Consider the following unbalanced oxidation-reduction reactions in aqueous solution: Ag+1aq2 + Li1s2¡ Ag1s2 + Li+1aq2 Fe1s2 + Na+1aq2¡ Fe2 + 1aq2 + Na1s2 K1s2 + H2O1l2¡ KOH1aq2 + H21g2 (a) Balance second reaction.

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Hi everyone for this problem, we're told to balance the following redox reaction in acidic condition. When we are balancing something in acidic condition, it means we're adding hydrogen ions and water to balance it out. Okay, so the first thing that we're going to want to do is separate this reaction that we were given into two half reactions. And so let's go ahead and do that now. So we have M. n. 0. 4 - is going to equal MN two plus and then r I minus is going to give us eye to. Okay, so that first step was separating it into two half reactions. The next thing that we're going to need to do is balance our atoms so we need to make sure our atoms equal each other on both sides for both of our half reactions. So taking a look at the first half reaction, We have one M. N. on both sides, but on the left side we have for oxygen's. So that means we need four oxygen's on the right side. So we're going to add water so that we can add Those oxygen's. So we're going to add four waters So that we can have four oxygen's on the right side. But by adding four waters, you can see here now that made us have eight hydrogen on the left side on the right side. So that means we need to add eight hydrogen to the right side. So now our atoms are balanced. You see we have eight hydrogen on both sides, we have four oxygen's on both sides and we have one Mn on both sides. For our first half reaction, let's do the same thing for our first one. We only have one eyed on on the left and two on the right. So that means we need to add a two here and so now our atoms are balanced. The next thing that we need to do is confirm that the total charges on both sides of our half reactions equal each other. And if they don't, we need to add electrons in order to balance those charges. So for the first half reaction, let's take a look at our total charge we have On the left side of our first half reaction we have a positive eight Plus a negative one Gives us a total charge of positive seven on the left side. And on the right side we have positive we just have a positive too. Okay, so We have positive seven On the left and positive two on the right. We need to subtract the highest Charge from the lowest charge. And when we do we have 7 -2 gives us five. So we're going to add five electrons to the side with the highest charge so that we can have the charges equal to each other. So plus five electrons here. So you see that we had a positive seven on the left. Now we're adding five electrons. So now that's going to make the left side's total charge too. So now we have to a total charge of two on the left side and a total charge of two on the right side. So our charges balance for the first half reaction. Let's go ahead and do the same thing for our second half reaction we have a total charge of negative two on the left side and a total charge of zero On the right side. Okay, so we're going to take the larger charge and subtract it from the smaller charge. So we have 0 -2 gives us positive two. So we're going to add two electrons to the side with the highest charge. So plus two electrons. Okay, so now you see our electrons, the charge is equal each other on both sides. We have a negative two charge on the left side of our second half reaction. And now a negative two charge on the right side of our second half reaction. Okay, so our charges balance so we can go ahead and say that our charges balance. The next thing that we need to do is make sure a number of electrons equal each other for both of our half reactions. So for the first half reaction, we have five electrons. For the second half reaction we have two electrons. This is a problem because they need to match each other. Okay, so in order to match each other, we're going to multiply both sides. Both half reactions to make the electrons equal the same number. So for our first half reaction, we can multiply it by two. And our second half reaction we can multiply it by five. Everything. And so when we do that, we'll get our electrons to match each other. Okay, so that gives us a new half react half reaction. So the new one for the first one is going to be 10 electrons plus H plus. So I'm doing this first half reaction here. I'm multiplying everything by two plus two. MN 04 - is going to yield two MN plus Plus eight H 20. Okay, so that's our first half reaction. Our second half reaction, we're multiplying everything by five. Okay, so we have five. We have 10 I minus is going to yield five I two Plus 10 electrons. And this was our second half reaction. So now that our electrons match were the final thing that we need to do. Or we're almost done. We need to cancel anything that appears on both sides. So our electrons 10 electrons on both sides is the only thing that appears on both sides. So that's the only thing that cancels now. We can write our final balanced half reaction by adding up everything on by adding both of our half reactions together. So we're going to add up the left side of both of our half reactions and the right side. So for the left side we get two MN 04 minus plus 16 H plus plus I minus. That's everything on the left side. For both of our half reactions yields to MN two plus plus eight H +20 Plus five i 2. Okay, so this is going to be our final balanced redox reaction in acidic condition. That's the end of this problem. I hope this was helpful.

Consider the following unbalanced oxidation-reduction reactions in aqueous solution: Ag+1aq2 + Li1s2¡ Ag1s2 + Li+1aq2 Fe1s2 + Na+1aq2¡ Fe2 + 1aq2 + Na1s2 K1s2 + H2O1l2¡ KOH1aq2 + H21g2 (a) Balance second reaction.

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Key Concepts

Oxidation-Reduction Reactions

Balancing Redox Reactions

Aqueous Solutions and Ionic Compounds