(b) Manganese(II) oxidation state chemistry

• The reactions of the manganese(II) ion:

  • Electron configuration of Mn²⁺ is [Ar]3d⁵

  • An aqueous solution of manganese(II) sulfate MnSO_4(aq) or manganese(II) chloride MnCl_2(aq) will do for most laboratory experiments investigating the chemistry of the manganese(II) state.

  • Manganese(II) salts are readily made by dissolving the carbonate, MnCO₃, in the appropriate dilute acid.

    • e.g. MnCO_3(s) + 2HCl_(aq) ===> MnCl_2(aq) + H₂O_(l) + CO_2(g)

    • H₂SO₄ for the sulfate, MnSO₄,  and 2HNO₃ for the nitrate, Mn(NO₃)₂.

    • The very pale pink hexaaquamanganese(II) [Mn(H₂O)₆]²⁺ is quite 'redox' stable in aqueous solution with respect to dissolved oxygen from air.

  • From manganese(II) solutions, the alkalis sodium hydroxide or ammonia, produce the hydrated manganese(II) hydroxide precipitate. There is no further reaction with excess of either alkali.

    • Mn²⁺_(aq) + 2OH^–_(aq) ===> Mn(OH)_2(s) 

    • (can be written as the neutral complex [Mn(OH)₂(H₂O)₄]⁰

    • the hydroxide is almost white if oxygen is excluded, but it gradually turns brown to form hydrated manganese(III) oxide.

    • then 4Mn(OH)_2(s)  + O_2(g) ===> 2Mn₂O_3(s)  + 4H₂O_(l)   

    • • the hydrated oxide Mn₂O₃can also be written as a hydroxy–oxide, MnO(OH)

    • Mn oxidised (II)==>(III) and ==>(IV) possibly to MnO₂ too?, O reduced (0)==>(–1)

    • VIEW more on ppts. with OH^–, NH₃ and CO₃^2–, & complexes with excess reagent

  • With manganese(II) ion solutions, alkaline aqueous sodium carbonate solutions produces a precipitate of manganese(II) carbonate.

    • Mn²⁺_(aq) + CO₃^2–_(aq) ===> MnCO_3(s) (white ppt.)

      • Like Mn(OH)₂ it readily oxidises to brown Mn₂O₃ or black MnO₂?

  • Oxidation of the manganese(II) ion

    • Acidified Mn²⁺ is not oxidised by hydrogen peroxide H₂O₂.

    • BUT alkaline Mn(OH)₂ + H₂O₂ gives brown Mn₂O₃ or MnO(OH), a hydrated manganese(III) oxide/hydroxide.

    • This again illustrates how redox potentials vary with pH i.e. change in relative stability of oxidation states for the Mn³⁺/Mn²⁺ half–cell potential.

  • The hexa–aqua manganese(II) ion readily forms complexes with polydentate ligands.

  • (i) [Mn(H₂O)₆]²⁺_(aq) + 3en_(aq) ===> [Mn(en)₃]²⁺_(aq) + 6H₂O_(l)   (en = H₂NCH₂CH₂NH₂)

    • K_stab = [[Mn(en)₃]²⁺_(aq)] / [[Mn(H₂O)₆]²⁺_(aq)] [en_(aq)]³

    • K_stab = 5.0 x 10⁵ mol^–3 dm⁹ [lg(K_stab) = 5.7]

    • Remember [H₂O] is not included in the equilibrium expression.

  • (ii) [Mn(H₂O)₆]²⁺_(aq) + EDTA^4–_(aq) ===> [Mn(EDTA)]^2–_(aq) + 6H₂O_(l)

    • K_stab = [[Mn(EDTA)₃]^2–_(aq)] / [[Mn(H₂O)₆]²⁺_(aq)] [EDTA^4–_(aq)]

    • K_stab = 1.0 x 10¹⁴ mol^–1 dm³ [lg(K_stab) = 14.0]

  • The higher K_stab value for EDTA reflects the greater entropy change. A simplistic, but not illegitimate argument, shows that in (i) a net gain of 3 particles, but in (ii) 5 more particles are formed.

• The electrode potential chart highlights the values for various oxidation states of manganese.

• Summary of some complexes–compounds & oxidation states of manganese compared to other 3d–block elements

(c) Manganese(III) oxidation state chemistry

• The chemistry of the manganese(III) ion

  • Electron configuration of Mn³⁺ is [Ar]3d⁴

  • The violet Mn(H₂O)₆]³⁺(aq) ion is unstable in aqueous solution, hence little reference to its reactions.

  • Give the two half-reaction E^ø data:

  • (i) Mn³⁺_(aq)  +  e^-  ===>  Mn²⁺_(aq)      (E^ø = +1.56V, so the manganese(III) ion is powerful oxidising agent)

  • (ii) O_2(g)  +  4H⁺_(aq)  +  4e^-  ===>  2H₂O_(l)   (E^ø = +1.23V, so is oxygen, but not as powerful as Mn³⁺_(aq))

  • So the Mn³⁺_(aq) ion will oxidise water to oxygen and be reduced to the +2 oxidation state - argument presented below.

  • (iii) 4Mn³⁺(aq)  +  2H₂O_(l)    ===>  4Mn²⁺_(aq)  +  O_2(g)  +  4H⁺_(aq)

  • To prove thermodynamic feasibility: E^ø_reaction = E^ø_(red) - E^ø_(ox) = (+1.56) - (+1.23) = 0.33V

  • Note how to balance equation (iii):

  • You add 4 x (i) to the reverse of equation (ii), a 4 electron change or 4 'units' of oxidation state to rise and fall,

  • the oxidation state decrease is (i) 4 Mn(+3) to 4 Mn(+2) and oxidation state increase (ii) 2O(-2) to 2O(0).

(d) Manganese(IV) oxidation state chemistry

• The chemistry of manganese(IV) in terms of manganese (IV) oxide

  • The only important manganese(IV) compound is the solid black oxide, MnO₂.

  • Manganese(IV) oxide is an excellent catalyst for the decomposition of hydrogen peroxide which is a useful way of making oxygen for school laboratory experiments. (See Gas Preparations)

    • 2H₂O_2(aq) ===> 2H₂O_(l) + O_2(g)

  • If a small quantity of manganese(IV) oxide is added to ice–cooled concentrated hydrochloric acid an anionic octahedral manganese(IV) chloro complex ion is formed.

    • If the mixture is filtered through glass wool the brown colour of the complex can be seen.

    • (i) MnO_2(s) + 4H⁺_(aq) + 6Cl^–_(aq) ===> [MnCl₆]^2–_(aq) + 2H₂O_(l)

    • The hexachloromanganate(VI) complex ion has an octahedral shape and co-ordination number of 6.

    • The overall charge on this manganese complex ion is 2- (+4 - 6x-1).

    • If the mixture is warmed, chlorine is formed as the complex decomposes to Mn(II) compounds.

    • (ii) [MnCl₆]^2–_(aq) ==> MnCl_2(aq) + 2Cl^–_(aq) + Cl_2(g)

    • The overall equation is

    • (iii) MnO_2(s) + 4H⁺_(aq) + 4Cl^–_(aq) ===> MnCl_2(aq) + Cl_2(g) + 2H₂O_(l)

    • So manganese(IV) oxide is acting as an oxidising agent i.e. the chloride ion is oxidised to chlorine.

    • Oxidation state changes: Mn from +4 to +2 (reduction) and Cl from -1 to 0 (oxidation).

(e) The chemistry of manganese(VI) oxidation state

• A solution of the tetrahedral (O-Mn-O bond angle 109.5^o) dark green manganate(VI) ion, MnO₄^2– can be made by strongly heating a mixture of manganese(IV) oxide, potassium hydroxide and potassium chlorate(V) in a crucible and extracting the manganese(VI) compound with water.

  • 3MnO₂ + 6OH^– + ClO₃^– ===> 3MnO₄^2– + 3H₂O + Cl^–

  • See for full redox analysis of the reaction

• However, the manganate(VI) ion, MnO₄^2– is unstable, especially in acid solution, and slowly undergoes disproportionation – i.e. a species in one oxidation state spontaneously and simultaneously changes into two species of different oxidation states – one higher and one lower in oxidation number. Adding dil. sulfuric acid to the crucible fusion extract will hasten the process.

• The green solution of the manganate(VI) ion changes to the purple manganate(VII) ion and a black precipitate of manganese(IV) oxide is formed.

• 3MnO₄^2–_(aq) + 4H⁺_(aq) ===> 2MnO₄^–_(aq) + MnO_2(s) + 2H₂O_(l)

• The equilibrium constant for the reaction, K, is ~10⁵⁸, so there ain't much chance of the green colour hanging around after acidification!

• The oxidation state changes are 3Mn(+6) ===> 2Mn(+7) + Mn(+4)

  • a total '18 units worth' of redox change, always check your oxidation number balancing before anything else! The oxidation states of H and O remain at +1 and –2 respectively.

• The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).

  • (i) MnO₄^–_(aq) + e^– ===> MnO₄^2–_(aq)  (E^Ø = +0.56V)

  • (ii) MnO₄^2–_(aq) + 4H⁺_(aq) + 2e^–  ===> MnO_2(s) + 2H₂O_(l)  (E^Ø = +1.70V, in acid solution)

    • a redox disproportionation reaction: Mn(VI) ==> Mn(VII) + Mn(IV) or Mn(+6) ==> Mn(+7) + Mn(+4)

  • (ii) has the more positive potential, so this will be the reduction half–cell reaction.

  • (i) has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.

  • E^Ø_reaction = E^Ø_reduction – E^Ø_oxidation = (+1.70) – (0.56) = +1.14V, well over 0V, therefore very feasible!

  • Incidentally:

  • Given the two half–cell reactions, you get the complete balanced equation by adding (ii) plus 2 x (i) reversed.

  • The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for (ii) in an alkaline media.

  • (iii)  MnO₄^2–_(aq) + 2H₂O_(l) + 2e^–  ===> MnO_2(s) + 4OH^–_(aq)  (E^Ø = +0.59V, in alkaline solution)

  • so, re–calculating gives

  • E^Ø_reaction = E^Ø_reduction – E^Ø_oxidation = (+0.59) – (0.56) = +0.03V, just over 0, therefore just feasible! but on the basis of an equilibrium argument, here, the far lower E^Ø_reaction, suggests the MnO₄^2– ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.

  • This is a good example of how change in pH can affect a standard reduction potential.

  • See also copper(I) chemistry for another example of disproportionation.

(f) The chemistry of the manganese(VII) oxidation state i.e. the manganate(VII) ion

• The tetrahedral deep purple manganate(VII) ion, MnO₄^–, can be considered as an intensely coloured and very stable complex ion (except in the presence of something that is readily oxidised!).

• Potassium manganate(VII), KMnO₄ is used to titrate (i) iron(II) ions, (ii) ethanedioates, (iii) hydrogen peroxide and (iv) nitrate(III) ions (old name 'nitrite').

• The titrations are done with dilute sulfuric acid present to prevent side reactions e.g. MnO₂ formation (brown colouration or black precipitate).

• The mineral acid must be dilute sulfuric acid because potassium manganate(VII) will oxidise hydrochloric acid (Cl^– ==> Cl₂) and nitric(V) acid is an oxidising agent itself, so use of either of these acids leads to inaccurate false titration results.

• The Mn²⁺ ions formed are almost colourless (very pale pink), so the end–point is the first permanent faint pink due to the first trace of excess of the brilliant purple manganate(VII) ion. 

• (i) MnO₄^–_(aq) + 8H⁺_(aq)  + 5Fe²⁺_(aq) ===> Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l)

• Theoretically, there are actually two simultaneous colour changes, both masked by the redox indicator change.

  • The purple manganate(VII) ion changes on reduction to the very pale pink manganese (II) ion,

  • and the pale green iron(II) ion changes on oxidation to the orange iron(III) ion,

  • However, in the dilute solution of the titration mixture, the first permanent pink colour does stand out from the pale orange of the iron(III) ion plus the very pale pink of the manganese(II) ion.

  • In the other examples (ii) to (iv) below, the reductants and oxidation products are colourless, so the colour of the very pale pink manganese(II) ion is visually overridden by the first drop of excess of bright purple potassium manganate(VII) at the end-point of the volumetric titration.

  • You do need excess dil. sulfuric acid in the titration and it will NOT act as an oxidising agent or a reducing agent to interfere with the quantitative and accurate volumetric redox titration.

  • e.g. acids you should NOT use for a potassium manganate(VII) redox titration:

    • conc. sulfuric acid is an oxidising agent, dilute is fine,

    • hydrochloric acid, manganate(VII) ion oxidises the chloride ion to chlorine,

    • nitric acid is an oxidising agent,

    • ethanoic acid is too weak to provide a high H⁺_(aq) concentration and many other weak organic acids are oxidised.

  • See iron chemistry part (16) for volumetric questions about this titration

• (ii) 2MnO₄^–_(aq) + 16H⁺_(aq)  +  5C₂O₄^2–(aq) ===> 2Mn²⁺_(aq)  +  8H₂O_(l) + 10CO_2(g)

  • this reaction speeds up as the titration proceeds because Fe³⁺ acts as a catalyst, this situation is known as auto–catalysis.

• (iii) 2MnO₄^–_(aq) + 6H⁺_(aq)  +  5H₂O_2(aq) ===> 2Mn²⁺_(aq)  +  8H₂O_(l)  + 5O_2(g)

• (iv) 2MnO₄^–_(aq) + 6H⁺_(aq)  +  5NO₂^–_(aq) ===> Mn²⁺_(aq)  +  5NO₃^–_(aq)  +  3H₂O_(l)

• See also fully worked examples of redox volumetric titration calculation questions on these titrations.

  • This set of questions also includes the full redox equation.

  • Constructing full inorganic redox equations from half–equations and redox titrations

  • and Balancing redox equations involving transition metal ions

• The autocatalysis of the ethanedioate/potassium manganate (VII) titration reaction by the Mn²⁺ ions is described under homogeneous catalysis in Appendix 6.

• Potassium manganate(VII), is strong enough to oxidise chloride ions. Running conc. hydrochloric acid onto the damp crystals is a handy way of making chlorine in the laboratory.

  • 2MnO₄^–_(aq) + 16H⁺_(aq)  + 10Cl^–_(aq) ===> 2Mn²⁺_(aq)  + 8H₂O_(l) + 5Cl_2(g)

  • This reaction is the reason that dilute sulfuric acid is used in potassium manganate(VII) titrations and not dil. hydrochloric acid, which would lead to inaccurate results.

• Potassium manganate(VII), is a strong enough oxidising agent for some organic synthesis reactions

  • e.g. acidified KMnO₄ oxidises methylbenzene to benzoic acid

  • C₆H₅CH₃ +  3[O]  == H⁺/MnO₄^-  ==>  C₆H₅COOH  +  H₂O