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Topic: Balancing Oxidation Reduction Reaction  (Read 6163 times)

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Offline XxslbabesxX

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Balancing Oxidation Reduction Reaction
« on: September 29, 2005, 07:06:08 PM »
I am having a hard time balancing this redox problem:

CH3OH + Cr2O7-2 ---> CH2O + Cr+3

First I found the oxidation #s then I separated them, balanced them, and added the necessary  H2O and H+ so that it looked like this.

CH3OH ---> CH2O  + 2H+

Cr2O7-2 + 14H+ ---> 2Cr+3 + 7H2O

When I went to find the change of electrons for each I found that Carbon had a change of 2, but Cr did not change at all because of the coefficient  2 before Cr+3.  2Cr+3 and Cr2O7 ended up equaling to six. I have no clue what to do from here. I would really appreciate a steer in the right direction. Thanks! :)
« Last Edit: September 29, 2005, 07:08:11 PM by XxslbabesxX »

Offline Borek

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Re:Balancing Oxidation Reduction Reaction
« Reply #1 on: September 29, 2005, 07:27:43 PM »
Add electrons to both half reactions to balance charge. Rest should be easy.
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Offline mike

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Re:Balancing Oxidation Reduction Reaction
« Reply #2 on: September 29, 2005, 07:58:16 PM »
Cr2O72- (Cr is +6)

Cr3+ (Cr is +3)
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Offline XxslbabesxX

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Re:Balancing Oxidation Reduction Reaction
« Reply #3 on: September 29, 2005, 09:03:49 PM »
Cr2O72- (Cr is +6)

Cr3+ (Cr is +3)

But the coeffiecient before Cr3+ is 2 so that means the number of electrons is 6 which is the same as Cr2O72-....or those that not matter?

Offline Borek

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Re:Balancing Oxidation Reduction Reaction
« Reply #4 on: September 29, 2005, 09:05:34 PM »
But the coeffiecient before Cr3+ is 2 so that means the number of electrons is 6 which is the same as Cr2O72-....or those that not matter?

There are 2*3+ and 2*6+, not 2*3+ and 6+ - you have two Cr on both sides of equation.
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Offline mike

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Re:Balancing Oxidation Reduction Reaction
« Reply #5 on: September 29, 2005, 09:16:15 PM »
one Cr6+ gains 3 electrons to form one Cr3+

two Cr6+ gains 6 electrons to form two Cr3+

etc...
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