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Chapter 21: Problem 32

(a) Write equations for the half-reactions that occur at the cathode and the anode when an aqueous solution of KCl is electrolyzed. Which chemical species is oxidized, and which chemical species is reduced in this reaction? (b) Predict the products formed when an aqueous solution of CsI is electrolyzed.

Short Answer

Expert verified
KCl electrolysis: Cl鈦� is oxidized, H鈧侽 is reduced. Products: H鈧� and Cl鈧�. CsI electrolysis: Products: H鈧� and I鈧�.

Step by step solution

01

Identify the Reaction Components for KCl Electrolysis

In the electrolysis of an aqueous solution of KCl, the major ions present are K鈦�, Cl鈦�, H鈧侽, H鈦�, and OH鈦� from water. These ions can participate in either oxidation or reduction reactions at the electrodes.
02

Write the Cathode Half-Reaction for KCl

At the cathode, reduction occurs. In an aqueous solution, water is usually reduced in preference to alkali metal ions like K鈦�. Thus, the cathode half-reaction involves the reduction of water: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \] This indicates water is reduced to hydrogen gas and hydroxide ions.
03

Write the Anode Half-Reaction for KCl

At the anode, oxidation takes place. Chloride ions (Cl鈦�) are oxidized instead of hydroxide ions because chloride has a lower oxidation potential. The half-reaction at the anode is: \[ 2Cl^- \rightarrow Cl_2 + 2e^- \]This indicates that chloride ions are oxidized to chlorine gas.
04

Determine Oxidized and Reduced Species in KCl Electrolysis

From the half-reactions, water (H鈧侽) is reduced (gaining electrons) at the cathode, and chloride ions (Cl鈦�) are oxidized (losing electrons) at the anode. Therefore, Cl鈦� is oxidized, and H鈧侽 is reduced.
05

Identify the Reaction Components for CsI Electrolysis

For the electrolysis of an aqueous solution of CsI, the major ions present are Cs鈦�, I鈦�, H鈧侽, H鈦�, and OH鈦�. These ions participate in electrode reactions similar to those in the KCl solution.
06

Write the Cathode Half-Reaction for CsI

As in KCl electrolysis, water undergoes reduction at the cathode in preference to cesium ions, so the cathode half-reaction is: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]Water is reduced to hydrogen gas and hydroxide ions.
07

Write the Anode Half-Reaction for CsI

Iodide ions (I鈦�) are oxidized at the anode rather than hydroxide due to their lower oxidation potential. The half-reaction is: \[ 2I^- \rightarrow I_2 + 2e^- \]Iodide ions are oxidized to iodine gas.
08

Predicting the Products for CsI Electrolysis

From the half-reactions, the products formed are hydrogen gas (H鈧�) at the cathode and iodine (I鈧�) at the anode.

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Key Concepts

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

Half-Reactions
In electrolysis, a half-reaction takes place at each electrode. These are the basic steps of the larger reaction. Each half-reaction shows either a gain or loss of electrons.
A typical setup involves two electrodes: a cathode and an anode.
At the cathode, reduction happens, meaning the substance gains electrons. At the anode, oxidation occurs, which means a substance loses electrons.
This split into two processes helps us understand what happens to electrons during the reaction.
Oxidation
Oxidation is when a species loses electrons. This always occurs at the anode during electrolysis.
In our examples, chloride ions (Cl鈦�) and iodide ions (I鈦�) undergo oxidation.
Here's how it works:
  • Chloride ions transform into chlorine gas (\(2Cl^- \rightarrow Cl_2 + 2e^-\)).
  • Iodide ions turn into iodine gas (\(2I^- \rightarrow I_2 + 2e^-\)).
Oxidation involves breaking down electrons from ions into gas forms, releasing electrons in the process.
Reduction
Reduction refers to the gain of electrons. This takes place at the cathode.
During reduction, molecules or ions gain electrons and transform into different compounds.
  • In our situations, water is reduced to hydrogen gas and hydroxide ions (\(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\)).
Reduction is essential because it balances out the electron transfer in the electrolysis process. It's the counterpart to oxidation.
Cathode
The cathode is where reduction occurs during electrolysis. At this electrode, substances gain electrons.
In both KCl and CsI electrolysis, water is reduced at the cathode in favor of the metal ions present.
This results in the formation of hydrogen gas and hydroxide ions:
  • \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\)
The cathode is vital for transforming materials, playing a key role in the completion of the circuit.
Anode
The anode is the electrode where oxidation happens. Here, species lose electrons.
In the electrolysis of both KCl and CsI solutions:
  • Cl鈦� ions at the anode form chlorine gas (\(2Cl^- \rightarrow Cl_2 + 2e^-\)).
  • I鈦� ions become iodine gas (\(2I^- \rightarrow I_2 + 2e^-\)).
The anode's role is crucial as it initiates the electron flow by removing electrons from the reaction species. It helps in forming gaseous products from ions.

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Most popular questions from this chapter

"Aerated" concrete bricks are widely used building materials. They are obtained by mixing gasforming additives with a moist mixture of lime, cement, and possibly sand. Industrially, the following reaction is important:\(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) \(3 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g})\) Assume the mixture of reactants contains \(0.56 \mathrm{g}\) of Al (as well as excess calcium hydroxide and water) for each brick. What volume of hydrogen gas do you expect at \(26^{\circ} \mathrm{C}\) and a pressure of \(745 \mathrm{mm} \mathrm{Hg} ?\)

Calcium reacts with hydrogen gas at \(300-400^{\circ} \mathrm{C}\) to form a hydride. This compound reacts readily with water, so it is an excellent drying agent for organic solvents. (a) Write a balanced equation showing the formation of calcium hydride from Ca and \(\mathrm{H}_{2}\). (b) Write a balanced equation for the reaction of calcium hydride with water (Figure 21.7 ).

Which of the first 10 elements in the periodic table are found as free elements in the Earth's crust? Which elements in this group occur in the Earth's crust only as part of a chemical compound?

A major use of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) is in steam boilers in power plants. (a) The reaction of hydrazine with \(\mathrm{O}_{2}\) dissolved in water gives \(\mathrm{N}_{2}\) and water. Write a balanced equation for this reaction. (b) \(\mathrm{O}_{2}\) dissolves in water to the extent of \(0.0044 \mathrm{g}\) in \(100 .\) mL of water at \(20^{\circ}\) C. What mass of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is needed to consume all of the dissolved \(\mathrm{O}_{2}\) in \(3.00 \times 10^{4} \mathrm{L}\) of water (enough to fill a small swimming pool)?

Place the following oxides in order of increasing basicity: \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{Al}_{2} \mathrm{O}_{3,} \mathrm{SiO}_{2}, \mathrm{SO}_{3}\)
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