Chemistry · Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number and balancing of redox reactions
¡) \) ii) acts as:
¡) \( \boldsymbol{H}_{2} \boldsymbol{O}_{2} \rightarrow \boldsymbol{H}_{2} \boldsymbol{O}+(\boldsymbol{o}) \) ii) \( \boldsymbol{H}_{2} \boldsymbol{O}_{2}+\boldsymbol{2} \boldsymbol{I}^{-} \rightarrow \boldsymbol{I}_{2}+\boldsymbol{2} \boldsymbol{O} \boldsymbol{H}^{-}, \boldsymbol{H}_{2} \boldsymbol{O}_{2} \) acts as:
- A. oxidising agent in both (i) and (ii)
- B. reducing agent in both (i) and (ii)
- C. oxidising agent in (i) and reducing agent in (ii)
- D. reducing agent in (i) and oxidising agent in (ii)
Step-by-step solution
In reaction (i), H2O2 → H2O + O (likely O2), the oxidation state of oxygen changes from -1 in H2O2 to 0 in O2 (oxidation), and to -2 in H2O (reduction). Thus H2O2 undergoes both oxidation and reduction, but considering the formation of O2, it acts as a reducing agent. In reaction (ii), H2O2 + 2I- → I2 + 2OH-, oxygen in H2O2 goes from -1 to -2 (reduction), while I- is oxidized to I2; hence H2O2 acts as an oxidizing agent.
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