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### Course: MCAT > Unit 9

Lesson 14: Electrochemistry- Electrochemistry questions
- Electrochemistry
- Redox reaction from dissolving zinc in copper sulfate
- Introduction to galvanic/voltaic cells
- Electrodes and voltage of Galvanic cell
- Shorthand notation for galvanic/voltaic cells
- Free energy and cell potential
- Standard reduction potentials
- Voltage as an intensive property
- Using reduction potentials
- Spontaneity and redox reactions
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential edited
- Nernst equation
- Using the Nernst equation
- Concentration cell
- Introduction to electrolysis
- Quantitative electrolysis
- Electrolysis of molten sodium chloride edited
- Lead storage battery
- Nickel-cadmium battery

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# Electrolysis of molten sodium chloride edited

Learn how electrolysis works with molten sodium chloride. See how a negative battery terminal provides electrons, causing a reduction half reaction. Understand that chloride anions oxidize to chlorine gas at the anode. Find out how to calculate the time needed to produce a certain amount of sodium using a constant current. Grasp the connection between moles of sodium and moles of electrons, and how to figure out the volume of chlorine produced.

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video. Created by Jay.

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video. Created by Jay.

## Want to join the conversation?

- At7:05, isn't the standard temperature 25° C?(5 votes)
- No, STP (standard temperature and pressure) is 0° C and 1 atm.

You might be mixing it up with the conditions for calculating standard free energy and cell potential (G=-nFE) where the standard conditions is 25° C.

It's really confusing but STP isn't the same as standard state conditions!

http://chemistry.about.com/od/gas2/f/What-Is-The-Difference-Between-Standard-Conditions-And-Standard-State.htm(9 votes)

- To make things easier, the electrodeposition equation would help. http://schoolbag.info/chemistry/mcat_2/89.html I hope this helps, just be careful with the value of n which isn't the number of electrons but the number of electrons per mole of substance. (The link explains it).(5 votes)
- When the moles of Na were calculated, how come, they weren't multiplied by 2 to account for the 2 moles of electrons transferred, when finding Q?(2 votes)
- If you look at the half reaction for Sodium (Na+ + e- --> Na(s)) there is a 1:1 mole ratio for moles of sodium to moles of electrons. For every mole of solid sodium produced 1 mole of electrons is consumed.(5 votes)

## Video transcript

- [Instructor] 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 Chlorine 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 drawn in some
liquid Sodium metal here forming on the electrode on the right, forming on the cathode. The other half reaction,
right at our other electrode, we know that their battery
draws electrons away from this electrode, so oxidation is occurring
at this electrodes. This would be the anode. So, 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, all right? 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 gonna start the same way, we're gonna 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 could 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. Zero times 10 to the sixth. All right, that's how many
grams of Sodium we have. If we divide that by 22.99,
the molar mass of Sodium, we get 4.35 times 10 to
the fourth Moles of Sodium. So how does that help us, right? 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. All right, 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
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. Right, 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, 'cause the Moles of electrons will cancel and we will get the charge, so we get 4.20 times 10 to the ninth, that would be Coulombs. That's how much charge we need to make this many Moles of Sodium. All right, 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. 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 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 into something that's a little bit easier
to comprehend, 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 are 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. This'd 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, two over one 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, 2x 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 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
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, 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 can use "Pivnert." 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, our temperature would be 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, all right? So that would be 21,750,
so we have 21,750 Moles. R is equal to 0.0821, 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, you will get 4.87 times
10 to the fifth liters, so you can plug it all into "pivnert," or you could use the shortcut way. Either way, you're gonna
get the same volume of Chlorine produced.