32– ions to an average of four at (+2.5) in the single S4O62– ion.

    • A bit of an awkward one in analysing sulphur and it is best to reason in terms of an average oxidation state of sulphur.

  • 2 x reduction half–cell, (i)  I2(aq) + 2e– ==> 2I–(aq)
    2 x oxidation half–cell, (ii) rev. 2S2O32–(aq) ==> S4O62–(aq) + 2e–
    added gives full equation 2S2O32–(aq) + I2(aq) ==> S4O62–(aq) + 2I–(aq)

  • This is used to quantitatively estimate iodine in aqueous solution. The indicator is a few drops of starch solution which forms a blue–black complex with iodine. The end–point is when the solution first becomes colourless with no remaining iodine to form the coloured complex. The iodine to be titrated may arise from a variety of reactions for analysis purposes, see 6.2, 6.6 and 6.12.

  • Section 6. Index of examples of constructing balanced ionic redox reaction equation from half–cell/EØ data

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Ex 6.11 The oxidation of chloride by acidified potassium manganate(VII)

Ex 6.12 The reduction of acidified dichromate(VI) with iodide ions

Ex 6.13 The oxidation of hydrogen sulphide by acidified potassium manganate(VII)

  • Half–cell reaction data:

    • (i) MnO4–(aq) + 8H+(aq)  + 5e– ==> Mn2+(aq) + 4H2O(l) (EØ = +1.52, reduction of the oxidising agent)

    • (ii) SO42–(aq) + 10H+(aq) + 8e– ==>  H2S(aq) + 4H2O(l)  (EØ = ?, oxidation of the reducing agent )

    • (ii) assumes that the sulphur is initially in a covalent molecular state and NOT sulphide ions, which would be readily protonated by the dilute sulphuric acid. It further assumes the sulphur is completely oxidised from –2 in H2S, to its maximum oxidation state of +6 as the sulphate(VI) ion.

    • So you have to balance up an 8 e–/8 ox. no. units reduction with a 5 e–/5 ox. no. units half–cell

  • 5 x oxi'n half–cell, (ii) rev. 5H2S(aq) + 20H2O(l) ==> 5SO42–(aq) + 50H+(aq) + 40e–
    8 x red'n half–cell, (i) 8MnO4–(aq) + 64H+(aq)  + 40e– ==> 8Mn2+(aq) + 32H2O(l)
    added – full redox equation 8MnO4–(aq) + 14H+(aq) + 5H2S(aq) ==> 8Mn2+(aq) + 5SO42–(aq) + 12H2O(l)

  • This is quite a tricky one to do with awkward numbers!

  • I'm not sure exactly what happens in practice, so above is theoretical, therefore in addition to the above 'construction' if the hydrogen sulphide is just oxidised to a sulphur precipitate, the equation would be ...

    • 2MnO4–(aq) + 6H+(aq) + 5H2S(aq) ==> 2Mn2+(aq) + 5S(s) + 8H2O(l)

    • 2 x Mn(VII) ==> 2 x Mn(II) of equation (i) (10 e– change), balanced by 5 x S(–2) to 5 x S(0) of equation (iii) reversed (below).

    • (iii) S(s) + 2H+(aq) + 2e– ==> H2S(aq) (EØ = +0.14V, less positive, so hydrogen sulphide gets oxidised)

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Ex 6.14 The conversion of chromium(III) to chromium(VI)

  • When hydrogen peroxide is added to an alkaline chromium(III) solution, oxidation occurs to give the yellow chromate(VI) ion CrO42– . 

    • 2Cr3+(aq) + 3H2O2(aq) + 10OH–(aq) ==> 2CrO42–(aq) + 8H2O(l)

    • Redox changes: oxidation 2Cr(+3) ==> 2Cr(+6), and for the corresponding ....

      • reduction 6 O(–1) in 3H2O2 ==> 6(–2) in 6 of the 8H2O

      • a total of 6 'units' oxidation state change, which I sometimes unofficially call 6 'electrons worth' of change!

    • Both Cr(VI) compounds and hydrogen peroxide and are oxidising agents e.g.

      • EØ = +1.33V for the half–cell reaction: Cr2O72–(aq) + 14H+(aq) + 6e– rev 2Cr3+(aq) + 7H2O(l)

      • Hydrogen peroxide is a stronger oxidising agent, and for the half–cell reaction:

      • EØ = +1.77 V  for H2O2(aq) + 2H+(aq) + 2e– rev 2H2O(l)

      • BUT, both of the above half-cell potentials are for acid–neutral conditions, so alternative half–cell potential data must be used for this Cr(III) to Cr(VI) conversion.

        • What we are looking at here is a case of a pH change producing different half–cell potentials and different species involved in a redox reaction.

      • For strongly alkaline conditions for the conversion of chromium(III) ion to the chromate(VI) ion the following half–cell potential data should be used involving the perhydroxyl ion (HO2– or HOO–):

        • (a) EØ = +0.88 V  for the half–cell reaction: HO2–(aq) + H2O(aq) + 2e– rev 3OH–(aq)

        • So alkaline hydrogen peroxide solution is still quite a strong oxidising agent.

        • Note the oxidant species is considered to be the perhydroxyl ion.

      • However, unlike the orange dichromate(VI) ion (Cr2O72–) in acid solution, the yellow chromate(VI) ion is a very weak oxidising agent in alkaline solution (in acid solution it reverts to the dichromate(VI) ion).

        • EØ = –0.11 V for the half–cell reaction (b): CrO42–(aq) + 4H2O(l) + 3e– rev Cr(OH)3(s) + 5OH–(aq)

      • From the standard electrode potentials (+0.88 V > –0.12 V) you can clearly see that the hydrogen peroxide can oxidise the chromium(III) ion/hydroxide to the chromate (VI) ion.

        • EØreaction = EØred – EØox = EØHOO–/OH–  –  EØCrO42–/Cr(OH)3 = +0.88 – 0.12 = + 1.00 V, a very feasible reaction!

        • Note that the chromium species EØ used in the argument involves chromium(III) hydroxide.

      • It is quite legitimate to do so, because when you add sodium hydroxide to a chromium(III) salt solution you get a green precipitate of chromium(III) hydroxide.

        • simple equation: Cr3+(aq) + 3OH–(aq) ===> Cr(OH)3(s)

      • When the hydrogen peroxide is added, this green precipitate is oxidised and dissolves to give a yellow solution of sodium chromate(III). The overall redox equation is:

        • 2Cr(OH)3(s)  +  3HO2–(aq)  +  OH–(aq) ===> 2CrO42–(aq)  +  5H2O(l)

      • The oxidation state change numbers and balancing are analysed below.

      • Some of these redox equations are quite tricky to work out – do a triple check ...

        • .... especially when combining two half–cell equations, which I've illustrated for this reaction ...

        • (i) balance the number of species in the equation with the oxidation state changes

          • 2Cr(+3) ==> 2Cr(+6) and 6 O(–1) in 3HO2– ==> 6 O(–2) in the oxygen's of the H2O's/CrO42- ion, overall a 6 'electron' change

        • (ii) check the ionic charge balance

          • on the left 3- + 1- = 4– is balanced by 2 x 2– = 4–

            • Note: if it involves both + and – ions, just obey the usual mathematical rules

        • (iii) double check the atom count

          • 2Cr + 13O + 13H on both sides of the equation finally!

        • Do all three and you shouldn't go wrong!

      • However on reflection, you do not get equation (c) by combing half–cell equations (a) and (b), BUT, using the perhydroxyl half–cell equation I've worked it through in one of my more 'nerdy' moments on my website to obtain equation (d). I've kindly left out the state symbols until the end for logic clarity! Check out the 'triple check' at each stage in deriving equation (d) for practice, that is well worth doing! Equation (d) is the most accurate depiction of the redox reaction that actually happens.

        • You may have to know how to carry out the reaction for your examination, but not all the details, BUT, you should understand all the principles used in this explanation and appreciate that changing the pH of an oxidant can change both its oxidising power and species involved.

3 x (a) for 6e change

3HO2– + 3H2O + 6e–

 ===>

9OH–
2 x (b) reversed for 6e change

2Cr(OH)3 + 10OH–

 ===>

2CrO42– + 8H2O + 6e–
= initial total, 6e cancel out

3HO2– + 3H2O + 2Cr(OH)3 + 10OH–

 ===> 2CrO42– + 8H2O + 9OH–

3H2O and 9OH– cancel out to give

3HO2– + 2Cr(OH)3 + OH–

 ===> 2CrO42– + 5H2O

leaving the final equation (d)!

2Cr(OH)3(s)  +  3HO2–(aq)  +  OH–(aq) ===> 2CrO42–(aq)  +  5H2O(l)

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7. Redox titration questions

  • Using the correct redox equation is obviously important when problem solving and performing calculations from redox titrations.

  • A set of problems involving some of these redox reactions, complete with worked out answers is available.

  • Redox titration questions

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