Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:
(i) P4 and F2 are reducing and oxidising agents respectively.
If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3.
P4(excess)+F2⟶+3PF3
However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.
P4+F2(excess)⟶+5PF5
(ii) K acts as a reducing agent, whereas O2 is an oxidising agent. If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is –2.
4K(excess)+O2→2K2−2O
However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is –1.
2K+O2(excess)⟶K2−1O2
(iii) C is a reducing agent, while O2 acts as an oxidising agent. If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2.
C(excess)+O2⟶+2CO
On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is +4.
C+O2(excess)⟶+4CO2
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:
(i) P4 and F2 are reducing and oxidising agents respectively.
If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3.
P4(excess)+F2⟶+3PF3
However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.
P4+F2(excess)⟶+5PF5
(ii) K acts as a reducing agent, whereas O2 is an oxidising agent. If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is –2.
4K(excess)+O2→2K2−2O
However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is –1.
2K+O2(excess)⟶K2−1O2
(iii) C is a reducing agent, while O2 acts as an oxidising agent. If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2.
C(excess)+O2⟶+2CO
On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is +4.
C+O2(excess)⟶+4CO2
