We're now going to take a look at how to balance an acidic redox reaction. Like I said before, it's going to 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. Alright. So step 1 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 two 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 dichromate ion gives us chromium 3 ion. We've done step 1, so let's go to step 2. Step 2 tells us to for each half reaction, balance elements that are not oxygen or hydrogen. So we have here 1 nitrogen, 1 nitrogen, so they're balanced. Here we have 2 chromiums but only 1 chromium, so put a 2 there. Step 3. For each half reaction balance the number of oxygens by adding water. So here in the first one we have 2 oxygens and here we have 3 oxygens. So I'm going to put 1 water on this side so that this side has 3 oxygens as well. For the other half reaction, we have 7 oxygens here, but no oxygens on the product side. So we need to put 7 waters. So now we've done step 3. Step 4, balance each half reaction, when it comes to hydrogen by adding H^{+}. So if we take a look here on the first half reaction we have 2 hydrogens within water. So this side on the left has 2 hydrogens. So to balance it out on the right so that it has 2 as well we put 2 H^{+}. Now, the other half reaction. We have 7 waters. So that's 7 times 2, that's 14 hydrogens on the product side, which means I'd have to put 14 H^{+} here. So now both sides have the same number of hydrogens. Now things can get a little bit tricky here, so pay very close attention to step 5. Balance the overall charge by adding electrons to the more positively charged side of each half reaction. Alright. So let's look at how we determine their charges. The Nitrite ion has a charge of minus 1. Water is neutral. So the overall charge on this side is minus 1. Look up the other side. We have minus 1 from the nitrate ion, but then we have 2 times plus 1, that's plus 2 from the H^{+}. So overall on this side, it is plus 1. Now we add electrons to the more positive side. We add the number of electrons necessary to make its charge on this right side, the same as on the left side. So since I'm starting out at plus 1, that would mean that I would need to add 2 electrons. 2 electrons make this side now equal to minus 1. Let's go to the other side. So if we look at the other side we have what? We have 2 minus 2 from the dichromide ion and then 14 times plus 1 is plus 14. So overall this side is plus 12. Then we have what? We have 2 times plus 3, this side is plus 6. Water is neutral so we don't have to account for a charge. Again, this side is plus 12, this side here is plus 6. I need to add enough electrons to the plus 12 side so that it comes out to be plus 6 as well because remember, add electrons to the more positive side. So for both sides to be plus 6, I would need to add 6 electrons to this side. So now doing that both sides will be plus 6. But here's the thing about step 5. If the number of electrons of both half reactions are different therefore, then multiply to get the lowest common multiple. So we have 2 electrons here and 6 electrons here. The lowest common multiple is 6, so I'd have to multiply this here by 3 so that this side also has 6 electrons just like the other half reaction. Now that we've done steps 1 through 5, step 6 is combine the half reactions and cross out reaction intermediates. Remember reaction intermediates are compounds that look the same with 1 look the same with 1 as a reactant and the other a product. So here, let me write them down now, this 3 gets distributed to everything inside of the, parentheses. So we're going to have 3 nitride ions plus 3 waters produce 3 nitrate ions plus 6 H^{+} plus 6 electrons. The other half reaction, I didn't have to multiply by anything so bring everything down. So dichromate + 14 H^{+} plus 6 electrons gives me 2 chromium 3 ions, plus 7 waters. Your electrons are always reaction intermediates because 1 will be a product, 1 will be a reactant. They must always completely cross out. Next, we see that we have water here and water here. All 3 of these waters cancel out with 3 from here. That leaves us 4 remaining. Then we have H^{+} here and H^{+} here. All 6 of these H^{+} cancel out with 6 from here which leaves us with 8. So at the end, our balanced redox reaction should come out as 3 NO_{2}^{−} + dichromate ion + 8 H^{+} gives me 3 nitrate ions + 2 dichromate ions or 2 chromium 3 ions + 4 H_{2}O. So this will be my balanced 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'll remember the order with enough of that.