In the following reaction between lead sulphide and hydrogen peroxide: \[\ce{PbS_{(s)} + 4H2O2_{(aq)} -> PbSO4_{(s)} + 4H2O_{(l)}}\] Which substance is reduced? Which substance is oxidised?

Reduction is the loss of oxygen. Hydrogen peroxide (H2O2) loses oxygen atoms as it converts into water (H2O). Therefore, Hydrogen peroxide (H2O2) is reduced. Oxidation is the gain of oxygen. Lead sulphide (PbS) gains four oxygen atoms to form lead sulphate (PbSO4). Therefore, Lead sulphide (PbS) is oxidised.

English

CBSEEnglish Medium Class 10

Question Papers

Question Papers1790

Textbook Solutions39241

MCQ Online Mock Tests19

Important Solutions20395

Concept Notes50

Time Tables18

Syllabus

In the following reaction between lead sulphide and hydrogen peroxide: PbS(s) + 4⁢H⁡2⁢O2⁢(aq) -> PbSO4⁢(s) + 4⁢H⁡2⁢O(l) (a) Which substance is reduced? (b) Which substance is oxidised?

Question

In the following reaction between lead sulphide and hydrogen peroxide:

\[\ce{PbS_{(s)} + 4H2O2_{(aq)} -> PbSO4_{(s)} + 4H2O_{(l)}}\]

Which substance is reduced?
Which substance is oxidised?

Short Answer

Solution

Reduction is the loss of oxygen. Hydrogen peroxide (H₂O₂) loses oxygen atoms as it converts into water (H₂O). Therefore, Hydrogen peroxide (H₂O₂) is reduced.
Oxidation is the gain of oxygen. Lead sulphide (PbS) gains four oxygen atoms to form lead sulphate (PbSO₄). Therefore, Lead sulphide (PbS) is oxidised.