Redox Reactions of Transition Elements (OCR A Level Chemistry A): Revision Note
Iron(II) & Iron(III) Ions
Oxidation of Fe2+ to Fe3+
A redox titration involves an oxidising agent being titrated against a reducing agent
Electrons are transferred from one species to another
In acid-base titrations indicators are used to show the endpoint of a reaction; however redox titrations using transition metal ions naturally change colour when changing oxidation state, so indicators are not always necessary
They are said to be 'self-indicating'
The redox reaction between iron(II) ions and manganate(VII) ions in acidic conditions is used as a basis for a redox titration
In this reaction:
Fe2+ is oxidised to Fe3+
MnO4- is reduced to Mn2+
Potassium manganate(VII) is commonly used which is an oxidising agent and is a deep purple colour
In acidic solutions it is reduced to the almost colourless manganese(II) ion by the Fe2+(aq)
The equation for the reaction is:
MnO4- (aq) + 8H+ (aq) + 5Fe2+ → Mn2+ (aq) + 5Fe3+ + 4H2O (aq)
purple colourless
Reduction of Fe3+ to Fe2+
An orange-brown solution of Fe3+(aq) ions can be reduced to pale green Fe2+(aq) ions by various reducing agents.
A potassium iodide solution is commonly used
The colour change can be masked by the formation of iodine which has a brown colour
In this reaction
Fe3+ is reduced to Fe2+
I- is oxidised to I2
The equation for this reaction is:
2Fe3+ (aq) + 2I- (aq) → 2Fe2+ + I2 (aq)
orange-brown pale-green brown
Chromium(III) & Dichromate Ions
Reduction of Cr2O72- to Cr3+
Aqueous dichromate (VI) ions, Cr2O72-, have an orange colour whilst aqueous chromium (III) ions, Cr3+(aq), have a green colour
Acidified Cr2O72- ions can be reduced to Cr3+(aq) ions by the addition of zinc
Zinc is a strong reducing agent and is capable of reducing both Cr2O72- to Cr3+ and Cr3+ to Cr2+
The equation for this reaction is:
Cr2O72- (aq) + 14H+ (aq) + 3Zn (s) → 2Cr3+ (aq) + 7H2O (l) + 3Zn2+ (aq)
orange green
With an excess of zinc, chromium(III) ions are reduced further to chromium(II), which is a pale blue colour
Zn (s) + 2Cr3+ (aq) → Zn2+ (aq) + 2Cr2+ (aq)
green pale blue
Examiner Tips and Tricks
Fe2+(aq) is a weaker reducing agent than zinc and will only reduce the dichromate to Cr3+
Standard electrode potentials can be used to compare the strength of reducing agents and you should be able to use E∅ to explain why certain redox reactions take place
Oxidation of Cr3+ to CrO42-
When transition metals in low oxidation states are in an alkaline solution, they are more easily oxidised than when in acidic solution
Hot alkaline hydrogen peroxide, H2O2, is a powerful oxidising agent which can be used to oxidise chromium(III) to chromium(VI), CrO42-
The equation for this reaction is
3H2O2 (aq) + 2Cr3+ (aq) + 10OH- (aq) → 2CrO42- (aq) + 8H2O (l)
dark green yellow
Chromium is oxidised from +3 in Cr3+ to +6 in CrO42-
Oxygen is reduced from -1 in H2O2 to -2 in CrO42-
Oxidation of CrO42- to Cr2O72-
Dilute sulfuric acid can be added to chromate (VI), CrO42- (aq), solution to produce a dichromate (VI), Cr2O72- (aq), solution
The equation for this reaction is
2CrO42- (aq) + 2H+ (aq) → Cr2O72- (aq) + H2O (l)
yellow orange
Examiner Tips and Tricks
Don't worry - you won't be expected to memorise and reproduce these full equations in exams.
You could be asked to construct and interpret redox equations from half equations and oxidation numbers though - so make sure you understand how to do it.
Reduction & Disproportionation of Copper Ions
Reduction of Cu2+ to Cu+
A pale blue solution of Cu2+ can be reduced to Cu+ by various reducing agents
A potassium iodide solution is commonly used
When excess iodide ions are present the following reaction occurs:
2Cu2+ (aq) + 4I- (aq) → 2CuI (s) + I2 (aq)
pale blue white precipitate brown
I- is oxidised to brown iodine, I2
Cu2+ is reduced to Cu+, forming a white precipitate
Disproportionation of copper(I) ions
When solid copper(I) oxide, Cu2O, reacts with hot dilute sulfuric acid, a brown precipitate of copper is formed together with a blue solution of copper(II) sulfate
In this reaction copper(I) ions, Cu+, have been simultaneously oxidised and reduced
As the same element has been reduced and oxidised, this reaction is disproportionation
Cu2O (s) + H2SO4 (aq) → Cu (s) + CuSO4 (aq)+ H2O (l)
Copper has been reduced from +1 in Cu2O to O in Cu
Copper has been oxidised from +1 in Cu2O to +2 in CuSO4
You've read 0 of your 5 free revision notes this week
Sign up now. It’s free!
Did this page help you?