Question

\[{K_2}C{r_2}{O_7}\] react with \[N{a_2}S{O_3}\] in acidic medium to give chromium (III) and sulphate ion. Write the balance ionic reaction.

Answer 1

Hint: \[{K_2}C{r_2}{O_7}\] is potassium dichromate and \[N{a_2}S{O_3}\]is sodium sulphite. The chromium is the central metal ion in potassium dichromate which is undergoing a change in oxidation state to chromium (III) ion and the sulfite in changing to sulphate ion.

Complete step by step answer:
Potassium dichromate is an inorganic salt composed of potassium, chromium and oxygen. Here chromium is the central metal ion. Chromium is a transition metal and which attains variable oxidation state after reaction.
Chromium is an element in the periodic table with atomic number \[24\]. Its electronic configuration is \[\left[ {Ar} \right]3{d^5}4{s^1}\] . Chromium is able to exist in a maximum of \[ + 6\] oxidation state.
Let us find the oxidation state of chromium in \[{K_2}C{r_2}{O_7}\]. Let the oxidation state of chromium is \[x\] and potassium dichromate is a neutral compound.
\[2{\text{ }} \times \] Valency of \[K\] + \[2{\text{ }} \times \] O.S. of \[Cr\] + \[7{\text{ }} \times \] valency of \[O\] = \[0\]
$2 \times ( + 1) + 2x + 7 \times ( - 2) = 0$
$2x = 12$
$x = 6$.
Thus chromium is in \[ + 6\] oxidation state and is undergoing reduction in acidic medium to generate chromium \[ + 3\] oxidation state.
Similarly let us find the oxidation state of sulphur in \[N{a_2}S{O_3}\]. Let the oxidation state of the central atom sulphur is \[y\].
\[2{\text{ }} \times \] Valency of \[Na\] + O.S. of \[S\] + \[3{\text{ }} \times \] valency of \[O\] = \[0\]
$2 \times \left( { + 1} \right) + y + 3\left( { - 2} \right) = 0$
$\Rightarrow$ $y = 4$
The oxidation state of sulphur in sulphate ion \[\left( {S{O_4}^{2 - }} \right)\] is
O.S. of \[S\] + \[4{\text{ }} \times \] valency of \[O\] = \[ - 2\]
$\Rightarrow$ $y + 4( - 2) = - 2$
$\Rightarrow$ $y = 6$.
Thus sulphur is going from \[ + 4\] to \[ + 6\] in an acidic medium i.e. an oxidation is occurring. Thus \[{K_2}C{r_2}{O_7}\] and \[N{a_2}S{O_3}\] will react to give a redox reaction. The corresponding half reactions are
Oxidation: $S{O_3}^{2 - } + {H_2}O \to S{O_4}^{2 - } + 2{H^ + } + 2{e^ - }$ ---(1)
Reduction: $C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O$ ---(2)
Multiplying equation (1) by \[3\] and adding to equation (2) gives the net balance ionic equation for the reaction.
$C{r_2}{O_7}^{2 - } + 8{H^ + } + 3S{O_3}^{2 - } \to 2C{r^{3 + }} + 3S{O_4}^{2 - } + 4{H_2}O$

Note:
This is an example of redox reaction which is normally encountered in inorganic quantitative analysis to determine the amounts or concentrations of \[{K_2}C{r_2}{O_7}\] and \[N{a_2}S{O_3}\] solutions. Potassium dichromate is generally favoured for strength determination because it is non hygroscopic in nature.