Question

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is?

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is D

6

Explanation for correct answer:-

Option (D) 6

Step 1: Oxidation of Ferrous ion

It is given to us that, Potassium dichromate (${\mathrm{K}}_{\mathrm{2}}\mathrm{C}{\mathrm{r}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{7}}$) reacts with the acidified $\mathrm{FeS}{\mathrm{O}}_{\mathrm{4}}\mathrm{.}{(\mathrm{N}{\mathrm{H}}_{\mathrm{4}}\mathrm{)}}_{\mathrm{2}}\mathrm{S}{\mathrm{O}}_{\mathrm{4}}.\mathrm{6}{\mathrm{H}}_{\mathrm{2}}\mathrm{O}$ (Mohr's salt) in the presence of Diphenyl-amine as an indicator.

The Ferrous ion gets oxidised to ${\mathrm{Fe}}^{3+}$from ${\mathrm{Fe}}^{2+}$by losing an electron due to the oxidation caused by the Potassium dichromate (${\mathrm{K}}_{\mathrm{2}}\mathrm{C}{\mathrm{r}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{7}}$)

Meanwhile, Chromium gains 3 electrons to form ${\mathrm{Cr}}^{3+}$from ${\mathrm{Cr}}^{6+}$and undergoes a reduction reaction.

The reactions are:

$$\begin{gathered} {\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\left[\mathrm{Oxidation}\right] \\ {\mathrm{Cr}}^{6+}+3{\mathrm{e}}^{-}\to {\mathrm{Cr}}^{3+}\left[\mathrm{Reduction}\right] \end{gathered}$$

The general reaction can be written as:

$\mathrm{F}{\mathrm{e}}^{\mathrm{2+}}+\mathrm{C}{\mathrm{r}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{7}}^{\mathrm{2-}}\to \mathrm{F}{\mathrm{e}}^{\mathrm{3+}}+\mathrm{C}{\mathrm{r}}^{\mathrm{3+}}$

Step 2: Balancing the reaction

$\mathrm{F}{\mathrm{e}}^{\mathrm{2+}}+\mathrm{C}{\mathrm{r}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{7}}^{\mathrm{2-}}\to \mathrm{F}{\mathrm{e}}^{\mathrm{3+}}+\mathrm{C}{\mathrm{r}}^{\mathrm{3+}}$

As we can see that the reaction is not yet balanced.

As we can also see that the reaction is taking place in an acidic medium, thus we need to add ${\mathrm{H}}^{+}$and ${\mathrm{H}}_2\mathrm{O}$ on both sides to balance the reaction.

We add $14{\mathrm{H}}^{+}$ and $7{\mathrm{H}}_2\mathrm{O}$ and multiply both the Ferrous ions with 6 to balance the reaction:

$6{\mathrm{Fe}}^{2+}+{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}\to 6{\mathrm{Fe}}^{3+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

Step 3: Writing the whole reaction

The whole reaction can also be written as:

$6{\left({\mathrm{NH}}_4\right)}_2\mathrm{Fe}{\left({\mathrm{SO}}_4\right)}_2+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+7{\mathrm{H}}_2{\mathrm{SO}}_4\to 6\left({\mathrm{NH}}_4\right)\mathrm{Fe}{\left({\mathrm{SO}}_4\right)}_2+{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+3{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+7{\mathrm{H}}_2\mathrm{O}$

Thus from the above reaction, we can very well see that 6 moles of Mohr's salt are required per mole of Potassium dichromate.

Therefore, 6 moles of Mohr's salt are required per mole of Potassium dichromate.