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Topic: Oxidation numbers (Read 6879 times)
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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
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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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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
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Borek
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Re: Oxidation numbers
«
Reply #3 on:
January 22, 2007, 03:05:28 PM »
Quote from: Kaleyrvt on January 22, 2007, 01:43:22 PM
1+ + x = 3 x 2-
1+ + x = 6-
1+ + x + 6- = 0
Finally, you better check your math...
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ChemBuddy
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Kaleyrvt
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Re: Oxidation numbers
«
Reply #4 on:
January 22, 2007, 03:29:41 PM »
I am confused
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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
g
o
o
g
l
e
and you'll get information and examples related with your post
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Kaleyrvt
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Posts: 57
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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
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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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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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