Please explain this.

Dear student
Please find the solution to the asked query:
a)Cl^- +NO $\to$   Cl₂ + N₂O
​Balancing the equation in acidic medium we add H+ to the side deficient in Hydrogen and water to the side deficient in Oxygen
So the equation becomes:2H ⁺+ ​2Cl^- +2NO $\to$   Cl₂ + N₂O + H₂O
Oxidising agent is the substance which undergoes reduction: NO[ NO undergoes reduction to form N₂O]
Reducing agent is the substance which undergoes oxidation : Cl- [Cl^- oxidises to Cl₂ in the reaction ]

b)PbO₂ + I₂  $\to$  Pb²⁺ + IO₃^-
Balancing the equation in acidic medium:
4H⁺ + 4​PbO₂ + I₂  $\to$  4Pb²⁺ + 2IO₃^- + 2H₂O
Oxidixing agent is PbO₂
Reducing agent is I2

c)Cr₂O₇^2- + S $\to$  Cr³⁺ + H₂SO₃
Balancing the equation in acidic medium​
8H⁺ +Cr₂O₇^2- + S $\to$  2Cr³⁺ + H₂SO₃ + 4H₂O
Oxidising agent is : ​Cr₂O₇^2-
Reducing agent is : S

d)BrO₃^-+ Mn²⁺ $\to$    Br₂ + MnO₂ 
Balancing the equation in acidic medium​​
8H⁺ +2BrO₃^-+ Mn²⁺ $\to$    Br₂ + MnO₂ ​+ 4H₂O​
Oxidising agent : ​BrO₃^-
Reducing agent : Mn²⁺

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