Question

Use the oxidation-number method to balance these ionic redox equations. a. \(\mathrm{MoCl}_{5}+\mathrm{S}^{2-} \rightarrow \mathrm{MoS}_{2}+\mathrm{Cl}^{-}+\mathrm{S}\) b. \(\mathrm{TiCl}_{6}^{2-}+\mathrm{Zn} \rightarrow \mathrm{Ti}^{3+}+\mathrm{Cl}^{-}+\mathrm{Zn}^{2+}\)

Short answer

Final balanced redox equations: a. \(\mathrm{MoCl}_5 + 2\mathrm{S}^{2-} \rightarrow \mathrm{MoS}_{2} + 5\mathrm{Cl}^{-} + S \) b. \(2\mathrm{TiCl}_{6}^{2-} + \mathrm{Zn} \rightarrow 2\mathrm{Ti}^{3+} + 12\mathrm{Cl}^{-} + \mathrm{Zn}^{2+}\)

Step-by-step solution

Identify the oxidation numbers

We start by assigning oxidation numbers to the elements involved in each equation. In this case, the oxidation numbers are: - Mo in \(\mathrm{MoCl}_{5}\): +6 (since the oxidation state of Cl is always -1) - S in \(\mathrm{S}^{2-}\): -2 - Mo in \(\mathrm{MoS}_{2}\): +4 (since the oxidation state of S is -2) - Cl in \(\mathrm{Cl}^{-}\): -1 - S in \(\mathrm{S}\): 0

Calculate the change in oxidation numbers

Calculate the change in oxidation numbers for Mo and S in the reaction by subtracting the initial oxidation numbers from the final ones: - For Mo: +4 - (+6) = -2 - For S: 0 - (-2) = +2

Balance the half-reactions for electrons

Balance the half-reactions based on the change in oxidation numbers, making sure that the number of moles of electrons gained equals the number of moles of electrons lost. - For Mo: \(\mathrm{MoCl}_5 \rightarrow \mathrm{MoS}_{2} + 2e^-\) - For S: \(2\mathrm{S}^{2-} + 2e^- \rightarrow S + S^{2-}\)

Balance the half-reactions for non-electron species

Now we need to balance the Cl species in the half reactions. Add \(\mathrm{10Cl}^-\) to both sides of the Mo half-reaction: \(\mathrm{MoCl}_5 + 5\mathrm{Cl}^{-} \rightarrow \mathrm{MoS}_{2} + 2e^- + 5\mathrm{Cl}^-\)

Combine balanced half-reactions

Combine the balanced half-reactions to obtain the final balanced redox equation: \(\mathrm{MoCl}_5 + 5\mathrm{Cl}^{-} + 2\mathrm{S}^{2-} + 2e^- \rightarrow \mathrm{MoS}_{2} + 2e^- + 5\mathrm{Cl}^{-} + S + \mathrm{S}^{2-}\) Simplify by removing the 2\(e^-\) on both sides and summing up the remaining terms on both sides to get the final balanced redox equation: \(\mathrm{MoCl}_5 + 2\mathrm{S}^{2-} \rightarrow \mathrm{MoS}_{2} + 5\mathrm{Cl}^{-} + S \) b. \(\mathrm{TiCl}_{6}^{2-}+\mathrm{Zn} \rightarrow \mathrm{Ti}^{3+}+\mathrm{Cl}^{-}+\mathrm{Zn}^{2+}\) Following the same steps as above:

Identify the oxidation numbers

- Ti in \(\mathrm{TiCl}_{6}^{2-}\): +4 - Zn in \(\mathrm{Zn}\): 0 - Ti in \(\mathrm{Ti}^{3+}\): +3 - Cl in \(\mathrm{Cl}^{-}\): -1 - Zn in \(\mathrm{Zn}^{2+}\): +2

Calculate the change in oxidation numbers

- For Ti: +3 - (+4) = -1 - For Zn: +2 - 0 = +2

Balance the half-reactions for electrons

- For Ti: \(\mathrm{TiCl}_{6}^{2-} + e^- \rightarrow \mathrm{Ti}^{3+} + 6\mathrm{Cl}^{-}\) - For Zn: \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^-\)

Balance the half-reactions for non-electron species

The Cl species is already balanced in the Ti half-reaction.

Combine balanced half-reactions

Multiply the Ti half-reaction by 2 to match the number of electrons in the Zn half-reaction: \(2\mathrm{TiCl}_{6}^{2-} + 2e^- \rightarrow 2\mathrm{Ti}^{3+} + 12\mathrm{Cl}^{-}\) Combine the balanced half-reactions: \(2\mathrm{TiCl}_{6}^{2-} + \mathrm{Zn} \rightarrow 2\mathrm{Ti}^{3+} + 12\mathrm{Cl}^{-} + \mathrm{Zn}^{2+} + 2e^-\) Simplify by removing the 2\(e^-\) on both sides to obtain the final balanced redox equation: \(2\mathrm{TiCl}_{6}^{2-} + \mathrm{Zn} \rightarrow 2\mathrm{Ti}^{3+} + 12\mathrm{Cl}^{-} + \mathrm{Zn}^{2+}\)

Key concepts

Oxidation Numbers

Oxidation numbers are a helpful way to keep track of the electrons in a chemical reaction, especially in oxidation-reduction, or redox, reactions. Each element in a compound is assigned an oxidation number, which represents the charge it would have if all the bonds were ionic. In our exercise, the role of oxidation numbers is crucial because they help determine which elements undergo oxidation (loss of electrons) and which undergo reduction (gain of electrons). This process involves:

• Identifying elements and calculating their oxidation states based on known oxidation state rules, such as oxygen usually being -2 and uncombined elements being 0.
• Recognizing the increase or decrease in these numbers to identify which species are oxidized and which are reduced in the reaction.

This method is foundational for balancing redox reactions because it allows us to better visualize and balance the loss and gain of electrons.

Half-Reaction Method

The half-reaction method is a systematic approach to balancing oxidation-reduction equations. It involves separating the overall reaction into two halves: oxidation and reduction processes. Each half-reaction shows the electrons lost (oxidation) or gained (reduction). Here's a quick breakdown of how this method is applied:

• First, identify the changes in oxidation numbers to determine oxidation and reduction.
• Next, write the two half-reactions. Make sure to include the electron terms explicitly in these equations.
• Balance each half-reaction for mass and charge, ensuring that the number of electrons lost is equal to the number of electrons gained.

By using the half-reaction method, we simplify complex redox equations by breaking them down into manageable parts, which makes balancing much more straightforward.

Electron Transfer

In every oxidation-reduction reaction, electron transfer is one of the key processes. Redox reactions involve the transfer of electrons between species, and this concept is central to understanding how these reactions proceed. Here’s how electron transfer plays a role:

• During oxidation, an element loses electrons, which are transferred to another substance. The substance losing electrons undergoes an increase in oxidation state.
• In reduction, an element gains electrons, resulting in a decrease in its oxidation state.
• The electrons lost in oxidation must equal the electrons gained during reduction to satisfy the law of conservation of charge.

Understanding electron transfer helps in visualizing the movement of electrons, ensuring that the equations are balanced not only for atoms and mass but also for charge.

Balance Ionic Equations

Balancing ionic equations, particularly redox equations, involves more than simply matching the number of atoms on each side. It requires ensuring that the charges are balanced as well. The steps generally include:

• Using oxidation numbers to figure out which atoms are losing and gaining electrons.
• Breaking the equation into oxidation and reduction half-reactions, indicating the electron transfer in each.
• Balancing the half-reactions separately for mass—using appropriate coefficients—and for charge by adding electrons where needed.
• Combining the balanced half-reactions, making sure electrons cancel each other out, and that both mass and charge are equal on both sides.

This multi-step process requires attention to both the atomic and ionic aspects of the reaction, ensuring that every species in the equation is thoroughly accounted for.