Question

Write the oxidation and reduction half-reactions for the redox equation. \(\mathrm{Pb}(\mathrm{s})+\mathrm{Pd}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Pd}(\mathrm{s})\)

Short answer

The oxidation and reduction half-reactions for the given redox equation are: Oxidation half-reaction: \(Pb(s) \rightarrow Pb^{+2}(aq) + 2e^-\) Reduction half-reaction: \(Pd^{+2}(aq) + 2e^- \rightarrow Pd(s)\)

Step-by-step solution

Determine the oxidation states of each element

First, let's assign oxidation states to each element in the given redox reaction: \(Pb(s)\): In its solid state, the oxidation state of Pb is 0. \(Pd(NO_3)_2(aq)\): Pd is in the ionic compound Pd(NO3)2 and has an oxidation state of +2 in this compound. The oxidation state of nitrogen in \(NO_3^-\) (nitrate ion) is +5, and that of oxygen is -2. \(Pb(NO_3)_2(aq)\): Pb forms an ionic compound with nitrate ions and has an oxidation state of +2 in this compound. \(Pd(s)\): In its solid state, the oxidation state of Pd is 0.

Identify the oxidation and reduction processes

Now that we've assigned oxidation states, we can determine which species are undergoing oxidation and reduction: Oxidation: Pb(s) goes from an oxidation state of 0 to +2 in \(Pb(NO_3)_2(aq)\). This means Pb is losing 2 electrons. Reduction: Pd(+2) in \(Pd(NO_3)_2(aq)\) goes from an oxidation state of +2 to 0 in Pd(s). This means Pd is gaining 2 electrons.

Write the half-reactions

Now let's write the half-reactions for oxidation and reduction: Oxidation half-reaction: \(Pb(s) \rightarrow Pb^{+2}(aq) + 2e^-\) Reduction half-reaction: \(Pd^{+2}(aq) + 2e^- \rightarrow Pd(s)\) These are the balanced half-reactions for the given redox equation as we have both mass and charge balanced in each of them.

Key concepts

Oxidation States

Oxidation states, also known as oxidation numbers, help us understand how electrons are transferred in redox reactions. They serve as a book-keeping method to track electron flow. Each atom in a compound is assigned an oxidation state based on specific rules:

• For an element in its pure form, like Pb(s) or Pd(s), the oxidation state is always 0.
• In ionic compounds, the sum of oxidation states of all atoms equals the overall charge of the compound.
• Common oxidation states include: oxygen as -2, hydrogen as +1, and halogens like chlorine as -1.

Understanding these rules allows us to identify which element is oxidized (loses electrons) and which is reduced (gains electrons) in reactions.

Half-Reactions

Half-reactions break down the redox process into two parts: one for oxidation and one for reduction. This helps clarify which substances lose or gain electrons:

• The oxidation half-reaction describes the process where an element loses electrons. For example, in the reaction where Pb(s) increases its oxidation state from 0 to +2.
• The reduction half-reaction involves gaining electrons, such as Pd gaining electrons and moving from an oxidation state of +2 to 0.

They are balanced separately for mass and charge, ensuring that the number of electrons lost in oxidation equals the number gained in reduction. Writing half-reactions ensures the conservation of charge and mass in the overall chemical process.

Chemical Equations

Chemical equations represent chemical reactions by showing reactants transforming into products. In redox reactions, they also indicate electron transfer:

• Reactants are shown on the left, and products on the right.
• Chemical equations must be balanced for both atoms and charge, reflecting the conservation of mass.
• In redox equations, balancing involves ensuring the number of electrons lost equals those gained, which is done through half-reactions.

For example, in the given reaction, Pb(s) transforms into Pb(NO3)2(aq) and Pd(NO3)2(aq) changes into Pd(s). By breaking it into half-reactions and balancing them, we achieve a clear understanding of how electron transfer drives the chemical changes.