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8.11 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?

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Answer-

These can be understood by the following examples-

$P_{4}$ is reducing agent and $Cl_{2}$ is an oxidizing agent

$P_{4}+10Cl_{2}(Excess)\rightarrow PCl_{5}$ [O.N of phosphorus +5] $\Rightarrow$ Higher O.S of P

$P_{4}(Excess)+6Cl_{2}\rightarrow 4PCl_{3}$ [O.N of phosphorus +3] $\Rightarrow$ Lower O.S of P

is an $O_{2}$ is a reducing agent and $C$ oxidising agent

$C+O_{2}\rightarrow CO_{2}$ (O is in excess) [O.N of C +4]

$C+O_{2}\rightarrow CO$ (C is in excess) [O.N of C +2]

$K$ is a reducing agent and $O_{2}$ is an oxidizing agent

$K+O_{2}\rightarrow K_{2}O$ (K is in excess) [O.N of O -2] (lower O.S.)

$K+O_{2}\rightarrow K_{2}O_{2}$ (O is in excess) [O.N of O -1] (lower O.S.)

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Gautam harsolia

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