Warning!!!!

Status
Not open for further replies.

vapspaz

Vaping Master
ECF Veteran
Verified Member
Oct 21, 2010
4,218
8,873
SC
Do NOT do a google search for CE4+.

It will make your head hurt. :blink:

My god does my little head hurt from reading this crap. And I just have to ask myself why? Why didn't you just close that page and move on dumbazz. :lol:


1. Write down oxidation reaction of 1 mole of Cr(NCS)6(4-)knowing that all N goes into NO3(-), all C goes into CO2, all S goes into SO4(2-) and chromium changes oxidation state to +3. Since it is acidic medium, you can put either H(+) or H2O to the left part of equation. H(+) won't help you to balance oxygens, but H2O will (do not forget to put H(+) to the right now):
Cr(NCS)6(4-)+54H2O-xe -->Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+). Now, the number x of electrons that were transfered from Cr(NCS)6(4-) can be determined if you equalize the total charge of the left and right parts of the equation:
-4-x*(-1)=3+6*(-1)+6*(-2)+108*(+1)
x=97
So, first half of the equation will look like this:
Cr(NCS)6(4-)+54H2O-97e -->Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+).
Second half is very simple: Ce(4+) +1e --> Ce(3+)
Multiply the second half by 97 and combine both halves:

Cr(NCS)6(4-)+54H2O+97Ce(4+) --> 97Ce(3+)+Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+).



This has been a public safety message from your local dumbazz Vapspaz. ;)
 

Briar

Ultra Member
ECF Veteran
Jul 28, 2009
2,350
2,558
64
A fool on the hill in Deposit, NY
:lol: :lol: :lol:

Thanks, I was just about to do the search!


Do NOT do a google search for CE4+.

It will make your head hurt. :blink:

My god does my little head hurt from reading this crap. And I just have to ask myself why? Why didn't you just close that page and move on dumbazz. :lol:


1. Write down oxidation reaction of 1 mole of Cr(NCS)6(4-)knowing that all N goes into NO3(-), all C goes into CO2, all S goes into SO4(2-) and chromium changes oxidation state to +3. Since it is acidic medium, you can put either H(+) or H2O to the left part of equation. H(+) won't help you to balance oxygens, but H2O will (do not forget to put H(+) to the right now):
Cr(NCS)6(4-)+54H2O-xe -->Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+). Now, the number x of electrons that were transfered from Cr(NCS)6(4-) can be determined if you equalize the total charge of the left and right parts of the equation:
-4-x*(-1)=3+6*(-1)+6*(-2)+108*(+1)
x=97
So, first half of the equation will look like this:
Cr(NCS)6(4-)+54H2O-97e -->Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+).
Second half is very simple: Ce(4+) +1e --> Ce(3+)
Multiply the second half by 97 and combine both halves:

Cr(NCS)6(4-)+54H2O+97Ce(4+) --> 97Ce(3+)+Cr(3+) + 6NO3(-) + 6CO2 + 6SO4(2-) + 108H(+).



This has been a public safety message from your local dumbazz Vapspaz. ;)
 

zoiDman

My -0^10 = Nothing at All*
Supporting Member
ECF Veteran
Apr 16, 2010
41,697
1
84,954
So-Cal
LMAO! I was hopping some one other than me caught those mistakes.

You weren’t in Equilibrium but it didn’t want to say anything. Or at least I don't think you were.

:blink:

Hey my thing was Hard Core Mathematics when I did my Under Graduate work. But if there was one thing I Remember in Chemistry, besides that the girl with the short skirt who wanted to sit next to me who was Smoking Hot, was Equilibrium.







;)
 

Secti0n31

Super Member
ECF Veteran
Feb 13, 2011
733
166
Ohio
I ordered one of the ego fitting ce4's from one of yer competitors just to try it. Considering I love ego batts and ce2 type cartos I figure why not? However in light of that I had to cut myself off after my last GV purchase until the next paycheck. This last run should have me with 6 tanks, 5 new 30ml bottles of juice in addition to the 300ish ml I currently have, a ce4+, a VV mod in the shop and 4 working ego's (2 genuine at 3.4v, two knockoffs at 3.7v). Not to mention the handfulls of boge and ce2 cartos, or all of the cr123 batteries.

Compared to the crap we vapers do/buy/learn/tweak, that oxidation reduction titration chemical reaction should be child's play.
 
Status
Not open for further replies.

Users who are viewing this thread