In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO 4 (s) must be converted into PbO 2 (s) and Pb(s). (b) How many coulombs of electrical charge are needed?

Hello. In this problem we are asked if 300 g of tin to nitrate has to be transformed into tin metal and tin four oxide in order to charge an experimental into nitrate battery. How much columns of electrical are needed. We'll begin by writing our two half reactions. So the oxidation half reaction and the reduction half reaction. We'll begin with the reduction half reaction. So we have 10 to nitrate then is reduced to form two metal. And then our oxidation half reaction. We have 10 to nitrate acting with water form or oxide, two nitrates For protons and two electrons. So we're going to combine these two half reactions. So we have two electrons being gained into being lost. So those canceled. The rest combines to form our overall reaction. We have two moles of tin to nitrate than reacting with water to form 10 metal plus 10 4 oxide plus four nitrates plus four your nines. And so we see that we have 10 to nitrate being reduced form 10 and we have 10 to nitrate being oxidized form 10 4 oxide. So half goes to form 10 half goes to form 10 4 oxide. So that means that 150 g. Then half of our 300 g of tin to nitrate. It's gonna be reduced to form 10. And the other half is going to be oxidized 10 4 oxide. In order to calculate the columns. Let's recall then Faraday Faraday is our charge and cool off Permal electrons. That will allow us to go from old electrons, two columns, which is what we're trying to find. So we'll begin with our 150 g 10 to nitrate. We will convert this to moles using the molar mass. The one mole of 10 to nitrate has a mass of two and 42 .7, grams. And then looking at our half reactions, we see that one mole of 10 to nitrate Is associated with two moles of electrons. And then we can make use of their day A mole of electrons and has a charge of 96,500 bombs. So verifying that our units cancel, we have grams of 10 to nitrate cancels molds of 10 to nitrate cancels molds of electron cancels and we're left with columns. This works out to 1.193 times 10 to the five columns. So this is the amount of columns then that is needed to transform our 300 g of 10 to nitrate into and 10 4 oxide to charge our experimental battery. Thanks for watching. Hope this helps

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 146b

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Chapter 19, Problem 146b

In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO₄(s) must be converted into PbO₂(s) and Pb(s). (b) How many coulombs of electrical charge are needed?

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Identify the balanced chemical reaction for the conversion of PbSO_4(s) to PbO_2(s) and Pb(s).

Determine the molar mass of PbSO_4 to convert the given mass (500.0 g) into moles.

Use stoichiometry to find the moles of electrons required for the conversion based on the balanced equation.

Calculate the total charge in coulombs using the relationship between moles of electrons and charge (1 mole of electrons = 96485 C).

Summarize the steps to ensure all conversions and calculations align with the stoichiometry of the reaction.

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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, which is fundamental in processes like charging a battery. In a lead storage battery, lead sulfate (PbSO4) is converted into lead dioxide (PbO2) and lead (Pb) through oxidation and reduction reactions, releasing or consuming electrons in the process.

Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis quantify the relationship between the amount of substance transformed at an electrode and the quantity of electric charge passed through the electrolyte. The first law states that the mass of a substance altered at an electrode is directly proportional to the total electric charge passed, which is crucial for calculating the charge needed to convert PbSO4 to PbO2 and Pb.

Coulombs and Charge Calculation

Coulombs (C) are the standard unit of electric charge in the International System of Units (SI). To determine the total charge required for the conversion of PbSO4 to PbO2 and Pb, one must calculate the moles of electrons involved in the reaction and then use Faraday's constant (approximately 96485 C/mol) to convert moles of electrons to coulombs.

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Electrolysis of a metal nitrate solution M(NO3)2(aq) for 325 min with a constant current of 20.0 A gives 111 g of the metal. Identify the metal ion M2+.

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In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO₄(s) must be converted into PbO₂(s) and Pb(s). (a) Does the reaction represent an electrolytic or galvanic cell?

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In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO₄(s) must be converted into PbO₂(s) and Pb(s). (c) If a current of 500 A is used, how long will it take?

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In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO4(s) must be converted into PbO2(s) and Pb(s). (b) How many coulombs of electrical charge are needed?

Verified Solution

Key Concepts

Electrochemical Reactions

Faraday's Laws of Electrolysis

Coulombs and Charge Calculation

In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO₄(s) must be converted into PbO₂(s) and Pb(s). (b) How many coulombs of electrical charge are needed?