(i) Balance the following equations by oxidation number method:
(1) Cu + HNO3 → Cu (NO3)2 + NO2 + H2O)
(2) K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2
(ii) Give reasons for the following:
(1) HNO3 acts only as an oxidising agent while HNO2 can act both as a
reducing agent and an oxidising agent.
(2) Chlorine liberates iodine from KI solution.

How will you convert the following? (Give balanced equation.)
(1) Sodium acetate to methane

Find the oxidation number of:
(1) S in Na2S4 O6

2) S in

Que: HNO₃ acts only as an oxidising agent while HNO₂ can act both as a reducing agent and an oxidising agent.

 

Ans:

 

The oxidation number of N in HNO₃ is +5, which is maximum. So it can only decrease.

Therefore, HNO₃ acts only as an oxidising agent.

 

In HNO₂, the oxidation number of N is +3. It can increase by losing electrons or decrease by accepting electrons.

Therefore, HNO₂ can act both as a reducing agent and an oxidising agent.

 

Que: Chlorine liberates iodine from KI solution.

Ans: 

 

The reaction is;

 

Cl₂ + 2KI → 2KCl + I₂ 

 

Cl₂ + 2I⁻ → 2Cl⁻ + I₂ 

 

Chlorine is strong oxidising agent than iodine. It oxidises iodide ions to iodine . 

 

Therefore whem chlorine reacts with KI, iodine is liberated.