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)`

Justify that the following reaction are redox reactions: a. `CuO(s)+H_

1.	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)`
Answer» (a) `CuO_((s)) + H_(2 (g)) to Cu_(s) + H_(2)O_(g)` Let us write the oxidation number of each element involved in the given reaction as : `overset(+2)(Cu)overset(-2)(O)_(s) + overset(0)(H_(2 (g))) to overset(0)(C)u_(s) + overset(+1)(H_(2)) overset(-2)(O)_(g)` Here , the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also , the oxidation number of H increases from 0 in `H_(2)` to +1 in `H_(2)O` i.e., `H_(2)` is oxidized to `H_(2)O` . Hence , this reaction is a redox reaction . (b) `Fe_(2)O_(3(s)) + 3 CO_((g)) to 2 Fe_((s)) + 3CO_(2 (g))` Let us write the oxidation number of each element in the given reaction as : `overset(+3)(F)e_(2)overset(-2)(O)_(3 (s)) + 3overset(+2)(C)overset(-2)(O)_((g)) to 2 overset(0)(Fe)_((s)) + 3 overset(+4)(C)overset(-2)(O_(2)(g))` Here , the oxidation number of Fe decreases from +3 in `Fe_(2)O_(3)` to 0 in Fe i.e., `Fe_(2)O_(3)` is reduced to Fe . On the other hand , the oxidation number of C increases from +2 in CO to +4 in `CO_(2)` i.e., CO is oxidized to `CO_(2)` . Hence , the given reaction is a redox reaction . (c) The oxidation number of each element in the given reaction can be represented as : `4 overset(+3)(B) overset(-1)(C)l_(3 (g)) + 3 overset(+1)(Li)overset(+3)(Al)overset(-1)(H)_(4(s)) to 2 overset(-3)(B_(2))overset(+1)(H)_(6 (g)) + 3 overset(+1)(Li)overset(-1)(C)l_((s)) + 3 overset(+3)(A)l overset(-1)(C)l_(3 (s))` In this reaction, the oxidation number of B decreases from +3 in `BCl_(3)` to `-3` in `B_(2)H_(6)` i.e., `BCl_(3)` is reduced to `B_(2)H_(6)` . Also , the oxidation number of H increases from `-1` in `LiAlH_(4)` to +1 in `B_(2)H_(6)` i.e., `LiAlH_(4)` is oxidized to `B_(2)H_(6)`. Hence, the given reaction is a redox reaction. (d) `2K_((s)) + F_(2 (g)) to 2K^(+) F_((s))^(-)` The oxidation number of each element in the given reaction can be represented as: `2 overset(0)(K)_((s)) + overset(0)(F)_(2 (g)) to 2 overset(+1)(K^(+)) overset(-1)(F^(-))_((s))` In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in `F_(2)` to `–1` in KF i.e., `F_(2)` is reduced to KF . Hence , the above reaction is a redox reaction (e) ` 4NH_(3 (g)) + 5 O_(2 (g)) to 4 NO_((g)) + 6 H_(2)O_((g))` The oxidation number of each element in the given reaction can be represented as: `4 overset(-3)(N) overset(+1)(H)_(3 (g)) + 5 overset(0)(O)_(2 (g)) to 4 overset(+2)(N) overset(-2)(O)_((g)) + 6 overset(+1)(H_(2)) overset(-2)(O)_((g))` Here, the oxidation number of N increases from –3 in `NH_(3)` to +2 in NO. On the other hand, the oxidation number of `O_(2)` decreases from 0 in `O_(2)` to` –2` in NO and `H_(2)O` i.e., `O_(2)` is reduced. Hence, the given reaction is a redox reaction.

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