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Chemistry: Post your doubts here!

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In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.
mHIO → nI2 + pHIO3 + qH2O
This equation can be balanced using oxidation numbers.
What are the values for n and p?
n p
a) 1 2
b) 2 1
c) 4 1
d) 4 2


the answer is B
I know how to figure out the oxidation numbers, but I need explanation on how to balance such equations using oxidation numbers... help please!!

You can try to create the 2 half equations

2HIO + 2H+ + 2e- --> I2 + 2H2O ------ (1)

HIO -+ 2H2O -> HIO3 + 4H+ + 4e- ------ (2)

Before combining the 2 equations, we need to make both equations have the same number of electrons
(2) x 2 + (1)

So ratio of I2 to HIO3 is 2: 1
 
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In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.
mHIO → nI2 + pHIO3 + qH2O
This equation can be balanced using oxidation numbers.
What are the values for n and p?
n p
a) 1 2
b) 2 1
c) 4 1
d) 4 2


the answer is B
I know how to figure out the oxidation numbers, but I need explanation on how to balance such equations using oxidation numbers... help please!!
Oxidation state of I in HIO, I2 and HIO3 is +1, 0 and +5 respectively.
Oxidation change from HIO to I2 is -2 (not -1 coz I-I)
Oxidation change from HIO to HIO3 is +4
Chem.png
Interchange the numbers, ie 2 for HIO3 and 4 for I2
2 : 4 ----> 1 : 2
n = 2 p = 1
 
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upload_2015-5-28_15-36-42.png
Doesn't Selenium also have 6 unpaired electrons??
So does chromium ... B is correct but then what about D?
 
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