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Topic: Someone help me understand this  (Read 8432 times)

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Offline Borek

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Re: Someone help me understand this
« Reply #15 on: November 26, 2013, 04:44:43 PM »
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:

Quote
Oxidation Half-Reaction: C-4 + 8e- :rarrow: C+4
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Offline dudebuddyguy

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Re: Someone help me understand this
« Reply #16 on: November 26, 2013, 05:04:39 PM »
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:

Quote
Oxidation Half-Reaction: C-4 + 8e- :rarrow: C+4

Hmm....I'm thinking it's gotta be the 8e-.

Would the answer be -8e-?

Offline Borek

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Re: Someone help me understand this
« Reply #17 on: November 26, 2013, 05:13:19 PM »
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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Offline dudebuddyguy

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Re: Someone help me understand this
« Reply #18 on: November 26, 2013, 05:17:15 PM »
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 (it'll automatically pull the balanced equation up.)
« Last Edit: November 26, 2013, 05:46:20 PM by dudebuddyguy »

Offline Borek

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Re: Someone help me understand this
« Reply #19 on: November 26, 2013, 05:55:13 PM »
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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Offline dudebuddyguy

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Re: Someone help me understand this
« Reply #20 on: November 26, 2013, 06:08:38 PM »
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
« Last Edit: November 26, 2013, 06:19:18 PM by dudebuddyguy »

Offline Borek

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Re: Someone help me understand this
« Reply #21 on: November 26, 2013, 06:28:24 PM »
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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Offline dudebuddyguy

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Re: Someone help me understand this
« Reply #22 on: November 26, 2013, 06:38:43 PM »
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.
« Last Edit: November 26, 2013, 08:22:27 PM by dudebuddyguy »

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