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Topic: redox reactions  (Read 3642 times)

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Offline kdpk

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redox reactions
« on: March 10, 2007, 11:03:15 AM »
fe(2+) + Cr2O7 (2-) -----------> Fe (3+) + Cr (3+)

9 (Fe (2+) ------------> Fe (3+) + e-)

9e- + 14H(+) + Cr2O7 (2-) ------------> Cr (3+) + 7H2O

=

9Fe(2+) -----> 9Fe(3+) + 9e-
9e- + 14H(+) + Cr2O7 (2-) ------------> Cr (3+) + 7H2O

And now I just cancell out?

i cant balance the charges no matter what i do pls help  ???

Offline Albert

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Re: redox reactions
« Reply #1 on: March 10, 2007, 01:41:21 PM »
9e- + 14H(+) + Cr2O7 (2-) ------------> Cr (3+) + 7H2O

The number of electrons is wrong: what are the oxidation numbers of chromium in the reagents and in the products?

Offline Borek

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Re: redox reactions
« Reply #2 on: March 10, 2007, 03:32:13 PM »
Problem is not in the number of electrons, but in the lost chromium. Wrong number of electrons is just a side effect of trying to balance charge :)
« Last Edit: March 10, 2007, 03:43:55 PM by Borek »
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Offline Sat87

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Re: redox reactions
« Reply #3 on: March 16, 2007, 10:08:17 PM »
Looks like your product is missing a Cr molecule
 


9Fe2++14H ++Cr2O72----------------------> 9Fe3++ Cr3++ 7H2O

         On the right we have                                                          On the left we have
Fe:9 H:14 Cr:2 O:7                                Fe:9 H:14 Cr:1 O:7

Check your initial equation again just to make sure its right

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