Question

In the following reaction between lead sulphide and hydrogen peroxide PbS(s) + 4H2O(aq) → PbSO4 (s) + 4H2O(l) a. Which substance is reduced b. which substance is oxydised.

Answer 1

I guess you mean the following equation:

PbS+2H202 →PbSO4+4H2O

Oxidation state of each element in a compound:

PbS-

Pb - +2

S - -2

2H202-

H - +1

O - -1

PbSO4-

Pb - +2

S - +6

O4 - -2

H20

H - +1

O - -2

As oxidation state of Pb in reactant side decreases from -2 to +6 in product side it is reducing agent and oxidation reaction.[ Oxidation: loss of electron ]

As oxidation state of oxide in reactant side increases from -1 to -2 in product side it is oxidizing agent and reduction reaction. [ Reduction: gain of electron ]

Answer 2

Answer:

$H_{2}O_{2}$ is reduced to $H_{2}O$ and $PbS$ is oxidized to $PbSO_{4}$.

Explanation:

• In $PbS$, the oxidation state of $Pb$ is +2, and that of $S$ is -2. So, the oxidation state of  $PbS$ is 0.
• In $PbSO_{4}$, the oxidation state of $Pb$ is +2, $S$ is +6, and $O$ is -2.
• The oxidation number of sulfur ($S$) is increased from -2 to +6. Hence, $PbS$ undergoes oxidation in the given reaction.
• In $H_{2}O_{2}$, each hydrogen atom has oxidation number +1 and each oxygen atom has oxidation number -1.
• In $H_{2}O$, each hydrogen has oxidation number +1 and the oxygen atom has oxidation number -2.
• The oxidation number of oxygen ($O$) is decreased from -1 to -2. Hence, $H_{2}O_{2}$  undergoes reduction in the given reaction.

Therefore, the substance being reduced is $H_{2}O_{2}$  and the substance being oxidized is $PbS$.

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