anonymous
  • anonymous
Using the half-reaction method, balance the following redox reaction occurring in acidic solution: NO3- + I2 ---> IO3- + NO2
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Frostbite
  • Frostbite
Okay a secure way to solve any redox problem is in the following order (unless very special cases) Oxidation-numbers (we want the number of electrons right) Charge. (In acid solution we use H+, base OH- Hydrogen. Oxygen. I suggest for you own good you try write each step and I correct you, in this way you get the most practice... I promise you I won't leave until we have come to conclusion.
anonymous
  • anonymous
Ok, Sounds good!
anonymous
  • anonymous
So first I have to separate the equation into two half reactions into oxidation and reduction portions.

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Frostbite
  • Frostbite
You don't have to, because we can handle them both using oxidation numbers. Are you familiar with that term?
anonymous
  • anonymous
hahaha no. :P Let me familiarize myself real quick.
anonymous
  • anonymous
Oxidation number is a measure of the charge of an atom.
Frostbite
  • Frostbite
True but in this case oxidation numbers is a way to keep track of the electrons in our redox reaction and we have a few rules that can help: In most cases: Hydrogen: +1 Oxygen -2 The sum of oxidation numbers have to equal the total charge.
Frostbite
  • Frostbite
So lets take as an example: NO3- Oxygen: -2 but we have 3 so it is 3*(-2) = -6 the total charge is -1 so N most have the oxidation number +5 as -1=-6+5.
Frostbite
  • Frostbite
Try NO2 for me :)
anonymous
  • anonymous
I2= 0 NO3=+5 IO3= +5 NO2=+4 Is this right?
Frostbite
  • Frostbite
As true as said.
Frostbite
  • Frostbite
well done. Now we analyse what oxidation numbers that change and use that to find out numbers to put in the redox reaction :)
anonymous
  • anonymous
Woah, woah, woah! Slow down there cowboy!
anonymous
  • anonymous
How do I go about doing that?
Frostbite
  • Frostbite
okay. We notice that the only oxidation numbers that change is N and I right?
Frostbite
  • Frostbite
Think I'm gonna do a worked example, you can watch here while we do the rest:
1 Attachment
Frostbite
  • Frostbite
Now we see that N go from +5 to +4 so it go 1 down. I go from 0 to +5 so it go 5 up. Now the numbers have to equal each other. So we say: 5 up* 2 (because we have I2) = 10 up Well then we need for I: 1 down * 10 = 10 down. that makes the number of electrons right and is secured.
Frostbite
  • Frostbite
Oh wait.... you have to use the half reaction method my god I'm sorry. did not see that.
Frostbite
  • Frostbite
call me stupid and forget everything I said! D;
anonymous
  • anonymous
No worries!! I was a little confused anyways! You are definitely NOT stupid! :D
Frostbite
  • Frostbite
But in this case.. enlighten me! :D
anonymous
  • anonymous
Ok, so I've been going over my notes to try and solve this problem and this is what I ended up with. NO3- + 2H ---> NO2 + H2O
anonymous
  • anonymous
Wait.... Something's wrong here..
anonymous
  • anonymous
oh my... *sigh* I guess I'll try this again!
anonymous
  • anonymous
8 H+(aq) + 10 NO3-(aq) + I2(s) ---> 2 IO3-(aq) + 10 NO2(g) + 4 H2O(l)
Frostbite
  • Frostbite
Got the same using my old method. ;)
anonymous
  • anonymous
Alright! Thank you so so much for your cooperation and help! Sorry about earlier, I'm just a sensitive person. :P
Frostbite
  • Frostbite
Not a problem and about the thing not being strong in chemistry... you can't fool me... you know it, just need to trust your self :)
anonymous
  • anonymous
Thank you :)

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