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Chapter 20: Problem 86

(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?

Short Answer

Expert verified
(a) An electrolytic cell is a device in which electrical energy is used to drive a non-spontaneous chemical reaction, causing a chemical change that would not otherwise occur on its own. (b) The electrode connected to the negative terminal of a voltage source in an electrolytic cell is the cathode. This is because the negative terminal provides electrons, making the connected electrode the site where reduction occurs. (c) The role of sulfuric acid in the electrolysis of water is to act as an electrolyte. It enhances the conductivity of water, as it dissociates into ions when mixed with water. These ions facilitate the movement of charge through the solution, allowing for more efficient electrolysis.

Step by step solution

01

Define an electrolytic cell

An electrolytic cell is a device in which electrical energy is used to drive a non-spontaneous chemical reaction. In other words, it is an electrochemical cell where electricity is utilized to cause a chemical change that would not otherwise occur on its own.
02

Determine whether the electrode connected to the negative terminal is an anode or cathode

The electrode connected to the negative terminal of a voltage source in an electrolytic cell is the cathode. In an electrolytic cell, reduction occurs at the cathode, which means electrons are gained by the species that are being reduced. Since the negative terminal of the voltage source provides electrons, the electrode connected to it becomes the site where reduction occurs, making it the cathode.
03

Explain the role of sulfuric acid in the electrolysis of water

The role of sulfuric acid in the electrolysis of water is to act as an electrolyte. Pure water is a poor conductor of electricity, but adding a small amount of sulfuric acid enhances its conductivity. This is because sulfuric acid dissociates into ions when mixed with water, which increases the concentration of ions in the solution. The presence of these ions facilitates the movement of charge through the solution, allowing electrolysis to occur more efficiently.

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Key Concepts

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

Understanding Electrochemistry
Electrochemistry is a branch of chemistry that studies the interrelation of electrical currents and chemical reactions. It involves analyzing both the chemical changes that produce electrical energy and the use of electrical energy to induce chemical changes.

An electrolytic cell, such as the one mentioned in the exercise, serves as a vital component in electrochemistry. It employs electrical energy to compel a non-spontaneous reaction, which means it drives a reaction that wouldn't occur naturally without an external voltage supply. Understanding the principles of electrochemistry is essential for a wide range of applications, including the recharging of batteries, electroplating, and the production of various chemicals through electrochemical processes.
Deciphering Electrode Reactions
Electrode reactions are fundamental to the function of electrolytic cells. An electrolytic cell consists of two electrodes: the anode and the cathode. Each electrode is a site where specific chemical reactions occur upon the introduction of electrical energy.

At the cathode, the reduction process takes place, where molecules or ions gain electrons. As pointed out in the solution, the electrode connected to the negative terminal of a power source in an electrolytic cell is the cathode, and thus, the site of reduction. Conversely, at the anode, oxidation occurs鈥攎olecules or ions lose electrons. These electrode reactions are guided by the flow of electrons from the external circuit into the cell and play a critical role in the electrolytic chemical transformation that occurs.
Chemical Electrolysis Explained
Chemical electrolysis is a process where electrical energy is utilized to cause a chemical reaction, specifically the separation of elements from their naturally occurring sources or compounds. In the context of the given problem, water electrolysis involves breaking down water (H2O) into hydrogen (H2) and oxygen (O2) gas.

As mentioned in the solution, sulfuric acid acts as an electrolyte in the electrolysis of water. Its dissociation into hydrogen and sulfate ions increases the ionic conductivity of water, facilitating the passage of electric current. This is crucial, as pure water has a low concentration of ions making it a poor conductor. By enhancing conductivity, sulfuric acid ensures that the electrical energy efficiently induces the desired chemical reaction during electrolysis.

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Most popular questions from this chapter

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q) \pm \mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-{ }^{-\rightarrow} \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q)-\cdots\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

(a) How many coulombs are required to plate a layer of chromium metal \(0.25 \mathrm{~mm}\) thick on an auto bumper with a total area of \(0.32 \mathrm{~m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in \(10.0 \mathrm{~s} ?(\mathrm{c})\) If the external source has an emf of \(+6.0 \mathrm{~V}\) and the electrolytic cell is \(65 \%\) efficient, how much electrical power is expended to electroplate the bumper?

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions. (Section 19.7) At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) : $$ \begin{aligned} \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-}--\rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-}--\rightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\mathrm{o}}=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(C y F e^{2+}\) by air? (b) If the synthesis of \(1.00\) mol of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ}\), how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2}\) ?

The capacity of batteries such as thetypical AA alkaline battery is expressed in units of milliamp-hours (mAh). An "AA" alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

Using the standard reduction potentials listed in Appendix \(\mathrm{E}\), calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q)-\rightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{Co}(s)+2 \mathrm{H}^{+}(a q)-\cdots \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)\) (c) \(10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q)-\cdots\) \(2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)\)
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