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Chapter 19: Problem 79

The products formed in the electrolysis of aqueous \(\mathrm{CuSO}_{4}\) are \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) Write equations for the anode and cathode reactions.

Short Answer

Expert verified
Cathode: \( \mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}(s) \); Anode: \( 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4\mathrm{e}^- \).

Step by step solution

01

Identify the Electrolyte Solutions and Electrolyte Components

Electrolysis involves breaking down the components of an electrolyte solution. Here, we have an aqueous solution of copper sulfate, \( \mathrm{CuSO}_4 \). In the solution, this dissociates into \( \mathrm{Cu}^{2+} \) ions and \( \mathrm{SO}_4^{2-} \) ions. Additionally, the water (\( \mathrm{H}_2\mathrm{O} \)) in the solution can also dissociate into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions.
02

Determine the Cathode Reaction

At the cathode, reduction occurs. The \( \mathrm{Cu}^{2+} \) ions in the solution are reduced to solid copper. This can be written as: \[ \mathrm{Cu}^{2+}(aq) + 2\mathrm{e}^- \rightarrow \mathrm{Cu}(s) \] Here, copper ions gain two electrons to become copper metal.
03

Determine the Anode Reaction

At the anode, oxidation occurs. In this scenario, water is more readily oxidized than sulfate ions. The oxidation of water at the anode forms oxygen gas, and can be described by the equation: \[ 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \] Thus, water molecules lose electrons and form oxygen gas.

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Key Concepts

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

Anode Reaction
The anode is where oxidation takes place during electrolysis. In the case of the electrolysis of aqueous copper sulfate \( \mathrm{CuSO}_4 \), it is crucial to identify which species undergoes oxidation. In our scenario:
  • The potential candidates for oxidation are the sulfate ion \( \mathrm{SO}_4^{2-} \) and water \( \mathrm{H}_2\mathrm{O} \).
  • However, water is more easily oxidized than the sulfate ions.
As a result, water at the anode leads to the production of oxygen gas. The balanced reaction occurring at the anode is given by:\[ 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \] In this reaction:
  • Each water molecule loses electrons.
  • Oxygen gas \( \mathrm{O}_2 \) is generated.
  • Four hydrogen ions \( \mathrm{H}^+ \) are produced.
Cathode Reaction
At the cathode, the process of reduction occurs. This means that ions gain electrons, resulting in the formation of a solid element. In the electrolysis of \( \mathrm{CuSO}_4 \) solution, the reduction process taking place involves the \( \mathrm{Cu}^{2+} \) ions:
  • Copper(II) ions, \( \mathrm{Cu}^{2+} \), are present in the solution.
  • These ions pick up electrons from the cathode.
This results in the formation of solid copper metal. The equation representing the cathodic reaction is:\[ \mathrm{Cu}^{2+}(aq) + 2\mathrm{e}^- \rightarrow \mathrm{Cu}(s) \]Breaking this down:
  • Two electrons are gained by each \( \mathrm{Cu}^{2+} \) ion.
  • This electron gain converts the ions into copper metal \( \mathrm{Cu} \).
Through this process, the beautiful reddish-brown copper metal appears at the cathode.
Electrolyte Solution
An electrolyte solution is a key component in the process of electrolysis. It consists of a solvent and an electrolyte, which is a substance that, when dissolved, results in an electrically conductive solution. For our particular example with aqueous \( \mathrm{CuSO}_4 \):
  • The solvent is water \( \mathrm{H}_2\mathrm{O} \).
  • The solute producing the electrolyte is copper sulfate \( \mathrm{CuSO}_4 \).
  • The resulting solution can conduct electricity due to the presence of free ions.
When copper sulfate dissolves in water:
  • \( \mathrm{CuSO}_4 \) dissociates into copper ions \( \mathrm{Cu}^{2+} \) and sulfate ions \( \mathrm{SO}_4^{2-} \).
  • The water itself dissociates slightly into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions.
These ions in the solution allow for the flow of electrical current, facilitating the electrochemical reactions at boththe anode and the cathode. This is why electrolyte solutions are crucial for carrying out electrolysis effectively.

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Most popular questions from this chapter

Predict the products formed in the electrolysis of an aqueous solution of \(\mathrm{CdSO}_{4}\)

Electrolysis of a solution of \(\mathrm{CuSO}_{4}\) (aq) to give copper metal is carried out using a current of 0.66 A. How long should electrolysis continue to produce \(0.50 \mathrm{g}\) of copper?

An old method of measuring the current flowing in a circuit was to use a "silver coulometer." The current passed first through a solution of \(\mathrm{Ag}^{+}(\mathrm{aq})\) and then into another solution containing an electroactive species. The amount of silver metal deposited at the cathode was weighed. From the mass of silver, the number of atoms of silver was calculated. Since the reduction of a silver ion requires one electron, this value equaled the number of electrons passing through the circuit. If the time was noted, the average current could be calculated. If, in such an experiment, 0.052 g of Ag is deposited during \(450 \mathrm{s},\) what was the current flowing in the circuit?

In the presence of oxgyen and acid, two half. reactions responsible for the corrosion of iron are $$ \begin{array}{c} \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \\ \mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{array} $$ Calculate the the standard potential, \(E^{\circ},\) and decide whether the reaction is product-favored at equilibrium. Will decreasing the pH make the reaction less thermodynamically product-favored at equilibrium?

A Write balanced equations for the follow. ing reduction half-reactions involving organic compounds. (a) \(\mathrm{HCO}_{2} \mathrm{H} \rightarrow \mathrm{CH}_{2} \mathrm{O}\) (acid solution) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) (acid solution) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \rightarrow\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (acid solution) (d) \(\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{4}\) (acid solution)
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