4.40a | Balance: Zn(s) + NO3−(aq) → Zn2+(aq) + N2(g) (in Acid)

Number 40 balance, each of the following equations, according to the half reaction method, okay. And then we have lettered a. So in this reaction we have to balance zinc solid, plus nitrate ion no3 minus, and that will yield zinc, two plus and then plus nitrogen, right n2. And they tell us that this is in an acidic solution. So I wrote down all the rules for you guys how to balance any redox reaction in acidic solution.

So it's a standard set of nine rules. You have to go in order. So let's let's have a go. Okay, so for the first thing we have to break this whole reaction into two half reactions. And the easiest way to think about this is that similar elements will yield the same elements right? So I need to turn this one reaction into two half reactions.

So let's see I'm going to link up the zinc solid that I have on the left side and that's going to go to the zinc like elements or like compounds go together. And then it makes sense that this ion right nitrate. O3, minus would link up with this one, mainly. Because they both have nitrogen in common, I don't really care about the oxygen at the moment because in step number three, we can always balance the oxygen. So just focus on other elements, just not oxygen and nitrogen at this oxygen and hydrogen at this point. So we have two half reactions.

We have zinc solid, and that will yield Zn, two plus and that's aqueous. And then we have the other uh, half reaction, no3 minus and that's aqueous yields and two and that's a gas. Okay, cool. So the first part all done. Let's move on now we just have to balance all the elements, except for hydrogen and except for oxygen. So don't, look at hydrogen don't. Look at oxygen in this step.

Look at all the other elements, I see two of them. I have my zincs on the top equation. And then I have my nitrogen on the bottom. They need to be balanced. So let's, just do these zincs first and notice how I like doing my step in both equations. So I'll do both equations for step two, and then I'll do both equations for step three.

And so on. And so forth. Okay, so let's balance the zinc.

Well, I have one zinc on my left side. And I got one zinc on my right side. So that's balanced already. I don't have to do anything with that one now going down to this half reaction, I got one nitrogen. But then on this side, I have two nitrogen, right, it's a n2. So I have to balance this.

That means that I have to put a number in front of my no3 to get me to two. What do you think? Yeah, I'm going to put a two here right?

Because two nitrogen then, and then I have. Two nitrogen. So now my nitrogen are balanced step number two is done now I'm going to balance my oxygen by adding h2o now, here's a quick, uh, trick just know that if you need to add one oxygen on any side, you will add it in terms of one h2o. So if you need to add two oxygen son one side, you'll, add two, h2o's, three, oxygen, three, hos. So let's, see, I don't have any oxygen in the first reaction.

So I don't care. But in the second one, I have an oxygen now how much total oxygen? Well, I have three coming from.

Here, right there's a three here, and then I have this big two. So how much total oxygen? Do I have? Yeah, I have six right. Two times three is six, and I have no oxygen on this side.

So I got to balance it. If I have 6 oxygen on this side, and I need to add 6 oxygen. I need to add 6, h2o. Okay. Cool now that step is balanced let's go to the next one. Now we balance hydrogen by adding h, plus.

And the rule of thumb is that if you need one hydrogen on any side, you will add it as one h, plus so two hydrogens, you'll. Add two h, pluses, three, hydrogens, three h, pluses. So I look on the top, I don't have any hydrogen on the left or the right side. So I skip over it.

But when I come down to here, I see that I have a hydrogen over here. Now, there's a 2, but now there's, a 6 in front of it. So 6 times 2 is 12 hydrogens on this side since I have none on this side, and I require 12 I'm going to add it as 12 h plus now that gets rid of the fourth step. Okay, looks like we're kind of halfway through now for the fun part, we have to. Balance the charges, and we balance the charges by adding electrons.

Electrons are always e negative. And you're always going to add the electrons to the more positive side. So I like to just make a barrier between the left and the right side for both of them. And I like to add up, the total charges to see what I'm dealing with so let's work with this equation.

First, I don't see a charge in the upper right-hand corner of zinc, which means that this element is neutral, it's, not positive nor negative. So it. Has a charge of zero.

And since there literally is just this one element here, my overall charge for the left-hand side would be zero now, let's do the same thing over here. But in this case, I do see a charge in the upper right-hand corner, it's, a two, plus there's, only one of them. So my overall charge would be a plus two. And those are the charges that you're going to be comparing we want to add the electrons to the more positive side out of the zero and the plus two, which one is more positive. The. Plus two, so I know that I'm going to have to add electrons on this side. But now the question is, how many?

Well, the answer is, how many does it take to go from a two to a zero? If you think of it on a lumber line? Yeah, they're. A difference of two numbers away. That's, how many electrons you add both sides are now going to be zero charge? Now we got to do the same thing for this.

Once I'll work from left to right. I see that I have a plus charge right so that's, a plus one, but I have 12 of them so 12 times a. Plus one there's, an overall plus 12 coming from the hydrogen. And now if I just do the nitrate, the nitrate has a negative one, but there's two of them. So two times a negative.

One is a negative two. And now you add these two together, just like there's addition here, there's addition up top here. So 12, plus a negative 2. Aka 12, minus 2 is a positive 10.

That's. The overall charge of the left side let's do this. But for both of them, I don't see any charges right?

I don't see any charges for both of these. So. They're, both neutral, so it's, a zero. So zero, plus zero is zero now, assess which one is the more positive side, the plus ten is.

So I know that I have to add electrons on this side, but how many? Well, how number how many numbers will it take me to get from a 10 all the way down to a zero 10? Right? So I know that I have to add 10 electrons.

This part is now done, here's. Your checkpoint guys just make sure that the electrons that you added are on opposite sides of your equation. Since they are here. One's on the reactant and one's on the product side, I'm all good. I can keep going, however, if you get to this point, and you see that you had like, you know, your two electrons on this side. And both electrons were on one side.

Go back. Do it over something went wrong. So before I um go to the sixth step I'm just going to make this a little clearer, I'm just going to clean up all my charges, because we don't need the math anymore that was only to figure out how many electrons we needed. So just to kind of. Make everything nice and neat. Okay.

Now, we're back I'll, just get rid of this minus one. Okay. So now we need to do step six, which is balance the electrons. So now I look at the actual numbers.

I have 10 electrons on this side right on this half reaction, and I have two electrons on the other side. I need to make them the same, and I need to make them the lowest number with multiplication. So what number does these have in common, or maybe I can multiply one to get to the other number.

What do you think yeah, I can totally just take this and times it by 5, right 2 times 5 will get me 10. But if I do that, I have to multiply the whole equation by 5. That's, totally, ok. And that means that all the coefficients in the front have to change you're multiplying, the 5 by all the coefficients.

So just for simplicity purposes, I'm just going to erase and write the new coefficient. But you can always just rewrite, um your equation again. So let's do it for this one? I had one zinc, right? But now one times five is. Now, five the same thing for this one, I had one zinc, there was no coefficient there, but one times five is now five. And now I go back over here.

I had 2. But now 2 times 5 is 10. So I'm just going to stick that in there. And now I can strip this away perfect. And now the electrons are balanced so that's done. Now, since they are the same, you can cancel like substances out on opposite sides, that's, why we have to get them to the same because we want to cancel them out. So since they're now, the same bye, bye.

They don't get placed into the actual ballast equation. Now I just look and see if I have anything else, that's the same. No, right? I have water on this side that I added, but I don't have any water on this side. So I can't do anything about that. So this is just rewriting we're done with this step. And now all we have to do is just rewrite as one equation.

It doesn't matter which one comes first on the reactant side, just make sure that you have all three of them. So maybe I'll just write. I don't know, I'll do the 12h plus first, and then possibly I'll say, the five, zinc solid, plus the two no3 minus, and that will yield the five zn2, plus aqueous I'm. Now, on the product side, the n2 gas, and then the six waters and that's it that is your full balanced equation. Pretty neat. Okay.

So guys, what'd, you think, hopefully this helped just make sure that you go through the steps and everything will be good. Okay, um. I hope you guys are having a great day and keep studying hard. Okay, I'm rooting for you guys and.

I'll? See you all in the next lesson, see you then bye.

Dated : 21-Mar-2022

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