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Chapter 20: Problem 90

(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

Expert verified
(a) An electrolytic cell is an electrochemical cell that uses an external voltage source to drive a non-spontaneous redox reaction, converting electrical energy into chemical energy. (b) The electrode connected to the negative terminal in an electrolytic cell is the cathode, where reduction occurs. (c) In the electrolysis of water, sulfuric acid increases the conductivity of the water by dissociating into ions, making the process more efficient. (d) Active metals like Al are obtained by electrolysis using molten salts rather than aqueous solutions to avoid competition with water for electrons and to improve efficiency due to high ionic conductivity.

Step by step solution

01

(a) Definition of an Electrolytic Cell

An electrolytic cell is a type of electrochemical cell that uses an external voltage source (like a battery) to drive a non-spontaneous redox reaction. In an electrolytic cell, electrical energy is converted into chemical energy, which can be used for purposes such as the decomposition of compounds or the plating of metals.
02

(b) Determining Electrode Type

The electrode connected to the negative terminal of a voltage source in an electrolytic cell is the cathode. This is because the negative terminal of the voltage source supplies electrons, which flow to the electrode. At the cathode, reduction occurs; this is the process in which a substance gains electrons. Conversely, at the anode (connected to the positive terminal), oxidation occurs, where a substance loses electrons.
03

(c) Role of Sulfuric Acid in the Electrolysis of Water

Adding a small amount of sulfuric acid to water for electrolysis helps increase the conductivity of the water. Pure water has very low electrical conductivity, which makes the electrolysis process inefficient. Sulfuric acid, being a strong acid, dissociates into ions when added to water. These ions increase the conductivity of the water, which allows the electrolysis process to occur more efficiently.
04

(d) Use of Molten Salts for Obtaining Active Metals

Active metals such as aluminum are obtained by electrolysis using molten salts rather than aqueous solutions because of the lower reduction potentials of active metals. When attempting to reduce the metal cation in an aqueous solution, water competes with the metal for electrons and gets reduced instead, producing hydrogen gas. The use of molten salts prevents this competition, as there is no water present in the electrolytic cell. Furthermore, molten salts have high ionic conductivity, which allows for more efficient electrolysis.

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Most popular questions from this chapter

(a) What is the definition of the volt? (b) Do all voltaic cells produce a positive cell potential?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) $2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ (b) $2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)$ (c) $2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (a) $\mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)$ (b) $\mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)$ (c) $\mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)$ (d) $2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+ 4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)$

(a) Aluminum metal is used as a sacrificial anode to protect offshore pipelines in salt water from corrosion. Why is the aluminum referred to as a "sacrificial anode"? (b) Looking in Appendix E, suggest what metal the pipelines could be made from in order for aluminum to be successful as a sacrificial anode.

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to $0.15 \mathrm{M}\(, at what concentration of \)\mathrm{I}^{-}$ would the cell have zero potential?
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