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Chlorine gas can oxidize zinc metal in a reaction that has been suggested as the basis of a battery. Write the half reactions involved. Label which is the oxidation half reaction and which is the reduction half-reaction.

Short Answer

Expert verified
Oxidation: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\); Reduction: \(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\).

Step by step solution

01

Define the Redox Process

In any redox (reduction-oxidation) reaction, one species is oxidized (loses electrons) and another is reduced (gains electrons). In this case, zinc (\(\text{Zn}\)) is oxidized and chlorine (\(\text{Cl}_2\)) is reduced.
02

Write the Oxidation Half-Reaction

The oxidation half-reaction for zinc is:\[\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\].Here, zinc loses two electrons to form zinc ions.
03

Write the Reduction Half-Reaction

The reduction half-reaction for chlorine is:\[\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\].Here, chlorine gas gains two electrons to form two chloride ions.
04

Label the Half-Reactions

- Oxidation Half-Reaction: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\)- Reduction Half-Reaction: \(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\)The zinc loses electrons, so it undergoes oxidation, while chlorine gains electrons and undergoes reduction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Half-Reaction
In chemistry, particularly when exploring redox reactions, the oxidation half-reaction plays a crucial role. This part of a redox reaction involves the loss of electrons from a substance. The substance that loses the electrons is said to be "oxidized." In the context of the zinc and chlorine reaction, zinc (\(\text{Zn}\) is the substance that undergoes oxidation. When zinc atoms lose two electrons, they transform into zinc ions (\(\text{Zn}^{2+}\) ). This change can be represented by the equation:
  • Oxidation Half-Reaction: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\)
The electrons released are then available for other substances in the reaction, such as chlorine in this case. It's important to remember that oxidation always involves a simultaneous reduction process where the lost electrons move to another atom or molecule.
Reduction Half-Reaction
In contrast to the oxidation process, the reduction half-reaction focuses on the gain of electrons. Reduction refers to the process where a substance gains electrons and is therefore said to be "reduced." As part of the zinc-chlorine reaction, chlorine gas (\(\text{Cl}_2\) undergoes reduction. The chlorine molecules gain two electrons, resulting in the formation of two chloride ions (\(2\text{Cl}^-\) ). This transformation is captured by the equation:
  • Reduction Half-Reaction: \(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\)
It's crucial to understand that the electrons consumed in the reduction process were those released during oxidation. This simultaneous process allows for the flow of electrons, which is a fundamental aspect of redox reactions.
Electrochemistry
Electrochemistry is a branch of chemistry that explores the relationship between electricity and chemical reactions. It plays a pivotal part in areas such as energy generation, battery design, and corrosion processes. The principles of electrochemistry are evident in redox reactions, where the transfer of electrons occurs.
This electron transfer is what powers electrochemical cells or batteries. In these cells, the oxidation and reduction reactions take place in separate compartments. The flow of electrons from the oxidation reaction to the reduction site generates an electric current.
The zinc-chlorine reaction is a primary example of how electrochemical principles apply in practical applications, offering potential for energy storage solutions.
Zinc-Chlorine Reaction
The zinc-chlorine reaction is a classic example of a redox reaction. This reaction serves as an illustrative model for understanding electron transfer processes that are fundamental in electrochemistry.
In this reaction, zinc metal interacts with chlorine gas and undergoes oxidation, while chlorine undergoes reduction. The overall chemical equation for this reaction is:
  • Overarching Reaction: \(\text{Zn} + \text{Cl}_2 \rightarrow \text{Zn}^{2+} + 2\text{Cl}^-\)
As zinc transforms into zinc ions and chlorine becomes chloride ions, an exchange of electrons occurs. This simple but powerful redox process can be harnessed in zinc-chlorine batteries, showing the practical interconnectivity of chemical reactions and electricity generation.

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Most popular questions from this chapter

Consider these half-reactions: $$ \begin{array}{lr} \hline \text { Half-reaction } & E^{\circ}(\mathrm{V}) \\ \hline \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(\mathrm{s}) & 1.52 \\ \mathrm{Pt}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}(\mathrm{s}) & 1.118 \\ \mathrm{Co}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{s}) & -0.277 \\ \mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}(\mathrm{s}) & -1.18 \\ \hline \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Co}(\mathrm{s})\) reduce \(\mathrm{Pt}^{2+}(\mathrm{aq})\) to \(\mathrm{Pt}(\mathrm{s})\) ? (f) Will \(\mathrm{Pt}(\mathrm{s})\) reduce \(\mathrm{Co}^{2+}(\mathrm{aq})\) to \(\mathrm{Co}(\mathrm{s})\) ? (g) Which ions can be reduced by Co(s)?

Explain how reactant-favored electrochemical reactions can be induced to make products.

The standard cell potential for the oxidation of \(\mathrm{Mg}\) by $$ \begin{aligned} \mathrm{Br}_{2} \text { is } 3.42 \mathrm{~V} \\ & \mathrm{Br}_{2}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \end{aligned} $$ (a) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for this reaction. (b) Suppose the equation is written with all coefficients doubled. Determine \(\Delta_{\mathrm{r}} G^{\circ}\) and \(E^{\circ}\) for this new equation.

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