(a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid? (d) Why are active metals such as Al obtained by electrolysis using molten salts rather than aqueous solutions?

Short Answer

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(a) An electrolytic cell is an electrochemical cell where an electric current drives a non-spontaneous redox reaction, causing the separation of elements or compounds. (b) The electrode connected to the negative terminal of a voltage source in an electrolytic cell is the cathode, where the reduction process occurs. (c) Sulfuric acid is added to water during electrolysis to increase the conductivity of the solution, making the process more efficient. (d) Active metals like Al are obtained using molten salts in electrolysis rather than aqueous solutions because molten salts allow for more efficient and effective separation of the metal, leading to a purer end product.

Step by step solution

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(a) Definition of an Electrolytic Cell)

An electrolytic cell is a type of electrochemical cell in which an electric current is passed through a solution containing ions, causing a chemical reaction to occur. The energy from the electric current is used to drive a non-spontaneous redox reaction leading to the separation of elements or compounds.
02

(b) Electrode at Negative Terminal)

In an electrolytic cell, the electrode connected to the negative terminal of the voltage source is the cathode. Here, reduction takes place, meaning that electrons are gained by the ions in the solution. This causes the ions to gain electrons and form neutral atoms or molecules, which then deposit onto the cathode or remain in the solution as reduced species.
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(c) Role of Sulfuric Acid in Electrolysis of Water)

During the electrolysis of water, a small amount of sulfuric acid (H2SO4) is often added to the water to increase the conductivity of the solution. Pure water has a low conductivity due to the low concentration of ions in the solution. Sulfuric acid is a strong electrolyte, meaning it dissociates into ions and improves the flow of electric current through the solution, making the process of electrolysis more efficient.
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(d) Reason for Using Molten Salts in Active Metal Electrolysis)

Active metals like aluminum are obtained through electrolysis using molten salts rather than aqueous solutions because they are highly reactive and tend to form stable compounds with water or other components of the solution. In molten salts, these metals exist as ions without the presence of water or other interfering species. This allows for a more efficient and effective separation of the active metal during the electrolysis process, producing a pure end product. Additionally, in molten salts, there is a higher concentration of the desired ions, which improves the overall efficiency and rate of the electrolysis reaction.

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Most popular questions from this chapter

(a) \(\mathrm{A} \mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of 7.60 \(\mathrm{A} .\) What mass of \(\mathrm{Cr}(s)\) is plated out after 2.00 days? (b) What amperage is required to plate out 0.250 mol Cr from a \(\mathrm{Cr}^{3+}\) solution in a period of 8.00 \(\mathrm{h} ?\)

In the Bronsted-Lowry concept of acids and bases, acid-base reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases? [Sections 20.1 and 20.2\(]\)

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Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. $$ \begin{array}{l}{\text { (a) } \mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s) \text { (acidic solution) }} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text { (acidic solution) }} \\ {\text { (c) } \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)(\text { acidic solution })} \\ {\text { (d) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (acidic solution) }} \\ {\text { (e) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (basic solution) }} \end{array} \\\ {\text { (f) } \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s) \text { (basic solution) }} \\ {\text { (g) } \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) \text { (basic solution) }} $$

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\) . (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

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