Name: Hydrogen Electrode
Uploaded: 2019-07-19
Duration: 3 min 7 s
Description: Every unit of measurement has a standard measurement associated with it. The same is with the electrode potential. Let's learn what the standard measurement...

Question

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.

Accepted Answer

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) $P_{4}$ and $F_{2}$ are reducing and oxidising agents respectively.
If an excess of $P_{4}$ is treated with $F_{2}$ , then $P F_{3}$ will be produced, where the O.N. of $P$ is $+ 3$ in $P F_{3}$
$2 P_{} (e x c e s s) + 3 F_{2} ⟶ {2 P F}_{3}$
O.N. of $P$ changes from $0$ to $+ 3$
However, if $P_{4}$ is treated with an excess of $F_{2}$ , then $P F_{5}$ will be produced, where the $O . N .$ of $P$ is $+ 5$ in ${PF}_{5}$ .
$P_{4} (e x c e s s) + F_{2} ⟶ {P F}_{5}$
O.N. of $P$ changes from $0$ to $+ 5$ .

(ii) $K$ acts as a reducing agent, whereas $O_{2}$ is an oxidising agent. If an excess of $K$ reacts with $O_{2}$ , then $K_{2} O$ will be formed, wherein the O.N. of $O$ is $- 2 i n K_{2} O$ .
$4 K (e x c e s s) + O_{2} ⟶ 2 K_{2} O$
The O.N. of oxygen changes from $0$ to $- 2$
However, if $K$ reacts with an excess of $O_{2}$ , then $K_{2} O_{2}$ will be formed, wherein the O.N. of $O$ is $- 1$ in $K_{2} O_{2} .$
$2 K + O_{2} (e x c e s s) ⟶ K_{2} O_{2}$
O.N. of oxygen atom changes from $0$ to $- 1$ .

(iii) $C$ is a reducing agent, while $O_{2}$ acts as an oxidising agent.
If an excess of $C$ is burnt in the presence of limiting amount of $O_{2}$ , then $C O$ will be produced, where the O.N. of carbon is $+ 2$ .
$C (e x c e s s) + \frac{1}{2} O_{2} ⟶ C O$
The O.N. of carbon changes from $0$ to $+ 2$
On the other hand, if $C$ is burnt in an excess of $O_{2}$ , then $C O_{2}$ will be produced, wherein the O.N. of $C$ is $+ 4$ .
$C + O_{2} (e x c e s s) ⟶ {C O}_{2}$
The O.N. of carbon changes from $0$ t0 $+ 4$ .

Important Questions on Redox Reactions

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