Balance the following redox reaction in an acidic solution.

Question

Cl₂ (g) + S₂O₃^2-(aq) → Cl^- (aq) + SO₄^2- (aq)

Accepted Answer

here. It says to balance the following Redox reaction in an acidic solution. So remember, Step one. We looked for elements that are not oxygen or hydrogen to determine our two half reactions. Here we have chlorine matched up with Corey, so that's one half reaction. So cl two gives me C l minus. Then we also have sulfur with sulfur. So as to 03 to minus gives me s 04 to minus Step two, we're gonna balance elements that are not oxygen or hydrogen. We have here to Corinne's, but then we only have one chlorine on the other side. So we're gonna put it to here. Then we're going to say we have to solve furs here, but only one over here. So we put it to here. Next balance. Oxygen's by adding water. We have no oxygen on the left half reaction. So move on to the one on the right. We have three oxygen's here, but then we have two times four here, so we have eight oxygen's on the product side. Alright, so eight oxygen's on this side. Three oxygen's on this side, which would mean I need to add five waters to this side that both sides have eight. Next balance, hydrogen. By adding h plus again, the half reaction on the left has no hydrogen is either. So move on to the one on the right. We have here five times to that's 10 hydrogen. So I'd have to put 10 h plus here on this side. Next, we have to balance out overall charge of each half reaction. Remember, at electrons to the more positively charged side of each half reaction here, call to has no charge. So the overall charge here is zero. Then we have two times minus one. So overall charge here is minus two. We add electrons, the more positive side. So the part the side with zero we had enough electrons to it. So it has the same overall charge as this side here. So to do that, I need to add two electrons. Alright, Next go to the other half reaction. We're going to say we have, uh, minus two year for our molecule or ion water has no charge for the overall charge. Here is minus two. Now we have two times minus two. That's minus four and then 10 times plus one is plus 10. So this idea is plus six. We need to add electrons had this more positive side. So that has the same overall charge as this side here. So to go from plus six to minus two would require that we add eight electrons. So I have eight electrons to this side. Realize that the number of electrons do not match. Here we have two electrons, but here we have eight. They have to be the same number of electrons. Lowest common multiple between them is eight. So I'd multiply everything here by four. So that be four c l two plus eight electrons. Here's me. Eight c l minus. Bring down the other half reaction. We don't have to multiply by anything, so it just comes down. Reactions come down. Everything on this side and they bring down the products. Okay. Remember to cross out your reaction intermediates. Things that look the same except ones are reacting in one's a product. Those will always be our electrons. They must always totally cancel out. There are no other reaction intermediates. So then bring down everything. Four l two plus s 203 to minus plus five h +20 gives me eight c l minus plus two sulfate ions plus 10 h plus. So this here represents my balanced redox reaction in an acidic solution.

Balance the following redox reaction in an acidic solution.Cl2 (g) + S2O32- (aq) → Cl- (aq) + SO42- (aq)

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Balance the following redox reaction in an acidic solution.
Cl₂ (g) + S₂O₃^2-(aq) → Cl^- (aq) + SO₄^2- (aq)