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For the electrolysis of \(\mathrm{CuSO}_{4}\) solution which is/are correct? (a) Cathode reaction: \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) using \(\mathrm{Cu}\) electrode (b) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2}++2 \mathrm{e}^{-}\)using \(\mathrm{Cu}\) electrode (c) Cathode reaction: \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) using Pt electrode (d) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}\) using \(\mathrm{Pt}\) electrode

Short Answer

Expert verified
Correct reactions are (a), (b), and (c).

Step by step solution

01

Understand the setup of electrolysis with CuSO4 solution

In electrolysis, ions in a solution are moved towards electrodes when an electric current is applied. For a solution of \( \mathrm{CuSO}_4 \), you have copper ions \((\mathrm{Cu}^{2+})\) and sulfate ions \((\mathrm{SO}_4^{2-})\) in the solution.
02

Identify the reactions at the cathode using Cu electrode

The cathode is the electrode where reduction occurs. For a Cu electrode, \( \mathrm{Cu}^{2+} \) ions from the solution will gain electrons to form solid copper, which deposits onto the electrode. The reaction is: \( \mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \). This confirms option (a) as correct.
03

Identify the reactions at the anode using Cu electrode

At the anode, oxidation takes place. For a Cu electrode, copper atoms from the electrode lose electrons to form \( \mathrm{Cu}^{2+} \) ions that go into the solution. The reaction is: \( \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \). This confirms option (b) as correct.
04

Examine the possibility of reactions with Pt electrode at the cathode

When using a Pt electrode, the primary reaction at the cathode under acidic conditions would involve hydrogen ions \((\mathrm{H}^+)\) being reduced to form hydrogen gas. The reaction \( 2 \mathrm{H}^+ + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2 \) is accurate and possible since Pt doesn't react with \( \mathrm{CuSO}_4 \). Therefore, option (c) is correct.
05

Evaluate reactions at the anode with Pt electrode

For a Pt anode in \( \mathrm{CuSO}_4 \) solution, typically oxygen gas is produced rather than copper forming \( \mathrm{Cu}^{2+} \). The reaction \( \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \) would not occur with Pt, as Pt is inert. Thus, option (d) is not correct.

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Key Concepts

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

Cathode Reaction
In the electrolysis of copper sulfate solution
  • The cathode is the site of reduction.
  • This is where positive ions gain electrons.
When using a copper (\( \text{Cu} \)) electrode as the cathode, the reaction occurring involves copper ions (\( \text{Cu}^{2+} \)) from the solution: \[\text{Cu}^{2+} + 2 \text{e}^{-} \rightarrow \text{Cu} \]This reaction signifies that these copper ions gain electrons to become solid copper
  • This process results in the deposition of copper on the cathode.
  • It effectively produces a pure copper layer on the electrode.
This type of reaction is advantageous for processes such as copper coating or purification. The stability and effectiveness of the copper electrode in capturing electrons make it an ideal choice for such electrodes.Cathode reactions are critical in shaping how electrolysis functions.
Anode Reaction
During the same electrolysis process, the anode reaction is equally significant.
  • The anode is the site of oxidation.
  • Here, negatively charged ions or elements lose electrons.
For a copper (\( \text{Cu} \)) electrode used as the anode, metallic copper oxidizes by losing electrons, which creates copper ions (\( \text{Cu}^{2+} \)): \[\text{Cu} \rightarrow \text{Cu}^{2+} + 2 \text{e}^{-} \]This transformation suggests that copper atoms on the anode surface dissolve into the solution as copper ions while releasing electrons.
  • This constant exchange is essential to maintaining the solution's copper ion content.
  • It also assists in balancing the ionic equation.
The copper electrolyte's dual role in both electrode reactions ensures the process' smooth continuation. Anode reactions are fundamental for ensuring proper functionality during electrolysis.
Electrode Materials
Electrode choice is vital in determining the reaction's fate in electrolysis. Copper (\( \text{Cu} \)) electrodes are commonly used due to their reactive nature in copper sulfate solutions.
  • Copper electrodes actively participate in both reduction and oxidation.
  • This participation generates efficient transfer and transformation of copper ions.
Conversely, platinum (\( \text{Pt} \)) electrodes are inert and serve a different role.
  • Being chemically stable, they do not participate in reaction.
  • These electrodes are often used if no chemical reaction should occur.
  • They are especially used for their conductive properties.
When used in copper sulfate solution, reactions remain electronic within the solution without altering the electrodes. Thus, the choice between copper and platinum electrodes depends heavily upon the desired outcome of electrolysis. Mastering the understanding of electrodes enables one to predict and control the course of electrolytic reactions effectively.

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Most popular questions from this chapter

The correct order of \(E_{M^{2+} / M}^{0}\) values with negative sign for the four successive elements \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) is \([2010]\) (a) \(\mathrm{Mn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Co}\) (b) \(\mathrm{Cr}>\mathrm{Fe}>\mathrm{Mn}>\mathrm{Co}\) (c) \(\mathrm{Fe}>\mathrm{Mn}>\mathrm{Cr}>\mathrm{Co}\) (d) \(\mathrm{Cr}>\mathrm{Mn}>\mathrm{Fe}>\mathrm{Co}\)

The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \(+1.97 \mathrm{~V}\) respectively. For which one of these metals the change in oxidation state form \(+2\) to \(+3\) is easiest? (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}\) (c) \(\mathrm{Fe}\) (d) \(\mathrm{Co}\)

The electrochemical series is the arrangement of various electrode systems in the increasing order of their standard reduction potentials. It has several important features. On moving from the top to the bottom in the series, tendency to gain electrons, i.e., to get reduced increases. The electrode systems having negative values of standard reduction potentials act as anode when connected to a standard hydrogen electrode, while those having positive values act as cathode. If \(E_{\mathrm{cu}}^{0}{ }_{\mathrm{c}}^{2+}=0.34 \mathrm{~V}\) and \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=0.8 \mathrm{~V}\), predict whether the reaction given below is feasible or not? $$ \mathrm{Cu}^{2}+(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) $$ (a) Not feasible (b) Feasible (c) Feasible on increasing the conc. of \(\mathrm{Ag}^{+}\)(aq). (d) easible at high temp.

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

Two electrochemical cells \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) and \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) are con- nected in series. What will be the net emf of the cell at \(25^{\circ} \mathrm{C} ?\) Given: \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}=-0.73 \mathrm{~V}\), \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}=+0.34 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \mid \mathrm{Fe}=-0.41 \mathrm{~V}\) (a) \(+1.85\) (b) \(-1.85 \mathrm{~V}\) (c) \(+0.83 \mathrm{~V}\) (d) \(-0.83 \mathrm{~V}\)

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