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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» Solution :The given statement can be verified by the FOLLOWING illustrations : (a) `UNDERSET(("reducing AGENT Excess"))(2C(s))+underset(("oxidising agent"))(O_(2)(g))tounderset(("Compound of lower O.S."))overset(+2)(2CO(g))""......(i)` `underset(("reducing agent"))(C(s))+underset(("oxidising angent Excess"))(O_(2)(g))tounderset(("Compound of higher O.S."))overset(+4)(CO_(2)(g))""......(ii)` In REACTION (i), the reducing agent, i.e., carbon is in excess, whereas in reaction (ii) the oxidising agent, i.e., `O_(2)` is in excess. Reaction (i) leads to the formation of CO (O.S. of C = + 2), while reaction (ii) leads to the formation of `CO_(2)` (O.S. of C=+4) (b) `underset(("reducing agent Excess"))(4Na(s))+underset(("oxidising agent"))(O_(2)(g))tounderset(("Compound of lower O.S."))(Na_(2)overset(-2)(O)(s))""......(i)` `underset(("reducing agent"))(2Na(s))+underset(("oxidising agent Excess"))(2O_(2)(g))tounderset(("Compound of higher O.S."))(Na_(2)overset(-1)(O_(2))(g))"".......(ii)` (c ) `underset(("reducing agent Excess"))(P_(4)(s))+underset(("oxidising agent"))(6Cl_(2)(g))tounderset("Compound of lower O.S.")(4overset(+3)(P)Cl_(3)(l))"".......(ii)` `underset(("reducing agent"))(P_(4)(s))+underset(("oxidising agent Excess"))(10Cl_(2)(g))tounderset("Compound or higher O.S.")(4overset(+5)(P)Cl_(5)(s))""........(ii)` |
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