Certainly! Let's analyze each of the given reactions to determine which one involves both oxidation and reduction.
(i) [tex]\( NH_3 (s) + H_2O (l) \rightarrow NH_4^+ (aq) + OH^- (aq) \)[/tex]
In this reaction, ammonia [tex]\( NH_3 \)[/tex] reacts with water to form ammonium ion [tex]\( NH_4^+ \)[/tex] and hydroxide ion [tex]\( OH^- \)[/tex]. In this process:
- [tex]\( NH_3 \)[/tex] gains a proton (H+) from water to form [tex]\( NH_4^+ \)[/tex].
- The water molecule loses a proton to form [tex]\( OH^- \)[/tex].
This is an acid-base reaction. No oxidation or reduction (transfer of electrons) is occurring in this reaction.
(ii) [tex]\( 2 H_2O_2 (l) \rightarrow 2 H_2O (l) + O_2 (g) \)[/tex]
For this reaction, consider the oxidation states:
- In [tex]\( H_2O_2 \)[/tex] (hydrogen peroxide), each hydrogen has an oxidation state of +1. Since the compound is neutral, total oxidation state of oxygen must be -2. With two oxygens, each oxygen has an oxidation state of -1.
- In [tex]\( H_2O \)[/tex] (water), hydrogen remains +1, but with two hydrogens, total oxidation state of oxygen must be -2. Thus, each oxygen in water has an oxidation state of -2.
- In [tex]\( O_2 \)[/tex] (oxygen gas), the oxygen molecule is in its elemental form, so each oxygen atom has an oxidation state of 0.
Here, the oxygen in [tex]\( H_2O_2 \)[/tex] changes from an oxidation state of -1 to -2 in [tex]\( H_2O \)[/tex] (reduction) and from -1 to 0 in [tex]\( O_2 \)[/tex] (oxidation). Thus, both oxidation and reduction are occurring simultaneously.
(iii) [tex]\( Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s) \)[/tex]
This reaction is a precipitation reaction where [tex]\( Ba^{2+} \)[/tex] and [tex]\( SO_4^{2-} \)[/tex] ions combine to form solid [tex]\( BaSO_4 \)[/tex]. Both [tex]\( Ba \)[/tex] and [tex]\( SO_4 \)[/tex] retain their oxidation states before and after the reaction (+2 for [tex]\( Ba \)[/tex] and -2 for [tex]\( SO_4 \)[/tex]). Hence, no change in oxidation states occurs. This is not a redox reaction.
Among the given options, the reaction [tex]\( 2 H_2O_2 \rightarrow 2 H_2O + O_2 \)[/tex] (ii) is the one where both oxidation and reduction occur.
Therefore, the correct answer is:
_(ii) [tex]\( 2 H_2O_2 (l) \rightarrow 2 H_2O (l) + O_2 (g) \)[/tex]_
