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Ch.20 - Electrochemistry
All textbooksBrown 15th EditionCh.20 - ElectrochemistryProblem 78b
Chapter 20, Problem 78b

During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (b) How many coulombs of electrical charge are transferred from Zn to MnO2?

Verified step by step guidance
1
insert step 1> Determine the molar mass of Zn (zinc) from the periodic table, which is approximately 65.38 g/mol.
insert step 2> Calculate the number of moles of Zn consumed using the formula: \( \text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} \).
insert step 3> Write the half-reaction for the oxidation of Zn at the anode: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \). This shows that each mole of Zn produces 2 moles of electrons.
insert step 4> Calculate the total moles of electrons transferred using the stoichiometry from the half-reaction: \( \text{moles of electrons} = 2 \times \text{moles of Zn} \).
insert step 5> Convert the moles of electrons to coulombs using Faraday's constant (\( F = 96485 \text{ C/mol} \)): \( \text{coulombs} = \text{moles of electrons} \times F \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrochemical Reactions

Electrochemical reactions involve the transfer of electrons between chemical species, typically occurring in galvanic cells like batteries. In an alkaline battery, zinc (Zn) is oxidized at the anode, releasing electrons that flow through the circuit to reduce manganese dioxide (MnO2) at the cathode. Understanding these reactions is crucial for calculating the charge transferred during the battery's operation.
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Faraday's Law of Electrolysis

Faraday's Law states that the amount of substance transformed at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the system. This law allows us to calculate the total charge (in coulombs) based on the moles of electrons involved in the reaction, which is essential for determining how much charge is transferred when zinc is consumed in the battery.
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Molar Mass and Moles

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole. For zinc, the molar mass is approximately 65.38 g/mol. By converting the mass of zinc consumed (4.50 g) into moles, we can determine the number of electrons transferred during its oxidation, which is necessary for calculating the total charge using Faraday's Law.
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Related Practice
Textbook Question

A voltaic cell is constructed that is based on the following reaction: Sn2+(aq) + Pb(s) → Sn(s) + Pb2+(aq) (a) If the concentration of Sn2+ in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell?

Textbook Question

During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO4(s). (a) What mass of PbO2(s) is reduced at the cathode during this same period?

Textbook Question

During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO4(s). (b) How many coulombs of electrical charge are transferred from Pb to PbO2?

Textbook Question

Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag2CrO4(s) → Li2CrO4(s) + 2 Ag(s) (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode?

Textbook Question

Heart pacemakers are often powered by lithium–silver chromate 'button' batteries. The overall cell reaction is 2 Li(s) + Ag2CrO4(s) → Li2CrO4(s) + 2 Ag(s) (b) Choose the two half-reactions from Appendix E that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions?

Textbook Question

In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries. The overall cell reaction for this relatively new battery is: 2 H2O(l) + 2 NiO(OH)(s) + Zn(s) → 2 Ni(OH)2(s) + Zn(OH)2(s) (b) What is the anode half-reaction?