Chromium Chemistry (Edexcel International A Level (IAL) Chemistry): Revision Note

Chromium - Reduction & Oxidation

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

• The half equations are arranged from the most negative E^θ at the top to the most positive E^θ 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 Cr₂O₇^2– (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. CrO₄^2– (aq) + 4H₂O (l) + 3e^– $\rightleftharpoons$ Cr(OH)₃ (aq) + 5OH^– (aq)     E^θ = – 0.13 V

  • 4. H₂O₂ (aq) + 2e^– $\rightleftharpoons$ 2OH^– (aq)     E^θ = + 1.24 V

• Half equation 4 has the most positive standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • H₂O₂ (aq) + 2e^– $\rightleftharpoons$ 2OH^– (aq)

• Half equation 3 has the least positive standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed

  • Cr(OH)₃ (aq) + 5OH^– (aq) $\rightleftharpoons$CrO₄^2– (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–	$\mathbf{\rightleftharpoons }$	6OH– (aq)
    2Cr(OH)3 (aq) + 10OH– (aq)	$\mathbf{\rightleftharpoons }$	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)₃ (aq) + 4OH^- (aq) + H₂O₂ (aq) $\mathbf{\rightleftharpoons }$ CrO₄^2- (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^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 5. Cr₂O₇^2– (aq) + 14H⁺ (aq) + 6e^– $\rightleftharpoons$ 2Cr³⁺ (aq) + 7H₂O (l)     E^θ = + 1.33 V

• Half equation 5 has the most positive standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • Cr₂O₇^2– (aq) + 14H⁺ (aq) + 6e^– $\rightleftharpoons$ 2Cr³⁺ (aq) + 7H₂O (l)

• Half equation 1 has the least positive / most negative standard electrode potential, E^θ, 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–	$\rightleftharpoons$	2Cr3+ (aq) + 7H2O (l)
    3Zn (s)	$\mathbf{\rightleftharpoons }$	3Zn2+ (aq) + 6e–

  •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

 Cr₂O₇^2- (aq) + 14H⁺ (aq) + 3Zn (s) $\mathbf{\rightleftharpoons }$ 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^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 2. Cr³⁺ (aq) + e^– $\rightleftharpoons$ Cr²⁺ (aq)     E^θ = – 0.41 V

• Half equation 2 has the most positive / least negative standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • Cr³⁺ (aq) + e^– $\rightleftharpoons$ Cr²⁺ (aq)

• Half equation 2 has the least positive standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed

  • Zn (s) $\rightleftharpoons$ 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–	$\rightleftharpoons$	2Cr2+ (aq)
    Zn (s)	$\rightleftharpoons$	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) $\mathbf{\rightleftharpoons }$ 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:

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 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(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