The solution of Pb(N3)2+Co(MnO4)3 -> CoO+MnO2+Pb3O4+NO it is from O.P.Tondon oxidation n reduction Q.29 (k)
The correct answer and explanation is :
Let’s balance the redox reaction step by step and provide a detailed explanation.
Given Unbalanced Reaction:
[
\text{Pb(N}_3\text{)}_2 + \text{Co(MnO}_4\text{)}_3 \rightarrow \text{CoO} + \text{MnO}_2 + \text{Pb}_3\text{O}_4 + \text{NO}
]
Step 1: Identify Oxidation and Reduction Half-Reactions
Oxidation Half-Reaction (Lead)
- Lead in Pb(N₃)₂: Oxidation state = +2 (Azide, N₃⁻, is -1 per azide ion).
- Lead in Pb₃O₄: The oxidation states are +2 and +4, as Pb₃O₄ consists of 2 Pb²⁺ and 1 Pb⁴⁺.
- So, Pb²⁺ is partially oxidized to Pb⁴⁺.
Reduction Half-Reaction (Manganese)
- Mn in Co(MnO₄)₃: Mn in MnO₄⁻ has an oxidation state of +7.
- Mn in MnO₂: Oxidation state of Mn = +4.
- So, Mn⁷⁺ is reduced to Mn⁴⁺.
Nitrogen Oxidation
- N in N₃⁻ (Azide ion) has an oxidation state of -1/3 per nitrogen.
- N in NO has an oxidation state of +2.
- So, nitrogen is oxidized from -1/3 to +2.
Step 2: Balance Half-Reactions
Oxidation Half-Reaction (Lead and Nitrogen)
[
3 \text{Pb}^{2+} + 2 \text{H}_2\text{O} \rightarrow \text{Pb}_3\text{O}_4 + 4 \text{H}^+ + 2 e^-
]
[
\text{N}_3^- \rightarrow \text{NO}
]
Reduction Half-Reaction (Manganese)
[
2 \text{MnO}_4^- + 2 \text{H}_2O + 2 e^- \rightarrow 2 \text{MnO}_2 + 4 \text{OH}^-
]
Step 3: Balance the Overall Reaction
The balanced equation is:
[
3 \text{Pb(N}_3\text{)}_2 + 2 \text{Co(MnO}_4\text{)}_3 + 2 \text{H}_2O \rightarrow 2 \text{CoO} + 6 \text{MnO}_2 + \text{Pb}_3\text{O}_4 + 6 \text{NO}
]
Explanation (300 Words)
This reaction involves both oxidation and reduction processes.
- Lead Oxidation:
Lead starts as Pb²⁺ in Pb(N₃)₂ and is partially oxidized to Pb³O₄, a mixed oxide where some Pb atoms reach +4 oxidation state. The oxidation occurs as Pb²⁺ loses electrons. - Manganese Reduction:
MnO₄⁻ (permanganate) initially has Mn in the +7 oxidation state, which is reduced to MnO₂ (Mn in +4 state). This reduction occurs by gaining electrons. - Azide (N₃⁻) Oxidation:
The azide ion (N₃⁻) consists of nitrogen in an average oxidation state of -1/3. It gets oxidized to NO (+2 oxidation state), releasing electrons in the process. - Cobalt Participation:
Cobalt in Co(MnO₄)₃ is already in a stable oxidation state, forming CoO as a byproduct. - Water Participation:
Water molecules help balance the reaction by supplying oxygen to form Pb₃O₄ and MnO₂, ensuring mass and charge balance.
Thus, the reaction follows redox principles, where Mn⁷⁺ and N₃⁻ undergo changes in oxidation states, leading to the final balanced equation.
I have provided an image illustrating the oxidation-reduction reaction, including oxidation states, electron transfers, and molecule involvement.