Question

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

Answer 1

oxygen is more electronegative so iodine must be +1

Answer 2

this reaction works in a same way as with Cl take a look at the pic

Answer 3

+1-->+5 Iodine change in oxidation state
+1--->0 and thats for iodine too
-2---->-6 thats for oxygen

Answer 4

oh wait i think u are getting something wron

Answer 5

oxygen is out

Answer 6

only iodine plays role here

Answer 7

so the difference in oxidation state for both iodine are -1 and +4

Answer 8

because its I2 we multiply it by 2 instead of 4 or am i wrong

Answer 9

(its diatomic)

Answer 10

okay wait a sec can u send me the link to the q itself?

Answer 11

sure here you go
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w10_qp_12.pdf
Q13

Answer 12

okay first of all oxidation state of I on reactant side is +1
on product side its 0 in I2 and +5 in HIO3

Answer 13

0 meaning gain 1 electron so its neutral so -1
+5 meaning lost 4 more electrons so +4
now don't pay much attention here
just try balancing the equation as we usually do in chemistry ok? i need to go now, when im back if u still can't figure out i'll help ok?

Answer 14

good luck with what your doing and thank you for your help and i think its 2 for n because the charge is shared between diatomic molecules

Answer 15

u don't even need to think of the charges to be honest, u just balance it the way we usually balance equations that's it

Answer 16

and yeah it's gonna be 2 and 1

Answer 17

The balanced equation is as follows;
\[5HIO \rightarrow 2I_{2} + HIO_{3} + 2H_{2}O \]