Question

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenyl-amine as an 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 conservation of the electron is used to determine the stoichiometry of the redox reaction. During the titration of Mohr’s salt and potassium dichromate, the analyte $\text{Fe}$is oxidised and the titrant $\text{C}{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{\text{2-}}$is reduced. Each $\text{C}{{\text{r}}^{\text{3+}}}$requires the three electrons per chromium ion.

Complete step by step answer:
- To perform the stoichiometry, it is necessary to have a balanced chemical reaction. Creating a balanced chemical reaction is the first step to solve this question.
- We provide that the potassium dichromate ${{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ reacts with the acidified $\text{FeS}{{\text{O}}_{\text{4}}}\text{.}{{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}.\text{6}{{\text{H}}_{\text{2}}}\text{O}$ in presence diphenyl-amine as the indicator.
- We know that the formula for Mohr’s salt is as$\text{FeS}{{\text{O}}_{\text{4}}}\text{.}{{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{.}\text{S}{{\text{O}}_{\text{4}}}$. Mohr’s salt is a mixture of ferrous sulphate and ammonium sulphate. During the oxidation reaction, ferrous sulphate of the Mohr’s salt is oxidized by the potassium dichromate${{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$.
- Let us write down the reaction associated with the titration of potassium dichromate with the Mohr’s salt in presence of diphenylamine.
- In a redox reaction, the ferrous $\text{(F}{{\text{e}}^{\text{2+}}}\text{)}$ is oxidized by dichromate ion ($\text{C}{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{\text{2-}}$). The $\text{(F}{{\text{e}}^{\text{2+}}}\text{)}$loses its electron to dichromate ion and itself undergoes the oxidation to form $\text{(F}{{\text{e}}^{\text{3+}}}\text{)}$ and the oxidation state of chromium changes from +6 to +3.
- Reactions are:
Oxidation reaction:$\text{F}{{\text{e}}^{\text{2+}}}\to \text{F}{{\text{e}}^{\text{3+}}}\text{+}{{\text{e}}^{\text{-}}}$
Reduction reaction:$\text{C}{{\text{r}}^{\text{6+}}}\text{+3}{{\text{e}}^{\text{-}}}\to \text{C}{{\text{r}}^{\text{3+}}}$
- The general reaction is as shown below,
$\text{F}{{\text{e}}^{\text{2+}}}\text{+C}{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{\text{2-}}\to \text{F}{{\text{e}}^{\text{3+}}}\text{+C}{{\text{r}}^{\text{3+}}}$
- The reaction is not balanced. Let us balance the reaction. Since the reaction takes place in an acidic medium add ${{\text{H}}^{\text{+}}}$and ${{\text{H}}_{\text{2}}}\text{O}$on either side of the reaction. The reaction can be balanced by adding 6${{\text{H}}^{\text{+}}}$and 7${{\text{H}}_{\text{2}}}\text{O}$on the left and right sides of the reaction respectively$\text{6F}{{\text{e}}^{\text{2+}}}\text{+C}{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{\text{2-}}\text{+14}{{\text{H}}^{\text{+}}}\to \text{6F}{{\text{e}}^{\text{3+}}}\text{+2C}{{\text{r}}^{\text{3+}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}$
This reaction can be also written as,$\text{6(N}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{Fe(S}{{\text{O}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{+}{{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{+7}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{6(N}{{\text{H}}_{\text{4}}}\text{)Fe(S}{{\text{O}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{+}{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{+C}{{\text{r}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{+3(N}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}$
Here, from a balanced chemical reaction
Therefore, n factor for $\text{C}{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{\text{2-}}$ is 6
n factor for $\text{F}{{\text{e}}^{\text{2+}}}$ is 1
-Hence, one mole of potassium dichromate ${{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$requires six moles of ferrous sulphate of Mohr’s salt. Thus, the total number of moles of Mohr’s salt required per mole of dichromate is six.
Therefore, the correct option is (D).

Additional information:
The coefficients from the balanced equation tell us about the ratio between the reactant and product. According to which the coefficient decides the number of moles of reactant requires the number of moles of the product required for the completion of the reaction.

Note: In redox reactions, the oxidation reactions can be balanced according to the medium. In acidic medium, the charges are balanced by adding ${{\text{H}}^{\text{+}}}$and${{\text{H}}_{\text{2}}}\text{O}$. However, for basic medium, the reactions can be balanced by considering the addition of $\text{O}{{\text{H}}^{\text{-}}}$on either side of the reaction.