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Advanced 3d block-transition metal chemistry

How to balance redox equations involving transition metal ions

Periodic Table - Transition Metal Chemistry - Doc Brown's Chemistry  Revising Advanced Level Inorganic Chemistry Periodic Table Revision Notes How to balance redox equations involving 3d block transition metal ions - oxidation state analysis of reactants and products

 Appendix 7 Balancing redox equations

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Oxidation state is revised in the context of balancing redox reactions and a set of guidelines is set out to ensure you arrive at the fully balanced complete ionic–redox equation derived from given half–cell reactions (half–reactions of the oxidation or the reduction). Examples covered include 1. The reaction between zinc metal and a silver salt solution, 2. The reaction between (a) acidified manganate(VII) ions with iron(II) ions and (b) acidified dichromate(VI) ions with iron(II) ions, 3. The reaction between acidified potassium manganate(VII) and hydrogen peroxide solutions aqueous, 4. The titration of ethanedioate with acidified potassium manganate(VII) solution, 5. The reduction of the vanadium(IV) oxo–cation by tin(II) salts, 6. The oxidation of chloride by acidified potassium manganate(VII) and 7. The reduction of acidified dichromate(VI) with iodide ions

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Appendix 7. Balancing redox equations

First, some basic definition reminders! (ox. st. = oxidation state)

OXIDATION is loss/removal of electrons from atom, ion or molecule – increase in oxidation state e.g.
  1. Fe ===> Fe2+ + 2e–
    • An iron atom loses 2 electrons to form the iron(II) ion e.g. in the initial chemistry of iron rusting or in an iron–acid reaction.
    • (ox. st. change from 0 to +2)
  2. Fe2+ ===> Fe3+ + e–
    • The iron(II) ion loses 1 electron to form the iron(III) ion, e.g. via chlorine or manganate(VII) oxidising agents.
    • (ox. st. change from +2 to +3)
 REDUCTION is gain/addition of electrons by an atom, ion, or molecule – decrease in oxidation state e.g.
  1. Cu2+ + 2e– ===> Cu
    • The copper(II) ion gains 2 electrons to form neutral copper atoms e.g. in electrolysis at the –ve cathode or when copper is displaced from its salt by a more reactive metal.
    • (ox. st. change from +2 to 0)
  2. Fe3+ + e– ===> Fe2+
    • The iron(III) ion gains an electron and is reduced to the iron(II) ion e.g. by adding zinc to acidified iron(III) salt solution.
    • (ox. st. change from +3 to +2)

Some simple equation analysis using oxygen and electron definitions

  1. iron(III) oxide + carbon monoxide ===> iron + carbon dioxide
    • Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(g)
    • In the blast furnace the iron(III) oxide is reduced to liquid iron (O loss),
    • the carbon monoxide is oxidised to carbon dioxide (O gain),
    • CO is the reducing agent (O remover/acceptor from Fe2O3),
    • and Fe2O3 is the oxidising agent (O donator to CO)]
    • to add ox states
  2. iron(III) oxide + aluminium ===> aluminium oxide + iron
    • Fe2O3(s) + 2Al(s) ===> Al2O3(s) + 2Fe(s)  (the Thermit reaction)
    • Iron(III) oxide is reduced and is the oxidising agent (its the oxygen loser i.e. oxygen donor),
    • and the aluminium is oxidised and is the reducing agent (its the oxygen gainer/acceptor/remover).
    • to add ox states
  3. copper + silver nitrate ===> silver + copper(II) nitrate
    • Cu(s) + 2AgNO3(aq) ===> 2Ag(s) + Cu(NO3)2(aq)
    • The nitrate ion NO3– is the spectator ion so the ionic–redox equation is
    • Cu(s) + 2Ag+(aq) ===> 2Ag(s) + Cu2+(aq)
    • in which copper atoms are oxidised by the silver ions by a two electron loss,
    • these electrons are transferred from the copper atoms to the silver ions,
    • so they are reduced by one electron gain each to silver atoms.
    • The silver ions are the oxidising agent (e– acceptor),
    •  and the copper atoms are the reducing agent (e– donor).
    • to add ox states

Oxidation states in more complex species

  1. vanadate(V) ion, VO43–, V is the +5 oxidation state, oxygen is -2

  2. manganate(VI) ion, MnO42–,  Mn in the +6 oxidation state, oxygen is -2

  3. manganate(VII) ion, MnO4–, (was called the permanganate ion), Mn in the +6 oxidation state, oxygen is -2

  4. tetrachlorocuprate(II) ion, [CuCl4]2– (Cu +2 ox. state, Cl is in a –1 ox. state)

  5. The dichromate(VI) ion, Cr2O72–

    • With 7 O's at (–2) ox. state, total 14–, the 2 Cr's must equal a total of +12 to give the 2– surplus charge on the ion, so the chromium is in the (+6) ox. state.
  6. Transition metal complex ions e.g. [CrCl2(H2O)4]+ ,
    • H2O is neutral, 2H (+1) and O (–2), so you can focus on the Cl and Cr.
    • The two chloride ligands are Cl– ions (ox. state –1), contributing a total charge of a 2–,
    • therefore the Cr must be in the +3 state to give an overall charge of a single + on the complex ion.
    • This complex might be called the tetraaquadichlorochromium(III) ion.
  7. Other examples of transition metal complexes
    • [TiCl6]2–, titanium is in a +4 ox. state (Cl is –1), (6 x –1) + (+4) = 2– overall charge on the complex ion.
    • [VO]2+(aq), or [VO(H2O)5]2+(aq), vanadium is in +4 ox. state (H is +1, O is –2),

BALANCING REDOX EQUATIONS

some guidelines applied to transition metal redox reactions in subsequent examples

  1. To state the obvious, use of the correct 'species' (e.g. usually two given/chosen as/from half–cell equation data)

  2. Make sure the 'species' direction change is correct – which is oxidised or reduced? (if not indicated, might have to decide from EØ data, the more +ve half–cell gets reduced)

  3. The ratio of half–cell equations in the full redox equation must be correct – the 'balance' must be based on oxidation number analysis ...

    • The total numerical increase in oxidation states must be equal to the total decrease in oxidation state.

  4. A useful check – the total ion charges should be the same on both sides of the equation (I find this a handy extra check especially with stray H+'s!)

  5. Still use the 'traditional atom count' – placed last because its not completely reliable with redox equations!

  6. All the half–cell equations are presented, by common convention now, as a reduction – electron gain, so one must be reversed!

  7. Apart from the examples here, where the emphasis is on redox reactions of transition metals, there are other detailed notes pages on oxidation number and redox analysis and redox equations with lots of help and examples explained.

triple check summary

1. The reaction between zinc metal and a silver salt solution

  • 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 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) sulfate to give a pink–brown coating of copper.

2. The reaction between acidified manganate(VII) ions and dichromate(VI) ions with iron(II) ions

  • (a) Manganate(VII) ions oxidising 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 sulfuric acid, as opposed to using other common mineral acids. Dil. sulfuric 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 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.

  • (b) Dichromate(VI) ions oxidising 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.

3. The reaction between acidified potassium manganate(VII) and hydrogen peroxide solutions

  • Half–cell reaction data:

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

    • (ii) O2(g) + 2H+(aq) + 2e– ==> H2O2(aq) (EØ = +0.68, H2O2 will act as reducing agent, EØ less positive)

  • Both are well known oxidising agents but in this situation hydrogen peroxide is the one to get oxidised (less +ve EØ).

  • Reduction: Mn (+7) is reduced to Mn (+2), 5e– gain, Mn +7 to +2 ox. state.

  • Oxidation: The O at (–1) in H2O2 is reduced to (–2) in H2O, O –1 to –2 ox. state.

  • Hydrogen (+1) of the H+ ions and the oxygen's (–2) of the MnO4– ion do not change oxidation state.

  • 5 x ox'n half–cell, (ii) rev'd  5H2O2(aq) ==> 5O2(g) + 10H+(aq) + 10e–
    2 x reduction half–cell, (i) 2MnO4–(aq) + 16H+(aq)  + 10e– ==> 2Mn2+(aq) + 8H2O(l)
    added full equation 2MnO4–(aq) + 6H+(aq) + 5H2O2(aq) ==> 2Mn2+(aq) + 5O2(g) + 8H2O(l)

  • Note the 16H+ on left and 10H+ on right result in just 6H+ on left after addition of the half–equations, so watch it!

  • The reaction can be used to quantitatively measure hydrogen peroxide concentrations. The end–point is the 1st permanent faint pink from a tiny excess of the potassium manganate(VII) solution from the burette.

4. The titration of ethanedioate with acidified potassium manganate(VII) solution

  • Half–cell reaction data:

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

    • (ii) 2CO2(aq/g) + 2e– ==> C2O42–(aq) (EØ = –0.49V based on (COOH)2)

  • Reduction: Mn (+7) is reduced to Mn (+2), 5e– gain, acts as the oxidising agent, electron acceptor.

  • Oxidation: Each ethanedioate ion loses two electrons to form carbon dioxide, acts as reducing agent.

    • (carbon's ox. state increases from +3 to +4, 1e– loss per C to get CO2 where C is +4)

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

  • 5 x ox'n half–cell, (ii) rev. 5C2O42–(aq) ==> 10CO2(aq/g) + 10e–
    2 x red'n half–cell, (i) 2MnO4–(aq) + 16H+(aq)  + 10e– ==> 2Mn2+(aq) + 8H2O(l)
    added – full equation 2MnO4–(aq) + 16H+(aq) + 5C2O42–(aq) ==> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
  • This reaction can be used to analyse samples of ethanedioc acid (oxalic acid) and ethanedioate salts (oxalates) or by starting with a very pure weighed samples of the acid or salt, you can standardise the potassium manganate(VII) solution. Self–indicating, first permanent pale pink.

5. The reduction of the vanadium(IV) oxo–cation by tin(II) salts

  • Half–cell reaction data:

    • (i) VO2+(aq) + 2H+(aq) + e– ==> V3+(aq) + H2O(l) (EØ = +0.34, reduction of the oxidising agent)

    • (ii) Sn4+(aq) + 2e– ==> Sn2+(aq) (EØ = +0.15, Sn2+ will act as a reducing agent, EØ less positive)

  • V at (+4) in VO2+ is reduced to V at (+3), 1e– gain per V,

  • Sn at (+2, tin(II) ion) is oxidised to Sn at (+4, tin(IV) ion), 2e– loss per Sn.

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

  • 1 x oxi'n half–cell, (ii) rev. Sn2+(aq) ==> Sn4+(aq) + 2e–
    2 x reduction half–cell, (i) 2VO2+(aq) + 4H+(aq) + 2e– ==> 2V3+(aq) + 2H2O(l)
    added – full equation 2VO2+(aq) + 4H+(aq) + Sn2+(aq) ==> 2V3+(aq) + 2H2O(l) + Sn4+(aq)

  • Tin(II) salts are used as reducing agents.

6. The oxidation of chloride 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) 1/2Cl2(aq) + e– ==> Cl–(aq)  (EØ = +1.36, lower EØ, so cannot act as an oxidising agent)

  • Both chlorine and potassium manganate(VII) are strong oxidising agents, but chlorine is the weaker, so chloride ions are oxidised to chlorine.

  • Oxidation: Chlorine as the chloride ions at (–1) lose electrons to give chlorine molecules at ox. state (0).

  • Reduction: Mn (+7) is reduced to Mn (+2), 5e– gain, acts as the oxidising agent.

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

  • 10 x oxi'n half–cell, (ii) rev. 10Cl–(aq) ==> 5Cl2(aq/g) + 10e–
    2 x red'n half–cell, (i) 2MnO4–(aq) + 16H+(aq)  + 10e– ==> 2Mn2+(aq) + 8H2O(l)
    added – full redox equation 2MnO4–(aq) + 16H+(aq) + 10Cl–(aq) ==> 2Mn2+(aq) + 8H2O(l) + 5Cl2(g/aq)

  • This reaction is used to prepare chlorine gas (green and toxic) by running conc. hydrochloric acid on to moistened potassium manganate(VII) crystals (old name potassium permanganate).

7. The reduction of acidified dichromate(VI) with iodide ions

  • Half–cell reaction data:

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

    • (ii) 1/2I2(aq) + e– ==> I–(aq)   (EØ = +0.54V, lower EØ, so cannot act as oxidising agent)

  • Oxidation: Iodide ions at (–1) lose electron to give iodine molecules at I(0).

  • Reduction: Each Cr at (+6) is reduced by gaining 3e– to give Cr at (+3), so the Cr2O72– ion is the oxidising agent.

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

  • 6 x oxidation half–cell, (ii) rev.  3I2(aq) + 6e– ==> 6I–(aq)
    1 x reduction half–cell, (i) Cr2O72–(aq) + 14H+(aq) + 6e– ==> 2Cr3+(aq) + 7H2O(l)
    added full redox equation Cr2O72–(aq) + 14H+(aq) + 6I–(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l)

  • This reaction can be used to quantitatively measure chromium(VI) in dichromates, Cr2O72–, or chromates, CrO42– (which change to dichromate(VI) on acidification, yellow ==> orange). Excess potassium iodide is added and the liberated iodine is titrated with standardised sodium thiosulfate solution (starch indicator, blue ==> colourless)

A level Revision notes for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Chemistry OCR GCE Chemistry Edexcel GCE Chemistry Salters Chemistry CIE Chemistry, WJEC GCE AS A2 Chemistry, CCEA/CEA GCE AS A2 Chemistry revising courses for pre–university students (equal to US grade 11 and grade 12 and AP Honours/honors level courses)

INORGANIC Part 10 3d block TRANSITION METALS sub–index:

10.1–10.2 Introduction to 3d–block Transition Metal chemistry

10.3 Chemistry of Scandium  *  10.4 Chemistry of Titanium

10.5 Chemistry of Vanadium  *  10.6 Chemistry of Chromium

10.7 Chemistry of Manganese  *  10.8 Chemistry of Iron

10.9 Chemistry of  Cobalt  *  10.10 Chemistry of Nickel

10.11 Chemistry of Copper  *  10.12 Chemistry of Zinc

10.13 Selected chemistry of other Transition Metals e.g. Ag and Pt

Appendix 1. Hydrated salts, acidity of hexa–aqua ions

Appendix 2. Complexes and ligands

Appendix 3. Complexes and isomerism

Appendix 4. Electron configuration and colour theory

Appendix 5. Redox equations, feasibility of reaction, Eø calculations

Appendix 6. Catalysis - types and effectiveness

Appendix 7. Redox equations - construction and balancing

Appendix 8. Stability constants of complexes and entropy changes

Appendix 9. Colorimetric analysis and determining a complex ion formula

Appendix 10 3d block – extended data table

Appendix 11 3d–block transition metal complexes, oxidation states & electrode potentials

Appendix 12 Hydroxide complex precipitate 'pictures', formulae and equations

Advanced Level Inorganic Chemistry Periodic Table Index: Part 1 Periodic Table history * Part 2 Electron configurations, spectroscopy, hydrogen spectrum, ionisation energies * Part 3 Period 1 survey H to He * Part 4 Period 2 survey Li to Ne * Part 5 Period 3 survey Na to Ar * Part 6 Period 4 survey K to Kr and important trends down a group * Part 7 s–block Groups 1/2 Alkali Metals/Alkaline Earth Metals * Part 8  p–block Groups 3/13 to 0/18 * Part 9 Group 7/17 The Halogens * Part 10 3d block elements & Transition Metal Series * Part 11 Group & Series data & periodicity plots * All 11 Parts have their own sub–indexes near the top of the pages

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