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)

To determine whether the given reactions are redox reactions, we need to analyze the changes in oxidation states of the elements involved. A redox reaction, or reduction-oxidation reaction, involves the transfer of electrons between species, resulting in changes in their oxidation states. Oxidation refers to the loss of electrons (increase in oxidation state), while reduction refers to the gain of electrons (decrease in oxidation state). Let’s break down each reaction step by step.

Reaction Analysis

(a) CuO(s) + H2(g) ? Cu(s) + H2O(g)

In this reaction, we can assign oxidation states:

• Cu in CuO is +2.
• O in CuO is -2.
• H in H2 is 0.
• Cu in Cu is 0.
• H in H2O is +1, and O is -2.

Here, copper (Cu) is reduced from +2 to 0, while hydrogen (H) is oxidized from 0 to +1. Thus, this reaction is a redox reaction.

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

Let’s examine the oxidation states:

• Fe in Fe2O3 is +3.
• O in Fe2O3 is -2.
• C in CO is +2.
• Fe in Fe is 0.
• C in CO2 is +4.

In this case, iron (Fe) is reduced from +3 to 0, and carbon (C) is oxidized from +2 in CO to +4 in CO2. Therefore, this reaction is also classified as a redox reaction.

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

For this reaction, we analyze the oxidation states:

• B in BCl3 is +3.
• Cl in BCl3 is -1.
• Li in LiAlH4 is +1.
• Al in LiAlH4 is +3.
• H in LiAlH4 is -1.
• B in B2H6 is -1.
• Li in LiCl is +1.
• Al in AlCl3 is +3.

Here, boron (B) is reduced from +3 in BCl3 to -1 in B2H6, indicating a gain of electrons, while aluminum (Al) remains at +3. This reaction is a redox reaction due to the change in oxidation state of boron.

(d) 2K(s) + F2(g) ? 2K+F–(s)

Let’s look at the oxidation states:

• K in K is 0.
• F in F2 is 0.
• K in K+ is +1.
• F in F– is -1.

In this reaction, potassium (K) is oxidized from 0 to +1, while fluorine (F) is reduced from 0 to -1. This confirms that it is a redox reaction.

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

Finally, we analyze the oxidation states:

• N in NH3 is -3.
• H in NH3 is +1.
• O in O2 is 0.
• N in NO is +2.
• O in H2O is -2.

In this case, nitrogen (N) is oxidized from -3 in NH3 to +2 in NO, while oxygen (O) is reduced from 0 in O2 to -2 in H2O. Thus, this reaction is also a redox reaction.

Summary of Findings

All five reactions involve changes in oxidation states, confirming that they are indeed redox reactions. Understanding these changes helps us grasp the electron transfer processes that define redox chemistry.