Question

A ${{6 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ mole }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ reacts completely with ${{9 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ mole }}{{\text{X}}^{{\text{n + }}}}$ to get ${\text{XO}}_{\text{3}}^{\text{ - }}$ and ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$. What is the value of n?

Answer 1

In the above question, we have to find the value of n, which is the charge present on atom X. Since, the ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ completely react with ${{\text{X}}^{{\text{n + }}}}$. So, first we will write a balanced equation for conversion of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ to ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ and from ${{\text{X}}^{{\text{n + }}}}$ to ${\text{XO}}_{\text{3}}^{\text{ - }}$. Then, we can compare the mole ratio and electron gain and loss ratio to find the value of n.

Complete step by step answer:
Since, ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$is converted to ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ and ${{\text{X}}^{{\text{n + }}}}$is converted to ${\text{XO}}_{\text{3}}^{\text{ - }}$. So, the first step is to write a balanced equation for the conversion.
${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ }} \to {\text{ C}}{{\text{r}}^{{\text{3 + }}}}$
First we will balance the chromate atom and oxygen atom by adding 7 water molecules on the right hand side.
${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ }} \to {\text{ 2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_2}{\text{O}}$
Since the oxidation state of chromate in ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ is 6 which changes to 3 in RHS . So, a chromate atom receives 3 electrons per atom and hence 6 electrons should be added. And to balance hydrogen atom we shall be adding 14 hydrogen ion and similarly for potassium
${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ + 6}}{{\text{e}}^ - }{\text{ + 14}}{{\text{H}}^ + }{\text{ }} \to {\text{ 2}}{{\text{K}}^ + }{\text{ + 2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_2}{\text{O}}$
Now, we have to balance the below reaction:
${{\text{X}}^{{\text{n + }}}}{\text{ }} \to {\text{ XO}}_{\text{3}}^{\text{ - }}$
Since the oxidation state of X changes from n to 5 so, there is loss of ${\text{5 - n}}$electrons. To balance oxygen atoms we will add 3 hydrogen molecules in the LHS.
${{\text{X}}^{{\text{n + }}}}{\text{ + 3}}{{\text{H}}_2}{\text{O }} \to {\text{ XO}}_{\text{3}}^{\text{ - }}{\text{ + (5 - n)}}{{\text{e}}^ - }$
To balance hydrogen atoms, we will add 6 hydrogen ions.
${{\text{X}}^{{\text{n + }}}}{\text{ + 3}}{{\text{H}}_2}{\text{O }} \to {\text{ XO}}_{\text{3}}^{\text{ - }}{\text{ + (5 - n)}}{{\text{e}}^ - }{\text{ + 6}}{{\text{H}}^ + }$
Now, we will compare the number of moles ratio and number of electron ratio to get the desired value of n:
$\dfrac{{{\text{No}}{\text{. of moles of }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}}{{{\text{No}}{\text{. of moles of }}{{\text{X}}^{{\text{n + }}}}}}{\text{ = }}\dfrac{{{\text{electron required for converting to C}}{{\text{r}}^{{\text{3 + }}}}}}{{{\text{electron required for converting to X}}{{\text{O}}^{{\text{3 - }}}}}}$
Substituting the values:
$\dfrac{{{{6 \times 1}}{{\text{0}}^{{\text{ - 3}}}}}}{{{{9 \times 1}}{{\text{0}}^{{\text{ - 3}}}}}}{\text{ = }}\dfrac{{{\text{5 - n}}}}{{\text{6}}}$
Simplifying:
$\dfrac{{{\text{36}}}}{{\text{9}}}{\text{ = 5 - n}}$
Rearranging:
$4 = 5 - n$
So, n =1

$\therefore$ The value of n is 1.

Note:
Potassium dichromate (${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$) is a common chemical reagent, most commonly used as an oxidizing agent. It is used for preparing strong cleaning solutions for glassware as well as etching materials. It is also used in leather tanning, photographic processing and many more.