Balance each redox reaction occurring in basic aqueous solution.
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Question

Balance each redox reaction occurring in basic aqueous solution.
b. Al(s) + MnO4-(aq) ¡ MnO2(s) + Al(OH)4-(aq)

Accepted Answer

Hello everyone. Today we have the following problem balance in basic solution. We have the following reaction here. So the first thing you wanna do is you want to split these two into half reactions. So our first reaction will be our first step. Our half reaction for our nitrogen dioxide will be nitrogen dioxide, any acquis form forming nitric oxide in the gaseous form. And our second half reaction will be our chromium in the solid phase To our chromium two plus in the Aquarius phase. The next thing I wanna do is you want to balance all non hydrogen and non oxygen elements. So what does that mean? So in balancing our oxygen and balancing our non groups, we look at our nitrogen for the first half reaction. We see that we have one nitrogen on both sides. And then for our chromium, we see we only have one chromium on each side. So we're done with that. Then we want to move on to balance our hydrogen and oxygen. And so first, we're going to balance our oxygen's by adding water. So we see here on the left for our first half reaction, we have to oxygen is currently present and we have one oxygen on the right. This means that in order to add that extra oxygen, we need to add one water to the right side of the equation here, We're just gonna add one liquid water for our second half fraction. We don't have to add any water because we have no oxygen's so it's gonna be kept the same, then we're going to start by balancing hydrogen is by adding protons. And so if we look at our reaction on the left with our nitrogen dioxide, we see on the left, we have zero hydrogen, but on the right, we have two, which means that we have to add two protons on our right hand side. Once again, for our second half reaction, we have no hydrogen. So we're going to keep it as is. And so after we're done balancing our hydrogen and oxygen, we're going to balance our charges and we do that by adding electrons. So this might get a little tricky. We see here that we have a negative charge here. We have a negative charge there and we have a positive charge with our hydrogen is here, there should be a two there to balance out the hydrogen. So if we have a positive 2 - this gives us an overall positive one charge. So we need to get this to zero. Essentially, what we're gonna do is we're gonna add one electronic to this side here. And that will bring this left side as neutral as is the case on the right side of the equation. Now, for the right hand side, we have a zero charge here on the left. However, we have a positive to charge here on the right, meaning that we have to add two electrons to the side on our right here to bring that charge to neutral. So what we're gonna do is we're gonna add these reactions together. So we're basically gonna put all the reactions on one side and all the products on the other side. On the left side, we're going to have our reactions. We're gonna have our nitrogen dioxide plus our protons plus our electrons and then our chromium and then the products that we're gonna have our nitric oxide, our water, Our two electrons and our chromium two plus Aquarius. And so now we have to essentially neutralize and add hydroxide to both sides so that we can cancel out the water. And this is because we're balancing in a basic solution. So there's a bunch of, of hydroxide ions floating around. And so how are we going to do this? Well, we said that we have to add hydroxide to both sides and we have to somehow cancel out water. So if we add hydroxide to the left here, then we're gonna add hydroxide to the right there. Never going to need to balance it essentially what we can do to balance this out is we can go ahead and we can multiply both sides by two or by four. And in doing that, We got our two hydroxide and then we get our Water, they're getting a coefficient of four and then we can cancel out our electrons. And so essentially what we worked up to is to find the final equation of our reaction. How is it such? And so with that, we have our final answer overall, I hope this helped. And until next time.

Balance each redox reaction occurring in basic aqueous solution. b. Al(s) + MnO4-(aq) ¡ MnO2(s) + Al(OH)4-(aq)

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