Balance the following redox reaction.

Question

Chemical equation for balancing a redox reaction involving Cl2 and KI.

Accepted Answer

Here we need to balance the following redox reaction. Now here's the thing, to correctly balance this, remember redox reaction involves reduction and oxidation. So you're only looking at the elements where oxidation numbers are changing. If an element's oxidation number doesn't change, it's not being oxidized or reduced and therefore should not be included within this balancing process. Alright. So we're gonna say here, if we look, CL is in its natural state, so so its oxidation number here is 0. Potassium is with iodide, potassium here is plus 1, iodide is minus 1. Potassium here is still plus 1, but now chlorine is minus 1. These are ionic compounds, so it's their charges are those numbers. Iodine here is in its natural state, so its oxidation number is 0. If you look carefully you'll see that potassium, its oxidation number stays the same throughout the whole process. So it's not being oxidized or reduced. It's chlorine and iodine whose oxidation numbers are changing, so they are what's important in terms of the redox reaction. So what we'd actually have is cl2 aqueous plus I minus aqueous gives me cl minus aqueous plus I 2 aqueous. This is what we care about. These are what that have oxidation numbers changing. So now we can break this up into its 2 half reactions. So chlorine with chlorine, iodine with iodine. So Cl 2 aqueous gives me Cl minus aqueous, I minus aqueous gives me I 2 aqueous. Alright. So now let's balance these half reactions. What's important here is to realize that the number of chlorines don't match on both sides. They have to when we're starting to balance out redox reactions. So I'd have to put a 2 here. Here there's 2 iodines here, but only 1 here, so I put a 2 here. Here. Now that I've balanced out my chlorine iodine, now I can start talking about adding electrons. Remember, we add electrons to the more positive side of each half reaction. Here, Cl 2 is neutral, so the total charge on this side is 0. Here, chlorine, chlorine actually is minus 1, but there are 2 of them. So the overall charge here is minus 2. Add electrons to the more positive side, enough electrons so that both sides have the same overall charge at the end. I need 0 to become minus 2, so I have to add 2 electrons. Let's go to the other side. Here we have 2 times minus 1, so the overall charge here is minus 2. Iodine here, it doesn't have a charge that we can see, so overall charge here is 0. Again adding electrons to the more positive side, I have to add 2 electrons to this side here, so that both sides are minus 2 at the end. Notice now that both number of electrons are the same, so I don't need to multiply anything to get the lowest common multiple. So just bring down everything and cancel out your electrons to get your balanced overall redox reaction. So electrons here cancel out, So we're gonna have now Cl2 aqueous plus 2 I minus aqueous gives me 2 c l minus aqueous plus I 2 aqueous as my balanced redox reaction.

Balance the following redox reaction.

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