Finding coefficients y and z in the iodate-peroxide redox reaction

[ChloeYip]

Consider the following chemical equation:

• 2IO₃^-+wH₂O₂+xH⁺=>I₂+yO₂+zH₂O

Which of the following is the correct combination of the reaction coefficient y and z?

• A) 4&5
• B) 5&4
• C) 5&6
• D) 6&5

How to obtain the answer? Thank you~

 

[Borek]

Try to balance the reaction equation.

 

[ChloeYip]

I know. I have tried simutaneous equation already, but i failed to do so.
Can you give me more concrete steps?
Thank you.

 

[Borek]

Simultaneous equations have problems with every system where hydrogen peroxide is being oxidized.

Hint: H₂O₂ → 2H⁺ + 2e^- + O₂

 

[epenguin]

A very easy example , should hit in eye that total electric charges left have to equal those right (charges somehow make things more obvious although the principle is not really different from anything without charges).

 

[ChloeYip]

i still don't understand...
how to solve with 4 variables...
i have tried to solve by charges and atoms, but have no ideas to get the answer
thanks

 

[epenguin]

I don't know what you have done. Simply, what is the total charge on the right, what is it on the left?

 

[epenguin]

I'll modify that. I'll say that the first step, not yet taken, doesn't come easier.

After that I see a difficulty. The answers you have given to choose from all seem wrong to me

 

[epenguin]

Ah on third thoughts I see it. This is really tricky and seems to be a test of chemical knowledge, not just stoichiometric balancing. Leads to something advanced and specialized.

But you haven't taken the first step. Do that and get a reasonable answer without worrying whether it fits the question options, then we can talk..

 

[ChloeYip]

to deal with the charge,
x should be 2
but the question is asking about y and z, it seems no use to find out x...
how they can be work out?
thanks

 

[Borek]

Can you write reduction half reaction for IO₃^-?

 

[epenguin]

to deal with the charge,
x should be 2
but the question is asking about y and z, it seems no use to find out x...
how they can be work out?
thanks

That's half the first step, can you complete it and write a reasonable reaction balance that makes sense? Never mind the question options, see what I said about them.

 

[Borek]

Come on, both of you, why do ignore the simplest approach - I gave two hints, follow them an you will get the correct answer in almost no time.

 

[ChloeYip]

i am not sure how to write the half equation...
is it like this?(just an attempt…):
2IO3-=>I2+(1/2)O2

what is the use for writting half equation?
what is the next step?
thanks~

 

[ChloeYip]

ah..
just forget to balance the charge with electron

correction:
2IO3-=>I2+(1/2)O2+6e-

 

[ChloeYip]

i really still don't have any idea on how to deal with it...

 

[Borek]

The problem is about balancing an equation and you approach it without knowing how to balance an equation?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Assume it is in acidic conditions, it will make things easier. Products will be water and iodine, that's how oxidizing acids behave.

 

[epenguin]

ah..
just forget to balance the charge with electron

correction:
2IO3-=>I2+(1/2)O2+6e-

Still needs correction - misled by way you have written maybe: there are not 6 e^- in 2IO₃^-. And where have five O gone?

(Where has Chloe Yip gone?)