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Copper can reduce silver ion to metallic silver, a reaction that could, in principle, be used in a battery. $$ \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) $$ (a) Write equations for the half-reactions involved. (b) Which half-reaction is an oxidation and which is a reduction? Which half- reaction occurs in the anode compartment and which takes place in the cathode compartment?

Short Answer

Expert verified
Oxidation: \( ext{Cu} \rightarrow ext{Cu}^{2+} \), reduction: \( ext{Ag}^+ \rightarrow ext{Ag} \). Oxidation at anode, reduction at cathode.

Step by step solution

01

Identify the Oxidation Half-Reaction

In the given reaction, inspect copper's role. Copper (Cu) is in its metallic form \(( ext{s})\), and it changes to copper ions \(( ext{Cu}^{2+})\). The half-reaction for this process is: \ \( ext{Cu}( ext{s}) \rightarrow ext{Cu}^{2+}( ext{aq}) + 2 ext{e}^- \). \ This is an oxidation because electrons are being lost.
02

Identify the Reduction Half-Reaction

Next, consider the role of silver ions. Silver ions \(( ext{Ag}^+)\) in solution gain electrons to form metallic silver \(( ext{Ag})\). The half-reaction for this is: \ \( 2 ext{Ag}^+( ext{aq}) + 2 ext{e}^- \rightarrow 2 ext{Ag}( ext{s}) \). \ This is a reduction because electrons are being gained.
03

Determine Oxidation and Reduction Sites

In electrochemistry, the anode is the site of oxidation, and the cathode is the site of reduction. Thus, \( ext{Cu}( ext{s}) \rightarrow ext{Cu}^{2+}( ext{aq}) + 2 ext{e}^- \) occurs at the anode, and \( 2 ext{Ag}^+( ext{aq}) + 2 ext{e}^- \rightarrow 2 ext{Ag}( ext{s}) \) happens at the cathode.

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Key Concepts

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

Oxidation-Reduction Reactions
Oxidation-Reduction reactions, often referred to as redox reactions, are fundamental to the field of electrochemistry. These reactions involve the transfer of electrons between substances, which results in changes to the oxidation states of the reactants. In the context of our exercise, consider the reaction where copper (Cu) interacts with silver ions (Ag鈦). During this reaction: - Copper loses electrons and is oxidized. - Silver ions gain electrons and are reduced. The oxidation of copper leads to the formation of copper ions (Cu虏鈦), while the reduction process turns silver ions into metallic silver (Ag). This exchange of electrons is the hallmark of redox reactions. By studying these reactions, we can gain insights into how energy transfer can be harnessed for technological applications, such as in batteries.
Half-Reaction Method
The half-reaction method is a powerful technique to balance redox reactions. In this method, the overall reaction is split into two component reactions, known as the half-reactions:- **Oxidation Half-Reaction**: Involves the loss of electrons. In our exercise example, copper (Cu) loses electrons to form copper ions (Cu虏鈦). The half-reaction is: \(\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\).- **Reduction Half-Reaction**: Entails the gain of electrons. Here, silver ions (Ag鈦) gain electrons to become metallic silver (Ag). The half-reaction is: \(2\text{Ag}^+(\text{aq}) + 2\text{e}^- \rightarrow 2\text{Ag}(\text{s})\).This method not only helps in balancing complex equations but also breaks down the reaction into more comprehensible parts. Each half-reaction emphasizes the electron transfer, highlighting the fundamental processes of oxidation and reduction separately. Understanding these half-reactions allows chemists to predict how different elements will behave in a chemical reaction.
Anode and Cathode
The concepts of anode and cathode are central to electrochemical cells. These terms refer to the electrodes where the half-reactions occur:- **Anode**: This is where oxidation takes place. In the given exercise, the reaction at the anode can be described as \(\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\). Essentially, the anode is the site of electron loss.- **Cathode**: This is the site of reduction. In our example, the cathode reaction is \(2\text{Ag}^+(\text{aq}) + 2\text{e}^- \rightarrow 2\text{Ag}(\text{s})\). Electrons are gained at the cathode.In an electrochemical cell, electrons flow from the anode to the cathode. This setup can be utilized in batteries where chemical energy is converted to electrical energy. Recognizing the roles of the anode and cathode is essential to grasp how electrochemical processes power devices and systems.

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Most popular questions from this chapter

Balance these redox reactions, and identify the oxidizing agent and the reducing agent. (a) \(\mathrm{FeO}(\mathrm{s})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})\) (b) \(\mathrm{P}_{4}(\mathrm{~s})+\mathrm{Br}_{2}(\ell) \longrightarrow \mathrm{PBr}_{5}(\ell)\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Co}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Co}^{3+}(\mathrm{aq})\) in acidic solution (d) \(\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{Cr}^{3+}(\mathrm{aq})\) in acidic solution (e) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s})\) in basic solution (f) \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (g) \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Explain how product-favored electrochemical reactions can be used to do useful work.

Write half-reactions for these changes: (a) Reduction of \(\mathrm{MnO}_{4}^{-}\) ion to \(\mathrm{Mn}^{2+}\) ion in acid solution (b) Reduction of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ion to \(\mathrm{Cr}^{3+}\) ion in acid solution (c) Oxidation of chlorine gas to \(\mathrm{ClO}^{-}\) ions (d) Reduction of hydrogen peroxide to water in acidic solution (e) Oxidation of nitrous acid to nitrate ions in acidic solution

You light a \(25-W\) light bulb with the current from a \(12-V\) lead-acid storage battery. Calculate how much energy the light bulb utilized after \(1.0 \mathrm{~h}\) of operation. Calculate how many coulombs passed through the bulb. Assume \(100 \%\) efficiency. \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s} .)\)

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