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Chapter 20: Problem 74

During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from Zn to \(\mathrm{MnO}_{2} ?\)

Short Answer

Expert verified
(a) Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of Zn consumed 脳 86.94 g/mol = \( \frac{4.50}{65.38} \) moles 脳 86.94 g/mol 鈮� 5.92 g (b) Coulombs transferred = moles of electrons transferred 脳 96,485 C/mol = \( ( \frac{4.50}{65.38} \) moles 脳 2) 脳 96,485 C/mol 鈮� 13294 C

Step by step solution

01

Convert the mass of Zn consumed to moles

To convert the mass of Zn consumed (4.50 g) to moles, we'll use the formula: Moles = mass / molar mass where the molar mass of Zn is 65.38 g/mol. Moles of Zn consumed = \( \frac{4.50}{65.38} \) moles
02

Find the number of moles of \(\mathrm{MnO}_{2}\) reduced

In the alkaline battery, the balanced half-reactions are as follows: Anode: \(Zn(s) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2e^-\) Cathode: \(MnO_{2}(s) + H_{2}O(l) + 2e^- \rightarrow MnOOH(s) + 2OH^-(aq)\) By comparing the anode and cathode half-reactions, we can see that the ratio of Zn to \(\mathrm{MnO}_{2}\) is 1:1. This means that the moles of \(\mathrm{MnO}_{2}\) reduced will be equal to the moles of Zn consumed. Moles of \(\mathrm{MnO}_{2}\) reduced = moles of Zn consumed
03

Convert moles of \(\mathrm{MnO}_{2}\) reduced to mass

To convert the moles of \(\mathrm{MnO}_{2}\) reduced to mass, we'll use the formula: Mass = moles 脳 molar mass where the molar mass of \(\mathrm{MnO}_{2}\) is 86.94 g/mol. Mass of \(\mathrm{MnO}_{2}\) reduced = Moles of \(\mathrm{MnO}_{2}\) reduced 脳 86.94 g/mol
04

Calculate the number of coulombs transferred

Now that we know the number of moles of Zn consumed, we can calculate the number of electrons transferred during the discharge. From the anode half-reaction, we know that for each mole of Zn consumed, 2 moles of electrons are transferred. Therefore: Moles of electrons transferred = moles of Zn consumed 脳 2 To find the number of coulombs transferred, we need to convert the moles of electrons to coulombs using Faraday's constant: Coulombs = moles of electrons transferred 脳 Faraday's constant where Faraday's constant is 96,485 C/mol. Coulombs transferred = moles of electrons transferred 脳 96,485 C/mol

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Key Concepts

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

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes in which the oxidation state of atoms changes. In alkaline batteries, these reactions facilitate the flow of electrons. They consist of two half-reactions: oxidation and reduction.

In the problem, zinc (Zn) undergoes oxidation at the anode, losing electrons to become zinc ions. This is expressed as:
  • Anode Reaction: \(Zn(s) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2e^-\)

At the cathode, manganese dioxide (\(MnO_2\)) is reduced, gaining electrons to form \(MnOOH\):
  • Cathode Reaction: \(MnO_{2}(s) + H_{2}O(l) + 2e^- \rightarrow MnOOH(s) + 2OH^-(aq)\)

Understanding these reactions helps conceptualize how the battery generates electricity through the movement of electrons.
Molar Mass Calculations
Molar mass calculations are essential for converting a substance's mass to moles, which plays a crucial role in stoichiometry.

To find the number of moles of a substance, you use the formula: \[\text{Moles} = \frac{\text{mass}}{\text{molar mass}}\] In the given exercise, zinc has a molar mass of 65.38 g/mol. With 4.50 g of zinc consumed, the moles of zinc are:
  • \(\frac{4.50}{65.38}\) moles of \(Zn\)

Similarly, for \(MnO_2\), knowing its molar mass is 86.94 g/mol allows us to convert the moles to mass. This step ensures accuracy in determining how much material is involved in the reaction.
Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. Within these cells, there are two electrodes: an anode and a cathode.

In an alkaline battery:
  • The **anode** is where oxidation occurs. For the battery in question, zinc serves as the anode, releasing electrons.
  • The **cathode** is where reduction occurs. Here, \(MnO_2\) gains electrons.

This setup allows for the consistent flow of electrons through the external circuit, powering devices. Understanding how these components interact is fundamental to grasping battery operation and effectiveness.
Faraday's Law of Electrolysis
Faraday's Law of Electrolysis connects the amount of substance that undergoes a reaction to the electrical charge applied. It states that the charge in coulombs is proportional to the amount of substance converted at the electrode.

Using Faraday's constant (96,485 C/mol), the number of moles of electrons transferred can be converted into electrical charge:
  • For every mole of \(Zn\) consumed, two moles of electrons are produced.
  • The relationship is expressed as: \[ \text{Coulombs} = \text{moles of electrons} \times 96,485 \text{C/mol} \]

This allows us to determine the overall charge transferred during the battery's discharge, linking the chemical nature of the reaction to its electrical output.

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Most popular questions from this chapter

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?

For a spontaneous reaction \(\mathrm{A}(a q) \rightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(a q),\) answer the following questions: (a) If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text { cell }}^{\circ} ?[\) Section 20.3\(]\)

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\) . (a) Why is an aqueous solution of MgCl_ not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96\(\%\) efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of \(97,000\) A for a period of 24 \(\mathrm{h} ?\)

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{znO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text { red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{V}\) . The overall cell potential is \(+1.35 \mathrm{V}\) . Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\) and \(\mathrm{Fe}^{3+}(a q) /\) \(\mathrm{Fe}^{2+}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\) .
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