Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)

a. S(s) + HNO3(aq) → H2SO3(aq) + N2O(g) (acidic solution)
b. BrO3-(aq) + N2H4(g) → Br-(aq) + N2(g) (acidic solution)
c. H2O2(aq) + ClO2(aq) → ClO20(aq) _ O2(g) (basic solution)

Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)

a. S(s) + HNO3(aq) → H2SO3(aq) + N2O(g) (acidic solution)

b. BrO3-(aq) + N2H4(g) → Br-(aq) + N2(g) (acidic solution)

c. H2O2(aq) + ClO2(aq) → ClO20(aq) _ O2(g) (basic solution)

Accepted Answer

Hello everyone here we have a question asking us to balance the following redox reaction under basic conditions and identify the oxidizing and reducing agents. We have chromium oxide acquis plus proxy die sulfate. Acquis forms chrome eight. Acquis plus sulfate acquis. So our first step is going to be separate the whole reaction into two half reactions. So that's going to be chromium oxide forms chrome eight and peroxide. A sulfate worms sulfate. And now our second step is to balance the non hydrogen and non oxygen elements. So our chromium is good. And on our proxy dot sulfate forming Sulfate we need to add a two in front of our sulfate. And our third step is to balance oxygen by adding water to the side that needs it. So we will add chromium oxide. We need to oxygen's so to water worms crow mate. And we are good with oxygen's with our other one. And now we need to balance our hydrogen by adding hydrogen ion to the side that needs hydrogen. So we'll have chromium oxide plus two water forms chrome eight plus four hydrogen. And we'll have proxy di sulfate forms to sulfate because that was good with hydrogen. So now we need to balance the charges by adding electrons to the more positive side. So we need three electrons on our first reaction and that's so it has a negative one charge. Like our chromite our diet our chromium oxide and for our proxy isil fight We need two electrons and now we need to balance electrons on two half reactions. So we need to multiply our first one by two And our 2nd 1 x three too. So they'll both have six electrons And we're going to multiply the other one x and losing electrons is oxidation. So our reducing agent is our chromium oxide and gaining electrons reduction. And so that is our oxidizing agent. So that is our peroxide. Isil fight. So that is our reducing agent and that is our oxidizing agent. Next we're going to combine our hydrogen with our hydroxide to form water and there is no hydrogen or hydroxide on our second reaction. So that's just going to stay proxy di sulfate plus six electrons forums six sulfate. And now we're going to cancel our common species. So we have four water on our left side and we have eight on our right. So that's going to reduce that to four. And our six electrons are going to cancel out. So our balanced redox reaction is going to be to chromium oxide. Acquis plus eight hydroxide. Acquis plus three Proxy. Di sulfate acquis forms too chromium to chrome eight. Aquarius plus four water liquid plus six sulfate queers. And that is our final balanced redox reaction. Thank you for watching. Bye

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.) H2O21aq2 + ClO21aq2 ¡ ClO2-1aq2 + O21g2 (basic solution)

Verified Solution

Key Concepts

Oxidation and Reduction

Oxidizing and Reducing Agents

Balancing Redox Reactions in Basic Solution