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Topic: Oxidation numbers  (Read 6879 times)

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Offline Kaleyrvt

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Oxidation numbers
« on: January 22, 2007, 01:43:22 PM »
Hey all-
Trying to figure out if I am doing this right-doing a chem course via correspondance
 Basically need to see if I am on the right track with these???
78)   Determine the oxidation numbers of the underlined elements in the following
formulas:

a) KClO3  =>   1+  +  x  =  3 x 2-
      1+  +  x  =  6-
      1+  +  x   +  6-  =  0
      Cl = x = 5+
       
b) NH4+    =>   x  +  4  x    1+
      x  +  4+           
      x  +  4+  =  0
      N = x =  4-
                  
c) Cr2O72-  =>   x  +  7 x 2-
      x  +  14-
      x  +  14- =  0
      Cr = x = 14+

d) MnO42-  =>   x  +  4 x 2-
      x  +  8-
      x  +  8- =  0
      Mn = x = 8+

Is this the right way to do this?? i have a feeling I am doing it wrong...Please give me any feedback possible
Thanks

Offline Dan

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Re: Oxidation numbers
« Reply #1 on: January 22, 2007, 02:10:02 PM »
a) is correct, you have the right idea, but for the others you must take into account the overall charge of the ion.

eg. for b), you say x + 4+ = 0. This is incorrect, it's not equal to 0, it should be something else...
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Offline DevaDevil

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Re: Oxidation numbers
« Reply #2 on: January 22, 2007, 02:12:16 PM »
And for c), don't forget there are 2 Cr atoms in the compound

Offline Borek

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Re: Oxidation numbers
« Reply #3 on: January 22, 2007, 03:05:28 PM »
1+  +  x  =  3 x 2-
1+  +  x  =  6-
1+  +  x   +  6-  =  0

Finally, you better check your math...
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Offline Kaleyrvt

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Re: Oxidation numbers
« Reply #4 on: January 22, 2007, 03:29:41 PM »
I am confused
 ???

Offline chiralic

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Re: Oxidation numbers
« Reply #5 on: January 22, 2007, 04:14:40 PM »
Hint: The neutral compound or element, the sum of the oxidation states is zero

Also check this link: http://www.chemguide.co.uk/inorganic/redox/oxidnstates.html (check section Examples of working out oxidation states)

You can use the keywords oxidation number or oxidation state on google and you'll get information and examples related with your post

Offline Kaleyrvt

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Re: Oxidation numbers
« Reply #6 on: January 22, 2007, 05:18:12 PM »
iS THIS BETTER??? Hopefully.. ???

b) NH4+    =>x  +  4 (1+) = 1+
       x  +  4+  = 1+         
       N = x =  -3



c) Cr2O7 2-  =>2 x  +  7(2-)= 2-
         2x  +  14- =2-
        Cr = x = +6

D)MnO4 2-  =>   x  +  4 x 2- = 2-
      x  +  8- =2-
      Mn = x = +6

Offline Borek

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Re: Oxidation numbers
« Reply #7 on: January 22, 2007, 05:28:26 PM »
Much better. Although I think it will be even better to use slightly different notation - instead of

2x + 7(2-) = 2-

put it as

2x + 7*(-2) = -2

This way it is not only OK chemically (assuming ON are OK) but it also makes a mathematical sense.
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Offline Dan

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Re: Oxidation numbers
« Reply #8 on: January 22, 2007, 07:48:31 PM »
Yes your answers are correct. * for multiply avoids confusion, as you are using x as an algebraic symbol.
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