Problem 68 - Chemistry Matter and Change

Question

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

Step-by-Step Solution

Answer

Both equations are balanced. Equation a: \( \mathrm{MoCl}_5 + 4 \mathrm{S}^{2-} \rightarrow \mathrm{MoS}_2 + 5 \mathrm{Cl}^{-} + 3 \mathrm{S} \). Equation b: \( \mathrm{TiCl}_6^{2-} + 2 \mathrm{Zn} \rightarrow \mathrm{Ti}^{3+} + 6 \mathrm{Cl}^{-} + 2 \mathrm{Zn}^{2+} \).

Step 1: Determine Oxidation Numbers for Reactants and Products (Equation a)

Identify the oxidation state of each element in the compounds for the first equation, \( \text{a. } \mathrm{MoCl}_5+\mathrm{S}^{2-} \rightarrow \mathrm{MoS}_2+\mathrm{Cl}^-+\mathrm{S} \).- Mo in \( \mathrm{MoCl}_5 \) and \( \mathrm{MoS}_2 \): Assume Mo is \(+5\) in \( \mathrm{MoCl}_5 \) and changes to \(+4\) in \( \mathrm{MoS}_2 \).- Cl in \( \mathrm{MoCl}_5 \) and \( \mathrm{Cl}^- \): Chlorine goes from \(-1\) in both compounds.- S in \( \mathrm{S}^{2-} \) and \( \mathrm{S} \): Sulfur goes from \(-2\) in \( \mathrm{S}^{2-} \) to 0 in \( \mathrm{S} \).

Step 2: Determine Oxidation Numbers for Reactants and Products (Equation b)

For the second equation, \( \text{b. } \mathrm{TiCl}_6^{2-}+\mathrm{Zn} \rightarrow \mathrm{Ti}^{3+}+\mathrm{Cl}^-+\mathrm{Zn}^{2+} \), determine these oxidation states:- Ti in \( \mathrm{TiCl}_6^{2-} \) to \( \mathrm{Ti}^{3+} \): Assume Titanium goes from \(+4\) to \(+3\).- Zn goes from 0 in metallic form to \(+2\) in \( \mathrm{Zn}^{2+} \).- Cl in \( \mathrm{TiCl}_6^{2-} \) and \( \mathrm{Cl}^- \): Chlorine remains at \(-1\).

Step 3: Identify Oxidation and Reduction Processes (Equation a)

For equation \(a\):- Oxidation: Sulfur goes from \(-2\) to 0.- Reduction: Molybdenum goes from \(+5\) in \( \mathrm{MoCl}_5 \) to \(+4\) in \( \mathrm{MoS}_2 \).

Step 4: Identify Oxidation and Reduction Processes (Equation b)

For equation \(b\):- Oxidation: Zinc goes from 0 to \(+2\).- Reduction: Titanium goes from \(+4\) to \(+3\).

Step 5: Balance the Oxidation and Reduction Equation (Equation a)

Balance the equation by ensuring equal change in oxidation numbers:- S oxidation: Two S atoms are needed because the increase is 2 per atom: \( 2(0 - (-2)) = 4 \).- Mo reduction: The decrease is 1 per atom change: \( 1(5 - 4) = 1 \).- Add Mo: Equation after balancing becomes \( 1 \mathrm{MoCl}_5 + 4 \mathrm{S}^{2-} \rightarrow 1 \mathrm{MoS}_2 + 5 \mathrm{Cl}^{-} + 3 \mathrm{S} \).

Step 6: Balance the Oxidation and Reduction Equation (Equation b)

Balance the equation by ensuring equal change in oxidation numbers:- Zn oxidation: Per atom, increase from 0 to \(+2\) is 2. Add a 1 to \( \mathrm{Zn} \).- Ti reduction: Decrease from \(+4\) to \(+3\) is 1 per atom. The equation becomes \( 1 \mathrm{TiCl}_6^{2-} + 2 \mathrm{Zn} \rightarrow 1 \mathrm{Ti}^{3+} + 6 \mathrm{Cl}^{-} + 2 \mathrm{Zn}^{2+} \).- Satisfy electron transfer: Ensure electrons lost and gained match to achieve \( 2 \mathrm{Zn} \) and \( 6 \) Cl.

Step 7: Verify Balance by Counting Atoms and Charges (Both Equations)

Check the atom balance and charge neutrality for both equations:- Equation \(a\): Each side has 3 sulfur, 5 chlorine, and 1 molybdenum.- Equation \(b\): Each side has 6 chlorine, 1 titanium, and 2 zinc.- Both equations balance for both atoms and charge. No extra charges appear on either side.

Key Concepts

Oxidation-Number Method

The Oxidation-Number Method is a systematic approach used to balance redox reactions. This method involves assigning oxidation numbers to each element in a chemical reaction to identify which elements are oxidized and which are reduced.

• Start by determining the oxidation states for each element in all reactants and products.
• Next, identify the changes in oxidation numbers to pinpoint the oxidation process (increase in oxidation state) and the reduction process (decrease in oxidation state).
• From there, you need to balance the changes in oxidation numbers to reflect an equal number of electrons lost and gained.

Balancing redox equations with this method ensures that the principle of charge conservation is maintained, as the total increase in oxidation number must equal the total decrease.

Balancing Chemical Equations

Balancing chemical equations is crucial for reflecting the conservation of mass and charge. Every chemical equation must have equal numbers of each type of atom on both sides of the reaction. When dealing with redox reactions, the process involves:

• Balancing each half-reaction separately for mass and charge.
• Ensuring that electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction.
• Combining the balanced half-reactions and simplifying to achieve a complete balanced equation.

In the provided equations, after determining the oxidation and reduction processes, adjustments were made to ensure that the total atoms and charges on each side were balanced.

Oxidation and Reduction Processes

Oxidation involves the loss of electrons, leading to an increase in oxidation state, while reduction is the gain of electrons, resulting in a decrease in oxidation state. Identifying these processes is fundamental in redox reactions:- In equation a:
- Sulfur is oxidized as it moves from an oxidation state of \(-2\) to 0.
- Molybdenum is reduced, going from \(+5\) to \(+4\).
- In equation b:
- Zinc experiences oxidation, changing from 0 to \(+2\).
- Titanium undergoes reduction, moving from \(+4\) to \(+3\).
Recognizing these shifts is essential for properly balancing the equations, as the electrons transferred must cancel one another out when the equations are balanced.

Oxidation State Determination

Determining oxidation states is the first step in analyzing a redox reaction. It involves determining the apparent charge an atom would have if all bonds were ionic.

• In most compounds, oxygen is assigned an oxidation state of \(-2\), and hydrogen \(+1\), unless specified otherwise.
• Assign oxidation states based on known tendencies of elements and charges in compounds.For instance, in \(\mathrm{MoCl}_5\), chlorine is \(-1\), allowing us to deduce that molybdenum must be \(+5\) to neutralize the charge.

Understanding these rules and applying them to each element facilitates the identification of oxidized and reduced species in a reaction, paving the way for balancing and verifying the equation's accuracy.