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.
Answer
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Hint: 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 oxidation state is in excess.
Following three illustration justify this –
Complete step by step solution:
(i) ${P_4}\left( {excess} \right) + {F_2}\xrightarrow{{}}\mathop P\limits^{ + 3} {F_3}$
When excess ${P_4}$ reacts with ${F_2}$then $P{F_3}$is produced in which $P$ has $ + 3$oxidation number.
Oxidising agent is ${F_2}$ and the reducing agent is ${P_4}$.
${P_4} + {F_2}\left( {excess} \right)\xrightarrow{{}}\mathop P\limits^{ + 5} {F_5}$
But if fluorine is in excess, $P{F_5}$ is formed in which $P$ has an oxidation number of $ + 5$.
(ii) $4K\left( {excess} \right) + {O_2}\xrightarrow{{}}2{K_2}\mathop O\limits^{ - 2} $
When excess $K$ reacts with oxygen then ${K_2}O$ is formed in which oxygen has an oxidation number of $ - 2$.
Oxidising agent is oxygen and the reducing agent is $K$.
$2K + {O_2}\left( {excess} \right)\xrightarrow{{}}{K_2}O_2^{ - 1}$
But if oxygen is in excess, then ${K_2}{O_2}$ is formed in which oxygen has an oxidation number of $ - 1$.
(iii) \[2\left( {excess} \right) + {O_2}\xrightarrow{{}}2\mathop C\limits^{ + 2} O\]
When an excess of carbon reacts with oxygen then $CO$ is formed in which carbon has $ + 2$ oxidation number.
$C + {O_2}\left( {excess} \right)\xrightarrow{{}}C{O_2}$
When an excess of oxygen is used $C{O_2}$ is formed in which carbon has $ + 4$ oxidation number.
Note: Oxidising agent: Species which oxidise the others but self-reduced are known as oxidising agent.
Reducing agent: Species which reduce the others but self-oxidise are known as reducing agents.
Following three illustration justify this –
Complete step by step solution:
(i) ${P_4}\left( {excess} \right) + {F_2}\xrightarrow{{}}\mathop P\limits^{ + 3} {F_3}$
When excess ${P_4}$ reacts with ${F_2}$then $P{F_3}$is produced in which $P$ has $ + 3$oxidation number.
Oxidising agent is ${F_2}$ and the reducing agent is ${P_4}$.
${P_4} + {F_2}\left( {excess} \right)\xrightarrow{{}}\mathop P\limits^{ + 5} {F_5}$
But if fluorine is in excess, $P{F_5}$ is formed in which $P$ has an oxidation number of $ + 5$.
(ii) $4K\left( {excess} \right) + {O_2}\xrightarrow{{}}2{K_2}\mathop O\limits^{ - 2} $
When excess $K$ reacts with oxygen then ${K_2}O$ is formed in which oxygen has an oxidation number of $ - 2$.
Oxidising agent is oxygen and the reducing agent is $K$.
$2K + {O_2}\left( {excess} \right)\xrightarrow{{}}{K_2}O_2^{ - 1}$
But if oxygen is in excess, then ${K_2}{O_2}$ is formed in which oxygen has an oxidation number of $ - 1$.
(iii) \[2\left( {excess} \right) + {O_2}\xrightarrow{{}}2\mathop C\limits^{ + 2} O\]
When an excess of carbon reacts with oxygen then $CO$ is formed in which carbon has $ + 2$ oxidation number.
$C + {O_2}\left( {excess} \right)\xrightarrow{{}}C{O_2}$
When an excess of oxygen is used $C{O_2}$ is formed in which carbon has $ + 4$ oxidation number.
Note: Oxidising agent: Species which oxidise the others but self-reduced are known as oxidising agent.
Reducing agent: Species which reduce the others but self-oxidise are known as reducing agents.
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