Identify the oxidant and reductant in the folowing rection : (a) 10 H^(+)(aq)+4Zn(s)+NO_(3)^(-)(aq)rarr4Zn^(2+)(aq)+NH_(4)^(+)(aq)+3H_(2)O(l) (b)I_(2)(g)+H_(2)S(g)+H_(2)S(g)rarrHI(g)+S(s)

Identify the oxidant and reductant in the folowing rection : (a) 10 H

1.	Identify the oxidant and reductant in the folowing rection : (a) 10 H^(+)(aq)+4Zn(s)+NO_(3)^(-)(aq)rarr4Zn^(2+)(aq)+NH_(4)^(+)(aq)+3H_(2)O(l) (b)I_(2)(g)+H_(2)S(g)+H_(2)S(g)rarrHI(g)+S(s)
Answer» Solution :(a) Writing the O.N of all the atoms above their symbols we have `10 overset(+1)H^(+)(aq)+4overset(0)Zn(s)+overset(+5)BO_(3)^(-)(aq) to 4 overset(+2)Zn^(2+)(aq)+ overset(-3)NH_(4)^(+)(aq)+3overset(+1)H_(2)O(l)` here there is no change in the O.N of H toms But hte O.N of Zn changes from zero in Zn to +2 in `Zn^(+)` therefore it is oxidised and hence Zn acts as a reductant The O.N of N decresses FORM +5 in `NO_(3)^(-) to -3 in NH_(4)^(+)` and therefore it is reduced and hence `NO_(3)^(-)` acts as the oxidant (b) Writing the O.N of all hte atoms above their symbols we have `I_(2)^(0)(g) + H_(2) overset(-2)H_(2)(g) to 2 overset(+1) overset(-1)I(g)+overset(0)S(s)` Here O.N of H does not change The O.N of `I_(2)` DECREASE frm zero in `I_(2)` to -1 in HI therefore `I_(2)` is REDUCING and hence it acts as an oxidant The O.N of S increases form -2 in `H_(2)S` to zero in S therefore `H_(2)S` is oxidation and hence it acts as th reductant