Classify each of the following unbalanced half-reactions as either an oxidation or a reduction. (d) SbCl₄^-(aq) → SbCl₆^-(aq)

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Hello. Everyone in this video, we're given four different reactions here and we're choosing the half reaction that is correctly classified. And oxidation reduction from the list below. So let's go ahead and recall that if we're losing electrons then we have an oxidation at hand. On the other hand, if we're gaining electrons, they were dealing with a reduction. All right. So basically in this problem we're just gonna do a lot of balancing. So for our first um reaction here, we have B I. O. Plus that yields R B. I solid and is said to be an oxidation. Well, like I said, we're just doing a lot of balancing here. So let's first go ahead and balance our oxygen's. We can see here on these are material side we have one atom of oxygen but on the right we have none. How we can add our oxygen atoms is by adding H 20. Of course HBO has one atom of oxygen. So because we need one oxygen, we'll just add one mole of H 20. And that's of course in its liquid state. So now we'll go ahead and balance out our high regions on the right side we just add our H 20. So there's two atoms of hydrogen but on the other material side we have none. So how we can add simply a hydrogen is by adding our hydrogen ions so we can add two moles of our H plus. So now we have our hydrants being balanced. And lastly last step we need to do for our balancing is to balance on our electrons. So our charges. So you can see here that on the right side, we have a neutral, everything's neutral here. But on the left side here we have well we have two plus ones and then one plus one, so two plus one is three. We'll need to make this side neutral by adding in three electrons. So we do so we have two H plus Reacting with three electrons as well as our original starting material. And this yields B I solid and H 20 liquid. So that's what we have here and we see here from Marseille material side that we're gaining these three electrons because we're getting electrons then this is a reduction and then we can see from the statement then that this is claimed to be an oxidation, so a is incorrect. So that's not one of the answers. Doing the same thing with half reaction be. So this is when our f minus yields are F two which is a gas. So we need to go ahead and balance out our flooring first. So on the right side we have two of the flooring on the left side we have one, we can go ahead and add a coefficient of two to balance out the florins. Next we need to go ahead and balance out our electrons. We'll see on the start material side we have two negative one um charges so that a negative two here. But the red side we have a neutral state. How can balance this out is by adding two electrons to the right. So you can see her then that we are losing electrons if we're losing electrons that we're dealing with an oxidation. So we see from a problem here, that statement or reaction B is supposed to be a reduction. So then B is also incorrect. Next we can do the same thing with half reaction see so we have here C O r foe to minus. And that yields C R O H three a solid. Alright, so first things first we're gonna go ahead and bounce on our auctions. We can bounce out our auctions again by adding H 20. So to the right side I'll go ahead and add one mole of our H 20. Next we'll go ahead and balance out our hydrogen. So on the left side we have no higher regions but on the right side we have a total of five. So we can add our high regions to the left side by adding in hydrogen cat ions. So here we'll go ahead and add five age pluses. Lastly we go ahead and bounce off the charges by adding electrons if needed. So on the right side here we have a neutral side for the left side here, we have a two miles with plus five. So to bounce it out to make it neutral we'll go ahead and balance out these positives with negatives which is in our case of course, is our electrons. So we'll add three electrons to the starting material side and that reacts with five moles of hydrant captains and one mole of c r 042 minus. Again that yield C r o H three, a solid and H 20 in its liquid state. So just growing down for more space here, we can see then that in this reaction from our sir material side we are gaining electrons ever gain electrons. We know that this is going to be a reduction. Let's go ahead and check what's stated here. So we can see that this reaction is supposed to be a reduction and that's definitely what we concluded. So C is one of our answers and I'll see if D is correct. So for half reaction Dean, we're starting off with our copper canyon and that yields are copper and solid state. So we just need to balance our charges in this case, we can see here on the left side we have a two plus charge but the right side is neutral. How we can make this side neutral is by adding a negative energy that's going to be our electrons. So go ahead and add two electrons here to the center material side. So we can see here, then there is a problem that from a certain materials we are gaining electrons again, if we're gaining electrons then we're dealing with a reduction. The problem states that this is an oxidation, so D. Is going to be incorrect. Making statements seem the one and true answer for this problem. Thank you all so much for watching.

Classify each of the following unbalanced half-reactions as either an oxidation or a reduction. (d) SbCl₄^-(aq) → SbCl₆^-(aq)

Verified Solution

Key Concepts

Oxidation and Reduction

Oxidation States

Half-Reactions

Classify each of the following unbalanced half-reactions as either an oxidation or a reduction. (d) SbCl4-(aq) → SbCl6-(aq)

Verified Solution

Key Concepts

Oxidation and Reduction

Oxidation States

Half-Reactions

Classify each of the following unbalanced half-reactions as either an oxidation or a reduction. (d) SbCl₄^-(aq) → SbCl₆^-(aq)