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Chapter 20: Problem 60

Describe what happens at the anode and the cathode in the electrolysis of KI (aq).

Short Answer

Expert verified
In the electrolysis of KI(aq), at the anode (positive electrode), iodide ions (I-) undergo oxidation to produce solid iodine (I2) and release two electrons: \[ 2I^- (aq) \rightarrow I_2(s) + 2e^- \]. At the cathode (negative electrode), hydrogen ions (H+) from water undergo reduction, gaining two electrons and forming hydrogen gas (H2): \[ 2H^+ (aq) + 2e^- \rightarrow H_2(g) \]. Overall, the process results in the formation of solid iodine (I2) and hydrogen gas (H2).

Step by step solution

01

Identify the anode and cathode in the electrolysis setup

In electrolysis, the anode is the positive electrode, and the cathode is the negative electrode. In this exercise, we are dealing with the electrolysis of KI (potassium iodide) in an aqueous solution.
02

Write the dissociation of KI in water

When KI dissolves in water, it dissociates into potassium ions (K+) and iodide ions (I-): \[ KI (aq) \rightarrow K^+ (aq) + I^- (aq) \]
03

Identify the reaction at the cathode

At the cathode, the reduction process takes place. In this case, H+ ions from water and K+ ions from KI will compete to undergo reduction. Since H+ ions from water have a higher reduction potential, it will be easier for them to undergo reduction at the cathode. The equation for the process at the cathode is: \[ 2H^+ (aq) + 2e^- \rightarrow H_2(g) \]
04

Identify the reaction at the anode

At the anode, the oxidation process takes place. In this case, I- ions from KI and OH- ions from water will compete to undergo oxidation. Since I- ions have a lower oxidation potential than OH- ions from water, they are more likely to undergo oxidation at the anode. The equation for the process at the anode is: \[ 2I^- (aq) \rightarrow I_2(s) + 2e^- \]
05

Describe the overall process in electrolysis of KI(aq)

In the electrolysis of KI in an aqueous solution, the anode and the cathode play crucial roles in the overall reaction. At the anode (positive electrode), iodide ions (I-) undergo oxidation, producing solid iodine (I2) and releasing two electrons. At the cathode (negative electrode), hydrogen ions (H+) from the water undergo reduction, gaining two electrons and forming hydrogen gas (H2). Overall, the electrolysis of KI in an aqueous solution results in the formation of solid iodine (I2) and hydrogen gas (H2).

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Key Concepts

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

Anode and Cathode
In electrolysis, the terms "anode" and "cathode" refer to two different electrodes involved in the electrical process. These are essentially the sites where the main chemical reactions occur. The anode is the positive electrode. Here, oxidation happens. This means that electrons are lost by ions or molecules.

The cathode, on the other hand, is the negative electrode, where reduction takes place. Reduction involves gaining electrons.

  • At the anode of an electrolysis cell, oxidation of iodide ions results in iodine formation.
  • At the cathode, reduction of hydrogen ions forms hydrogen gas.
Understanding the roles of the anode and cathode is crucial for grasping the concept of electrolysis effectively.
Potassium Iodide
Potassium iodide (KI) plays a fundamental role in the electrolysis process. It is a chemical compound made of potassium (K) and iodine (I) atoms. When dissolved in an aqueous solution, it dissociates into ions.

  • Potassium ions (\( ext{K}^+ \)) and iodide ions (\( ext{I}^- \)) are formed when KI dissolves in water.
  • The dissociation reaction can be represented as: \[ KI (aq) \rightarrow K^+ (aq) + I^- (aq) \]
Here, iodide ions (\( ext{I}^- \)) head to the anode for oxidation whereas hydrogen ions are more favorable for reduction at the cathode.
It is this dissociation that facilitates the electrochemical reactions in the electrolysis of KI.
Oxidation and Reduction
The concepts of oxidation and reduction are central to electrochemical reactions in electrolysis.
- **Oxidation** refers to the loss of electrons. During this process, the substance that loses electrons is known as the reducing agent. At the anode, iodide ions (\( 2I^- \)) lose electrons to form iodine (\( I_2 \)). - **Reduction** involves the gain of electrons. It occurs at the cathode where the substance that gains electrons is termed the oxidizing agent. In this scenario, hydrogen ions (\( 2H^+ \)) gain electrons to form hydrogen gas (\( H_2 \)).

  • Oxidation equation: \[ 2I^- (aq) \rightarrow I_2(s) + 2e^- \]
  • Reduction equation: \[ 2H^+ (aq) + 2e^- \rightarrow H_2(g) \]
Remembering "OIL RIG" ("Oxidation Is Loss, Reduction Is Gain") can help in recalling these important processes.
Electrochemical Reaction
An electrochemical reaction involves the movement of electrons between substances through an external circuit. This reaction is fundamental for electrolysis, a process that occurs in an electrolyte solution.
In the electrolysis of potassium iodide (KI), the reaction occurs in an aqueous solution and includes two half-reactions鈥攐ne at the anode and one at the cathode.

  • A complete electrochemical reaction includes the dissociation of potassium iodide and ends with the formation of iodine and hydrogen gas.
  • These reactions are enabled by the passage of electric current, allowing chemical changes at each electrode.
Overall, by using electrical energy, electrochemical reactions can convert compounds into simpler elements or compounds, offering insights into the interaction of electricity and chemistry.

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Most popular questions from this chapter

Recycling Explain how recycling aluminum conserves energy

Explain why the products of the electrolysis of brine and the electrolysis of molten sodium chloride are different.

A battery is assembled using tin and mercury, which have the following reduction half-reactions: $$ \begin{array}{l}{\mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}} \\\ {\mathrm{Hg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}}\end{array} $$ a. Write a balanced equation for the cell's reaction. b. What is oxidized and what is reduced? Identify the oxidizing agent and the reducing agent. c. Which reaction occurs at the anode? At the cathode? d. What is the cell potential? Use Table 20.1 e. If sodium sulfate solution is in the salt bridge, in which direction do the sulfate ions move?

$$ \mathrm{Hg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(\mathrm{l}) \text { and } \mathrm{Cr}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}) $$

Where does oxidation take place in an electrolytic cell?
See all solutions

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