Question

Write the oxidation and reduction half-reactions represented in each of these redox equations. Write the halfreactions in net ionic form if they occur in aqueous solution. a. \(\mathrm{PbO}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Pb}(\mathrm{s})\) b. \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq})\) c. \(\mathrm{Sn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{SnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\)

Short answer

For the given redox equations, the oxidation and reduction half-reactions are as follows: a. Oxidation half-reaction: \(4NH_{3}(g) \rightarrow 2N_{2}(g) + 12H^{+} + 12e^{-}\) Reduction half-reaction: \(PbO(s) + 2H^{+} + 2e^{-} \rightarrow Pb(s) + H_{2}O(l)\) b. Oxidation half-reaction: \(S_{2}O_{3}^{2-}(aq) \rightarrow S_{2}O_{4}^{2-}(aq) + 2e^{-}\) Reduction half-reaction: \(I_{2}(s) + 2e^{-} \rightarrow 2I^{-}(aq)\) c. Oxidation half-reaction: \(Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}\) Reduction half-reaction: \(2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\)

Step-by-step solution

Problem a: Identify Oxidized and Reduced Species

In the given equation: \(\mathrm{PbO}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Pb}(\mathrm{s})\) Pb is initially in the form of PbO and finally ends up in a solid-state as Pb. NH3 is consumed in the reaction and forms N2. Notice that the oxidation state of Pb in PbO is +2, while in its elemental form it's 0. Thus, \(Pb^{2+}\) is being reduced to Pb. Meanwhile, the oxidation state of Nitrogen in NH3 is -3 while in N2 it is 0. Thus, N in NH3 is being oxidized to N2.

Problem a: Write Oxidation and Reduction Half-Reactions

Oxidation half-reaction: The oxidation half-reaction is the transformation of NH3 to N2, with N going from an oxidation state of -3 to 0. Balancing the reaction: \(4NH_{3}(g) \rightarrow 2N_{2}(g) + 12H^{+} + 12e^{-}\) Reduction half-reaction: The reduction half-reaction is the transformation of PbO to Pb, with Pb going from an oxidation state of +2 to 0. Balancing the reaction: \(PbO(s) + 2H^{+} + 2e^{-} \rightarrow Pb(s) + H_{2}O(l)\)

Problem b: Identify Oxidized and Reduced Species

In the given equation: \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq})\) Iodine (I2) is consumed and forms NaI, while Na2S2O3 is consumed and forms Na2S2O4. The oxidation state of iodine in I2 is 0, while in NaI it's -1. Thus, I2 is being reduced to \(I^{-}\). The oxidation state of S in Na2S2O3 is +2, while in Na2S2O4 it's +2.5. Thus, S is being oxidized.

Problem b: Write Oxidation and Reduction Half-Reactions

Oxidation half-reaction: The oxidation half-reaction is the transformation of S in \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) to S in \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\). Balancing the reaction: \(S_{2}O_{3}^{2-}(aq) \rightarrow S_{2}O_{4}^{2-}(aq) + 2e^{-}\) Reduction half-reaction: The reduction half-reaction is the transformation of I2 to \(I^{-}\), with I going from an oxidation state of 0 to -1. Balancing the reaction: \(I_{2}(s) + 2e^{-} \rightarrow 2I^{-}(aq)\)

Problem c: Identify Oxidized and Reduced Species

In the given equation: \(\mathrm{Sn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{SnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) Sn is consumed and forms SnCl2, while HCl is consumed and forms H2. The oxidation state of Sn in its elemental form is 0, while in SnCl2 it's +2. Thus, Sn is being oxidized to \(Sn^{2+}\). The oxidation state of H in HCl is +1, while in H2 it's 0. Thus, H is being reduced from \(H^{+}\) to H2.

Problem c: Write Oxidation and Reduction Half-Reactions

Oxidation half-reaction: The oxidation half-reaction is the transformation of Sn to SnCl2, with Sn going from an oxidation state of 0 to +2. Balancing the reaction: \(Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}\) Reduction half-reaction: The reduction half-reaction is the transformation of HCl to H2, with H going from an oxidation state of +1 to 0. Balancing the reaction: \(2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\)

Key concepts

Oxidation States

Understanding oxidation states is crucial in redox chemistry. They help us keep track of the electrons during chemical reactions, especially in redox processes where the transfer of electrons leads to changes in oxidation state. To find the oxidation state of an element in a compound, remember that the sum of oxidation states must equal the charge of the molecule or ion. For neutral compounds, this sum is zero.

In the problem provided, we saw how the oxidation state of lead (Pb) changes from +2 in PbO to 0 in pure Pb, signifying a reduction, as it gained electrons. Similarly, nitrogen (N) in ammonia (NH₃) changes its oxidation state from -3 to 0 when it forms nitrogen gas (N₂), indicating an oxidation as it loses electrons. These examples exemplify the concept: oxidation state decreases during reduction and increases during oxidation.

Balancing Redox Equations

Balancing redox equations ensures that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This balance is fundamental to obey the law of conservation of mass and charge. For instance, in the given solution, the oxidation half-reaction for ammonia involves losing a total of 12 electrons, while the reduction half-reaction for lead oxide gains only 2 electrons. To balance these half-reactions, we have to ensure that the total number of electrons lost equals the electrons gained by multiplying the reduction half-reaction by 6, allowing us to add these half-reactions together to give the overall balanced redox equation.

Tips for Balancing Redox Equations:

• Start by identifying the oxidized and reduced species.
• Use the changes in oxidation states to determine the number of electrons transferred.
• Balance the atoms involved in the redox change first, followed by other elements.
• Finally, ensure the total charge is balanced on both sides by adding electrons where necessary.

Net Ionic Equations

Net ionic equations show the actual chemical change without spectator ions, which do not participate in the reaction. These equations highlight the essence of redox reactions by focusing on the species that exchange electrons. In the problems solved, the net ionic form represents only the components undergoing a change in oxidation state, stripped of any ions not directly involved in the redox process.

For example, in the oxidation of Na₂S₂O₃ to Na₂S₂O₄, the sodium ions are spectators and don't appear in the net ionic equation. The benefit of using net ionic equations is that it simplifies complex reactions, making it easier to identify the actual redox changes at play, which is a vital skill in understanding chemical reactivity and stoichiometry.