Question

When ${\text{S}}{{\text{O}}_2}$ is passed through an acidified solution of ${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}$, then chromium sulphate is formed. Change in oxidation state of Cr is from.
A. ${\text{ + 4}}\,{\text{to}}\,{\text{ + 2}}$
B. ${\text{ + 6}}\,{\text{to}}\,{\text{ + 3}}$
C. ${\text{ + 7}}\,\,{\text{to}}\,{\text{ + 2}}$
D. ${\text{ + 5}}\,{\text{to}}\,{\text{ + 3}}$

Answer 1

Oxidation state is the charge on the ion. When an ion gets or loses electrons its oxidation number changes. The potassium dichromate is an oxidizing agent. Potassium dichromate gets reduced.

Complete step by step answer:
${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}$ is known as potassium dichromate.
Potassium dichromate is a strong oxidizing agent. So, it gains electrons in a reaction.
Determine the oxidation state of chromium in potassium dichromate as follows:
The oxidation state of oxygen is $- 2$ and potassium is $+ 1$.
So,
$\Rightarrow \left( { + 1 \times 2} \right) + \left( {x \times 2} \right) + \left( { - 2 \times 7} \right) = 0$
$\Rightarrow + 2 + 2x - 14 = 0$
$\Rightarrow 2x\, = \, + 12$
$\Rightarrow x\, = \, + 6$
So, the oxidation state of chromium is$+ 6$.
The reaction of sulphur dioxide with potassium dichromate is as follows:
${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}\, + \,{{\text{H}}^ + }\,\, \to \,{\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}\, + \,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{O}}$
Determine the oxidation state of chromium in chromium sulphate as follows:
The oxidation number of sulphate is $- 2$.
$\Rightarrow \left( {x \times 2} \right) + \left( { - 2 \times 3} \right) = 0$
$\Rightarrow 2x = + 6$
$\Rightarrow x\, = \, + 3$
So, the oxidation state of chromium is$+ 3$.
So, the oxidation number of chromium is changing from${\text{ + 6}}\,{\text{to}}\,\,{\text{ + 3}}$.

**Therefore, option (B) ${\text{ + 6}}\,{\text{to}}\,{\text{ + 3}}$ is correct.

Note: **
Dichromate is of orange colour. The product of reaction of potassium dichromate with sulphur dioxide is of green colour. The product is chromium sulphate. The green colour appears due to the reduction of chromium during the reaction. The reaction of the reduction of potassium dichromate in an acidic medium is as follows: ${\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }\, + \,14\,{{\text{H}}^{\text{ + }}}\, + 6{{\text{e}}^ - }\, \to \,2\,{\text{C}}{{\text{r}}^{3 + }}\, + \,{\text{7}}\,{{\text{H}}_2}{\text{O}}$. The oxidation number of chromium is changing from ${\text{ + 6}}\,{\text{to}}\,{\text{ + 3}}$ in acidic medium.