Balance each redox reaction occurring in basic aqueous solution. c. Cl₂(g) → Cl^–(aq) + ClO^–(aq)

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Accepted Answer

Hey everyone welcome back. So let's get started with this video. So here they want us to balance the following redox reaction in basic solution. And they asking what is the coefficient in front of O H minus. Okay, so the first step here is to separate these into two half reactions so that we're going to have BR two, going to be our 03 minus. Then pr two going to be our minus. So then here we have er being oxidized and reduced. So it's a disproportion ation reaction. So let's go ahead and write the half reactions down. So then we're going to have BR two, going to BR. And then the other half reaction is BR two going to be our minus. Okay, so then the next step is to balance elements except oxygen and hydrogen. So then here we have to be our so then on the other side we're going to add to here. So now the bros are balanced, So on the other one there's two on the left, so then we're going to put a two here, so now the bros are balanced. Okay, so the next we're going to balance out oxygen's by adding water to the side that needs it. So then here we have two times three that is six oxygen's on the product side. So then we're going to add six waters to the react inside. On the other side, there is no oxygen. So then it's fine. And then next we're going to balance out hydrogen is by adding H plus to the side that needs it. So then here, since we added six water six times to that is 12. So then we're going to add 12 H plus to this side. Once again, this one does not have hydrogen, so it's fine. So then our next step is to balance out the overall charge by adding electrons to the more positive side. So let's go ahead and solve for the overall charge. So then here we have 2 - and 12 plus. So then the overall 12 Plus the two That is going to equal a positive 10 charge. So in the overall charge here is plus 10 on the other side. Both of these are in their natural state, so it's going to be zero. So we're going to add electrons to the more positive side. And in order to make these equal, we're going to have to make This 1 - zero. So then we're going to add 10 electrons because 10 electrons added to the positive 10 will give us zero. So that both sides are not going to be equal. So now let's look at the other side here, we have two and it's minus so that the overall charge on the product side is going to be negative too. The reacting side is its natural state. So zero. So once again we're going to add electrons to the more positive side. So then here we're going to add two electrons. So now this one is going to have overall charge is negative two as well. Okay, so the electron swim must match for both half reactions. Therefore the common number is going to be 10. And we're going to multiply this one but times five. So that now this one is going to be 10 electrons when we multiply it. And this is also 10 electrons. Okay, so let's go ahead and bring everything down. So we have six H 20. Cause B R two going to to B R O three minus plus 12 H plus plus 10 electrons. And then for the other one, don't forget to multiply by five. So we're going to have five br two plus 10 electrons going to 10 B r minus. Okay, so now we cross out any intermediates, recall? The electrons must always totally cancel out. So then let's go ahead and start with the electrons. So 10 and 10 those cancel out. And that is it. That's all we can cancel out. So then let's go ahead and bring everything down again. So in these pr too, so you can actually add up. So then we're going to have five plus one, six PR 2. Then bring down the water six H 20. Going to to B R 03 minus plus 12 H plus plus 10 B R minus. Okay, so then this represents our balanced reaction an acidic solution. So that this is the acidic sit. Okay, but we want basic solution so we're going to look at how many H. Plus ions there is and add O. H. S. To both sides. So then here we have 12 h. Plus. So we're going to add 10 0. H. To both sides. Sorry not 10 we have 12. So we look at how many H. Plus is we have and we're gonna add the same number. So then we have 12. So we're going to add 120 H -2 both sides. So then let's go ahead. So then if we have 12 0. H. And then we're adding it to H. H. O. H. is the same as H. 20. Which is water. So then we're going to have here 12 H. 20. So then let's bring everything down 10 B. R minus. And then the two b. R. 3 -. And then here we're adding those 12. So then six ft off two plus six H. 20. Closer 12 0. H. We added. Okay so then now let's go ahead, we have water on both sides so that we can actually cancel out. So then we're gonna cancel these six out and we're going to subtract six from here and we're going to stay with six more. So then once again let's rewrite this And it's going to be six p. R. Two plus 12 0. H minus. Going to to b. r. 0. 3 minus plus six H. +20. Plus 10 B. R minus. You can see here how these coefficients can be divisible by two. So let's go ahead and divide everything by two. So We're gonna divide by 1/2 and we end up with three. We are too Plus 60 H -. Going to be our of three minus plus three H +20 plus five B r minus. And this is going to be in basic solution. Okay. And our question is, what is the coefficient in front of O H minus? And the coefficient in front of OH - is six. Okay, thank you for watching. I hope this helped and I'll see you in the next video.

Balance each redox reaction occurring in basic aqueous solution. c. Cl2(g) → Cl–(aq) + ClO–(aq)

Verified Solution

Balance each redox reaction occurring in basic aqueous solution. c. Cl₂(g) → Cl^–(aq) + ClO^–(aq)