Verified Solution • 2025–26 Session moderate

QJEEM23Ch9975: JEE Main 2023 Chemistry - Class 12 JEEM23Ch9975 Solution: acts as a reducing agent in :

H2O2\mathrm{H}_{2} \mathrm{O}_{2} acts as a reducing agent in :

Multiple Choice Options

  • A Mn2++2H2O2→MnO2+2H2O\mathrm{Mn}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
  • B 2Fe2++2H++H2O2→2Fe3++2H2O2 \mathrm{Fe}^{2+}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O}
  • C Na2S+4H2O2→Na2SO4+4H2O\mathrm{Na}_{2} \mathrm{S}+4 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}
  • D 2NaOCl+H2O2→2NaCl+H2O+O22 \mathrm{NaOCl}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}

Step-by-Step Verified Solution

H2O2\mathrm{H}_{2} \mathrm{O}_{2} act as a reducing agent.

2Na+1O−2Cl+1+H2O2⟶2NaCl+H2O+O22 \stackrel{+1}{\mathrm{Na}}\stackrel{-2}{\mathrm{O}} \stackrel{+1}{\mathrm{Cl}}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}

Cl\mathrm{Cl} from (+1) state changes to Cl−1\mathrm{Cl}^{-1} .

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