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Whenever a reaction between an oxidsing agent nad a reducing aent is carried out a compound of lower oxidation state I is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if oxidising agent is in excess justify this statement giving three illustrations |
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Answer» <P> Solution :(i)C is reducing agent while `O_(2)` is n oxidising agent if excess of carbon is burnt in a limitedsupport of `O_(2)` CO formed in which the OXIDATION state of C is +2 if however excess of `O_(2)` is used the initial formed CO gets oxidised to `CO_(2)`in oxidation state of C is +4`2C(s) +O_(2)(g)rarr 2CO(g),C(s)+O_(2)(g)rarrCO_(2)(g)` (ii) `P_(4)` is a reducing agent while `CI_(2)`is usedthe initally formed `PCI_(3)` react further to form `PCI_(5)`in which the oxidation state of p is +5 `P_(4)(s)+6CI_(2)(g)rarr4overset(+3)PCI_(3), P_(4)(s)10 CI_(2)rarr4 overset(+5)PCI_(5)` (iii) NA is a reducing agent while `O_(2)` is an oxidising agent when excessof Na is used sodium oxide formed in the oxidation state of O is -1 which is higher than -2 `4NA(s)+O_(2)(g)rarrNa_(2)O(s), 2NA(s)+2O_(2)(g)rarrNa_(2)O_(2)(s)` |
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