For each of the following balanced oxidation–reduction
reactions,...

Question

For each of the following balanced oxidation–reduction
reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total
number of electrons transferred in each reaction.
(a) 2 MnO4-1aq2 + 3 S2 - 1aq2 + 4 H2O1l2 ¡ 3 S1s2 +
2 MnO21s2 + 8 OH-1aq2

Accepted Answer

Hi everyone. So I asked to identify each of the oxidation numbers for each element in the balanced redox reaction below. We're gonna look at each reactant and product individually. Well C. R. O. T minus a quiz and this contains oxygen, this is not a peroxide or a super oxide. I was gonna have an observation number of -2. And then chromium We need to calculate this overall charge of -1 on the compound. It's gonna be equal to the oxidation number of chromium plus two Times -2 because we have two oxygen and we have a negative to oxidation number for oxygen. So the oxidation number for chromium is gonna be plus three O. H minus a quiz oxygen which is not a peroxide or a super oxide. So it's gonna have an oxidation number of negative two and then we have hydrogen which we need to calculate have a negative one charge on the overall compound. It's gonna be equal to oxidation number of hydrogen plus negative two. For the oxidation number of oxygen number hydrogen We get plus one S two oh eight two -A. Careers get oxygen which is not a peroxide or super oxide. We're gonna get -2 for the oxidation number. And for sulfur We need to calculate this have overall charge of -2 on the compound. It's gonna be equal to two. The oxidation number of sulfur because there's two sulfur atoms plus eight. I was negative too because there's eight oxygen atoms and has a negative two oxidation number For Sulfur get Plus seven. Then we have c. r. 0. 4 two minus agreeance. You have oxygen just not a peroxide or super oxide. We get -2 for the oxidation number and we have chromium. If we need to calculate we have a negative to overall charge on the compound, let's go to the extension number of chromium plus four times negative two. There's four oxygen atoms and has an oxidation number of negative two of chromium get plus six H 20. You have oxygen which is not a peroxide or super oxide. We get -2. And for hydrogen we need to calculate of overall charge of zero because it is a neutral compound, It's equal to two. How's the oxidation of hydrogen? There's two hydrogen plus negative two. For the oxidation number of oxygen for hydrogen get plus one sl four two minus square. You have oxygen just not a peroxide or super oxide. We get -2 again. And for sulfur we need to calculate this negative to overall charge on the compound is equal to the oxidation number of sulfur last four times negative two because there's four oxygen and has an oxidation number of negative two. For Salford get plus six chromium, We're gonna get plus three plus six offer from plus seven plus six oxygen and from negative two and I get to hydrogen Plus one plus 1 plus one. Thanks for watching my video and I hope it was helpful

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