Question

State, giving reasons, whether the substances printed in bold letters have been oxidised or reduced.

(a) $PbO+CO\to Pb+C{O}_2$

(b) $H_{2}S+C{l}_2\to 2HCl+S$

Answer 1

(a) This is an oxidation-reduction process, where both oxidation and reduction has taken place simultaneously.

PbO is Lead (II) Oxide that is reduced to Lead (Pb) and Carbon monoxide or CO is oxidized to $C{O}_2$ or Carbon dioxide.

Hence, the oxidized substance is Carbon dioxide or $C{O}_2$ is the oxidized substance, while Lead (Pb) is the reduced substance in the chemical reaction.

Here, Carbon Monoxide acts as the reducing agent and Lead Oxide acts as the oxidizing agent



(b) In the given reaction,

Oxidised substance- $H_{2}S$

Reduced substance- $C{l}_2$

Oxidising Agent- $C{l}_2$

Reducing Agent- $H_{2}S$

Oxidation in the reaction is of:-
S going from oxidation number -2 to 0

Reduction in the reaction is of:-
Cl going from oxidation number 0 to -1

As Cl goes from oxidation number 0 to -1 which is a reduction reaction, so $C{l}_2$ is the oxidizing agent .