Chromium (Edexcel A Level Chemistry): Revision Note

Reduction & Oxidation of Chromium Species

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

• We will use zinc and hydrogen peroxide as oxidising agents

• The half equations are arranged from high negative E^Θ at the top to high positive E^Θ at the bottom

  • The best reducing agent is the top right species (Zn (s))

  • The best oxidising agent is the bottom left species (Cr₂O₇^2-(aq))

Oxidation from +3 to +6

• The two half equations we need to consider are 3 and 4

• Chromiums oxidation number changes from +6 to +3 in half equation 3

• The E^Θ value for half equation 3 is more negative than the E^Θ for half equation 4

  • Cr(OH)₃ (aq) is the best reducing agent

  • H₂O₂ (aq) is the best oxidising agent

• We can obtain the overall equation by reversing half equation 3 and combining it with equation 4

  • When adding half equations remember to multiply them so each has the same number of electrons

 2Cr(OH)₃ (aq) + 4OH^- (aq) + 3H₂O₂ (aq) → 2CrO₄^2- (aq) + 8H₂O (l)

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

Reduction from +6 to +3

• The two half equations we need to consider are 1 and 5

• Chromiums oxidation number changes from +6 to +3 in half equation 3

• The E^Θ value for half equation 1 is more negative than the E^Θ for half equation 5

  • Zn is the best reducing agent

  • Cr₂O₇^2- is the best oxidising agent

• We can obtain the overall equation by reversing half equation 1 and combining it with equation 5

  • When adding half equations remember to multiply them so each has the same number of electrons

 Cr₂O₇^2- (aq) + 14H⁺ (aq) + 3Zn (s) → 2Cr³⁺ (aq) + 7H₂O (l) + 3Zn²⁺ (aq)

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

Reduction from +3 to +2

• The Cr³⁺ ion can be further reduced by zinc

• The two half equations we need to consider are 1 and 2

• Chromiums oxidation number changes from +3 to +2 in half equation 3

• The E^Θ value for half equation 1 is more negative than the E^Θ for half equation 2

  • Zn (s) is the best reducing agent

  • Cr³⁺ (aq) is the best oxidising agent

• We can obtain the overall equation by reversing half equation 1 and combining it with equation 2

  • When adding half equations remember to multiply them so each has the same number of electrons

 2Cr³⁺ (aq) + Zn  (s) → 2Cr²⁺ (aq) + Zn²⁺ (aq)

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

The Dichromate(VI) - Chromate(VI) Equilibrium

• The chromate CrO₄^2- and dichromate Cr₂O₇^2- ions can be converted from one to the other by the following equilibrium reaction

2CrO₄^2- (aq) + 2H⁺ (aq) ⇌ Cr₂O₇^2- (aq) + H₂O (l) 

• Chromate(VI) 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