Give the preparation of potassium dichromate and state its uses.
Give the structures of chromate and dichromate ions. How can chromates and dichromates be interconverted?
Give the chemical reactions showing the oxidizing nature of potassium dichromate.

(N/A) Preparation: The yellow-coloured sodium chromate is obtained by the fusion of chromite ore $[FeCr_{2}O_{4}]$ with sodium carbonate in the presence of excess air.
$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$
The yellow-coloured solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate $Na_{2}Cr_{2}O_{7} \cdot 2H_{2}O$ can be crystallized out.
$2Na_{2}CrO_{4} + 2H^{+} \rightarrow Na_{2}Cr_{2}O_{7} + 2Na^{+} + H_{2}O$
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore prepared by treating the solution of sodium dichromate with potassium chloride.
$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$
Thus,orange-coloured crystals of potassium dichromate are obtained.
Uses: It is used as a primary standard in volumetric analysis because of its non-hygroscopic nature.
It is used in the leather industry and for the formation of azo compounds.
It is used as an oxidizing agent in organic chemistry.
It is used to measure chemical oxygen demand $(COD)$.
Structures: $CrO_{4}^{2-}$ (chromate) is tetrahedral,whereas $Cr_{2}O_{7}^{2-}$ (dichromate) consists of two tetrahedra sharing one corner with a $Cr-O-Cr$ bond angle of $126^{\circ}$.
Interconversion: Depending upon the $pH$ of a solution,the chromate ion and dichromate ion are interconvertible as:
$2CrO_{4}^{2-} + 2H^{+} \rightarrow Cr_{2}O_{7}^{2-} + H_{2}O$ (Yellow to Orange)
$Cr_{2}O_{7}^{2-} + 2OH^{-} \rightarrow 2CrO_{4}^{2-} + H_{2}O$ (Orange to Yellow)
Oxidizing Nature: In acidic medium,the dichromate ion acts as a very good oxidizing agent. The half-reaction is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$ $(E^{\circ} = +1.33 \ V)$
Acidified dichromate solution oxidizes iodides to iodine,sulphides to sulphur,$Sn(II)$ to $Sn(IV)$,and $Fe(II)$ to $Fe(III)$. The half-reactions are:
$(i) \ 6I^{-} \rightarrow 3I_{2} + 6e^{-}$
$(ii) \ 3H_{2}S \rightarrow 6H^{+} + 3S + 6e^{-}$
$(iii) \ 3Sn^{2+} \rightarrow 3Sn^{4+} + 6e^{-}$
$(iv) \ 6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^{-}$