Balancing Redox Reactions: Acidic Solutions Example 2

Jules Bruno
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we're not gonna take a look at how to balance a acidic Redox reaction. Like I said before, it's gonna require a brand new approach to balance these types of reactions. So follow the steps listed below, and you'll be able to balance Redox reactions. Now if we take a look here, it says in this example, balance the following Redox reaction if it is found to be in an acidic solution, All right, so Step One tells us that we have to break the full Redox reaction into two half reactions. Now remember, we focus on the elements that are not oxygen or hydrogen to determine the to half reactions. So nitrogen is an element that is not oxygen or hydrogen, and it's found here and here. That gives us one half reaction. Don't worry about the state that they're in. We'll put those in later, after we're done balancing. And then what else? We see chromium here. Chromium is not oxygen or hydrogen, so that's our other half reaction. So die chromite ion gives us chromium three ion. We've done Step one, so let's go to step two. Step to tells us two for each half reaction balanced elements that are not oxygen or hydrogen. So we have here one nitrogen, one nitrogen, so they're balanced. Here we have two chromium, but only one chromium, so put it to their Step three for each half. Reaction balance the number of oxygen's by adding water. So here in the first one, we have to oxygen's. And here we have three oxygen's. So I'm gonna put one water on this side so that this side has three oxygen's as well. For the other half reaction, we have seven oxygen's here, but no oxygen's on the product side, so we need to put seven waters. So now we've done Step three step forward balance each half reaction when it comes to hydrogen by adding H plus. So if we take a look here on the first half reaction, we have two hydrogen within water, so this side on the left has to hydrogen. So to balance it out on the right, so that it has to is while we put two h plus, now the other half reaction. We have seven waters, so that's seven times to that's 14 hydrogen on the product side, which means I'd have to put 14 h plus here. So now both sides have the same number of hydrogen Now, things could get a little bit tricky here, so pay very close attention to step fine. Balance the overall charge by adding electrons to the mawr. Positively charged side of each half reaction. All right, so let's look at how we determine their charges. The nitrite ion has a charge of minus one. Water is neutral. So the overall charge on this side is minus one. Look at the other side. We have minus one from the nitrate ion, but then we have two times plus one. That's plus two from the H plus is so overall on this side it is plus one. Now we add electrons for the more positive side. We add the number of electrons necessary to make its charge on this side on this right side, the same as on the left side. So since I'm starting out at plus one, that would mean that I would need to add two electrons. Two electrons would make this side now equal to minus one. Let's go to the other side. So if we look at the other side, we have what we have tu minus two from the dye Chromite ion and then 14 times plus one is plus 14. So overall this side is plus 12. Okay, then we have what we have two times plus three decide is plus six. Water is neutral, so we don't have to account for a charge again. This side is plus 12. This side here is plus six. I need to add enough electrons to the plus 12 side so that it comes out to be plus six as well. Because remember at electrons the more positive side. So for both sides to be plus six, I would need to add six electrons to this side. So now doing that, both sides will be plus six. But here's the thing about step five if the number of electrons of both half reactions are different, differ than multiply to get the lowest common multiple. So we have two electrons here in six electrons here, the lowest common multiple of six. So I'd have to multiply this here by three so that this side also has six electrons just like the other half reaction. Now that we've done steps one through five steps, sixes combined, the half reactions and cross out reaction Intermediates. Remember reaction. Intermediates are compounds that look the same with one look the same with one as a reactant and the other a product. So here, let me write them down. Now this three gets distributed toe everything inside of the parentheses. So we're gonna have three nitrite ions plus three waters produce three nitrate ions plus six h plus plus six electrons. The other half reaction. I didn't have to multiply by anything. So bring everything down. So die chrome eight plus 14 age plus plus six electrons gives me two chromium three ions plus seven waters your height. Your electrons are always reaction intermediates because one will be a product when will be reacted, they must always completely cross up. Next we see that we have water here and water here. All three of these waters cancel out with three from here. That leaves us four remaining. Then we have hpe plus here in H plus here. All six of these h plus cancel out with six from here, which leaves us with eight. So at the end, are balanced. Redox reaction should come out as three no tu minus plus die chromite ion. So when you write that down die chromite ion plus eight h plus gives me three nitrate ions plus to die crow made ions to chrome and three ions plus four h two out So this will be my balance Redox reaction within an acidic solution. I know the steps can be a little bit daunting, But again, if you want to be able to balance a Redox reaction, these are the steps that you need to employ in order to answer the question correctly. So just remember, keep practicing and you remember the order with enough of that.