Question

4MnO4 + 3N2O + 2OH -> 6NO2 + 4MnO2 + H2O In the above reaction, the oxidation state of nitrogen changes from v to ? How many electrons are transferred in the reaction? 

          4MnO4 + 3N2O + 2OH -> 6NO2 + 4MnO2 + H2O
In the above reaction, the oxidation state of nitrogen changes from v to ?
How many electrons are transferred in the reaction?
4mno43n2o2oh 6no24mno2h2o in the above reactionthe oxidation state of nitrogen changes from vto how many electrons are transferred in the reaction 79534

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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4MnO4 + 3N2O + 2OH -> 6NO2 + 4MnO2 + H2O In the above reaction, the oxidation state of nitrogen changes from v to ? How many electrons are transferred in the reaction?
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Transcript

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00:02 In this question we are given this equation and we are asked first of all about the oxidation state of nitrogen and how it changes.
00:10 So to answer that we need to assign oxidation numbers and specifically only for the hydrogen.
00:17 So in our first compound on the reactant side we have n2h4.
00:21 Well we know that the oxidation number of hydrogen is always positive one except when hydrogen is in a metal hydride and this is not a metal hydride.
00:30 So hydrogen is positive one and there are four of those for a total of positive four.
00:39 Well we know for a compound, a neutral compound, when we add the oxidation numbers together they have to add up to zero and there are two nitrogen here.
00:48 So we need to determine what x is so that when we add it to four we get zero.
00:55 Well solving this equation for x gives us negative two.
00:59 The oxidation number of nitrogen is negative two...
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