Justify that the following reaction are redox reactions:
a. `CuO(s)+H_(2)(g)rarr Cu(s)+H_(2)O(g)`
b. `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g)`
c. `4BCl_(3)(g)+3LiAlH_(4)(s) rarr 2B_(2)H_(6)(g)+3LiCl(s)+3AlCl_(3)(s)`
d. `2K(s)+F_(2)(g)rarr 2K^(o+)F^(Θ)(s)`
e. `4NH_(3)(g)+5O_(2)(g)rarr 4NO(g)+6H_(2)O(g)`

Explanation

`overset(+2-2)(CuO)(s)+overset(0)(H_(2))(g)rarr Cu(s)+overset(+1-2)(H_(2)O)(g)`
Here, `O` is removed from `CuO`, therefore, it is reduced to `Cu` while `O` is added to `H_(2)` to form `H_(2)O`. Therefore, it is oxidised. Further, oxidation number of `Cu` decreases from `+2` in `CuO` to `0` in `Cu`, but that of `H` increases from `0` in `H_(2)` to `+1` in `H_(2)O`. Therefore, `CuO` is reduced to `Cu` but `H_(2)` is oxidised to `H_(2)O`. thus, this is a redox reaction.
b. `overset(+3)(Fe_(2))overset(-2)(O_(3))(s)+overset(+2)(3CO)(g)rarr 2 overset(0)(Fe)(s)+3overset(+4)(CO_(2))(g)`
Here oxidation number of Fe decreases from `+3` in `Fe_(2)O_(3)` to `0` in `Fe` while that of `C` increases from `+2` in `CO` to `+4` in `CO_(2)`. Further, oxygen is removed from `Fe_(2)O_(3)` and added to `CO`. Therefore, `Fe_(2)O_(3)` is reduced while `CO` is oxidised. Thus, this is redox reaction.
c. `overset(+3)(4Br)overset(-1)(Cl)(g)+overset(+1)(Li)overset(+3)(Al)overset(-1)(H_(4))(s)rarr 2overset(-1)(B_(2))overset(+1)(H_(6))(g)+overset(+1)(3LiCl)(s)+3overset(+3)(Al)overset(-1)(Cl_(3))(s)`
Here, oxidation number of `Br` decreases from `+3` in `BCl_(3)` to `-3` in `B_(2)H_(6)` while that of H increases from `-1` in `LiAlH_(4)` to `+1` in `B_(2)H_(6)`. Therefore, `BCl_(3)` is reduced white `LiAlH_(4)` is oxidised. Further, H is added to `BCl_(3)`, but is removed from `LiAlH_(4)`. Therefore, `BCl_(3)` is reduced white `LiAlH_(4)` is oxidised. Thus, it is a redox reaction.
d. `2K(s)+F_(2)(g)rarr 2K^(o+)F^(Θ)(s)`
Here, each `K` atom has lost one electron to form `K^(o+)` while `F_(2)` has gained two electrons to form two `F^(Θ)` ions. Therefore, `K` is oxidised while `F_(2)` is reduced. Thus, it is a redox reaction.
e. `4overset(-3)(N)overset(+1)(H_(3))(g)+5overset(0)(O_(2))(g)rarr overset(+2)(4N) overset(-2)(O)(g)+6 overset(+1)(H_(2))overset(-2)(O)(g)`
Here, oxidation number of `N` increases from `-3` in `NH_(3)` to `+2` in `NO` while that `O` decreases from `0` in `O_(2)` to `-2` in `NO` or `H_(2)O`.
Therefore, `NH_(3)` is oxidised while `O_(2)` is reduced.
Further, `H` bas been removed from `NH_(3)` but added to `O_(2)`. Therefore, `NH_(3)` has been oxidised while `O_(2)` is reduced.
Thus, this is a redox reaction.