Question

Write the oxidation and reduction half-reaction represented in each of these redox equations. Write the half-reactions in net ionic form if they occur in aqueous solution. a. \(P b O(s)+N H_{3}(g) \rightarrow N_{2}(g)+H_{2} O(1)+P b(s)\) b. \(I_{2}(s)+N a S_{2} O_{3}(a q) \rightarrow N a_{2} S_{2} O_{4}(a q)+N a I(a q)\) c. \(\operatorname{Sn}(s)+2 H C l(a q) \rightarrow \operatorname{Sn} C l_{2}(a q)+H_{2}(g)\)

Short answer

For the given redox equations, the oxidation and reduction half-reactions in net ionic forms are as follows: a. Oxidation: \(2NH_{3}(g) \rightarrow N_{2}(g) + 6e^{-}\) Reduction: \(PbO(s) + 2e^{-} \rightarrow Pb(s) + \frac{1}{2}O_{2}(g)\) b. Oxidation: \(S_{2}O_{3}^{2-}(aq) \rightarrow S_{2}O_{4}^{2-}(aq) + 2e^{-}\) Reduction: \(I_{2}(s) + 2e^{-} \rightarrow 2I^{-}(aq)\) c. Oxidation: \(Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}\) Reduction: \(2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\)

Step-by-step solution

Identify the species being oxidized and reduced

The oxidation states of all the atoms in this reaction are as follows: - Pb: +2 in PbO, 0 in Pb; - N: -3 in NH3, 0 in N2; - O: -2 in PbO and H2O. It's essential to observe that N is oxidized from -3 to 0 and Pb is reduced from +2 to 0.

Write the half-reactions

Considering the atoms and their oxidation states, we can write the half-reactions as: - Oxidation half-reaction: \(2NH_{3}(g) \rightarrow N_{2}(g) + 6e^{-}\) - Reduction half-reaction: \(PbO(s) + 2e^{-} \rightarrow Pb(s) + \frac{1}{2}O_{2}(g)\) ##Reaction b## b. \(I_{2}(s)+NaS_{2}O_{3}(aq) \rightarrow Na_{2}S_{2}O_{4}(aq)+NaI(aq)\)

Identify the species being oxidized and reduced

The oxidation states of all the atoms in this reaction are as follows: - I: 0 in I2, -1 in NaI; - S: +2 in both Na2S2O3 and Na2S2O4. It's important to observe that I is reduced from 0 to -1 and S is oxidized from +2 to +2.5.

Write the half-reactions in net ionic form

Considering the atoms and their oxidation states, we can write the half-reactions in net ionic form as: - Oxidation half-reaction: \(S_{2}O_{3}^{2-}(aq) \rightarrow S_{2}O_{4}^{2-}(aq) + 2e^{-}\) - Reduction half-reaction: \(I_{2}(s) + 2e^{-} \rightarrow 2I^{-}(aq)\) ##Reaction c## c. \(Sn(s)+2HCl(aq) \rightarrow SnCl_{2}(aq)+H_{2}(g)\)

Identify the species being oxidized and reduced

The oxidation states of all the atoms in this reaction are as follows: - Sn: 0 in Sn, +2 in SnCl2; - H: +1 in HCl, 0 in H2; - Cl: -1 in both HCl and SnCl2. It's important to observe that Sn is oxidized from 0 to +2, and H is reduced from +1 to 0.

Write the half-reactions in net ionic form

Considering the atoms and their oxidation states, we can write the half-reactions in net ionic form as: - Oxidation half-reaction: \(Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}\) - Reduction half-reaction: \(2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\)

Key concepts

Understanding Oxidation Half-Reaction

In redox reactions, the oxidation half-reaction is where a substance loses electrons. Think of oxidation as the charge of an atom becoming more positive. For example, when nitrogen is contained in ammonia, \( NH_3(g) \), it undergoes oxidation by losing electrons to transform into nitrogen gas, \( N_2(g) \). In our example reactions, the species that is oxidized can be figured out by comparing the oxidation states:

• Reaction (a): Nitrogen's oxidation state increases from \(-3\) in \( NH_3 \) to \( 0 \) in \( N_2 \).
• Reaction (b): Sulfur's oxidation state increases slightly from \(+2\) in \( Na_2S_2O_3 \) to higher values in \( Na_2S_2O_4 \).
• Reaction (c): Tin's oxidation state rises from \( 0 \) in \( Sn \) to \(+2\) in \( SnCl_2 \).

The general trend is oxidation increases the oxidation state due to electron loss. If you carefully watch how the electrons are transferred, it becomes easier to identify and balance the equation.

Grasping Reduction Half-Reaction

The reduction half-reaction is the other side of a redox process, where a substance gains electrons. Reduction results in a decrease in the oxidation state or a more negative charge. To keep it simple, when iodine, \( I_2(s) \), gains electrons, it gets reduced to iodide ions, \( I^- \), which you might find in table salt as sodium iodide, \( NaI \).In our sample reactions:

• Reaction (a): Lead is reduced; its oxidation state decreases from \(+2\) in \( PbO \) to \( 0 \) in pure lead \( Pb \).
• Reaction (b): Iodine is reduced as it goes from \( 0 \) in \( I_2 \) to \(-1\) in \( I^- \).
• Reaction (c): Hydrogen is reduced from \(+1\) in \( HCl \) to \( 0 \) in \( H_2 \).

Reduction is about gaining negativity—that means adding electrons and lowering the oxidation state, counterbalancing the oxidation half-reaction.

Explaining Net Ionic Equations

Net ionic equations simplify reactions by focusing on the species undergoing change, eliminating spectator ions that don't change in the reaction. It's like peeling away what's unnecessary to get to the heart of the reaction. Consider these practical net ionic scenarios:

• In reaction (a), the net ionic form stresses the electron shifts, explicitly showing nitrogen oxidation and lead reduction.
• Reaction (b)'s net ionic equation highlights sulfur's oxidation and iodine's reduction, dismissing sodium ions that don't alter in the process.
• Reaction (c) cleanly presents tin's oxidation and hydrogen's reduction by discarding the unchanged chloride ions.

Net ionic equations help you concentrate on the essential electron exchanges, making balancing redox equations a touch easier. They're all about breaking down complex equations into understandable steps, enabling a clearer grasp of the chemical artistry involved.