Write unbalanced oxidation and reduction half-reactions for the f...

Question

Write unbalanced oxidation and reduction half-reactions for the following processes. . 
(a) Mn(s) + NO3-(aq) -> Mn2+(aq) + NO2(g)

Accepted Answer

Hey everyone welcome back. So let's get started with this video. So here they want us to identify the reduction half reaction and the oxidation have. Okay, so then first we need to write the two half reactions to figure out which one is with. Okay, so then here we have c. 00. two and then here we also have Ceo. So then this is going to be one of us have reactions. Here we have the r here we have pR as well. So then this is going to compose second interaction. So then our first one is going to be C O. Two. Going to see, oh you're correct. The other half reaction is going to be we are too going to H B R. Oh okay. So then the next step is going to look at oxidation numbers for each element. Let's get started with this one. So then we know that oxygen has a charge of two. So that we're going to solve for the charge of C. O. So then you say X. There's two, oxygen's. So then two, The charge is -2 And the overall charge of this compound is zero. So when we solve for X, we're solving for C. O. And that is a charge positive for. So then ceo is a charge of plus four. And here And went to a charge of plus three. So as you can see the value decreased, a decrease because they gained an electron. So this one is going to be our reduction. So then this is our word half reaction. So then I took out the other one. So then here we have H B R. O. We know H is a plus one Pr here we're solving for it. So it's X. And then once again we know Oh is negative. two Overall charge of this compound is zero. So then we sulfur X, which is B. R. And we get plus one. So then here the charge of pr is plus one. Well here it's in its natural state as its in di atomic form. So it's zero. So here the charged increased. So since it's increase, that means it lost electrons. So then this is going to be oxidation. This is our oxidation half reaction. Okay, thank you for watching. I hope this helped. And I'll see you in the next video.

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