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प्रश्न
\[\ce{H2O2}\] is a better oxidising agent than water. Explain.
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उत्तर
\[\ce{H2O2}\] is a better oxidizing agent than water. \[\ce{H2O2}\] acts as an oxidizing agent in acid as well as in alkaline medium.
\[\ce{H2O2 + 2H^{+} + 2e^{-} ->[Acid] 2H2O; E^Θ = 1.77 V}\]
\[\ce{H2O2 + OH^{-} + 2e^{-} ->[Alkaline] 3OH-; E^Θ = 0.88 V}\]
Oxidation state of oxygen changes from – 1 to – 2. Oxidising nature of \[\ce{H2O2}\] can be interpreted on account of possession of labile oxygen.
\[\ce{H2O2 -> H2O + O}\]
When water acts as an oxidizing agent, it is reduced to \[\ce{H2}\]. Water reacts with number of active metals whose electrode potential is less than – 0.83 V.
\[\ce{\underset{Reduction}{2Na} + \underset{Oxidant}{2H2O} -> 2NaOH + H2}\]
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| \[\ce{H, O}\] | \[\ce{D2O}\] | |
| Melting point / K | 373.0 | 374.4 |
| Enthalpy of vapourisation at (373 K)/kJ mol–1 | 40.66 | 41.61 |
| Viscosity/centipoise | 0.8903 | 1.107 |
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