 ## Analytical Chemistry

Learn the toughest concepts covered in your Analytical Chemistry class with step-by-step video tutorials and practice problems.

16. Electroanalytical Techniques

# Fundamentals of Electrolysis

Electrolysis
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concept

## Fundamentals of Electrolysis 4m
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So when we talk about electrolysis, we're gonna say that electrolysis deals with passing an electrical current through a substance in order to produce chemical changes now, because we're using outside energy, we're gonna say that processes dealing with electrolysis are non spontaneous. Remember non spontaneous things do not occur naturally and therefore require this outside energy in order to drive the reaction forward. Now, here we're going to say a good example of electrolysis, we have the passing of electric current through water, this helps to generate the standard components of water. So here we have water as a liquid, We drive electricity through it. And as a result we produce oxygen, gas and hydrogen gas. This is not a natural thing. We need that outside energy and form of the current in order to force the reaction to occur. Some of these concepts we've seen before. But just to revisit, remember we deal with electrical current. We're gonna say here the units for electrical currents are in amperes Or amps. So if we got were given 15 amps we'd say an amp is equal to columns per second. So this would be equal to 15 columns per second. And remember when we're talking about current, we can use the variable I I hear, which is current equals Q divided by T. So cue here would be our charge here. This would be our time. If we're talking about moles of electrons, we can say the moles of electrons within a reaction are determined by moles of electrons equals here, current times time divided by Faraday's constant. Remember current is cool ums per second time. Here will be in seconds. Faraday's constant has the units of columns over moles of electrons. So if you see columns cancel out seconds cancel out. And that's how you left with moles of electrons at the end. Now with non spontaneous processes, we should realize that our self potential will be negative, it'll be less than zero. And we have to take into account that with non spontaneous processes, our current is not negligible. It's a necessary part of the whole process. Because of this, we're gonna have to take into account three other terms we're gonna have a comic potential which we're gonna say is e it's the voltage necessary to overcome resistance. Are when the current i is flowing. Remember here would say I can also equal E over R. So I again, as our current e here is our voltage and our here is our resistance. We're gonna also have another term over potential. This is the voltage required to overcome the activation energy for reaction under given electrode, it's a non spontaneous process. So the activation energy has to be overcome in order to drive the reaction forward. Next we have concentration polarization occurs when there is a difference in the concentration of reactant on the surface of the electrode when compared to the solution itself. So the concentration of ions could be a very different number when it's close to the metal surface of our electrode or nerd electrode as opposed to within the bulk solution itself. Now, we're gonna say here, electrolysis is made more difficult by these different values because here your potential here is subtracting from our total and our over potentials are subtracting from our total our concentration polarization is found here between these two. So all these concepts help to make our overall cell potential more negative. And remember, the more negative and smaller cell potential becomes, the more non spontaneous process, the more current you need to supply in order to drive the reaction forward. So remember when we're talking about electrolysis, we're basically running current through a substance to elicit a response. And we have to do this because the whole process in itself is not a spontaneous one.
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example

## Fundamentals of Electrolysis Calculations 1 2m
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So here it states that aluminum can be electro plated at the cathode of an electrolysis cell by the half reaction. A L three plus plus three electrons reduces down to aluminum solid. Here, it says, how much time would it take for 825 mg of aluminum to be plated at a current of 4.1 and peers. So here we're using current to drive electrons onto the metal ions to create a precipitate a solid. Since the current is being used in electrolysis, this is a non spontaneous process here, we're gonna have to figure out time. I don't specify what units time will be in, so we'll just out for seconds. What we're gonna do first is realize that 4.1 amperes really means that we have 4.1 columns for one second. So it's our goal to isolate seconds here at the end, that means I'm gonna have to find a way of canceling out columns by using these grams these milligrams. So we're gonna start out with 825 mg of aluminum. And the first thing we're gonna do is convert those milligrams into grams. So one mg is 10 to the negative three g. So milligrams cancel out Now we have grams. Next I'm gonna change those grams into moles. So one mole of aluminum is 26.982g. Then we're going to stay here according to my equation for every one mole of aluminum solid. We have three moles of electrons involved. So one mole of aluminum involves three moles of electrons Moles of electrons are part of Faraday's constant. So it's 96,045 columns per one mole of electrons. Now that we have these columns, we can finally use the amps that we had originally. So take those 4.1 Coombs and put them on the bottom And 1/2 on top. So here at the end, what we'll get is we'll get 2.2 times 10 to the three seconds. So that's the approximate time it would take to basically plate that many milligrams of aluminum with a current of 4.1 being applied. Now that we've seen this example. Move on to example two. You can attempt it on your own, but if you get stuck, don't worry, come back and see how I approach the same exact question.
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example

## Fundamentals of Electrolysis Calculations 1 5m
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All right. So in this question it says in the electrolysis of molecular iodine to iodide ions for 0.15 moller of sodium iodide solution containing 4.2 times 10 to the negative for Mueller iodine at a ph of six with a partial pressure of oxygen gas equal to 1.25 bar, calculate the voltage needed to drive the reaction. Alright, so we have two half reactions given to us. And remember because this is part of electrolysis, this represents a non spontaneous process. And we're gonna say normally under a spontaneous process, the cathode, which is positive here should be the larger cell potential. But again, because we're under non spontaneous conditions, everything is reversed. Now the cathode, we'll have the smaller cell potential and the anodes will have the larger one. So that way when we do catholic minus a note at the end, we'll get a self potential. That's negative, illustrating that we have a non spontaneous process that's happening. Alright, so what we're gonna do here is we're gonna calculate 1st, Let's calculate what the the standard cell potential the potential self potential of the cathode is. So that equals the standard cell potential of the cathode minus 0.5916 volts, divided by the number of electrons transferred times log of products. Overreact ints. So here my product here is um a liquid. So we're gonna ignore that. That's gonna be one over Here. We have a gas and it has a pressure and bar so that's pressure of 02. Okay. And then we're gonna stay here times H plus And the coefficient years 4 to the 4th Plug in our values. That's negative .41V -15916V divided by number of electrons transferred. Alright, so the number of black trans transferred have to be the same in both half reactions. The second one is only a two. So we have to multiply this by two. Okay, so we're gonna multiply that by two. Alright, so now it's still four here, so it's four electrons, four moles of electrons times log of One over here. We want the pressure in bars, so that's 1.25 times H plus two. The fourth. Since you know ph you know what H plus is because H plus is 10 to the negative ph That's 10 to the -6 to the 4th. So here that would be negative .41V. And then here, if you do this all correctly it'll give you -35V. So that comes out to negative .76V. Now we do the same thing for the anodes. So now it's standard cell potential of the an ode minus 0.5916 volts divided by n times log of we multiplied everyone by two. So the number of electrons would be the same in both half reactions. So now that's actually I -2. The 4th Divided by I 2 to the 2nd. So that's 0.54 volts minus 0.5916 volts two by two by four electrons times log of I minus to the four cell, I minus is 40.15 Moeller to the 4th divided by concentration of I two. So 4.2 times 10 to the negative four squared. So that's gonna give me 0.54 volts minus 0.5 volts, which gives me positive 0.49 volts. Now we just say that my cell potential now equals cathode minus an ode. So that's negative 0.76 volts minus 0.49 volts Equals -1.25V. So since the self potential is negative, it's non spontaneous process, which makes sense because electrolysis is a non spontaneous process. This means that I have to run a current that high in order to drive the reaction forward. Remember under non spontaneous conditions we have to use outside energy in order to have the reaction occur. In this case we need to apply or use 1.25V of outside energy. So this whole process can occur. Now that we've seen this example. Take a look at the practice question left at the bottom. If you get stuck, don't worry, come back and see how I approach that same practice question 4
Problem

During electrolysis the concentration of I2 increases to 8.3 x 10 -3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed?

Electrolysis Calculations
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example

## Electrolysis Calculations 1 3m
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Hey guys. So here we're talking about the concept of electrical current. So when we say electrical current, another term for it would be amps or amperes. We're gonna stay here. Current or amps or amperes, it's really just charge over time. We're gonna say here charge, we use C to represent columns and time here would be in seconds. So, knowing this is key to get answering these questions. So for the first one it says gold can be plated out of a solution containing gold three based on the following half reaction. So here we have gold three ion reacting with three moles of electrons to give us one mole of gold solid here were asked what massive gold is plated out by a 41 minute flow of 6.8 and peers or current. Alright, so they're asking us to figure out the mass of gold, realize here when they say 6.8 A. That represents amperes. So this really represents 6.8 columns per one second. Because again, amperes amps is columns per second now, because this has seconds in it. That means that we have to convert the minutes given to us in two seconds as well. So we're gonna start out with the minutes, we have 41 minutes, We're gonna stay here for every one minute. It comes out to 60 seconds. Now that we have seconds we can cancel out the seconds in the amps. So 6.8 columns per one second. Yeah. And where do we see columns? We see columns with Faraday's constant. Faraday's constant, which is f So Faraday's constant is 96,485 columns per one mole of electrons. So we want to cancel out cooling. We're gonna put 96,004, columns on the bottom so they can cancel out and that's equal to one mole of electrons. Now, if we take a look at this equation for every one mole of gold, there are three electrons involved. So we're gonna say for every one mole of gold solid, There are three moles of electrons involved. So moles of electrons cancel out now that I have electrons that can change that into g. So for every one mole of gold, its mass on the periodic table is 196.97 g for gold. So we have at the end is g of gold, which comes up to 11.38 g of gold. So that will be our final answer here and remember. Current is just charged over time. And other words for it. Amps and peers or just a
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example

## Electrolysis and Current Calculations 1 2m
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here, it states a solution of manganese five is used to plate out manganese and an electrochemical cell. If a total of 1.13 g of mag unease is plated out in a total time of 1600 seconds. What was the electrical current used? So remember, electrical current is the same thing as amps or amperes. Remember here, this is just cool um per second. So we need to find columns and divide them by the total number of seconds. We already know the total number of seconds in the question. We have 1600 seconds total. So 1600 seconds will go on the bottom. All we have to do now is figure out the amount of columns we have to do that. We're gonna start out with the 1.13 g of manganese given. So we have 1.13 g of manganese. We're gonna stay here for every one mole of magazines. Were told that its mass is this 54.94 g. So those cancel apt. Now we're gonna stay here for every one mole of manganese. How many moles of electrons do we have involved where we're going from? Plus five to neutral manganese. That means we're dealing with five moles of electrons. So moles of magnesium cancel out now we have five moles of electrons. And now that we have moles of electrons, we can use thurday's constant. So that's one mole of electrons is 96,485 columns. So moles of electrons cancel out now have columns. And when I punch this in this gives me 9,922.47 columns. So take that and plug it up here. So we divide that by 1600. That gives us 6.20 amps or amperes. So that will be our final answer for the amount of electrical current.
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example

## Electrolysis Calculations 1 4m
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