Balance each redox reaction occurring in basic aqueous solution. a. H₂O₂(aq) + ClO₂(aq) → ClO₂^–(aq) + O₂(g)

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hey everyone in this example, we need to consider the redox reaction below and we need to balance this reaction under basic conditions. So what we're going to refer to is because this is redox, we need to separate this whole reaction into half reactions. So looking at our first reactant, we have our reaction our bromate an eye on so that's sorry B r oh three minus. And this is going to produce our an ion bromide. For our second half reaction we have hydro sign. So that's end to H four gas. Which sorry, that should say G for gas which produces our product and two gas. So this is nitrogen gas. So we need to take note of the non hydrogen and non oxygen atoms. So for our first half reaction, that's booming and we need to make sure that grooming is balanced on both sides. We have one mole of roaming here and one mole of roaming here. So this brooding is balanced for our second half reaction. We're looking at the non oxygen and non hydrogen atoms that would be nitrogen here. We have two moles of nitrogen and two moles of nitrogen on the other side. So we would say that nitrogen is also balanced in this half reaction. And so now we're going to move forward and focus on balancing our oxygen. And we should recall that one mole of oxygen is equivalent to one mole of water. And so we can balance out oxygen by using water. So what we would say is that for our first half reaction we have three moles of oxygen here and we are missing oxygen on the product side. So we would go ahead and bounce out the oxygen by saying that we have for our bromate, we produce our bromide an ion plus three moles of water to bounce out our oxygen. So now we would have three oxygen's on the product side, and three oxygen's on the reactant side for this first half reaction. And now for our second half reaction to bounce out the hydrogen rather, we should recall that one mole of hydrogen is equivalent to one mole of H plus proton. And so we can use H plus to bounce out hydrogen. So it's clear that we have formals of hydrogen on the product or on the reactant side here. And we don't have any hydrogen on the product side. So that's where we will add our H Plus. So what we should have is our hide, resign so and two H 4, sorry, produces nitrogen gas Plus four moles of H plus and we should have a grease labels here for water and H Plus. Now, we also need to make sure we have hydrogen balanced for our first half reaction because it's bounced for our second half reaction. But when we added the three moles of water to balance out the oxygen, we changed the hydrogen here. And so we need six moles of hydrogen here. So we would go ahead and expand this equation so that we also on the reactant side. Sorry, let's just make room here. So on our reactant side we want to add six H plus, so plus six H plus. And this would also have an Accu label so that now we have both hydrogen and oxygen balanced. And this applies to both of our half reactions now. So our next step is to balance out our charges. And for our first half reaction, we would recognize that on our product side we have a net charge of minus one coming from the bromide and i on here and then on our reactant side we have a net charge of we have a coefficient of six in front of this Plus. So that means plus six minus one would be minus five. So we want to go ahead and get a net charge of minus one on this side by canceling out this plus six charge here and we would do so by adding six electrons. So what we should have is now the bromate, an eye on plus six H plus plus six electrons yields our bromide, an ion Plus our three moles of water. Likewise here, for our second half reaction, we have a net charge on the product side of plus four because we have that coefficient of four. And here we have a net charge of zero because this is a neutral compound. So we need to cancel out this plus four charge by adding four electrons. So we would do that below. We would have our hydro zayn, Oh sorry, that's a gas. So hydrogen gas yields are nitrogen gas Plus four h plus plus for electrons to balance out that charge. And it's clear that we've added electrons to the more positive sides of both of our half reactions. Here, we have a plus four charge. So we added four electrons here, we had a plus six charge. So we added six electrons. And now our next step is to go ahead and make sure that we have the same electrons for both half reactions. Here we have six electrons and here we have four electrons. So we need to think of a factor that can make these electrons the same and multiply by that factor. So for our first half reaction we would multiply this equation by two. And then for our second half reaction, we would multiply this equation by three to get our electrons to equal each other. And this would give us a total of 12 electrons for both of our reactions. So we can rewrite these reactions. So now we should have two moles of our bromate Plus 12 moles of each plus Plus 12 electrons, yields 12 moles of the bromide. An ion Plus six moles of water. Likewise, for our second reaction, we multiplied by three. So we would have three moles of hydrogen gas, Yields three moles of sorry, nitrogen gas Plus moles of H plus plus 12 electrons. So now our electrons are matching on both of our half reactions. And now we want to cancel out what is the same but on opposite sides of our equations. So now that our electrons match on both equations, we can cancel them out as well as our age plus. So we can get rid of that on the reactive side here as well as on the product side for a second half reaction. And then the electrons here and then for our second half reaction as well. And this would give us our overall reaction here where we would add these two reactions so that we now have just one equation And this would give us two moles of roommate Plus three moles of hydrogen. And this is a gas, Yields two moles of our bromide, an ion. And just a correction here. This should be two, not 12 because we only multiplied the two to the bromide an eye on here. So then this is added to six moles of water Plus our three moles of nitrogen gas. And so for our final answer, this is going to be our balanced redox reaction that is balanced under basic conditions. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Balance each redox reaction occurring in basic aqueous solution. a. H₂O₂(aq) + ClO₂(aq) → ClO₂^–(aq) + O₂(g)

Verified Solution

Key Concepts

Redox Reactions

Balancing Redox Reactions in Basic Solution

Half-Reaction Method

Balance each redox reaction occurring in basic aqueous solution. a. H2O2(aq) + ClO2(aq) → ClO2–(aq) + O2(g)

Verified Solution

Key Concepts

Redox Reactions

Balancing Redox Reactions in Basic Solution

Half-Reaction Method

Balance each redox reaction occurring in basic aqueous solution. a. H₂O₂(aq) + ClO₂(aq) → ClO₂^–(aq) + O₂(g)