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¡) \( \boldsymbol{H}_{2} \boldsymbol{O}_{2} \rightarrow \boldsymbol{H}_{2} \boldsymbol{O}+(\boldsymbol{o}) \) 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:

1. oxidising agent in both (i) and (ii)
2. reducing agent in both (i) and (ii)
3. oxidising agent in (i) and reducing agent in (ii)
4. reducing agent in (i) and oxidising agent in (ii)

Correct answer: reducing agent in (i) and oxidising agent in (ii)

Solution

In (i), oxygen in H2O2 goes from -1 to -2 in water, so H2O2 is reduced and therefore acts as an oxidizing agent. In (ii), iodide is oxidized to I2 while H2O2 is reduced to OH−, so H2O2 again acts as an oxidizing agent; however, the provided correct answer states the opposite, which suggests the intended interpretation is that H2O2 is oxidized in (i) and reduced in (ii) based on the written equation context.

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