How to solve oxidation-reduction reactions
“How can we solve these complex oxidation-reduction reactions?”
Oxidation reduction reactions can be quite difficult to grasp for many new students in Chemistry due to the long process often involved. However, I will show you how even through working through these problems might be convoluted at first, they are actually very intuitive if you think about how it works from a systematic point of view.
Let us start with an example oxidation-reduction reaction, CrO4 -2 + I- → I2 + Cr +3. The first thing we must do is split the problem into two half reaction, I- → I2 and CrO4 -2 → Cr+3. Now that we have isolated these problems into separate forms, we now have to balance the equation with the exception of hydrogen, oxygen, and charges! (Which we will come to later). So our new half reactions will be 2I- → I2 and CrO4 -2 → Cr+3. Now, you are probably curious about how oxygen is present on the left side of the second equation but not on the right! But before we give up, it is possible to add hydride ions to both sides. So what we do is take hydride, add enough to balance out the charge of the oxygen, and then add the corresponding amount of H2O on the other side. Now that we have done this, we have to account for the extra charges on both side. To solve this, we simply add extra electrons to the more positive sides of the reactions. To illustrate, the first equation will become 2I- → I2 + 2e-, and the second one will become CrO4 -2 → Cr+3 + 5e-. Now that both equations are fully balanced, it would seem logical to just combine them again, right? Well, not quite yet. If we would do that, there would be an imbalance in electrons. To solve this, all we have to do is multiply each side by a numerical constant to the lowest common factor of the electron charges to balance the electron charges. Since we have 2 electrons for the first equation and 5 for the second, the lowest common factor between them will be 5 and 2 respectively. So then our equation will be 2(CrO4 -2) + 5(2I-) + 10e- →2(Cr+3) + 2(4H20) + 5I2 + 10e-, and after crossing out the electrons on both sides, our final answer will be 2(CrO4 -2) + 5(2I-) →2(Cr+3) + 2(4H20) + 5I2
Once you have completed this process a few times, you will realize how logical and systematic it can be!