Electrolysis of molten sodium chloride
Example quantitative electrolysis problem using molten sodium chloride. Created by Jay.
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- what is the difference between electode and anion(6 votes)
- In terms of electrolysis, an electrode refers to either one of the metal bars used during electrolysis whereas anion is the ion (negatively charged species) present in the solution. For example, Cu^2+ is an ion (cation) whereas Cu metal is the electrode used in electrolysis.(4 votes)
- thanks so much for your video!
can you create more videos of electrolysis ?especially for ionic solutions i am not clear about the overvoltage! thanks in advance.(7 votes)
- Is it realistic to ask how much Cl2 is produced at STP given molten NaCl would be at a far higher temperature and/or pressure than 0oC and 1.0atm? Or was STP just chosen for simplicity?(4 votes)
- The Cl₂ would be produced at a different temperature or pressure than STP. But Cl₂ is a gas. You could have calculated its volume at any reasonable combination of P and T.
But we know that 1 mol Cl₂ has a volume of 22.4 L.
STP is a convenient choice of conditions, because it is directly related to an easily-memorized amount (1 mol) of the gas.(6 votes)
- Can you just use the equation Moles = Current * Time / Faraday's Constant * moles electrons for the kind of problem presented around5:03?(6 votes)
- why do sodium ions gain two electrons instead of one?(2 votes)
- They don't! They gain only one electron.(8 votes)
- Aren't positive ions called cations? Then why does he say "we have liquid sodium ions and liquid chlorine anions"?(3 votes)
- Yes, Na⁺(l) ions are sodium ions and also sodium cations.
It is understood that sodium ions are cations because sodium is a metal.
But Cl⁻(l) ions are chloride ions and also chloride anions.
It is understood that chloride ions are anions because chlorine is a nonmetal.(4 votes)
- How does electrons from chlorine reaches sodium?
Because upto positive terminal its okay as positive terminal is attracting the electrons but what after that , DOES these pass through the electrolyte and if so why aren't they repelled by the negative terminal of the battery?(3 votes)
- Why doesn't sodium react with chlorine gas again ?(2 votes)
- It doesn't react therefore so it can be as a gas(2 votes)
- Why do we write Sodium chloride as "NaCl", why not "ClNa"?(1 vote)
- In ionic compounds, we generally write the cation (positive ion) first and the anion (negative ion) last.(3 votes)
- I am struggling to understand why a concentrated solution of sodium chloride produces Cl bubbles faster than a dilute solution of sodium chloride that produces Oxygen bubbles.
Let's say I add 6 moles of water to 2 moles of sodium chloride to form dilute sodium chloride solution.If I electrolyse this solution, I will eventually get 3 moles of Oxygen gas since 6 moles of oxide ions will react to form Oxygen.
On the other hand, let's say I add 2 moles of water to 6 moles of sodium chloride to form concentrated sodium chloride solution.If I electrolyse this solution, I will eventually get 3 moles of Chlorine gas since 6 moles of chloride ions will react to form Cl2.
I don't understand why it is scenario 2 is faster since the same moles of gas (3 moles) is produced in both scenarios. In the first scenario, the water is in high abundance, so lots of it is delocalised to form oxygen and in the second scenario, the chloride ions are in high abundance so there are lots of Chloride ions that delocalise. Since the same amount of both gases are produced (3 moles each), what makes Cl's reaction occur faster than the O's reaction?(2 votes)
- [Voiceover] Here's a simplified diagram for the electrolysis of molten sodium chloride. So if you melt solid sodium chloride, you get molten sodium chloride. So you get sodium ions, liquid sodium ions, and you get liquid chloride anions. So that's what we have here, we have sodium ions and chloride anions. Remember, in an electrolytic cell, the negative terminal of the battery delivers electrons. In this case, delivers electrons to the electrode on the right. And when those liquid sodium ions come in contact with those electrons, we get a reduction half-reaction, so let me write, this is a reduction half-reaction. Sodium ions gain electrons, and are reduced to form liquid sodium metal. So liquid sodium metal is one of our products. And this would form at the cathode. Remember, reduction occurs at the cathode. Let me draw in some liquid sodium metal here forming on the electrode on the right, forming on the cathodes. The other half-reaction, at our other electrode, we know that our battery draws electrons away from this electrode. So oxidation is occurring at this electrode. This would be the anode. So chloride anions, liquid chloride anions are oxidized to chlorine gas, so we lose two electrons. So chloride anions are oxidized to chlorine gas at the anode, and so we'd have bubbles of chlorine gas forming at this electrode. This is an important reaction in industry. So to form sodium metal, this is a good way to do it. How long does it take to produce 1.00 times 10 to the third kilograms of sodium, so that's a lot of sodium, with a constant current of 3.00 times 10 to the fourth amps. So we have another quantitative electrolysis problem here. This one's a little bit harder than the one we did before, but we're going to start the same way. We're going to start with the definition of current. So, current is equal to charge over time, so Q over t, where charge is in coulombs and time is in seconds. So right now we know the current is three times 10 to the fourth amps. So 3.00 times 10 to the fourth, is equal to the charge over the time. We want to know the time. We want to know how long it takes to make that much sodium. So we're trying to find the time. If we could find the charge, then we could find the time using this equation. So let's do that, let's find the charge, and we know we can start with the mass of sodium here. So we're given 1.00 times 10 to the third kilograms, we need to get that into grams. So what is 1.00 times 10 to the third kilograms in grams? So of course, 1.00 times 10 to the sixth grams. Next, we can find the moles of sodium that we're trying to produce. So if we have grams of sodium, this is grams of sodium, how do we figure out moles of sodium? We could divide by the molar mass, so we divide by the molar mass of sodium, which is 22.99, and that would be grams per mole. If we divide by the molar mass, the grams cancel out, and we would get moles of sodium. So let's get out the calculator. So we have 1.00 times 10 to the sixth, that's how many grams of sodium we have. If we divide that by 22.99, the molar mass of sodium, we get, this would be, let's see, 4.35 times 10 to the one, two, three, four. So 4.35 times 10 to the fourth, let's write that down, 4.35 times 10 to the fourth moles of sodium. So how does that help us? If we have moles of sodium, how does that help us find our charge? Well, we can relate the moles of sodium to the moles of electrons. So let's go back up here to our half-reaction where we have our moles of sodium. So right here, so we're trying to make 4.35 times 10 to the fourth moles of sodium. So how many electrons do we need to do that? Well, the mole ratio is two to two, so you need the same number of moles of electrons, you need 4.35 times 10 to the fourth moles of electrons, and that helps us out. So let me go back down here, we have some more room, and let me write down 4.35. 4.35 times 10 to the fourth, so because of our mole ratios we know this is the same number of moles of electrons that we need. We're trying to find charge, we're trying to find this charge here, and we know we can go from moles of electrons to the total charge using Faraday's constant. Faraday's constant is the charge carried by one mole of electrons. And we know that's 96,500. So there's a charge of 96,500 coulombs for every one mole of electrons. So if we multiply these together, we'll get charge, because the moles of electrons will cancel, and we will get the charge, so let's do that. We have 4.35 times 10 to the fourth, and we're going to multiply that by Faraday's constant, which is... So, times, 96,500, and so we get let's see, that'd be 4.2 times 10 to the, let's see, one, two, three, four, five, six, seven, eight, nine, so 4.20 times 10 to the ninth, let's write that down. 4.20 times 10 to the ninth, that would be coulombs, right? That would be coulombs, that's how much charge we need to make this many moles of sodium. So we have the charge, and so now we can plug that back into here, and solve for the time. So let's do that. We have 3.00 times 10 to the fourth, that was our current, and current is equal to charge over time. So our total charge is 4.20 times 10 to the ninth, and that's all over the time. So to solve for the time, we just need to divide this number, so let's do that, so that's 4.20 times 10 to the ninth, we're going to divide that number by 3.00 times 10 to the fourth, and that should give us our time in seconds, so our time in seconds this would be 1.4 times 10 to the one, two, three, four, five, so 1.4 times 10 to the fifth. So the time in seconds is 1.40 times 10 to the fifth. So we have our time, I guess we could leave the answer like that. But let's convert that to something that's a little bit easier to comprehend. First, let's convert that into minutes. So if you have 1.40 times 10 to the fifth seconds, how many minutes is that? We could just divide that by 60. So if we divide that by 60, how many minutes? That's 2,333 minutes. How many hours is that? We could just divide that by 60, so we could figure out how many hours, so 38.9 hours, approximately, so it would take us approximately 38.9 hours so it would take us 38.9 hours to make, let's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium. We can also figure out how many liters of chlorine gas are produced when we make our sodium. So let's say we're at STP here, so standard temperature and pressure. Remember, standard temperature is zero degrees C, and standard pressure is one atmosphere. So we've already found that we're trying to make 4.35 times 10 to the fourth moles of sodium. So how many moles of chlorine gas will we make? Well, look at our mole ratios here. So this would be a two to one mole ratio. So we can set up a proportion, so we could do sodium to chlorine, so this would be a two to one mole ratio, so for every two moles of sodium that are produced, one mole of chlorine gas is produced. So 2/1 is equal to 4.35 times 10 to the fourth, that's how many moles of sodium we were making, over X, where X represents the moles of chlorine that we would make. So two X is equal to 4.35 times 10 to the fourth, so X is equal to 4.35 times 10 to the fourth divided by two. So if we find 4.35 times 10 to the fourth, we need to divide that by two, and we get 21,750 moles of chlorine, so this would be 21,750 moles of chlorine gas that are produced. We're trying to find liters of chlorine gas that are produced, so one thing we could do, just to make our lives easy, would be to say, we know that one mole of an ideal gas occupies 22.4 liters at STP. So if we have this many moles, and we assume this is an ideal gas, we could find out how many liters that is by multiplying that number by 22.4. So 21,750 moles... if we multiply that by 22.4, then we should get the volume in liters. So let's do it, let's take that number and multiply it by 22.4. And we get 4.87, times 10 to the one, two, three four, five, so we get 4.87 times 10 to the fifth liters of chlorine. So there's another way to do it, if you don't want to take the shortcut, you can actually plug everything into PV is equal to nRT, so you could use piv-nert, so PV is equal to nRT, we're at STP, so standard pressure is one atmosphere, and temperature is zero degrees C, or 273.15 Kelvin. So we can plug those in, so our pressure is one atmosphere, one atmosphere, our temperature would be 273.15, so 273.15 Kelvin. We're trying to find the volume, we're trying to find liters, so we're trying to find V, n would be equal to the number of moles, so that would be 21,750, so we have 21,750 moles. R, R is equal to .0821... And I didn't leave room for the units in there. But the units would cancel out to give us liters for the volume, and if you solve this, if you use your calculator to solve this, you will get the same answer. You will get 4.87 times 10 to the fifth liters, so you plug it all into piv-nert or you could use the shortcut way. Either way, you're going to get the same volume of chlorine produced.