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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?

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Question

In the following reaction between lead sulphide and hydrogen peroxide:

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

  1. Which substance is reduced?
  2. Which substance is oxidised?
Short Answer
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Solution

  1. Reduction is the loss of oxygen. Hydrogen peroxide (H2O2) loses oxygen atoms as it converts into water (H2O). Therefore, Hydrogen peroxide (H2O2) is reduced.
  2. Oxidation is the gain of oxygen. Lead sulphide (PbS) gains four oxygen atoms to form lead sulphate (PbSO4). Therefore, Lead sulphide (PbS) is oxidised.
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Chapter 1: Chemical Reactions and Equations - Exercise 2 [Page 61]

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Lakhmir Singh Chemistry [English] Class 10
Chapter 1 Chemical Reactions and Equations
Exercise 2 | Q 21. | Page 61

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