Chromium Chemistry (Edexcel International A Level Chemistry)

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Chromium - Reduction & Oxidation

  • For chromium we need to consider the following standard electrode potential values

reduction-and-oxidation-of-chromium-species-table

  • The half equations are arranged from the most negative Eθ at the top to the most positive Eθ at the bottom
    • The strongest oxidising agent is the reactant from the electrode reaction with the most positive standard electrode potential
      • In this case, the most positive value is + 1.33 V, which means that Cr2O72– (aq)  is the strongest oxidising agent
    • The strongest reducing agent is the product from the electrode reaction with the most negative standard electrode potential
      • In this case, the most negative value is – 0.76 V, which means that Zn (s) is the strongest reducing agent 

Oxidation of Cr from +3 to +6 by H2O2 

  • The two half equations we need to consider are 3 and 4
  • Chromium is being oxidised from an oxidation number of +6 to +3 in half equation 3
  • So, the two half equations we need to consider are:
    • 3. CrO42– (aq) + 4H2O (l) + 3e rightwards harpoon over leftwards harpoon Cr(OH)3 (aq) + 5OH (aq)     Eθ = – 0.13 V
    • 4. H2O2 (aq) + 2e rightwards harpoon over leftwards harpoon 2OH (aq)     Eθ = + 1.24 V
  • Half equation 4 has the most positive standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • H2O2 (aq) + 2e rightwards harpoon over leftwards harpoon 2OH (aq)
  • Half equation 3 has the least positive standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Cr(OH)3 (aq) + 5OH (aq) rightwards harpoon over leftwards harpoonCrO42– (aq) + 4H2O (l) + 3e
  • We can obtain the overall equation by combining the reduction equation (half equation 4) and the oxidation equation (the reversed half equation 3)
    • Remember: When combining half equations, they must have the same number of electrons in both equations
      3H2O2 (aq) + 6e  begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  6OH (aq)
      2Cr(OH)3 (aq) + 10OH (aq) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 2CrO42– (aq) + 8H2O (l) + 6e
    •  The equal number of electrons on both sides of the equation cancel out and 6OH (aq) cancel out on both sides to give the overall equation: 

 2Cr(OH)(aq) + 4OH- (aq) + H2O2 (aq) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style CrO42- (aq) + 8H2O (l)

  • This reaction is carried out in alkaline conditions due to the presence of OH- ions in the equation

Reduction of Cr from +6 to +3 by Zn

  • Chromium is being reduced from an oxidation number of +6 to +3 in half equation 5
  • So, the two half equations we need to consider are:
    • 1. Zn2+ (aq) + 2e rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 5. Cr2O72– (aq) + 14H+ (aq) + 6e rightwards harpoon over leftwards harpoon 2Cr3+ (aq) + 7H2O (l)     Eθ = + 1.33 V
  • Half equation 5 has the most positive standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • Cr2O72– (aq) + 14H+ (aq) + 6e rightwards harpoon over leftwards harpoon 2Cr3+ (aq) + 7H2O (l)
  • Half equation 1 has the least positive / most negative standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Zn (s) → Zn2+ (aq) + 2e 
  • We can obtain the overall equation by combining the reduction equation (half equation 3) and the oxidation equation (the reversed half equation 1)
    • Remember: When combining half equations, they must have the same number of electrons in both equations
       Cr2O72– (aq) + 14H+ (aq) + 6e  rightwards harpoon over leftwards harpoon  2Cr3+ (aq) + 7H2O (l)
      3Zn (s) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  3Zn2+ (aq) + 6e 
    •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

 Cr2O72- (aq) + 14H+ (aq) + 3Zn (s) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 2Cr3+ (aq) + 7H2O (l) + 3Zn2+ (aq)

  • This reaction is carried out under acidic conditions due to the presence of H+ in the equation

Reduction of Cr from +3 to +2 by Zn

  • Chromium is being reduced from an oxidation number of +3 to +2 in half equation 2
  • So, the two half equations we need to consider are:
    • 1. Zn2+ (aq) + 2e rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 2. Cr3+ (aq) + e rightwards harpoon over leftwards harpoon Cr2+ (aq)     Eθ = – 0.41 V
  • Half equation 2 has the most positive / least negative standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • Cr3+ (aq) + e rightwards harpoon over leftwards harpoon Cr2+ (aq)
  • Half equation 2 has the least positive standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Zn (s) rightwards harpoon over leftwards harpoon Zn2+ (aq) + 2e 
  • We can obtain the overall equation by combining the reduction equation (half equation 2) and the oxidation equation (the reversed half equation 1)
    • Remember: When combining half equations, they must have the same number of electrons in both equations
      2Cr3+ (aq) + 2e rightwards harpoon over leftwards harpoon  2Cr2+ (aq) 
      Zn (s) rightwards harpoon over leftwards harpoon  Zn2+ (aq) + 2e 
    •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

2Cr3(aq) + Zn (s) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 2Cr2+ (aq) + Zn2+ (aq) 

  • As this reaction is a further step from the previous reduction this reaction is also carried out under acidic conditions

Colour changes of chromium 

  • Another approach to transition metal chemistry is to look at the reactions of transition metal ions and complexes with various reagents
  • This information is sometimes presented in the form of a reaction scheme:

5-3-2-colour-changes-of-chromium


Reaction scheme outlining the colour changes associated with the reactions of different chromium species 

 

  • You would be expected to know:
    • The formula and corresponding state, (aq) or (s), of the transition metal species
    • The colour of the transition metal species
    • The reagents and conditions required to convert one transition metal species into another
    • How to write the equation for the conversion of one transition metal species into another
  • For example:
    • A green solution of hexaaquachromium(III) ions, [Cr(H2O)6]3+ (aq), will react with dilute NaOH (aq) to form a grey-green precipitate of Cr(H2O)3(OH)3 (s)
    • [Cr(H2O)6]3+ (aq) + 3OH- (aq) → [Cr(H2O)3(OH)3] (s) + 3H2O (l) 

Dichromate(VI) & Chromate(VI) Equilibrium

  • The chromate CrO42- and dichromate Cr2O72- ions can be converted from one to the other by the following equilibrium reaction

2CrO42- (aq) + 2H(aq) ⇌ Cr2O72- (aq) + H2O (l) 

  • Chromate(IV) ions are stable in alkaline solution, but in acidic conditions the dichromate(VI) ion is more stable
  • Addition of acid will  push the equilibrium to the dichromate
    • This results in a colour change from yellow to orange
  • Addition of alkali will remove the H+ ions and push the equilibrium to the chromate
  • This is not a redox reaction as both the chromate and dichromate ions have an oxidation number of +6
    • This is an acid base reaction

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Sonny

Author: Sonny

Expertise: Chemistry

Sonny graduated from Imperial College London with a first-class degree in Biomedical Engineering. Turning from engineering to education, he has now been a science tutor working in the UK for several years. Sonny enjoys sharing his passion for science and producing engaging educational materials that help students reach their goals.