- 6/4/2025
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00:00Hello everyone. Welcome to this exciting lecture series on electrochemistry. In this video,
00:09we are going to explain Hittor's rule which is related to mobility of iron in an electrolytic
00:15solution. As we study electrolysis, we often think about ions simply moving towards opposite
00:24electrodes, cations to the cathode and anions to the anode. But in reality, they don't
00:31all travel at the same speed. Some minds move faster and some minds move slower. And this
00:37difference affects the concentration of ions near each electrode during the electrolysis.
00:43This is where Hittor's rule comes into play. Hittor was a scientist who studied how the
00:49concentration of ions changed near the electrodes while a current passed through an electrolyte.
00:55When an electric current passes through an electrolytic solution, the ions begin to move towards the
01:01respective oppositely charged electrodes. Cations move towards the cathode and the anions move
01:07towards the anode. However, it is important to note that not all ions move at the same speed.
01:14Some ions travel faster while others move more slowly depending on their nature and size.
01:21Let us say a cation is moving away from the anode. The rate at which its concentration decreases
01:27around the anode is directly related to how fast that cation is moving away. The same principle
01:34applies to the anion moving from the cathode. Hittor observed this behavior in a lab setting
01:41and proposed a rule based on his findings. His rule helps us understand how much of each ion
01:48is being transported in a given time. Hittor's rule states that the loss of concentration around
01:57an electrode is proportional to the speed of ions moving away from it. So this is a very simple
02:04definition of the Hittor's rule which is related to mobility of ions. This rule is especially
02:11important in electrochemical calculations like determining transport numbers and help us
02:16understand the distribution of ions during the process of electrolysis.
02:23Let us visualize a simple electrolysis cell setup where we have A as anode, this one is A, and C as
02:35the cathode. These two are the electrodes through which current enters and exits the cell. Now imagine drawing
02:44two imaginary vertical planes labeled as AA, this one A, A dash, and second one is B, B dash. It divides
02:53the cell into the three components. This one is the anode compartment, first one, middle compartment, and third one is the cathode compartment.
03:05In this diagram, plus sign is used to represent a cation, which is positively charged ion, and minus sign represents the anion, which is a negatively charged particle, or the ion.
03:27As the electric current flows into the cell, the cations move toward the cathode, and the anions move in the opposite direction toward the anode.
03:37This movement is key to understanding how ions redistribute during electrolysis, and why Hittor's rule is important in tracking these changes across the compartment.
03:48Before the process of electrolysis begins, let us consider the initial conditions of the electrolyte solution in the cell.
04:01There are total 13 ion pairs redistributed across the compartments of the cell. Here we can see that these are the three ion pairs in the cell.
04:16In this scenario are position one. So first one is the position one, this one. In position one, which is our starting point, the anode compartment
04:28have four electron, four pairs. These are the four pairs, positive and negative, positive and negative, and two others. And also the cathode compartment have four pairs.
04:40There are five ion pairs in the middle compartment. So total there are 13 ion pairs. Four plus five plus four is equal to 13.
04:52Now, let us analyze three different scenarios of ion movement when electrolysis starts. First one is that ions move, anions move alone.
05:03Second one is that both anions and cations move at the same rate. And third one is that cations move faster than the anions. We will look into these scenarios in detail in the coming slides.
05:17Now, let us go towards the case one, which is anions move alone. In this case, we assume that only anions are mobile and they move towards the anode. Here we can see that in the second scenario, which is the position two, we see that anions move towards the anode. Two negative charges move towards the anode under the influence of electric current.
05:46As the electrolysis begins, two anions move from the cathode compartment. Here, from the cathode compartment, two anions have moved towards the anode compartment. After this movement, the new ion distribution reaches where we label as the position two. So this one is the new position.
06:03Here we can see that there are four anions. Here we can see that there are four anions in this compartment and four anions in this compartment. But in position two, here are the six anions and here are the two anions. The two anions have moved from this compartment all the way from here to into the anode compartment.
06:22At the electrodes, two cations are discharged at the cathode. Two cations are being discharged at the cathode and the two anions are being discharged at the anode. So the number of discharge ion on both sides is equal to two. Two ions are discharged here and two ions are being discharged here.
06:48The concentration in the cathode here. The concentration in the cathode compartment decreases by two electrons or the two ion pairs. We can see that the anode concentration remains unchanged. In the position one, we have four ion pairs and in position two, we also have four ion pairs. The extra two anions have been discharged from the anode.
07:15However, because only anode. However, because only anions moved, they left behind the repaired cation in the cathode compartment. This results in the decrease in a concentration by two ions in the cathode compartment. Meanwhile, the anode compartment remains unchanged in concentration because the arriving anions replenish the anions that were discharged.
07:40Moving toward the second scenario where both anions and cations move at the same rate. In the second case, we now assume that both anions and cations are equally mobile. That is, they move at the same speed toward the respective electrodes. So what happens here? Two cations or four cations move away from the anode compartment.
07:58toward the cathode compartment toward the cathode compartment toward the cathode compartment. And four anions move from the cathode compartment to the anode compartment. This results in a new ion distribution, which we refer to as position three. At the electrodes, four cations get discharged.
08:16And four anions are being discharged. And four anions are being discharged at the anode. So again, the total number of discharged ion is eight, where four ions are discharged at each electrode. Now, since both types
08:46the cathode compartment have lost two ion pairs, the cathode compartment have lost two ion pairs or four ions and the anode compartment have also lost four ions or two ion pairs. This means that both compartments see a reduction of two ion pairs. So the concentration decreases equally in the anode and the cathode compartment.
09:08This case beautifully illustrates Hitor's idea that when ions move at equal rates, the fall in concentration is the same on both sides.
09:25In this third case, we consider an interesting and very realistic scenario. The cations move faster than the anions, specifically twice as fast.
09:36So what happens during the electrolysis in this case? As current passes through the solution, two cations move toward the cathode, but only one anion move toward the anode.
09:49This gives us a new arrangement of ions referred to as position four. So here we can see that from the cathode compartment, one anion have moved from here all the way to here, and two cations have moved from anode compartment to the cathode compartment.
10:03The total number now, let us talk about the discharge of electrons or the ions at the electrodes. In the anode compartment, three anions are being discharged. And in the cathode compartment, three cations are being discharged. So the total number of discharged ion is again the same, that is three.
10:28Anode compartment. Anode compartment. Anode compartment decreases by two ion pairs. Here we can see that at in the position one, we had the four electron pairs, but in the position four, we have two ion pairs.
10:41And the cathode compartment. And the cathode compartment decreases by one electron pair. Before it were four electron pair ion pairs. And now there are three ion pairs.
10:53It is evident from the above consideration that ions are always discharged in equivalent amounts on the opposite selectors. If we can see from all of the scenarios that if in the scenario two, two charges are being discharged in the anode compartment, then same number of positive charges are being discharged in the cathode compartment.
11:18In the scenario three. In the scenario three, four positive charges were being discharged. And in the anode compartment in the scenario three, again, four negative charges were being discharged. In this case, which is the case three, three negative charges and three positive charges are being discharged. So in every case, equivalent amount of ions are being discharged on the opposite electrodes.
11:45Specifically, in the fourth scenario. Specifically, in the fourth scenario, we have seen that the anode compartment, ion pairs are being decreased by two amount and the cathode compartment, one ion pair is decreased. So anode compartment concentration decreases more rapidly than the cathode compartment.
12:08This concentration changes. So what does this show? The fall in concentration is proportional to the speed of ions moving away from it, which was exactly according to the Hittor's rule. And importantly, even if the ions move at their different speeds, the electrodes still discharge equivalent amount of ions in terms of total charges.
12:32The concentration changes. The concentration changes near electrodes happen because anions and cations move at different speeds.
12:39Faster moving ions leave their compartment more quickly, causing causing a greater decrease in the concentration. So unequal speeds of ions are the real reason behind these changes during the electrolysis. So here we can see that the concentrations have been changed due to more speed of cations than the anions.
12:50Cathode compartment have more concentration of ions than the anions. Cathode compartment have more concentration of ions than the anode compartment, which is entirely due to the speed of cations or the anions.
13:11Cathode compartment. Based on our findings, we conclude that loss in concentration around an electrode, whether it is a cathode or anode, will be proportional to the speed of ions moving away from it. That is exactly what Hittor's rule says.
13:34Mathematically. Mathematically, delta c or the concentration around c anode compartment will be proportional to positive ions or the velocity of positive ions and the concentration change in the cathode compartment will be proportional to the velocity of negative ions.
13:50Mathematics. Here in these equations, V plus corresponds to speed of cations and V minus corresponds to speed of anions. We can write the expression as fall of concentration around anode divided by fall of concentration around cathode will be equal to speed of cations divided by speed of anions and denoting these speeds by their symbols V plus divided by V minus.
14:18Mathematics. When we drive this correlation, we made a key assumption that the discharged ions simply move away or deposit without reacting with the electrode material itself. Essentially, we assume the ions don't chemically combine with the electrodes. But in the real situation, this assumption does not always hold true. Often, the ions may actually react or combine with the electrode material
14:47rather than just depositing on it. When this happens, instead of seeing a drop in ion concentration near the electrode, we observe an increase in the concentration. This is because the ions accumulate as they react, which changes the local concentration dynamics.
15:08So, it is important to remember that while Hittor's rule provide a good baseline, real electrochemical systems can show more complex behavior depending on the interaction between ions and the electrodes.
15:23During the process of electrolysis, the electric current does not just flow freely. It is actually carried by charged particles called ions. Specifically, both the positively charged ions called cations and the negatively charged ions called anions. They both contribute to carrying the current.
15:45Now, the transport number. Now, the transport number, which is also known as the Hittor's number, is a way to express how much of the total current each type of ions carry. In other words, it tells us the fraction or percentage of the total current that is transported by either the cation or the anion.
16:07So, if V plus represents the speed of migration of cations and V minus of that of anions, then the transport number will be of the cation will be equal to V plus divided by V plus and V minus.
16:22So, here below we have the total velocity of both of the ions and the fraction of it can be calculated by dividing the transport velocity of cations to the velocity of total ions.
16:35Similarly, the transport number of anions can be calculated by using this equation, which is V minus, which is the speed of migration of anions divided by V plus and V minus.
16:50This concept helps us understand how the current is split between the two types of ions during the electrolysis.
16:57To get a sense of this, imagine the speeds at which these ions move.
17:02If we represent the speed of cations as V plus and the speed of ions represented by V minus, this speed will influence how much of the current each ion carries.
17:14Later on, we will also see how these speeds relate mathematically to the transport number, which help us to quantify their concentration contributions precisely.
17:29So, the transport number can be represented by T, T plus for the cations and T minus for the anions.
17:42So, moving further in the calculations, T plus will be equal to V plus divided by V plus plus V minus and T minus which is the transport number for the anions will be equal to this equation.
17:55So, changing the equation, T plus divided by T minus will be equal to V plus divided by V minus.
18:02When we divide both of these equations, the denominator got cancelled out and further simplifying the equation, we get the value as T plus and T minus will be equal to 1.
18:14If the speed ratio is V plus and V minus is denoted by R, then we can have R is equal to T plus over T minus which is equal to T plus divided by 1 minus T minus.
18:27This shows the concentration of positive charges rather than the negative transport number.
18:34And also for the negative transport number, if we try to find the negative transport number of anions, then we can simply use this equation.
18:45Here, R represents the speed ratio of cation to anion.
18:49So, that is all.
18:53I hope you have learned something new and got benefited from this lecture.
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