(k) redox reactions and accompanying colour changes for:

(i) interconversions between Fe²⁺ and Fe³⁺

(ii) interconversions between Cr³⁺ and Cr₂O₇ ^2–

(iii) reduction of Cu²⁺ to Cu⁺ and disproportionation of Cu⁺ to Cu²⁺ and Cu

{Fe²⁺ can be oxidised with H⁺/MnO₄ ^– , Fe³⁺ reduced with I^–, Cr³⁺ can be oxidised with H₂O₂/ OH^– & Cr₂O₇ ^2– reduced with Zn/H⁺, Cu²⁺ can be reduced with I^–.}

{In aqueous conditions, Cu⁺ readily disproportionates.}

{Learners will not be required to recall equations but may be required to construct and interpret redox equations using relevant half-equations and oxidation numbers}

(l) interpretation and prediction of unfamiliar reactions including ligand substitution, precipitation, redox.

There are various definitions of Oxidation and Reduction involving electrons, Oxygen and Hydrogen.

In this case, however, we will stick to:

Oxidation = increasing Oxidation Number; Reduction = reducing Oxidation Number.

Changing the oxidation number of a Transition Metal complex will change its colour.

This alone might be interesting but the main interest is using that colour change quantitatively.

In an Acid-Base titration we use indicators to show when a reaction has reached its end-point and then use this information to calculate something about the unknown acid if we have made up a standard solution of base (or unknown base if we have made up a standard solution of acid).

But with Transition Metal complexes changing colour during redox reactions we can use their own colour change to indicate when the reaction is over.

You may, or may not, recall that Dichromate (Vi) is Cr₂O₇^2- (a) and that it is orange.

And also that it reduces to green Cr³⁺ (b) ions when acting as an oxidising agent.

Cr₂O₇^2- + 14 H⁺ + 6e^- --> 2Cr³⁺ + 7H₂O

Since dichromate is a good oxidising agent a range of reducing agents would be able to reduce it to Cr³⁺ ions, although the syllabus only mentions Zinc.

For this to happen zinc would have to be oxidised to is stable Zn²⁺ ion.

Zn --> Zn²⁺ + 2e^-

Putting the two half-equations together gives:

3 Zn + Cr₂O₇^2- + 14 H⁺ -> 3Zn²⁺ + 2Cr³⁺ + 7H₂O

However, since zinc is a rather powerful reducing agent the Chromium (III) ions can be reduced to Chromium (II)

2Cr³⁺ + Zn --> Zn²⁺ + 2Cr²⁺

You could tell if you had taken the reaction this far since Cr²⁺ is light blue, whereas Cr³⁺ is distinctly green

To oxidise Chromium (III) would require an oxidising agent stronger than dichromate.

The syllabus suggests alkaline Hydrogen Peroxide (H₂O₂/OH^-)

However, this will oxidise to Chromate (VI) (CrO₄^2-) rather Dichromate

3H₂O₂ + 6e^- --> 6OH^- & 16 OH^- + 2 Cr³⁺ --> 2 CrO₄^2- + 8H₂O + 6e^-

2Cr³⁺ + 3H₂O₂ + 10 OH^- -> 2CrO₄^2- + 8 H₂O

So the final solution will be closer to yellow than the orange of dichromate

Although, it is easy to reverse the reactions by changing the pH.

2Cr³⁺ + 3H₂O₂ + 10 OH^- ⇌ 2CrO₄^2- + 8 H₂O

Adding more Alkali (OH^-) would force the equilibrium to the right (to reduce OH^- concentration)

Adding more Acid would remove OH^- and force the equilibrium to the left (to raise OH^- concentration)