Chemistry - Hydrogen and Its compound Question with Solution | TestHub

The oxidation state of oxygen in H₂O₂ is -1. H₂O₂ acts as an oxidizing agent when its oxygen is reduced to -2 (e.g., in H₂O). H₂O₂ acts as a reducing agent when its oxygen is oxidized to 0 (e.g., in O₂).

Let's analyze each reaction:

(i) $2{\mathrm{FeSO}}_4+{\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3+2{\mathrm{H}}_2\mathrm{O}$Fe²⁺ (oxidation state +2) is oxidized to Fe³⁺ (oxidation state +3). H₂O₂ (oxygen oxidation state -1) is reduced to H₂O (oxygen oxidation state -2).

Therefore, H₂O₂ acts as an oxidizing agent.

(ii) ${\mathrm{MnO}}_2+{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{\oplus }\to {\mathrm{Mn}}^{2+}+{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$MnO₂ (manganese oxidation state +4) is reduced to Mn²⁺ (oxidation state +2). H₂O₂ (oxygen oxidation state -1) is oxidized to O₂ (oxygen oxidation state 0).

Therefore, H₂O₂ acts as a reducing agent.

(iii) $\mathrm{NaOCl}+{\mathrm{H}}_2{\mathrm{O}}_2\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$NaOCl (chlorine oxidation state +1) is reduced to NaCl (chlorine oxidation state -1). H₂O₂ (oxygen oxidation state -1) is oxidized to O₂ (oxygen oxidation state 0).

Therefore, H₂O₂ acts as a reducing agent.

(iv) ${\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}$Na₂SO₃ (sulfur oxidation state +4) is oxidized to Na₂SO₄ (sulfur oxidation state +6). H₂O₂ (oxygen oxidation state -1) is reduced to H₂O (oxygen oxidation state -2).

Therefore, H₂O₂ acts as an oxidizing agent.

(v) ${\mathrm{Pb}}_3{\mathrm{O}}_4+{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{HNO}}_3\to 3\mathrm{Pb}({\mathrm{NO}}_3{)}_2+{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$

Pb₃O₄ (lead oxidation state +8/3, a mixed oxide of PbO and PbO₂) is oxidized to Pb(NO₃)₂ (lead oxidation state +2). This is incorrect. Pb₃O₄ contains Pb(II) and Pb(IV). In this reaction, Pb(IV) is reduced to Pb(II). H₂O₂ (oxygen oxidation state -1) is oxidized to O₂ (oxygen oxidation state 0).

Therefore, H₂O₂ acts as a reducing agent.

(vi) $2{\mathrm{Fe}}^{3+}+{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}\to 2{\mathrm{Fe}}^{2+}+{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$Fe³⁺ (oxidation state +3) is reduced to Fe²⁺ (oxidation state +2). H₂O₂ (oxygen oxidation state -1) is oxidized to O₂ (oxygen oxidation state 0).

Therefore, H₂O₂ acts as a reducing agent.

Reactions where H₂O₂ acts as an oxidizing agent (x): (i), (iv)

So, $x=2$.

Reactions where H₂O₂ acts as a reducing agent (y): (ii), (iii), (v), (vi)

So, $y=4$.

The value of $\frac{x}{y}=\frac{2}{4}=0.5$.