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

4 Cl 2 (g) + S 2 O 3 2- (aq) + 5 H 2 O (l) → 8 Cl - (aq) + 2 SO 4 2- (aq) + 10 H + (aq)

Balance the following redox reaction in an acidic solution. Cl 2 (g) + S 2 O 3 2- (aq) → Cl - (aq) + SO 4 2- (aq)

Here it says to balance the following redox reaction in an acidic solution. So remember, step 1, we look for elements that are not or hydrogen to determine our 2 half reactions. Here we have chlorine matched up with chlorine, so that's one half reaction. So c l 2 gives me c l minus. Then we also have sulfur with sulfur. So s203 2 minus gives me s04 2 minus. Step 2, we're gonna balance elements that are not oxygen or hydrogen. We have here 2 chlorines, but then we only have 1 chlorine on the other side, so we're gonna put a 2 here. Then we're going to say we have 2 sulfurs here, but only 1 over here, so we put a 2 here. Next, balance oxygens by adding water. We have no oxygens on the left half reaction so move on to the to the one on the right. We have 3 oxygens here, but then we have 2 times 4 here so we have 8 oxygens on the product side. Alright? So 8 oxygens on this side, 3 oxygens on this side, which would mean I need to add 5 waters to this side so that both sides have 8. Next, balance hydrogens by adding h plus. Again, the half reaction on the left has no hydrogens either, so move on to the one on the right. We have here 5 times 2, that's 10 hydrogens. So I have to put 10 h+ here on this side. Next, we have to balance out overall charge of each half reaction. Remember, add electrons to the more positively charged side of each half reaction. Here, c l two has no charge, so the overall charge here is 0. Then we have 2 times minus 1, so overall charge here is minus 2. We add electrons to the more positive side, so the part the side with 0. We add enough electrons to it, so it has the same overall charge as this side here. So to do that I need to add 2 electrons. Alright. Next, go to the other half reaction. We're gonna say we have, minus 2 here for our molecule or ion. Water has no charge so the overall charge here is minus 2. Now we have 2 times minus 2, that's minus 4. Mhmm. And then 10 times plus 1 is plus 10. So this side here is plus 6. We need to add electrons to this more positive side so that it has the same overall charge as this side here. So to go from plus 6 to minus 2 would require that we add 8 electrons. So we add 8 electrons to this side. Realize that the number of electrons do not match. Here we have, 2 electrons but here we have 8. They have to be the same number of electrons. Lowest common multiple between them is 8, so I multiply everything here by 4. So that'd be 4 c l 2 plus 8 electrons gives me 8 c l minus. Bring down the other half reaction, we didn't have to multiply by anything so it just comes down. Reactants come down, so everything on this side, and then bring down the products. Remember to cross out your reaction intermediates, things that look the same except one's a reactant and one's a product. Those will always be your electrons. They must always totally cancel out. There are no other reaction intermediates, so then bring down everything. 4 c l 2+2o32minusplus5h2o gives me 8 c l minus plus 2 sulfate ions plus 10 h plus. So this here represents my balanced redox reaction in an acidic solution.