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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.
Ok, Sounds good!
So first I have to separate the equation into two half reactions into oxidation and reduction portions.
You don't have to, because we can handle them both using oxidation numbers. Are you familiar with that term?
hahaha no. :P Let me familiarize myself real quick.
Oxidation number is a measure of the charge of an atom.
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.
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.
Try NO2 for me :)
I2= 0 NO3=+5 IO3= +5 NO2=+4 Is this right?
As true as said.
well done. Now we analyse what oxidation numbers that change and use that to find out numbers to put in the redox reaction :)
Woah, woah, woah! Slow down there cowboy!
How do I go about doing that?
okay. We notice that the only oxidation numbers that change is N and I right?
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.
Oh wait.... you have to use the half reaction method my god I'm sorry. did not see that.
call me stupid and forget everything I said! D;
No worries!! I was a little confused anyways! You are definitely NOT stupid! :D
But in this case.. enlighten me! :D
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
Wait.... Something's wrong here..
oh my... *sigh* I guess I'll try this again!
8 H+(aq) + 10 NO3-(aq) + I2(s) ---> 2 IO3-(aq) + 10 NO2(g) + 4 H2O(l)
Got the same using my old method. ;)
Alright! Thank you so so much for your cooperation and help! Sorry about earlier, I'm just a sensitive person. :P
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 :)
Thank you :)