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Question 3

Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

(d) 2K(s) + F2(g) → 2K+F(s)

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

Answer

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

Let us write the oxidation number of each element involved in the given reaction as: 

+2  -2        0             0             +1    -2

Cu O(s) +  H2(g)  →  Cu(s)  +  H2  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 H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.

 

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Let us write the oxidation number of each element involved in the given reaction as:

+3   -2          +2  -2            0          +4  -2

Fe2 O3(s) + 3C  O(g) → 2Fe(s) + 3C O2(g)

Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.



 

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

Let us write the oxidation number of each element involved in the given reaction as:

 +3  -1            +1   +3   -1            -3   +1         +1  -1             +3   -1

4B Cl3(g) + 3  Li   Al   H4(s) → 2B2 H6(g) + 3Li   Cl(s)   + 3 Al  Cl3 (s)

In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

 

(d) 2K(s) + F2(g) → 2K+F(s)

Let us write the oxidation number of each element involved in the given reaction as:

0               0                    +1   -1

2K(s)   +   F2(g)    →    2KF(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 F2 to – 1 in KF i.e., F2 is reduced to KF.

Hence, the above reaction is a redox reaction.

 

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

Let us write the oxidation number of each element involved in the given reaction as:

  -3   +1            0              +2  -2         +1    -2

4 N  H3(g) + 5 O2(g) → 4N  O(g) + 6HO(g)

Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

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