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). (c) If a current of 500 A is used, how long will it take?

Hello everyone. So in this problem we're gonna talk about the electrochemical process of analyzing titanium. So here we're given to chemical reactions went on the catholic side one on the outside here we're being asked to calculate for the time it takes to produce a dark green titanium oxide film with a thickness of 10.3 or 0.13 micro meters and a square sheet of titanium metal with an edge length of 20 centimeters. The density Of titanium is given as well as the current issues. That's .25 amps. So first, what we can do is utilize these two chemical reactions here we see that some of the products and materials can cancel out. For example, of course our four h plus, since they're on opposite sides of the arrows, those two can actually cancel out as well as our four electrons. So we do this and we take the overall net reaction. This is what we just have left so far starting materials, we have one mole of titanium solid as well as two moles of H liquid. This yields into two moles over H two gas as well as one mole of our titanium oxide. Alright, so now that we have this, we can go ahead and calculate for the volume of titanium oxide film is an ammeter cute. So here again we're calculating for volume and centimeter cubed. Alright, so as you recall the equation for volume is length, times width, times height. So we are already given the height and that is 0.13 micro meters. And then we're seeing that the edge length is 20 centimeters. So that's going to be our length And with So what we're gonna do first is taking the 0.13 micro meters. We're gonna go ahead and convert this into centimeters. I'll do a step by step process. So first I'll take the micro meters and two m. So for every one micro meter this is about 10 to the negative six m. And now that we're in meters can go and go into centimeters. So for every 10 to the negative two m that we have sprays this and put this more clearly. Okay, so we have meters down here then we have one centimeter. Okay, you can see here that the unit cancelation for micro meters cancels as well as the meter. Alright, so that's for one of the parts and then here, of course already said that they link them with is the scene. So we'll just have this at 20 cm and we'll go ahead and simply cube this or square this. So, once I put everything into my calculator, I see that the numerical value that I get is 5.12. I apologize. That's five points to times 10 to the negative three centimeter. Cute. The cube comes from because we have one centimeter here and two centimeters here, of course one and two. That is total of three. All right, so we did get the volume of titanium oxide film and centimeter cubes. Now we can go hot, go on to calculate for the moles of titanium oxide. Let's go ahead, scroll down again. For more space here. So again, calculating for the moles Of TIO two. Alright, so here we're first cackling for the molar mass of our molecule. So for titanium that is 47. g per mole. All these values can be found in european arctic table. And then for auction here, since we have two atoms of auction, each contributes a weight of 16.0 g per mole. We simply just double this value. So once I put that into my calculator I get the value of 79.867 g per mole. Alright, now, using this as a conversion factor for calculating now the moles of titanium oxide, we're starting off with our volume first that is 5.2 times 10 to the negative. Three centimeter. Cute. I'm just gonna go ahead, scroll down. Okay, we're gonna go ahead and use the density of titanium oxide. So here we have 4.23 g for every one centimeter cubes. And now that we have our units in terms of grams, we can go ahead and use our molar mass. So for every one mole of T I 02 we have 79. g. Alright, so again, let's go ahead and check out this unit conversion here, the centimeter cubes will be canceled as well as our grams of titanium oxide. So once I put everything into the calculator here we get the value of 2.7541 times 10 to the negative four are units of course is just going to be what's the remainder? And that is just the molds of titanium oxide and last thing we have to calculate here before we get our final answer is going to be the time. Of course. So here we're finding calculating for time. Alright, so for time we have 18, which is just amps this equals to one column, one. Cool, um per second. So it's just for seconds and then our Faraday's constant Is equal to 964, columns per mole of electrons. So not calculating for time here, we can go ahead and use the moles of titanium oxide that we have. So that is 2.7541 times 10 to the negative four moles of titanium oxide. Here we can go ahead and put this in terms of electrons. So for every one mole I have titanium oxide, we know that we have four moles of electrons because of our two chemical equations on top here, they're going to go ahead and use the Faraday's constant. So for every one mole Of electrons, we have 9000 or 96,485 columns. So I just put C for columns there. Finally using our conversion here That for every 2nd We have 0.25 columns and now we can go from Are seconds and two minutes and we just know that of course for every one minute that we have we have 60 seconds. Alright, so I'm just gonna go ahead and put everything into my calculator here, and once I do so I get the numerical value, if we just round to use two significant figures that he goes to 7.1 units being minutes, let's go ahead and go back to our dimensional analysis here. So we see that the molds of titanium oxide cancels our molds of electronic council. I realized that we did make a little mistake here for this unit conversion, since we have columns on top, of course we need our columns on the bottom for one second. There we go, that makes more sense. Okay, so our columns will cancel, our second will cancel and we do get the remainder of minutes for our units, so the time or T is equal to 7.1 minutes and this right here is going to be my final answer for this problem. Thank you all so much for watching

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 146c

Chapter 19, Problem 146c

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?

Verified step by step guidance

First, write the balanced chemical equation for the conversion of PbSO4(s) to PbO2(s) and Pb(s). The overall reaction involves the reduction of PbSO4 to Pb and the oxidation of PbSO4 to PbO2.

Calculate the molar mass of PbSO4 using the atomic masses of lead (Pb), sulfur (S), and oxygen (O). This will help in determining the number of moles of PbSO4 present in 500.0 g.

Determine the number of moles of electrons required for the conversion. For each mole of PbSO4, two moles of electrons are needed: one for the reduction to Pb and one for the oxidation to PbO2.

Use Faraday's law of electrolysis, which states that the amount of substance transformed in an electrochemical reaction is directly proportional to the quantity of electricity (in coulombs) passed through the substance. The formula is: \( Q = n \times F \), where \( Q \) is the total charge in coulombs, \( n \) is the number of moles of electrons, and \( F \) is Faraday's constant (approximately 96485 C/mol).

Finally, calculate the time required using the formula \( t = \frac{Q}{I} \), where \( t \) is the time in seconds, \( Q \) is the total charge calculated in the previous step, and \( I \) is the current in amperes (500 A in this case).

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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 an electrochemical cell. In the context of a lead storage battery, the conversion of PbSO4 to PbO2 and Pb involves oxidation and reduction processes, where lead ions undergo changes in oxidation states. Understanding these reactions is crucial for calculating the amount of charge transferred during the charging process.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis relate the amount of substance transformed at an electrode to the quantity of electric charge passed through the cell. The first law states that the mass of a substance altered at an electrode is proportional to the total electric charge. This principle is essential for determining how long it will take to convert a specific mass of PbSO4 into PbO2 and Pb, given a constant current.

Current and Time Relationship

The relationship between current (I), charge (Q), and time (t) is described by the equation Q = I × t. This relationship allows us to calculate the time required for a specific charge to flow through the circuit. In this scenario, knowing the current (500 A) and the total charge needed to convert the lead sulfate is necessary to determine how long the charging process will take.