# Balancing?

Discussion in 'DAT Discussions' started by Dencology, May 28, 2008.

1. ### Dencology 2+ Year Member

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Guys i am going crazy over this. i think the answer is not right. but can you check it?

MnO4-(aq) + H2C2O4(aq) _> Mn2+(aq) + CO2(g) + H2O(l) + H+(aq)

choose the correct order of coefficients:

a. 2,2,5,6,8,10
b. 2,5,6,2,10,8
c, 2,10,8,5,2,6,
d. 10,8,4,5,6,8,
e. 1,5,1,10,4,2,

ans. E. i

my solution:

2(5e-+ 8H+ +Mno4-__> Mn2+ + 4H2o) 1st half reaction

5(H2C2O4__> 2CO2 + 2H+ +2e-) 2nd half reaction

10e- + 16H+ +2MnO4- + 5H2C2O4 --> 2Mn2+ + 8H2O + 10CO2 + 10H+ + 10e-

now simplyfining:

6H+ + 2MnO4- + 5H2C2O4--> 2Mn2+ + 8H2O + 10CO2

so the coefficient is 6,2,5,2,8,10. but it is not part of the choices. what is wrong?

2. ### dentalplan 2+ Year Member

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first off ihave 2 questions. where did u get this question? And secondly, is the question presented exactly as you wrote it, i.e, is
MnO4-(aq) + H2C2O4(aq) _> Mn2+(aq) + CO2(g) + H2O(l) + H+(aq) already written down for you?

Ok anyway, You have the balanced redox reaction correct. I did it out and I go the exact same thing. However
MnO4-(aq) + H2C2O4(aq) _> Mn2+(aq) + CO2(g) + H2O(l) + H+(aq)
is chemically impossible, methinks.

If you just look at it, they have H+ on the right side. In our balanced reaction, we have H+ on the left. Just that right there spells something suspicious. So i'm thinking whoever wrote this question was either high, or some other kind of drug, cuz i don't know why E would be the right answer. You said they put E as the right answer, right? I mean E even includes 2 species with coefficients of 1. Thats messed up right there because in our correct version of the balanced equation we don't even have any species with a coefficient of 1.

So i'm just wondering where you got this question, cuz i'm almost certain there are some errors here. do they give any explanation in the book you're using?

3. ### Zerconia2921 Bring your A-game! 2+ Year Member

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I agree while doing the half reactions for H2C2O4 I was not getting the correct e-

For example:

H2C2O4 ---> CO2 H2c2o4 carbon has a +6 charge but for CO2 it has a +4 charge. The electrons should cancel out and to do so they need to be on opposite sides of the reactions. The first half reaction is correct for MnO4 to Mn2+ . Double check if the reaction is written correctly.

Once you get the right number of electrons that cancel out, then look at the number of oxygen atoms. Then depending on what factor used to cancel the electrons add the same amount of water on the other side to balance oxygen. Lastly, balance it off with H+ if your in acidic solution or OH- if your in a basic.

4. ### Dencology 2+ Year Member

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i agree with all of you. i also went crazy over this. but after all we know how to balance them.

5. ### dentalplan 2+ Year Member

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he/she has everything balanced correctly. the matter at hand pertains to the ****tiness of the sample problem. we're tryin to figure out if there is an error on their part.

im basically 100% sure there is. but i hate getting too cocky lol cuz then its an ugly fall , lol.

6. ### Zerconia2921 Bring your A-game! 2+ Year Member

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Ohhhhhhhhhhh righty there somone drinking too much red bull

7. ### nixon13 7+ Year Member

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Easier way to do this is look at the species with Mn. That is the 1st and 3rd coefficient. You know those two are going to be the same coefficient so eliminate the answers with differect coefficients for the 1st and 3rd species. Took me about a second to do, rather than trying different possibilities.

8. ### doc toothache 10+ Year Member

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+7.............(+5e)............. +2
MnO4-(aq) + H2C2O4(aq) _> Mn2+(aq) + CO2(g) + H2O(l) + H+(aq)
.....................I.................................I
......................(+3)..........(-1e)..........(+4)

(Carbon goes from +3 to +4).
To balance equation loss of electrons is multiplied by 5, giving 10H on left side. The hydrogens on the right side are then balanced to obtain 10H. With 10 C on the left side, multiply CO2 by 10. The equation becomes:

MnO4-(aq) + 5H2C2O4(aq) _> Mn2+(aq) + 10 CO2(g) + 4H2O(l) + 2 H+(aq)