The d- and f-Block Elements
Potassium dichromate and permanganate
Write preparations and redox equations for K₂Cr₂O₇ and KMnO₄.
Two transition-metal compounds are especially important in the laboratory, both strong oxidising agents.
Potassium dichromate, K₂Cr₂O₇ (orange), is made from the ore chromite, FeCr₂O₄, in three steps:
- Fuse chromite with sodium carbonate in air: this gives yellow sodium chromate, Na₂CrO₄.
- Acidify with sulphuric acid: the chromate becomes sodium dichromate, Na₂Cr₂O₇.
- Add KCl: the less soluble potassium dichromate crystallises out.
Chromate (yellow) and dichromate (orange) turn into each other when the pH changes. Add acid and yellow chromate goes orange; add alkali and orange dichromate goes yellow. In both, chromium is +6.
In acid, dichromate takes 6 electrons and becomes green Cr³⁺:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
It oxidises iodide to iodine, iron(II) to iron(III), and hydrogen sulphide to sulphur.
Potassium permanganate, KMnO₄ (deep purple), is made from pyrolusite, MnO₂:
- Fuse MnO₂ with KOH and an oxidising agent (such as KNO₃) to get dark green potassium manganate, K₂MnO₄.
- This manganate disproportionates in neutral or acid solution, or is oxidised electrolytically, to give permanganate.
In acid, permanganate takes 5 electrons and becomes almost colourless Mn²⁺:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
It oxidises iron(II) salts, sulphur dioxide and oxalic acid. Its purple colour vanishes as it reacts, so it acts as its own indicator in titrations.
To balance a full equation, match the electrons: one dichromate needs six Fe²⁺ (each gives one electron); one permanganate needs five.
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