Consider the following unbalanced oxidation-reduction
reactions i...

Question

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 third reaction.

Accepted Answer

Hi everyone for this problem, we're told to balance the following redox reaction in acidic condition. When we're in acidic condition, that means we are adding hydrogen ions and water in order to balance our reaction. The first thing that we're going to need to do is write out our half reactions. Okay, so we have PBO. two is going to this is a solid is going to form PB two plus and we have cl minus is going to form Cl two gas. Okay, so now we have our half reactions, which is our first step. Our second step is going to be to balance our atoms. So let's take a look at our first half reaction. For our first half reaction we have one lead on both sides, but on the left side we have two oxygen's. And so in order to balance the oxygen's we're going to need to add two water molecules to the right. So when we add two water molecules to the right now we have two oxygen's. But by adding two water molecules to the right, we added four hydrogen atoms to the right side. So that means we need to do the same to the left side. So we're going to add four H plus. So now our atoms are balanced for the first half reaction, we have four hydrogen on both sides and we have two oxygen's on both sides. All right, so the next thing we're going to do is the same thing for the second half reaction. So the second half reaction we have one chlorine on the left and two on the right. So that means we need to put it to here. So now our atoms are balanced for both of our half reactions. The next thing that we're going to do is balance our charges. We need to make sure that our charges match on both sides for both of our half reactions. So let's take a look at our first half reaction. For the left side we have a total charge of plus four Is equal to on the right side. We have a total charge of plus two. These total charges don't match. And so we need to make them match. We're going to do that by adding We're going to subtract the two numbers from each other. Subyract the highest from the lowest and then add that number of electrons to the side with the highest number. So 4 -2 is two. So we need to add two electrons to the side with the highest charge. So when we add two electrons to the side with the highest charge, that's going to bring the left sides charge to positive two. Ok, because we have a positive four charge minus minus two electrons is going to give us a positive too. Alright, so our charges now balance each other for the first half reaction. For the second half reaction, our total charge on the left side is - is equal to our total charge on the right side is zero. So we're going to take the highest number and subtract it by the lowest number. So zero minus negative two gives us positive two. So we're going to add positive two to the side with the highest charge and that is the right side. So we're going to add two electrons here by adding two electrons here. Now you see that Our charge on the right equals our charge on the left -2 is equal to negative two. So now our charges are balanced. Okay, so now the next thing we need to do is balance our electrons. We need to make sure a number of electrons equal each other for both of our half reactions. And they do our first half reaction. We have two electrons. Our second half reaction. We have two electrons. So we don't need to do anything here with balancing our electrons. So now we can go ahead and cancel anything. We see that shows up on both sides and that is two electrons. So those two electrons show up on both sides so we can cancel. Now the last thing we need to do is add up both of our half reactions and that's going to give us our final balanced redox reaction. So for our left side we're going to add up both of everything on both sides. Everything on the left side of both of our half reactions. And so that is going to give us PBO two solid plus four H plus plus two C. L minus. This is a quiz and that is going to give us PB two plus Aquarius plus two H liquid plus C. L two gas. So we went ahead and added up both of our half reactions, the left side and the right side. And this is 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 third reaction.

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