Identify the oxidising agent (oxidant) in the following reactions
(a) Pb3O4 + 8HCl ⎯→ 3PbCl2 + Cl2 + 4H2O
(b) 2Mg + O2⎯→ 2MgO
(c) CuSO4 + Zn ⎯→ Cu + ZnSO4
(d) V2O5 + 5Ca ⎯→ 2V + 5CaO
(e) 3Fe + 4H2 O ⎯→ Fe3O4 + 4H2
(f) CuO + H2⎯→ Cu + H2O
(a) Pb3O4 is the oxidising agent.
Here, in Pb3O4 the oxidation state of Pb is +6. And in PbCl2, the oxidation state is +2. Since the oxidation state is decreased thus, Pb3O4 is the oxidant.
(b) O2 is the oxidising agent.
Because O2 is an element and the oxidation state of any element is 0. Thus, the oxidation state of oxygen is 0. Now, in MgO, the oxidation state of oxygen is -2.
Since, the oxidation state of oxygen is decreased from 0 to -2. Thus, O2 is the oxidising agent.
(c) CuSO4 is the oxidising agent.
The oxidation state of Cu in CuSO4 is +2. On the reactants side, the oxidation state of Cu changes to 0, as it is in its elemental form. On the other hand, the oxidation state on Zn changes from 0 to +2. Therefore, Cu has been Reduced and Zn has been oxidised. Cu has gained electrons and hence, CuSO4 is the Oxidising agent.
(d) V2O5 is the oxidising agent.
The oxidation state of Calcium goes from 0 (in its elemental state) to +2 on the products side. On the other hand, the oxidation state of V changes for +5 to 0. Therefore, we can see that V has been reduced (and has gained electrons ). Therefore, V2O5 is the oxidising agent.
(e) H2 O is the oxidising agent.
The oxidation state of Oxygen on the reactants side is -2 whereas on the products side is -3. This gain of electrons leads us to believe that H2 O is the oxidising agent in the above reaction.
(f) CuO is the oxidising agent.
The oxidising state of Cu has changed from +2 on the reactants side to 0 on the products side. It has been reduced and hence, CuO is the oxidising agent in this reaction.
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