Explanation:
According to this equation:
Cr2O72− + 3HNO2 + 5H+ → 2Cr3+ + 3NO3− + 4H2O
The half-reactions are:
Oxidation)
[tex]3\text{ N}^{III}=>\text{ 3 N}^V+\text{ 6e}^{-1}[/tex]Reduction)
[tex]2\text{ Cr}^{VI}+\text{ 6e}^{-1}=>\text{ 2 Cr}^{III}[/tex]HNO2 is the reducing agent
Cr2O7- is an oxidizing agent
Answer: HNO2 is the reducing agent
[tex]3\text{ N}^{III}=>\text{ 3 N}^V+\text{ 6e}^{-1}[/tex]