- K2Cr2O7 is more popular in the laboratory because it is not deliquescent,Sodium dichromate is handled as its di-hydrate Na2Cr2O7·2H2O.It is around twenty times more soluble in water than the potassium salt. (Why??)
- Acidified Potassium dichromate(Orange) converts into (green) Cr3+ during its redox titrations, Hence, Equivalent mass of K2Cr2O7 is M/6, where M is its molar mass.
- Potassium dichromate can be used only in ACID solutions,since it will be converted to chromate in basic solution (non-redox hydrolysis) :Cr2O72- + 2OH- –> 2CrO42- + H2O; (Chromate will be converted back to dichromate,if the solution is acidified.)
- When an alkali is added to an orange-red solution containing dichromate ions,a yellow solution is obtained due to the formation of chromate ions.For example: K2Cr2O7 + K2CO3 → 2 K2CrO4 + CO2
- KMnO4: Acidic medium:MnO4- + 8H+ + 5e- = Mn2+ (colourless) + 4H2O; Neutral medium:2H2O + MnO4− + 3e−⟶ MnO2↓ (brown) + 4OH−; Strongly alkaline medium:MnO4− + e− ⟶ MnO42− (green);
- Potassium permanganate is a STRONGER oxidizing agent than potassium dichromate in acidic solutions,Check their reduction potentials (E0MnO4/Mn2+ = 1.51 V ; E0K2Cr2O7/Cr3+ = 1.33 V) Still DICHROMATE is widely used in redox titrations because of several advantages over permanganate.
- KMnO4 act as SELF-INDICATOR during its redox titrations, In permanganate titrations,a slight faint persisting pink colour signals the endpoint of the titration because of the intense colour of the excess oxidizing agent potassium permanganate.
- A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, Though the dichromate solutions are intensely orange coloured solutions and a single drop of it imparts yellow colour to a colourless solution, it can’t be used as a self-indicator like KMnO4. This is because its reduction product (Cr3+) is green which hinders in the visual detection of end point by observing dichromate colour, therefore an indicator such as sodium diphenylamine is used.
- Dichromate converts primary alcohols into aldehydes and, under more strong conditions, into carboxylic acids. In contrast, potassium permanganate tends to give carboxylic acids as the sole products
- K2Cr2O7 can be obtained pure, is stable up to its melting point,and can be considered a primary standard. Its solutions are also very stable, do not suffer decomposition when exposed to light and less readily reduced by organic matter than KMnO4. However, solutions of permanganate are not very stable, especially towards light, and its solutions need frequent standardization to ensure that no appreciable decomposition has occurred. [Primary standards:1. are available to a high degree of purity. 2. high relative formula mass. 3. stable in the environment (air/water). 4. suitably reactive]
- Potassium permanganate is also almost impossible to be obtained in a sufficiently pure form (there is almost always some manganese(IV) oxide present),
and hence KMnO4 is not a primary standard. For the standardization of KMnO4 solutions, reduction by oxalic acid or Mohr’ salt is often used.
- K2Cr2O7 can be used in the presence of hydrochloric acid or sulphuric acid,
It does not oxidize aqueous HCl. (E° Cl2/Cl-) = +1.36V
- KMnO4 cannot usually be used in the presence of hydrochloric acid,
since HCl will be oxidized by KMnO4:: 2 KMnO4 + 16 HCl = 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2
(For acidification of KMnO4 solution, only dilute H2SO4 is suitable)
- With both KMnO4 & K2Cr2O7, HNO3 & Conc. H2SO4 are not suitable to make the solution acidic, as these are strong oxidants and hence may oxidise some reducing agents during titrations.
- Potassium permanganate decomposes at 230 °C to potassium manganate and manganese dioxide, releasing oxygen gas:
2 KMnO4 → K2MnO4 + MnO2 + O2
- When heated strongly, K2Cr2O7 decomposes with the evolution of oxygen.
4K2Cr2O7 → 4K2CrO4 + 2Cr2O3 + 3O2
- Treatment with cold sulphuric acid gives red crystals of chromic anhydride (CrO3):
K2Cr2O7 + 2H2SO4 → 2CrO3 + 2 KHSO4 + H2O
On heating with concentrated acid, oxygen is evolved:
2 K2Cr2O7 + 8H2SO4 → 2K2SO4 + 2Cr2(SO4)3 + 8 H2O + 3O2
- The colour in KMnO4 arises from an electronic transition, but it is actually not a d-d transition, since the Mn in this compound has no d electrons.
It arises from a LMCT charge transfer process within the molecule, in which photons promote an electron from the highest energy molecular orbital in one of the Mn– O bonds to an empty d orbital on the manganese.
This promotion is equivalent to the energy of a yellow photon, so yellow light is absorbed leaving us to see pink , the complementary color.