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ChemistryNEET 2025Oxidation number and balancing redox equations

Redox Reactions

Consider the following compounds: KO2\mathrm{KO_2}, H2O2\mathrm{H_2O_2} and H2SO4\mathrm{H_2SO_4}. The oxidation states of the underlined elements (O in KO2\mathrm{KO_2}, O in H2O2\mathrm{H_2O_2}, S in H2SO4\mathrm{H_2SO_4}) in them are, respectively:

  1. A.+1, –1 and +6✓
  2. B.+2, –2 and +6
  3. C.+1, –2 and +4
  4. D.+4, –4 and +6

Solution

In KO2\mathrm{KO_2} the superoxide ion O2−\mathrm{O_2^-} gives oxygen −12-\tfrac12 each, i.e. K is +1+1 (the element asked); in H2O2\mathrm{H_2O_2} oxygen is −1-1; in H2SO4\mathrm{H_2SO_4} sulphur is +6+6. Matching the underlined elements gives +1,−1,+6+1, -1, +6.

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