Explain PEM electrolysis step by step with arrow pushing and lewis dot structure...

Without looking at the text, describe the steps for drawing Lewis structures.

Question 108 is a Louis. Simple question for Adam's, not molecules. You're asked to do this in a group and have everybody draw the Louis symbols for elements hydrogen through neon on the periodic table, so you should get something that looks like this. Helium has one valence. Electrons are hydrogen has one valence electron helium has to each represented with a small circle or a dot lithium. Now going to the second period just has one Valence electron beryllium has to. Boron has three. Carbon has four, and we begin to add them around one of the time until we have included one on all four sides. And then we double up. Nitrogen has five, Oxygen has six, Florian has seven and Neon has eight. The formal charge on all of these is going to be equal to zero because a formal charge of anything but zero occurs on Lee with bonding where electrons are being shared unequally because there is no bonding involved. No Covalin bonding. There's no sharing of electrons. There's no unequal sharing of electrons. Therefore, all formal charges are zero

Explain how to draw an ionic. Do this with an example. So let's take potassium iodide as an example. So what I have is potassium iodine. I know that each of these comes from potassium, which has a single valence electron iodine, which has seven. That there is a transfer of electrons from the metal to the non metal. And so potassium becomes a positively charged I on with no electrons in the village shell where it started at. So it does have a full Bayliss shell. It's just not what we show here. Um, so it had the positive potassium eye on, and then the iodine now has gained eight electrons. We show that with the eight dots around it, and we're gonna drop brackets around this symbol and that a charge to demonstrate, um that this is a charge I on. And this is how we show and I ionic move a structure. Sometimes we have Lewis structures where there is more than one of one of the ions. So, for example, if I had sodium oxide, um, what I would do is the same thing. The sodium would still become positively charged, but there are two of them. And then for the oxide, I would have all eight electrons, and then this would have a two minus charge. And so this is the way that it formed this mutual compound is I have to positive charges for every minus two charge of the oxygen.

If we have Ah, barium, boron hydroxide plus water. We can show the reaction using Lewis structures. So be O H three plus water. So water will donate electrons to boron. More hydrogen will be released and we will form B O H four plus h plus. And so B O H three is our Lewis acid. An H 20 is our Lewis base.

Hi. Are you ready to draw some Lewis structures and then use those Lewis structures to help us identify the Lewis acid and the Lewis base? In a reaction, Let's get started. I want to start off with boric acid. So drawing the Louis structure for boric acid. Boron has three valence electrons. Each oxygen brings in six, and each hydrogen will bring in one giving us a Louis structure that looks like this for the boric acid. The boric acid is reacting with water and according to the balanced equation, there actually to water molecules. So I want to draw the Louis structure. For each of those, there's one water and our second water, right. If we recall the octet rule tells us that elements, most elements, they're definitely exceptions. But most elements would like to have eight valence electrons. That is the definition of the octet rule. So when we look at Boron right now, boron on Lee has six valence electrons. When these substances react, the bond in one of our water molecules is going to break between the oxygen and the hydrogen, giving us an H plus and an O. H minus. This h minus that lone pair of electrons from where the bond broke are going to be donated and attached to the boron in this empty position. Since the O. H is donating a lone pair of electrons, it is known as the Louis Space Louis basis. Donate electrons, so just be identified as our base. The boric acid, on the other hand, is accepting the lone pair of electrons, so it is known as the Lewis acid. So we have a bond breaking in the water between the hydroxide and the hydrogen ion, and that is being bonded to the bar. Um, so then you might ask, what happens to this other hydrogen, this proton, this long proton? That's by itself. It, in turn, is going to attach to the lone pair on the other water molecule. When it does this, we can see that the water is still acting as a lone pair donor, so the water is still acting as the Lewis base, so each of these water molecules acted as a Lewis base, donating alone pair of electrons. The bore the boric acid, acted as lewis acid and the H plus ion. It was also a lone pair, except er so it acted as an acid. Let's see what the end result of this is. So on our product side swing, hammer, boron. But now our boron is going to have a full octet as it's gonna be surrounded by four oxygen's that are then attached to correspondingly four hydrogen. Okay, alright. And our other product is going to be and this is going to have a negative charge back up a second year that's gonna have a negative charge or other product is gonna be a hydro me. Um, I on or h 30 plus. All right, So there's your Lewis structures and how you use those Lewis structures to identify your Lewis acid and your Lewis base. Thanks for watching. I hope you found this helpful. Have a great day.