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An aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in water is electrolysed using Pt electrodes.The products at the cathode and anode are respectively: (a) \(\mathrm{H}_{2}, \mathrm{SO}_{2}\) (b) \(\mathrm{O}_{2}, \mathrm{NaOH}\) (c) \(\mathrm{H}_{2}, \mathrm{O}_{2}\) (d) \(\mathrm{O}_{2}, \mathrm{SO}_{2}\)

Short Answer

Expert verified
The products at the cathode and anode are respectively H2 and O2, so the correct answer is (c) H2, O2.

Step by step solution

01

Identify the Electrolysis Reactions

Electrolysis is the process by which ionic substances are decomposed into simpler substances when an electric current is passed through them. Cathode (negative electrode) reaction: Cations are attracted to the cathode, where they undergo reduction (gain electrons). Anode (positive electrode) reaction: Anions are attracted to the anode, where they undergo oxidation (lose electrons).
02

Determine the Cathode Reaction

In aqueous solutions, cations like Na+ (from Na2SO4) compete with the H+ ions from water to get reduced. Since Na+ has a higher reduction potential than H+, H+ ions are preferentially reduced to hydrogen gas (H2).Cathode reaction: 2H2O(l) + 2e- → H2(g) + 2OH-(aq)
03

Determine the Anode Reaction

At the anode, the sulfate ions (SO4^2-) or water will undergo oxidation. It's more energetically favorable for water to be oxidized than sulfate ions.Anode reaction: 2H2O(l) → O2(g) + 4H+(aq) + 4e-
04

Identify the Correct Products

The correct products of the electrolysis of an aqueous solution of Na2SO4 using Pt electrodes are H2 at the cathode and O2 at the anode.

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Key Concepts

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

Electrolysis Reactions
Electrolysis is a fascinating process widely used in various industries, including the extraction and purification of metals, and even in the production of chemical compounds. At its core, electrolysis involves the conversion of electrical energy into chemical change. The process takes place in an electrolytic cell, where an electric current passes through an aqueous solution, or a molten salt, causing chemical reactions at the electrodes.

During electrolysis, positive ions, or cations, migrate to the cathode (the negative electrode) where they gain electrons through a process called reduction. Conversely, negative ions, or anions, move towards the anode (the positive electrode) and lose electrons in a process called oxidation. This movement and subsequent reaction at the electrodes are referred to as electrolysis reactions.

These reactions are not random; they follow specific chemical and physical rules, and the type of substance being electrolysed as well as the type of electrodes used can influence what products are formed. For instance, in the electrolysis of aqueous sodium sulfate ()
Reduction and oxidation
Reduction and oxidation, commonly known as redox reactions, are chemical processes that involve the transfer of electrons between substances. In the context of electrolysis, these reactions occur at the two electrodes and are responsible for the transformation of the ions in solution into new substances.

Let's simplify these concepts:
  • Reduction is the gain of electrons by an atom or ion. It occurs at the cathode in an electrolytic cell, where electrons from the external circuit enter the aqueous solution and are acquired by the cations. This is what happens to the hydrogen ions ()
Standard electrode potentials
Understanding standard electrode potentials is integral for predicting and explaining the outcomes of electrolysis. These potentials, measured in volts, are a measure of the driving force behind an electrochemical reaction and indicate how easily an atom or ion can gain or lose electrons. The more positive the standard electrode potential of a species, the more likely it is to be reduced (gain electrons). Conversely, a more negative potential implies a greater tendency to lose electrons (oxidation).

In electrolysis, the standard electrode potentials help us determine which ions are preferentially reduced or oxidized. For example, in the solution of sodium sulfate ()

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Most popular questions from this chapter

The electrolytic decomposition of dilute sulphuric acid with platinum electrode, cathodic reaction is : (a) Reduction of \(\mathrm{H}^{+}\) (b) Oxidation of \(\mathrm{SO}_{4}^{2-}\) (c) Reduction \(\mathrm{SO}_{3}^{2-}\) (d) Oxidation of \(\mathrm{H}_{2} \mathrm{O}\)

Given the listed standard electrode potentials, what is \(E^{\circ}\) for the cell : \(4 \mathrm{BiO}^{+}(a q)+3 \mathrm{~N}_{2} \mathrm{H}_{5}^{+}(a q) \longrightarrow 4 \mathrm{Bi}(s)+3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+7 \mathrm{H}^{+}(a q)\) \(\mathrm{N}_{2}(g)+5 \mathrm{H}^{+}(a q)+4 e^{-} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q), E^{\circ}=-0.23 \mathrm{~V}\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow \mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l), \quad E^{\circ}=+0.32 \mathrm{~V}\) (a) \(+0.55\) (b) \(+0.34\) (c) \(+1.88\) (d) \(+0.09\)

For the electrochemical cell \(\operatorname{Pt}(s) \mid \begin{aligned}&\mathrm{H}_{2}(g)\left|\mathrm{H}^{+}(1 M) \| \mathrm{Cu}^{2+}(1 M)\right| \mathrm{Cu}(s), \text { which one of the } \\\&1 \text { atm }\end{aligned}\) following statements is true ? (a) \(\mathrm{H}_{2}\) liberated at anode and \(\mathrm{Cu}\) is deposite at cathode. (b) \(\mathrm{H}_{2}\) liberated at cathode and \(\mathrm{Cu}\) is deposite at anode. (c) Oxidation occurs at cathode. (d) Reduction occurs at anode.

The electric conduction of a salt solution in water depends on the : (a) Size of its molecules (b) Shape of its molecules (c) Size of solvent molecules (d) Extent of its ionization

How many Faradays are required to reduce \(0.25 \mathrm{~g}\) of \(\mathrm{Nb}(\mathrm{V})\) to the metal? (Atomic weight : \(\mathrm{Nb}=93 \mathrm{~g}\) ) (a) \(2.7 \times 10^{-3}\) (b) \(1.3 \times 10^{-2}\) (c) \(2.7 \times 10^{-2}\) (d) \(7.8 \times 10^{-3}\)

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