Write unbalanced half-reactions for the following net ionic...

Write unbalanced half-reactions for the following net ionic equations: (a) $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{IO}_{4}^{-}(a q)$ (b) $\mathrm{NO}_{3}^{-}(a q)+\mathrm{SO}_{2}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{NO}_{2}(g)$

Write unbalanced half-reactions for the following net ionic equations:
(a) $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{IO}_{4}^{-}(a q)$
(b) $\mathrm{NO}_{3}^{-}(a q)+\mathrm{SO}_{2}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{NO}_{2}(g)$

Key Concepts

Redox Reactions

Redox reactions are chemical reactions that involve the transfer of electrons between species. In these reactions, one species undergoes oxidation (loses electrons) while another undergoes reduction (gains electrons). This concept is central to understanding how the reactants are transformed into products in many chemical processes.

Half-Reaction Method

The half-reaction method involves splitting a redox reaction into two separate expressions—one for oxidation and one for reduction. Each half-reaction is balanced individually for atoms and charge, which simplifies the process of balancing overall redox equations, whether they are in acidic, basic, or neutral media.

Oxidation States

Oxidation states (or oxidation numbers) are assigned to elements within compounds and ions to keep track of electron transfer in redox processes. Determining the oxidation state of each element helps identify which species are oxidized and which are reduced, serving as a guide to writing the correct half-reactions.

Unbalanced Half-Reactions

Writing unbalanced half-reactions involves formulating separate expressions for the oxidation and reduction processes without yet balancing atoms or charge. This initial step is crucial for understanding the individual electron transfer processes before applying further steps to balance the overall reaction.

Transcript

00:01 Okay, so this question is asking us to find the unbalanced half reactions of these certain reactions.

00:06 So how you can find an unbalanced half reaction is you actually take the overall reaction that you're given, and you want to find the oxidation number and see which elements are being reduced and which are being oxidized, because then it'll allow you to figure out which is the oxidation half reaction, which is the reduction half reaction.

00:40 Okay, so if we take a look here, we have oxygen, which we can automatically say is negative 2.

00:47 So this is negative 8 total.

00:49 Since this is still a negative charge overall, then we can say that mn.

00:55 Inganese is a positive 7.

00:59 And then if we take a look at the manganese over here, we have oxygen.

01:02 Which is a negative 2, so it's a negative 4 total, since this is a neutral chart, and this would be a positive 4.

01:10 So went from a positive 7 to a positive 4, meaning it has been reduced.

01:20 Now looking at iodine, so we have oxygen, again, negative 2, so it's negative 6.

01:25 But since this is still a negative, then that would mean that iodine would be a positive 5.

01:33 Take a look over here, oxygen, negative 2, negative 8 total, but since then since then it would be a positive 5.

01:38 Still a negative, it means it would be a positive 7.

01:42 So iodine went from a positive 5 to a positive 7, meaning it has been oxidized.

01:51 So our half reactions would be, for the reduction half reaction, it would be mn 04 negative, nucleus to mno2.

02:09 Solid, since these are unbalanced, we can leave it like that and then for the oxidation half reaction i'm actually going to do it down here the oxidation half reaction we have i have 3 negative aqueous to i have 4 negative aqueous since they're unbalanced we can leave it like that so that is our answer for the first part and then we're going to do the exact same thing for the second part so the second part, we're going to do the exact same thing.

02:56 We're going to take a look at the equation, take a look at the reaction, and see what is being oxidized and what is being reduced, and then those would be our half reactions.

03:26 Okay, great.

03:27 So we have n -o -3 -aquias plus s -o -2, aqueous plus n -o -2.

03:44 Yes.

03:45 So if i look at nitrogen, you see, it's a next to oxygen, oxygen is down negative 2.

03:51 This is negative 6, but there's still that negative charge.

03:55 So then that would be a positive 5...

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