Problem 44 - Chemistry The Central Science

Question

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \text { or } \mathrm{Br}_{2}(l)} \\ {\text { (b) } \mathrm{Zn}^{2+}(a q) \text { or } \mathrm{Cd}^{2+}(a q)} \\ {\text { (c) } \mathrm{Cl}^{-}(a q) \text { or } \mathrm{ClO}_{3}(a q)} \\ {\text { (d) } \mathrm{H}_{2} \mathrm{O}_{2}(a q) \text { or } \mathrm{O}_{3}(\mathrm{g})}\end{array} $$

Step-by-Step Solution

Answer

The stronger oxidizing agents for each pair of substances are: (a) Cl₂(g), (b) Cd²⁺(aq), (c) ClO₃⁻(aq), and (d) O₃(g).

Step 1: (a) Cl₂(g) or Br₂(l)

To compare the oxidizing abilities of Cl₂(g) and Br₂(l), we will check their standard reduction potentials in Appendix E. The reaction for chlorine gas is: \[ \mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(a q) \] The \( E° \) for this reaction is \( +1.36 \ \text{V} \). The reaction for bromine liquid is: \[ \mathrm{Br}_{2}(l) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(a q) \] The \( E° \) for this reaction is \( +1.07 \ \text{V} \). Since Cl₂(g) has a higher standard reduction potential than Br₂(l), we can conclude that Cl₂(g) is the stronger oxidizing agent.

Step 2: (b) Zn²⁺(aq) or Cd²⁺(aq)

For this comparison, we will check the standard reduction potentials for Zn²⁺(aq) and Cd²⁺(aq). The reaction for zinc ions is: \[ \mathrm{Zn}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s) \] The \( E° \) for this reaction is \( -0.76 \ \text{V} \). The reaction for cadmium ions is: \[ \mathrm{Cd}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s) \] The \( E° \) for this reaction is \( -0.40 \ \text{V} \). Comparing the standard reduction potentials, Cd²⁺(aq) has a higher value than Zn²⁺(aq), which means Cd²⁺(aq) is the stronger oxidizing agent.

Step 3: (c) Cl⁻(aq) or ClO₃⁻(aq)

Let's compare the standard reduction potentials for Cl⁻(aq) and ClO₃⁻(aq). We already found the reaction for Cl⁻(aq) in part (a). However, oxidation is required in this case, so we need to reverse the reaction and change the sign of \( E° \): \[ \mathrm{2Cl}^{-}(a q) \rightarrow \mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \] with \( E° = -1.36 \ \text{V} \). The reaction for ClO₃⁻(aq) is: \[ \mathrm{2ClO}_{3}^{-}(a q) + 12 \mathrm{H}^{+}(a q) + 10 \mathrm{e}^{-} \rightarrow \mathrm{Cl}_{2}(g) + 6 \mathrm{H}_{2}\mathrm{O}(l) \] The \( E° \) for this reaction is \( +1.50 \ \text{V} \). Since ClO₃⁻(aq) has a higher standard reduction potential than Cl⁻(aq), we can conclude that ClO₃⁻(aq) is the stronger oxidizing agent.

Step 4: (d) H₂O₂(aq) or O₃(g)

Lastly, we will compare the standard reduction potentials for H₂O₂(aq) and O₃(g). The reaction for hydrogen peroxide is: \[ \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2 \mathrm{H}^{+}(a q) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) \] The \( E° \) for this reaction is \( +1.77 \ \text{V} \). The reaction for ozone is: \[ \mathrm{2O}_{3}(g) + 2 \mathrm{e}^{-} \rightarrow 3 \mathrm{O}_{2}(g) \] The \( E° \) for this reaction is \( +2.07 \ \text{V} \). Comparing the standard reduction potentials, O₃(g) has a higher value than H₂O₂(aq), so we can conclude that O₃(g) is the stronger oxidizing agent.

Key Concepts

Standard Reduction Potential

Standard reduction potential is a measurement that tells us how easily a substance gains electrons in comparison to a standard hydrogen electrode (SHE). This potential is represented by the symbol \( E^° \), and is measured in volts (V). A higher \( E^° \) value indicates a greater tendency for a substance to gain electrons, thus being a stronger oxidizing agent.
For example, in the case of chlorine gas \( \mathrm{Cl}_{2}(g) \), it has an \( E^° \) of \(+1.36 \ \text{V}\), meaning it is more likely to gain electrons and act as an oxidizing agent compared to bromine liquid \( \mathrm{Br}_{2}(l) \) with an \( E^° \) of \( +1.07 \ \text{V}\).

When deciding which substance is a stronger oxidizing agent, always compare their standard reduction potentials. The greater the potential, the stronger the oxidizing ability.

• Measured against the standard hydrogen electrode.
• The higher the \( E^° \) value, the more potent the oxidizing agent.
• Helps to predict the direction of redox reactions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons from one substance to another. Every redox reaction is composed of two half-reactions: reduction, where a substance gains electrons, and oxidation, where a substance loses electrons.

For instance, in our exercise, when chlorine gas \( \mathrm{Cl}_{2}(g) \) gains two electrons, it is reduced to two chloride ions \( \mathrm{Cl}^{-}(aq) \). This process shows the reduction half-reaction of the redox process. The corresponding oxidation half would involve a different substance that donates electrons.

Recognizing redox reactions is critical because they are foundational for many chemical processes, such as energy production in batteries and metabolic pathways in living organisms.

• Composed of two processes: reduction and oxidation.
• Essential for energy transformations.
• Involves electron transfer between compounds.

Oxidation and Reduction

Oxidation and reduction are two halves of a redox reaction. Oxidation is the loss of electrons, while reduction is the gain of electrons. Remember this by using the mnemonic: "OIL RIG" - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

When looking at the oxidation of a substance in redox reactions, it is important to know which component is providing the electrons. In our example, \( \mathrm{Zn}^{2+}(aq) \) undergoes reduction by gaining electrons to become metallic zinc \( \mathrm{Zn}(s) \), while a different substance would be oxidized during this process.

The ability for a substance to oxidize or reduce another compound is defined by its standard reduction potential. Therefore, substances with a high standard reduction potential will be excellent at undergoing reduction reactions and will thus be good oxidizing agents. Understanding these processes gives insight into both natural and human-made systems.

• Oxidation = Electron loss.
• Reduction = Electron gain.
• Central to processes like corrosion and respiration.