• The 'simple' hexaaquachromium(VI) cation, [Cr(H₂O)₆]⁶⁺, cannot exist in aqueous media.

  • In fact, I doubt if the 'simple' Cr⁶⁺ ion can exist in any ionic compound, the high charge would 'theoretically draw 'over' the electron cloud of the negative anion to form a covalent bond and it would be a very acidic complex.

  • Note that chromium(VI) oxide, CrO₃ and chromium(VI) fluoride, CrF₆, (both compounds have Cr in +6 oxidation state), are covalent compounds, despite the relatively large electronegativity difference between the metal and non–metal.

    • BUT, being very electronegative, they can stabilise chromium compounds-ions in the higher +6 oxidation state.

  • If the oxidation state of the central metal ion is over +3, it appears that deprotonation via proton transfer to water is so facilitated that in most cases (there may be exceptions?) all protons are 'theoretically' lost to give the oxyanion.

    • ie the theoretical [Cr(H₂O)₆]⁶⁺ ends up in reality as Cr₂O₇^2– or CrO₄^2– depending on the pH of the aqueous solution.

    • The reason for this situation is that the high charge density of the 'theoretical' central metal ion, gives it a high polarising power.

      • You can theoretically conceive the situation of imagining the central metal ion pulling on the electrons of the M–OH₂ and M–OH ligand bonds in two stages to leave a M=O bond in the oxyanion

        • ie considering one co–ordinated water molecule (and ignoring the charge on intermediate complexes) ...

          • M–OH₂  ==proton loss==>  M–OH  ==proton loss==>  M=O

  • There is, theoretically, always an equilibrium between the chromate(VI) ion and the dichromate(VI) ion which can be expressed in two ways.

    • (i) 2CrO₄^2–_(aq) +  2H⁺_(aq) Cr₂O₇^2–_(aq) + H₂O_(l)      (adding acid to a chromate(VI) salt solution)

    • (ii) Cr₂O₇^2–_(aq) +  2OH^-_(aq) 2CrO₄^2–_(aq) +  H₂O_(l)  (adding alkali to a dichromate(VI) salt solution)

    • Therefore from Le Chatelier's principle, high pH (alkaline) favours the formation of the chromate(VI) ion and low pH (acid) favours dichromate(VI) ion formation.

• Oxidation of a chromium(III) ion to a chromium(VI) ion:

• When hydrogen peroxide is added to an alkaline chromium(III) solution, oxidation occurs to give the yellow chromate(VI) ion CrO₄^2– . 

  • 2Cr³⁺_(aq) + 3H₂O_2(aq) + 10OH^–_(aq) ===> 2CrO₄^2–_(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.

    • E^Ø = +1.33V for the half-cell reaction: Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– 2Cr³⁺_(aq) + 7H₂O_(l)

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

    • E^Ø = +1.77 V  for H₂O₂(aq) + 2H⁺(aq) + 2e^– 2H₂O(l)

    • BUT, both of the above are for acid–neutral conditions, so different 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) E^Ø = +0.88 V  for the half–cell reaction: HO₂^–(aq) + H₂O(aq) + 2e^– 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 (Cr₂O₇^2–), the yellow chromate(VI) ion is a very weak oxidising agent in alkaline solution (in acid solution it reverts to the dichromate(VI) ion).

      • E^Ø = –0.11 V for the half–cell reaction (b): CrO₄^2–(aq) + 4H₂O(l) + 3e^– 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 in alkaline conditions.

      • E^Ø_reaction = E^Ø_red – E^Ø_ox = E^Ø_HOO–/OH–  –  E^Ø_{CrO42–/Cr(OH)3} = +0.88 – 0.12 = + 1.00 V, a very feasible reaction!

      • Note that the chromium species used in the E^Ø 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).

    • So the reaction can better written as (though derivation not shown, see equation (d) at the end):

    • (c) 2Cr(OH)₃(s)  +  3H₂O₂(aq)  +  4OH^–(aq)  ===>  2CrO₄^2–(aq)  +  8H₂O(l)

    • The oxidation state change numbers and balancing are as above.

    • 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 3H₂O₂ ==> 6 O(–2) in 6 of the oxygen's of the H₂O's, 6e change

        • The total oxidation state increases must equal the total decrease in oxidation states.

      • (ii) check the ionic charge balance

        • on the left 4– is balanced by 2 x 2– = 4–, if it involves both + and – ions, just obey the usual mathematical rules

      • (iii) double check the atom count

        • 2Cr + 16O + 16H 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) even though it is a legitimate equation,

      • 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), set out below.

      • 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!

      • I think 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.

• As mentioned already, when the resulting yellow solution from above is acidified with dilute sulfuric acid, the orange dichromate(VI) ion Cr₂O₇^2–  is formed.

• The equilibrium is pH dependent. From 'Le Chatelier's Principle':

  • in more acidic solution, more H⁺, decrease pH ==> more orange (net change L to R) or in

  • more alkaline, less H⁺ (removed by OH^–), increase pH <= more yellow (net change R to L).

  • 2CrO₄^2–_(aq) + 2H⁺_(aq) Cr₂O₇^2–_(aq) +  H₂O_(l) (no change in ox. state)

• The dichromate(VI) ion is reduced in two stages by a zinc and dilute hydrochloric/sulfuric acid mixture.

  • Cr(VI, +6) ==> Cr(III, +3): Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– 2Cr³⁺_(aq) + 7H₂O_(l)

    • orange (+6) ==> green (+3),  E^Ø = +1.33V 

  • Cr(III, +3) ==> Cr(II, +2): Cr³⁺_(aq) + e^– Cr²⁺_(aq)

    • green (+3) ==> blue (+2), E^Ø = –0.41V, so Cr(II) is readily oxidised by dissolved oxygen and can only be retained in an inert atmosphere.

  • Note the  E^Ø_{Zn(s)/Zn2+(aq)} is –0.76V, so the reducing power of zinc is sufficient to effect either of the two chromium oxidation state reduction changes.

    • The full redox equations for the reactions which happen on the surface of the zinc are:

    • Cr₂O₇^2–_(aq) + 3Zn_(s) + 14H⁺_(aq) 2Cr³⁺_(aq) + 3Zn²⁺_(aq) + 7H₂O_(l)

    • 2Cr³⁺_(aq) +  Zn_(s) 2Cr²⁺_(aq) + Zn²⁺_(aq)

    • You will see hydrogen formed as a by–product of the zinc–acid reaction but the reductions take place on the surface of the zinc.

    • The reductions occur by electron transfer on the surface of the zinc.

• Potassium dichromate(VI), K₂Cr₂O₇,  can be crystallised to high purity standard without water of crystallisation, and is a valuable 'standard' redox volumetric reagent.

  • e.g. It can used to titrate iron(II) ions in solution acidified with dilute sulfuric acid, using a redox indicator like barium diphenylamine sulfonate (sulfonate) which is less readily oxidised than iron(II) ions, but once all the iron(II) ions are oxidised the indicator is oxidised to a blue colour.

  • The iron(III) ions formed affect the indicator to give an inaccurate end point so phosphoric(V) acid is also added at the start to complex the Fe³⁺ ions as they form.

  • Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) ===> 2Cr³⁺_(aq) + 6Fe³⁺_(aq) + 7H₂O_(l)

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

    • The orange dichromate(VI) ion changes on reduction to the green chromium(III) ion,

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

    • so without the indicator I'm not sure exactly how the colour change you would really observe would pan out, but the problem is got round by using a special redox indicator!

    • One of the most common indicator is sodium diphenylaminesulfonate, which turns from colourless to purple with first tiny excess of dichromate at the end-point.

    • You must have excess dil. sulfuric acid in the titration flask mixture - you can use dil. hydrochloric acid, because the dichromate(VI) ion is not a powerful enough oxidising agent to oxidise chloride ion to chlorine.

  • See also fully worked examples of redox volumetric titration calculation questions, and ...

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

  • and Balancing redox equations involving transition metal ions

• The dichromate(VI) ion in acid solution is a strong oxidising agent – examples of oxidising action:

• See above for oxidation of iron(II) ions.

• It oxidises iodide ions to iodine.

• Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6I^–_(aq) ===> 2Cr³⁺_(aq) + 3I_2(aq) + 7H₂O_(l)

  • The released iodine can be titrated with standard sodium thiosulfate solution using starch indicator.

  • 2S₂O₃^2–_(aq)  +  I_2(aq)  ===>  S₄O₆^2–_(aq) + 2I^–_(aq) (black/brown/blue ==> colourless endpoint)

  • This reaction between the released iodine and sodium thiosulfate can be used to estimate oxidising agents like dichromate(VI) ions.

    • The iodine is titrated with standardised sodium thiosulfate (e.g. 0.10 mol dm^–3) using a few drops of starch solution as an indicator. Iodine gives a blue colour with starch, so, the end–point is very sharp change from the last hint of blue to colourless.

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

• Soluble chromate(VI) salts give yellow solutions, but lead(II) ions give a yellow ppt. of lead(II) chromate(VI) and silver ions a dark red ppt. of silver chromate(VI).

• Pb²⁺_(aq) + CrO₄^2–_(aq) ===> PbCrO_4(s)   and 

  • 2Ag⁺_(aq) +  CrO₄^2–_(aq) ===> Ag₂CrO_4(s)

  • Note that a few drops of potassium chromate(VI) is used as an indicator when titrating chloride solutions with silver nitrate solution in neutral solution.

    • The solubility product for the white precipitate. of silver chloride is

    • K_sp = [Ag⁺_(aq)][Cl^–_(aq)] = 2 x 10^–10 mol²dm^–6

  • and is exceeded before the solubility product of silver chromate(VI) because of the relatively high concentration of chloride ions prior to reaching the end-point.

    • K_sp = [Ag⁺_(aq)]²[CrO₄^2–_(aq)] = 3 x 10^–12 mol³dm^–9

    • until all the chloride is precipitated.

    • The next drop of silver nitrate causes the precipitation of brownish–red silver chromate, so the end point is the formation of the dark red precipitate.

CHROMIUM(II) oxidation state chemistry:

• The blue hexaaquachromium(II) ion, [Cr(H₂O)₆]²⁺_(aq), can be formed by reducing chromium(III) salt solutions with zinc and hydrochloric acid but it is rapidly oxidised back to violet-green chromium(III) ions by dissolved oxygen unless protected by an inert atmosphere.

• See Redox Electrode Potential Chart and given the half-cell potentials ...

• (i) O_2(g)  +  4H⁺_(aq)  +  4e^-  ===>  2H₂O_(l)   (E^ø = +1.23V, will act as the oxidising agent, and reduced)

• (ii) Cr³⁺_(aq) +  e^-  ===>  Cr²⁺_(aq)   (E^ø = -0.42V, will act as the reducing agent, and oxidised)

• E^ø_reaction = E^ø_reduction  -  E^ø_oxidation

• E^ø_reaction = (+1.23)  -  (-0.42)  =  +1.65V  (very positive, so very feasible!)

• for the reaction:

  • O_2(g)  +  4H⁺_(aq)  +  4Cr²⁺_(aq)  ===>  4Cr³⁺_(aq)  +  2H₂O_(l

  • Note the balancing from the 4 electron change equation.

  • One of (i) balances four of (ii) reversed.

  • Redox equation triple check!

    • Check on oxidation number changes,  the increase must numerically balance the decrease in oxidation states,

    • Check the total ion charge is same on both sides of equation,

    • Finally, do the usual check on the atom sums of each element are the same on each side of the equation.