Question

Given, $F{{e}^{3+}}$ is reduced to $F{{e}^{2+}}$ by using:
(A) ${{H}_{2}}{{O}_{2}}$ in the presence of NaOH
(B) $N{{a}_{2}}{{O}_{2}}$ in water
(C) ${{H}_{2}}{{O}_{2}}$ in the presence of ${{H}_{2}}S{{O}_{4}}$
(D) $N{{a}_{2}}{{O}_{2}}$ in the presence of ${{H}_{2}}S{{O}_{4}}$

Answer 1

Hint: Recollect the concept of redox reactions. Here the question says, ferric ion is reduced to ferrous ion. So, reduction reaction is taking place and ferric ion is gaining one electron to form ferrous ion. Take a look at each of the reactants given in the options and then choose the reactant which can act as a reducing agent.

Complete answer:
- Iron is a transition metal which can exist in two oxidation states.
- One is ferrous ion, $F{{e}^{2+}}$ which has +2 oxidation state and another one is ferric ion, $F{{e}^{3+}}$ which has +3 oxidation state.
- Now, in the question, ferrous ions are getting reduced to ferric ions.
- Reduction is the process which involves gain of electrons.
- For the reduction process to take place, free electrons should be present in the reaction medium which also indicates that the reacting species should donate electrons and therefore, the reactant should be a reducing agent.
- A reducing agent is the reagent which reduces other reactants by itself getting oxidized.
- Hydrogen peroxide acts as a reducing agent in the basic medium and reduces $F{{e}^{3+}}$ to $F{{e}^{2+}}$. The reaction is given as,
\[2F{{e}^{3+}}+{{H}_{2}}{{O}_{2}}+2O{{H}^{-}}\to 2F{{e}^{2+}}+{{O}_{2}}+2{{H}_{2}}O\]
- $N{{a}_{2}}{{O}_{2}}$ in water will also act as a reducing agent since hydrogen peroxide and sodium hydroxide is formed when $N{{a}_{2}}{{O}_{2}}$ is added to water.
\[N{{a}_{2}}{{O}_{2}}+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}{{O}_{2}}\]
\[2F{{e}^{3+}}+{{H}_{2}}{{O}_{2}}+2O{{H}^{-}}\to 2F{{e}^{2+}}+{{O}_{2}}+2{{H}_{2}}O\]

Therefore, the answer is both options (A) and (B).

Note:
Remember sulphuric acid is a strong oxidizing agent so in acidic medium, reduction will not take place. Only in the basic medium, hydrogen peroxide will get oxidized to oxygen and water giving rise to two free electrons and therefore, hydrogen peroxide will act as a reducing agent in the basic medium.