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. Justify the above statement with three examples.

Asked by Pragya Singh | 1 year ago |  55

Solution :-

When an oxidizing agent and a reducing agent react, a lower oxidation state compound is formed if the reducing agent is in excess, and a higher oxidation state
compound is formed if the oxidizing agent is in excess. As an example, consider the following:

(i)$$P_{ 4 } \;and\; F_{ 2 }$$​ are reducing and oxidizing agent respectively.

In an excess amount of $$P_{ 4 }$$​ is reacted with $$F_{ 2 }$$​, then $$PF_{ 3 }$$​ would be produced, where the oxidation no. of P is +3.

$$P_{ 4 } \; _{(excess)}\; + \; F_{ 2 } \; \rightarrow \; PF_{ 3 }$$

If $$P_{ 4 }$$​ is reacted with excess of $$F_{ 2 }$$​, then $$PF_{ 5 }$$ would be produced, where the oxidation no. of P is +5.

$$P_{ 4 } \; + \; F_{ 2 } \; _{(excess)} \; \rightarrow \; PF_{ 5 }$$

(ii) K and $$O_{ 2 }$$​ acts as a reducing agent and oxidizing agent respectively.

If an excess of K reacts with $$O_{ 2 }$$​, it produces $$K_{ 2 }O$$. Here, the oxidation number of O is -2.

$$4 \; K \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; 2 \; K_{ 2 }O^{ -2 }$$

If K reacts with an excess of $$O_{ 2 }$$​, it produces $$K_{ 2 }O_{ 2 }$$​, where the oxidation number of O is –1.

$$2 \; K \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; K_{ 2 }O_{ 2 }^{ -1 }$$

(iii) C and O_{ 2 }O2​ acts as a reducing agent and oxidizing agent respectively.

If an excess amount of C is reacted with insufficient amount of $$O_{ 2 }$$​, then it produces CO, where the oxidation number of C is +2.

$$C \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; CO$$

If C is burnt in excess amount of $$O_{ 2 }$$​, then $$CO_{ 2 }$$ is produced, where the oxidation number of C is +4.

$$C \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; CO_{ 2 }$$

Answered by Abhisek | 1 year ago

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