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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Duzzy on May 27, 2012, 05:41:50 PM

Title: Oxidation/eletrons
Post by: Duzzy on May 27, 2012, 05:41:50 PM

The reaction that occurs is: 6H+ + 5H2C2O4 + 2 MnO4
- ® 10 CO2 + 2Mn2+ + 8H2O.
A number of electrons needed for oxidation of one H2C2O4 molecule is:
A. 1, B. 2, C. 3, D. 5, E. 10

no idea where to start and where to begin. *delete me*

Title: Re: Oxidation/eletrons
Post by: Dan on May 28, 2012, 03:09:45 AM

Tart by identifying the oxidation state of Mn on both sides of the equation.

Title: Re: Oxidation/eletrons
Post by: Duzzy on May 28, 2012, 03:51:12 AM

+7

Title: Re: Oxidation/eletrons
Post by: Dan on May 28, 2012, 04:21:30 AM

Yes, +7 on the reactants side, now what about the product side? What is the oxidation state of Mn²⁺?

Title: Re: Oxidation/eletrons
Post by: Duzzy on May 28, 2012, 06:45:12 AM

so +7 on product side and +2 on state side.

Title: Re: Oxidation/eletrons
Post by: Dan on May 28, 2012, 09:36:57 AM

+7 on the reactants side, +2 on products side.

So, how many electrons do you need to go from +7 to +2?

Title: Re: Oxidation/eletrons
Post by: Duzzy on May 28, 2012, 09:47:17 AM

5

Title: Re: Oxidation/eletrons
Post by: Borek on May 28, 2012, 10:26:51 AM

Can you try to think by yourself, or do we have to spoon feed you all the time?

This is only partially correct answer. Yes, 5 electrons per each Mn atom present in the reaction, but in the balanced equation there are two atoms of Mn on each side. This is enough information to calculate numbers of electrons lost by the oxidizer, now just divide it by the number of the oxidized molecules.

Title: Re: Oxidation/eletrons
Post by: Duzzy on May 28, 2012, 02:21:16 PM

Ahhhh I get it now. Thanks!