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Chapter 9: Problem 113

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The product at the anode and cathode are (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}, \mathrm{H}_{2}\)

Short Answer

Expert verified
The products are \( \mathrm{O}_{2} \) at the anode and \( \mathrm{H}_{2} \) at the cathode (option a).

Step by step solution

01

Determine components of the solution

Identify the ions present in the dilute aqueous solution of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \). The solution contains \( \mathrm{Na}^{+} \), \( \mathrm{SO}_{4}^{2-} \), and \( \mathrm{H}_{2} \mathrm{O} \) molecules, which dissociate into \( \mathrm{H}^{+} \) and \( \mathrm{OH}^{-} \) ions.
02

Identify possible reactions at the cathode

At the cathode, where reduction occurs, possible reductions are:1. \( \mathrm{H}_{2} \mathrm{O}(l) + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(g) + 2\mathrm{OH}^{-}(aq) \) (reduction of water to hydrogen gas)2. \( \mathrm{Na}^{+}(aq) + \mathrm{e}^{-} \rightarrow \mathrm{Na}(s) \) (reduction of sodium ions)In aqueous solutions, water is more easily reduced than sodium, thus \( \mathrm{H}_{2} \) gas forms at the cathode.
03

Identify possible reactions at the anode

At the anode, where oxidation takes place, possible oxidations are:1. \( 2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) + 4\mathrm{e}^{-} \) (oxidation of water to oxygen gas)2. \( 2\mathrm{SO}_{4}^{2-}(aq) \rightarrow \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(aq) + 2\mathrm{e}^{-} \) (formation of peroxodisulfate ion)In a dilute solution, the oxidation of water is more feasible. Therefore, the product at the anode is \( \mathrm{O}_{2} \) gas.
04

Conclusion based on both electrodes

The reactions at the electrodes result in the formation of hydrogen gas at the cathode and oxygen gas at the anode. Compare this conclusion with the provided options: The correct option is (a) \( \mathrm{O}_{2}, \mathrm{H}_{2} \).

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Key Concepts

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

Aqueous Solution
An aqueous solution is any solution where water is the solvent.
When \({Na}_{2} \mathrm{SO}_{4}\) (sodium sulfate) is dissolved in water, it dissociates into its constituent ions.
This includes sodium ions \(\mathrm{Na}^{+}\), sulfate ions \(\mathrm{SO}_{4}^{2-}\), hydrogen ions \(\mathrm{H}^{+}\), and hydroxide ions \(\mathrm{OH}^{-}\).
The dissociation of these ions is crucial to conduct electricity, which is a characteristic feature of electrolyzed solutions.
Therefore, a thorough understanding of what makes up an aqueous solution sets the stage for understanding the electrochemical reactions occurring at the electrodes.
Reduction Reactions
Reduction reactions are half of the fundamental processes in electrolysis.
At the cathode, reduction reactions occur where gain of electrons transforms ions to neutral atoms or molecules.
In the electrolysis of aqueous sodium sulfate, one possible reduction reaction is the conversion of water to hydrogen gas: \[ \mathrm{H}_{2} \mathrm{O}(l) + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(g) + 2\mathrm{OH}^{-}(aq) \]
This reaction indicates that water is more readily reduced than the sodium ions in the solution.
As such, hydrogen gas bubbles form and rise from the cathode.
Oxidation Reactions
Oxidation reactions are integral to electrolysis and take place at the anode.
In these reactions, the loss of electrons occurs.
In the electrolysis of a sodium sulfate solution, the common oxidation reaction at the anode is the conversion of water to oxygen gas: \[ 2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) + 4\mathrm{e}^{-} \]
It is more favorable for water to undergo oxidation than for sulfate ions to form peroxodisulfate, especially in a dilute solution.
This results in oxygen gas being produced.
Sodium Sulfate
Sodium sulfate is a salt that plays a passive but essential role during electrolysis in aqueous solutions.
It dissociates completely in water to yield sodium ions \(\mathrm{Na}^{+}\) and sulfate ions \(\mathrm{SO}_{4}^{2-}\).
These ions are not directly involved in the main reduction or oxidation reactions during electrolysis but contribute to the overall ionic conductivity of the solution.
Understanding its role helps grasp the greater context of eaction conditions required for specific outcomes in electrolysis.

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Most popular questions from this chapter

In the reaction \(4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 4 \mathrm{Fe}^{3+}+6 \mathrm{O}_{2}^{2-}\) which of the following statements is incorrect? (a) metallic iron is reducing agent (b) \(\mathrm{Fe}^{3+}\) is an oxidizing agent (c) metallic iron is reduced to \(\mathrm{Fe}^{31}\) (d) redox reaction

The values of standard oxidation potentials of following reactions are given below: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=0.762 \mathrm{~V}\) \(\mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{e} ; E^{\circ}=0.440 \mathrm{~V}\) \(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=-0.345 \mathrm{~V}\) \(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+2 \mathrm{e}^{-} ; E^{\circ}=-0.800 \mathrm{~V}\) Which of the following is most easily reduced? (a) \(\mathrm{Fe}^{2+}\) (b) \(\mathrm{Ag}^{+}\) (c) \(\mathrm{Zn}^{2+}\) (d) \(\mathrm{Cu}^{2+}\)

Oxidation numbers of carbon in \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}\) and diamond respectively are (a) \(+3,4\) and \(+4\) (b) \(+3,-4\) and zero (c) \(+6,+4\) and zero (d) \(+6,+4\) and \(+4\)

A standard hydrogen electrode has zero electrode potential because (a) hydrogen is easiest to oxidize (b) its electrode potential is assumed to be zero (c) hydrogen atom has only one electron (d) hydrogen is the lightest element

A certain current liberates \(0.504 \mathrm{~g}\) of hydrogen in 2 hour. The amount (or mass) of copper deposited at the cathode during the electrolysis of \(\mathrm{CuSO}_{4}\) (aq) solution by the same current flowing for the same time is (atomic masses \(\mathrm{H}=1.0, \mathrm{Cu}=63.5\) ) (a) \(16.0 \mathrm{~g}\) (b) \(32.0 \mathrm{~g}\) (c) \(8.6 \mathrm{~g}\) (d) \(1.60 \mathrm{~g}\)
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