Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: dudebuddyguy on November 25, 2013, 02:47:18 PM
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Oxidation half-reactions and Reduction half-reactions.
I've read over the explanation given:
- Assign oxidation numbers to each symbol.
- Determine which element has lost electrons from reactant to product. (The reactant charge will be larger, for example, –3 to 0).
- Determine which element has gained electrons from reactant to product. (The product charge will be larger. Ex +2 to –1).
- Write out each half reaction.
- Balance the half reactions so that electrons lost in oxidation = electrons gained in reduction.
And I'm still confused. Hopefully someone can "dumb it down a shade" (so to speak) and help me understand this.
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Have you tried to apply it to any reaction? Say,
Cl2 + H2SO3 + H2O :rarrow: HCl + H2SO4
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Have you tried to apply it to any reaction? Say,
Cl2 + H2SO3 + H2O :rarrow: HCl + H2SO4
Well the one question on my coursework is
CH4 + O2 -> CO2 + H2O
But I'm totally confused when I reach the 4th step (writing half reactions). I'm a little confused on the 2nd and 3rd step
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CH4 + O2 -> CO2 + H2O
So I've been watching a few videos on this, and I think I'm starting to get it, and it seems to be a bit easier than I first thought.
So, adding in the oxidation state numbers of everything, I get:
C(-4) H(+1) + O2(0) :rarrow: C(+4) O(-2) + H(+1) O(-2)
Carbon starts out with a -4 charge, and on the right, ends up with a +4 charge, so it ended up losing 4 electrons, correct?
Hydrogen starts out with a +1 charge, and ends up with a +1 charge, so I disregard that altogether for the oxidation/reduction equations?
O2 starts out with 0 charge, and ends up turning into two Oxygen molecules, each with a -2 charge. (-2) + (-2) = 0, so I disregard this as well?
So the oxidation half-reaction would be C(-4) :rarrow: C(+4) .
uh..+4e- ?
EDIT: Wait a mo, the Oxygen would be the reduction half-reaction. So (-2) + (-2) = (-4), so it gained +4e-?
EDIT 2: No wait, +2e-, because you don't add the two -2's together, they count as one -2.
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Carbon starts out with a -4 charge, and on the right, ends up with a +4 charge, so it ended up losing 4 electrons, correct?
No. After it lost 4 electrons its ON (oxidation number) was 0, and it changed further.
O2 starts out with 0 charge, and ends up turning into two Oxygen molecules, each with a -2 charge. (-2) + (-2) = 0, so I disregard this as well?
(-2) + (-2) is not zero.
So the oxidation half-reaction would be C(-4) :rarrow: C(+4)
Yes.
uh..+4e- ?
No. See above. Imagine you owe me $4 (so you are at -$4). How many bucks do you need to have $4 after you pay your debt?
EDIT: Wait a mo, the Oxygen would be the reduction half-reaction. So (-2) + (-2) = (-4), so it gained +4e-?
Yes.
EDIT 2: No wait, +2e-, because you don't add the two -2's together, they count as one -2.
No. There are two atoms per molecule, each gaining two electrons, so the reaction goes like
O2 + 4e- :rarrow: 2O(-2)
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I know I've got it correct now, let me explain the +4e-. After watching YouTube videos, that seems to be the common way to write down if electrons have been lost or gained.
I didn't see anyone using -4e-, so I naturally assumed that all electron statements, be it gain or loss, use "+"
EDIT: Wait...I'd need $8. What does 8 have to do with it?
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Initially ON was -4, at the end it is +4. It is exactly the same situation. You need to remove 4 electrons to get to ON zero, then you need to remove another 4 electrons to get to ON +4. 8 in total.
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Initially ON was -4, at the end it is +4. It is exactly the same situation. You need to remove 4 electrons to get to ON zero, then you need to remove another 4 electrons to get to ON +4. 8 in total.
Ohh yeahhhh. Damn my math skills. So ON would be +8e- (because + in chemical equations = loss, and - = gain, for some odd reason)
Seriously, why is chemistry so ass-backwards? Oxidation means loss of electrons, and reduction (which you think would be loss) means a gain of electrons. I FEEL LIKE I'M TAKING CRAZY PILLS. (obligatory Zoolander reference)
EDIT: O, not ON
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Ohh yeahhhh. Damn my math skills. So ON would be +8e- (because + in chemical equations = loss, and - = gain, for some odd reason)
Seriously, why is chemistry so ass-backwards? Oxidation means loss of electrons, and reduction (which you think would be loss) means a gain of electrons. I FEEL LIKE I'M TAKING CRAZY PILLS. (obligatory Zoolander reference)
EDIT: O, not ON
Nothing is backward here, perhaps you math ;) Trick is, electron charge is -e, not e. Blame physicists for that, it was their idea that the current flows in the opposite direction to the electron flow.
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Ohh yeahhhh. Damn my math skills. So ON would be +8e- (because + in chemical equations = loss, and - = gain, for some odd reason)
Seriously, why is chemistry so ass-backwards? Oxidation means loss of electrons, and reduction (which you think would be loss) means a gain of electrons. I FEEL LIKE I'M TAKING CRAZY PILLS. (obligatory Zoolander reference)
EDIT: O, not ON
Nothing is backward here, perhaps you math ;) Trick is, electron charge is -e, not e. Blame physicists for that, it was their idea that the current flows in the opposite direction to the electron flow.
Yes, my math is a tad backwards :P But you and Dan have both been most helpful and patient with me and I appreciate that :) I do admit that my original intention on joining this forum was to get quick and easy answers so I could quickly complete my online course, but those aren't my intentions any more. Onto the topic at hand;
So just "8e-" then. Most "teachers" on YouTube put "+" before the electron count which was a tad confusing. Should a + be placed before the electrons in any specific situations?
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Not sure what you refer to. It is nothing unusual to see the half reaction written as
Fe :rarrow: Fe2+ + 2e-
+2e- means just two electrons between the products (well, they almost never get separated and freely floating, but we have to list them somehow in the half reaction equation).
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Oxidation means loss of electrons, and reduction (which you think would be loss) means a gain of electrons.
Easy way to remember it is that reduction is reducing the charge. Adding electrons reduces the charge. Therefore reduction is adding electrons.
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Oxidation means loss of electrons, and reduction (which you think would be loss) means a gain of electrons.
Easy way to remember it is that reduction is reducing the charge. Adding electrons reduces the charge. Therefore reduction is adding electrons.
OILRIG Oxidation is Losing, Reduction is Gaining. Learned that on YouTube :P
Not sure what you refer to. It is nothing unusual to see the half reaction written as
Fe :rarrow: Fe2+ + 2e-
+2e- means just two electrons between the products (well, they almost never get separated and freely floating, but we have to list them somehow in the half reaction equation).
Ohhh, I see now. So the people on YouTube were just using + in their reaction formulas just like you explained, to show electrons between products.
So with my original equation, the correct answer would be
Original Formula: CH4 + O2 :rarrow: CO2 + H2O
Adding in Oxidation States: C-4 H+1 + O20 :rarrow: C+4 O-2 + H+1 O-2
Oxidation Half-Reaction: C-4 + 8e- :rarrow: C+4
Reduction Half-Reaction: O2 + 4e- :rarrow: 2O-2
Hopefully I've done everything properly :D
EDIT: After looking at my Oxidation and Reduction reactions, I'm starting to think I need to switch them around, so that the oxygen formula is the Oxidation reaction, and the Carbon formula is the reduction.
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You are almost there, just check your oxygen half reaction.
For the reaction equation to be balanced you need to have not only atoms balanced, but also a charge. What is total charge on the left? What is total charge on the right? (Please remember there is no such thing as C4- nor C4+, although these are reasonable notational tools at this moment).
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You are almost there, just check your oxygen half reaction.
For the reaction equation to be balanced you need to have not only atoms balanced, but also a charge. What is total charge on the left? What is total charge on the right? (Please remember there is no such thing as C4- nor C4+, although these are reasonable notational tools at this moment).
Judging from my knowledge gained from YouTube videos, I need to balance each individual chemical.
Carbon has 1 on each side, so that's balanced.
Hydrogen, which is not part of the half reactions, has 4 atoms on the left, 2 on the right. So I'd have to add 2 Hydrogen atoms on the right (but in which reaction? I have no clue lol)
Oxygen had O2 on the left, and 2O2 on the right. So we need to add another Oxygen atom to the left to balance it.
How to write all this out? I'm not sure. Thing is, my coursework says nothing about balancing the half-reactions. As long as I have the reactions written out, I'm good to go. But because this could appear on my final exam in the future, I should probably learn this.
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Sorry, I wrote oxygen, but I meant carbon. The oxygen half reaction:
Reduction Half-Reaction: O2 + 4e- :rarrow: 2O-2
is OK, but there is a problem here:
Oxidation Half-Reaction: C-4 + 8e- :rarrow: C+4
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Sorry, I wrote oxygen, but I meant carbon. The oxygen half reaction:
Reduction Half-Reaction: O2 + 4e- :rarrow: 2O-2
is OK, but there is a problem here:
Oxidation Half-Reaction: C-4 + 8e- :rarrow: C+4
Hmm....I'm thinking it's gotta be the 8e-.
Would the answer be -8e-?
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Why don't you try to put them on the right side?
At the moment on the left you have -4 (from carbon) and -8 (from electrons) - that is -12 in total. And only +4 on the right.
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Why don't you try to put them on the right side?
At the moment on the left you have -4 (from carbon) and -8 (from electrons) - that is -12 in total. And only +4 on the right.
I see it now!
C(-4) + C(+4) :rarrow: 8e-
Math is fun :P
EDIT: WAIT! I'm completely wrong! I know what to do :D
C-4 + 8e- :rarrow: 3C+4
-4 + 8 = 12 :rarrow: 3C x +4 = 12
EDIT 2: So I used this wonderful automatic balancing tool, and apparently C{-4} + 8e = C{+4} was incorrect. The correct balancing was C{-4} = 8e- + C{+4}. No additional charges needed!
Click here to visit the balancing calculator at WebQC.org (http://www.webqc.org/balance.php?reaction=C%7B-4%7D%3De%2BC%7B%2B4%7D) (it'll automatically pull the balanced equation up.)
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The correct balancing was C{-4} = 8e- + C{+4}. No additional charges needed!
I told you to put them on the right, didn't I?
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The correct balancing was C{-4} = 8e- + C{+4}. No additional charges needed!
I told you to put them on the right, didn't I?
I mis-read what you said lmao.
I'm onto the next equation which has Chlorine ions (hooray).
Mn + Cl2 -> MnCl7
Oxidation numbers: Mn{+2} + Cl{-2} = Mn{+2} + Cl{-7}
Reduction: Cl{-2} + -5e = Cl{-7}
The Manganese however hasn't lost or gained anything. It's sat at a solid +2. So the coursework noted that some questions may not have an Oxidation reaction, and this is obviously an example of this.
Grade me and let me know how I did (out of 10) :P
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Mn + Cl2 -> MnCl7
Oxidation numbers: Mn{+2} + Cl{-2} = Mn{+2} + Cl{-7}
I am afraid none of these is OK.
You have elemental forms on the left. http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method - rule number 2.
Then, on the right - start with the rule number 5, followed by the rule number 6.
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Mn + Cl2 -> MnCl7
Oxidation numbers: Mn{+2} + Cl{-2} = Mn{+2} + Cl{-7}
I am afraid none of these is OK.
You have elemental forms on the left. http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method - rule number 2.
Then, on the right - start with the rule number 5, followed by the rule number 6.
Update: Ok, so I've done everything properly (I think), so I'm done with that.
Now I just need to figure out WHICH REACTIONS ARE SPONTANEOUS AND NON-SPONTANEOUS without using that damned Gibbs Free Energy equation BECAUSE MY MATH SKILLS SUCK.