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Balancing Redox Reactions Using Oxidation Number Method 4,846 views
So here are some tips, and tricks for balancing Redox reactions using the oxidation number method. So here we have a couple of examples. So the first thing you want to do is identify all the things that are changed, and so label their oxidation numbers.
So usually the trick is, they are not Oxygen, usually 90 percent of the time. So circle all these. The Manganese +2, and then the Iodine would be +7 for its oxidation number. Then this Iodine would be +5. Then the Manganese would be +7.
So what we so is we mash up. So the Manganese basically went from +2 to +7. So went by 5. Then that means that the Iodine, if you notice went down by 2. Now, how do you know that those are the ones that changed? Well, the oxidation number's changed. If you take a look at an element and the oxidation number is identical, then you know that it didn't get oxidized, and it did not get reduced.
So you take a look here, and you look for the lowest common multiple between these two numbers. So 5 and 2, the lowest common multiple is 10. So that means I need to multiply the Manganese by 2, and the Iodine by 5. So that means I can get 2 atoms of Manganese, and I needed 5 atoms of Iodine on each side in my equation.
So the easy trick is, just put the coefficients there. So I have 2MN2+, don't forget the charges. Then I have 2MnO4-. Then plus let's do that here. I have O4- plus 5IO3-.
Then you just follow the regular steps. So you balance out your oxygens by adding water. So since I have this here, I have 23 Oxygens on the right side. 20 on the left side, so I need 3 waters on the left side. Then balance that to your Hydrogens by adding H+. So since I have 6 Hydrogens from the waters, I need 6H+ on the right side. That's the quick way of balancing a Redox reaction. This is acidic of course, because I have H+.
Let's do another example. So same thing. Label all the oxidation numbers. So I have a couple of metals here. So I have the Chromium which is +6, Manganese is 0, because it's by itself +2 for this Manganese, +3 for the Chromium here. That means that the Manganese nice and easy, went up by 2, and Chromium went down by 3. So lowest common multiple 3. 3 times 3 is 6. So I have a 3 here, and a 2 here. That means I need 3 atoms of Manganese here, and 2 atoms of Chromium here. It's good to write them out, because this makes it nice, and easy.
So let's think of 2 atoms of Chromium. On the Dichromate, we already have 2 atoms of Chromium, so I don't need a coefficient there. Then let's do the Chromium again on the right side. So I need a 2 in front of the Cr3+ to make it so that I have 2 atoms of Chromium. Then Manganese, 3Mn plus 3Mn2+.
Don't forget the charges, because some tips are, my students tend to forget the charges, and then it makes the balancing of the charges and the atoms a little difficult.
So now we add water to balance out the Oxygens. So I need 7H20 on the right side because I have 7O on the left side. Next add H+ to balance out the Hydrogens. So I need 14 on the left side. Now if you double-check, we have 2Chromiums on both sides. 3Manganese on both sides. We have 7Oxygens on both sides, and we have 14Hydrogens on both sides. The atoms of each element match up.
Then the charges, +14 minus 2, that's +12 on the left side. Then on the right side; 2 times +3, and then 3 times +2, that's 12 also. So that also matches up. So for an acidic solution, that would be our final answer.
Just to recap; label all oxidation numbers that changed. So if you're not sure, you can always take a guess to figure out which one changed. That's number one.
Number two; the lowest common multiple, and then figure out the number of atoms of each of those elements that got oxidized or reduced. Then three; use water to balance out Oxygens. Then you add H+ to balance out the Hydrogens. Then if you want to make it basic, then you'd add an equal number of Hydroxides to both sides, and then group the waters together, and cancel them out.
So hopefully this three step method will make it so it's easier. And you have a lot of easy tips for balancing Redox equations using the oxidation method.
The trick is lowest common multiple, once you find that, then everything else should proceed nice and quick, and easy. So hopefully this helps you out with balancing Redox equations. Have a good one.