Balance the half-reactions in Problem 19.42, assuming that they o...

Question

Balance the half-reactions in Problem 19.42, assuming that they occur in acidic solution. <QUESTION REFERENCES PROBLEM 19.42>

Accepted Answer

Hello everyone in this video we're given these four different half reactions and they're occurring in acidic solution surgeries trying to balance them out. So keyword here in this problem is acidic solution. So it's safe to assume that there's going to be HT O and H plus kind of just floating around in an acidic solution. So those can be used as our source of oxygen atoms and hydrogen atoms. Taking a look at the first reaction over here. This one right here. So let's kind of maybe just rewrite us out. So this is the first reaction where we have I to which is in a solid state to give us I minus which is in its state. So first thing I wanna do is just bounce out our ions or our elements that we have. So we can see here on the left side that we have to Adams and then this right side we just have one. So we just completed a coefficient of two. So now we can go ahead and bounce out the charges. So on the right side we have two negative charges but this side is neutral. How we can a neutralize This is by adding two electrons. So to this we'll add two electrons. And if we do so then the final one will be I two solid state plus two electrons here and that gives us two mol of i minus and ions. And this would be our first past half reaction. All right now moving on to second one again. Right this out so we have C. U plus to give us see you solid. So the right and left side are all balanced out in terms of our atoms, What we need to do is just simply add in electron. We need to balance our charges. So again on the left side here we need to draw an arrow instead. Okay, that definitely helps. Alright, so on the left side here we have a positive one charge but on the right side we have a neutral state to kind of a neutralize this. What we can do is add an electron where will I electron is simply by adding it to the left side. So if we add to the left side this will neutralize the left side. Now the charges for the left and right are all balanced. So this is going to be our second balance equation. Alright, moving on to our 3rd. Alright, so this is MN. Give us an and two plus. Alright, so right off the bat, I can see here that our auction atoms are missing on the right side here we have four. So how can add our oxygen atoms is by adding H20. So we'll add in our case four moles of H 20. Now we can see that we have four atoms of auction on the right and four on the left. Now we can see that we have eight hydrogen is on the right but none on the left but adding purely our hydrogen sources. We can add our H plus carry ons. So I do. So LBMN 04 -, we're gonna add eight H plus cat ions which is a quiz and this will give us Mn two plus, which again is a quiz plus four H +20 in its liquid state. Now we need to go ahead and bounce off the charges. So we see here on the right that we have a two plus charge. But on the left we have plus one or minus one and plus eight. So that's seven. So we need to balance it out by adding five electrons on the left side to go these two positive charges. So we do. So then the final reaction will be we simply just add in five electrons to the left side. So our starting material side and everything on the product side or the right side will stay the same. Alright, so that's the third passed out half reaction. Now, the final one, let's just scroll down to give us more space. Alright, the last one here You have PBO two To give us PB two plus. Alright, so first things first scouting in again, our oxygen source on the left side, we have two items of auction. So on the right I automatically add two moles of H 20. Now we see that we have four hydrogen is total. So the left side, I will add that four moles of H plus and everything on the right side. The product side will stay the same. Alright, last thing to do is go ahead to bounce out our charges. So the right side we have a plus two charge. But the left side we have a plus four charge because we have four moles of H plus. All right. So how can balance this out is by adding two electrons to the star material side. So we have PB. 0. 2 reacting with four H plus is so at this moment then we have a plus four charge. But once we add in those two negative electrons that bonds out to being two plus on the bright side we have P two plus and two moles of H 20. So H 20. Is of course neutral. And then we have this two plus charge. So now is everything is balanced for our fourth reaction and that's it. So remember to use our H 20 as an oxygen source and our H plus as purely the hydrogen source. And we always need to balance out our electrons are charges. All right, thank you so much for watching this video. Hopefully this helps

Balance the half-reactions in Problem 19.42, assuming that they occur in acidic solution.

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