NEET 2024 Question Paper, Download Free PDF All SET Q, R, S, T Exam Paper

Potassium Dichromate - K2Cr2O7 - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:24 PM | #NEET

Syllabus

Quick Facts

8 Questions around this concept.

Solve by difficulty

MEDIUM

Which one of the following statements is correct when $SO_2$ is passed through the acidified $K_2Cr_2O_7$ solution?

The solution turns blue.

The solution is decolorized.

SO₂ is reduced.

Green Cr₂(SO₄)₃ is formed.

View Solution

Concepts Covered - 1

K2Cr2O7

Potassium dichromate is a very important chemical used in leather industry and as an oxidant for preparation of many azo compounds. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of chromite ore (FeCr₂O₄) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows:

$4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+7 \mathrm{O}_{2} \rightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+8 \mathrm{CO}_{2}$

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na₂Cr₂O₇.2H₂O can be crystallised.

$2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{Na}^{+}+\mathrm{H}_{2} \mathrm{O}$

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.

$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}$

Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.

$\begin{array}{l}{2 \mathrm{CrO}_{4}^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O}} \\ {\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH^{-}} \rightarrow 2 \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}}\end{array}$

The structures of chromate ion, CrO₄^2– and the dichromate ion, Cr₂O₇^2– are shown below. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr–O–Cr bond angle of 126°. Sodium and potassium dichromates are strong oxidising agents; the sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis. In acidic solution, its oxidising action can be represented as follows:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$

Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions are noted below:

$\begin{array}{l}{6 \mathrm{I} \rightarrow 3 \mathrm{I}_{2}+6 \mathrm{e}^{-}} \\ {3 \mathrm{H}_{2} \mathrm{S} \rightarrow 6 \mathrm{H}^{+}+3 \mathrm{S}+6 \mathrm{e}^{-}}\end{array}$ $\begin{array}{l}{3 \mathrm{Sn}^{2+} \rightarrow 3 \mathrm{Sn}^{4+}+6 \mathrm{e}^{-}} \\ {6 \mathrm{Fe}^{2+} \rightarrow 6 \mathrm{Fe}^{3+}+6 \mathrm{e}^{-}}\end{array}$

The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g.,

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$