Balance the following redox reaction in a basic solution

Question

H₂O₂ (aq) + ClO₂(aq) → ClO₂^-(aq) + O₂ (g)

Accepted Answer

In this practice question, we're told to balance the following redox reaction in a basic solution. Alright. So following step 1, we know we look for elements that are not oxygen or hydrogen first. So here we have chlorine with chlorine. So that's gonna give us one half reaction which is c l o 2 gives me c l o two minus. Now all we have left is this compound and this compound. So by default that's gonna represent our other half reaction. So we have h two zero two gives us o 2. Now that we have that, next we balance elements that are not oxygen or hydrogen first. Here we have 1 chlorine and 1 chlorine, so that's already balanced. Now balance the number of oxygens by adding water. We have 2 oxygens on this side, 2 oxygens on this side. So it's already balanced. Let's go to the other socket. We have 2 oxygens here and 2 oxygens here, still balance 2. Alright. So move on to hydrogen. We balance hydrogens by adding h plus. In the first half reaction neither side has hydrogen. On the other half reaction, we have 2 hydrogens here, but none on this side. So put 2 h plus. Now the hard part, figuring out overall charge for each side of the two half reactions. C l o 2 here has no charge, so its overall charge on this side is 0. Here we have one negative one charge, so the overall charge is minus 1. Remember, add electrons to the more positive side so that it has the same overall charge as this side. So how do we go from 0 to minus 1? How many electrons do we need to add? We need to add 1 electron. Let's go to the other side. H202 which is hydrogen peroxide has no charge, so it's 0 for the overall side. Then here it's 2 times plus 1, so that's plus 2. Again, add electrons to the more positive side, so that equals this side. How do we go from a plus 2 charge to a, zero charge? We add 2 electrons. Notice your electrons do not match. When they don't match, we have to get to the lowest common multiple. So we'd have to multiply this equation by 2. Everything within it gets multiplied by 2. So 2 c l o 2 plus 2 electrons gives me 2 cl0 2 minuteus. Then we're going to say here for our other half reaction, we have h two zero two gives us o 2 plus 2 h+plus2electrons. Now cancel out reaction intermediates. The electrons, remember, are always your reaction intermediates. They're always supposed to totally cancel out. Now, bring down everything because those are our only reaction intermediates in this question. So we have 2 c l o 2 +2 gives me 2 c l o two minus plus o two plus 2 h positive. At this point we've done all the steps necessary to, balance it in acidic solution. But again, we're balancing it in basic solution, So we add additional step 7. We add o h minus to both sides based on the amount of h plus remaining. And remember, when you have h+ohminus together, they can combine to give us water. So bring down everything. 2cl02 +2+2ohminus gives us 2cl0two-+2 h two o. So here, h two o is only found on one side, so we don't have to cancel lock those h two o molecules on the product side at all. So this would be our final balanced equation in a basic solution.

Struggling with GOB Chemistry?

Balance the following redox reaction in a basic solution
H₂O₂ (aq) + ClO₂(aq) → ClO₂^-(aq) + O₂ (g)

Watch next

Balancing Redox Reactions: Basic Solutions practice set

Struggling with GOB Chemistry?

Balance the following redox reaction in a basic solutionH2O2 (aq) + ClO2 (aq) → ClO2- (aq) + O2 (g)

Watch next

Balancing Redox Reactions: Basic Solutions practice set

Balance the following redox reaction in a basic solution
H₂O₂ (aq) + ClO₂(aq) → ClO₂^-(aq) + O₂ (g)