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Topic: Redox Half Reactions again  (Read 19121 times)

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hihi123

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Redox Half Reactions again
« on: May 20, 2005, 02:23:58 AM »
This is the same type of problem as my previous post:

KIO4(aq) + KI(aq) +HCl(aq) ----> KCl(aq) + I2(s) + H2O(L)

a) write the oxidation half reaction
b) write the reduction half reaction
c) write the balanced equation

I dont see which one is being oxidized! I think that I is being reduced. Please *delete me*

Offline Donaldson Tan

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Re:Redox Half Reactions again
« Reply #1 on: May 20, 2005, 03:16:22 AM »
the forum rules require you to show that you have attempted the problem. nevertheless, i will still give u some hints.

1. write the ionic equation for this reaction
2. all iodine in the reactant side is converted to I2 in the product side.
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hihi123

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Re:Redox Half Reactions again
« Reply #2 on: May 20, 2005, 03:29:44 AM »
ok, when i write out the ionic eq, i get this:

K+1 + I+6 +O4-2 + K+1 + I-1 + H+1 + Cl-1 -----> K+1 + Cl-1 + I2 + H2O

I just dont see how the I's combine to form I2. Also, it looks like I is the only one changing, so how can I write both two half reactions? I know I am doing something wrong, I just don't know what. Any help would be appreciated.

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Re:Redox Half Reactions again
« Reply #3 on: May 20, 2005, 03:35:11 AM »
cancel the common ions on both left and right hand side. you will realise that IO4- is reduced to I2 and I- is oxidised to I2. This should be sufficient for you to write the half equations.
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Re:Redox Half Reactions again
« Reply #4 on: May 20, 2005, 04:15:15 AM »
ok, when i write out the ionic eq, i get this:

K+1 + I+6 +O4-2 + K+1 + I-1 + H+1 + Cl-1 -----> K+1 + Cl-1 + I2 + H2O

It is not a net ionic reaction! Why did you split ion IO4(-) into some non existing entities like I(+6) and O4(-2)?

Regardless of the fact that such entities doesn't exist, what will be the charge of IO4 if it will be made of I(+6) a (O4)(-2)?

Are you sure it was IO4(-) from the start, not IO3(-)?
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hihi123

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Re:Redox Half Reactions again
« Reply #5 on: May 20, 2005, 04:25:38 AM »
The original eq was KIO4 + KI +HCl ----> KCl + I2 +H2O  so I dont think it was IO3 from the start.... unless I'm completing missing something which is quite possible.

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Re:Redox Half Reactions again
« Reply #6 on: May 20, 2005, 04:47:39 AM »
honestly, i never seen IO4 before. the unbalanced ionic equation (in my opinion) is:

IO3- + I- + H+ => I2
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hihi123

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Re:Redox Half Reactions again
« Reply #7 on: May 20, 2005, 04:55:17 AM »
Well, the reaction may not work very well in real life, but on paper, my teacher wants me to make it work...

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Re:Redox Half Reactions again
« Reply #8 on: May 20, 2005, 06:25:39 AM »
on paper, the reduction equation is:
(1) 2IO4- + 16H+ +14e => I2 + 8H2O

and the oxidation equation is:
(2) 2I- => I2 + 2e

to balance the number of electrons, one set of equation (1) must be coupled with 7 sets of equation (2), ie. (1) + 7(2):

14I- + 2IO4- + 16H+  => 8H2O + 8I2

« Last Edit: May 20, 2005, 04:35:44 PM by geodome »
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hihi123

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Re:Redox Half Reactions again
« Reply #9 on: May 20, 2005, 08:05:01 AM »
Thank you so much for your help, i think i understand what you did, except on the final balancing of the eqs, why can't you combine all of the I2s?

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Re:Redox Half Reactions again
« Reply #10 on: May 20, 2005, 01:32:47 PM »
if i directly combine the two equations, the number of electrons won't balance. I must have an equal number of electrons on the reactant and product sides. Equation (1) has 14 electron on the reactant side, whereas Equation (2) has 2 electrons on the product side. I need to multiply Equation (2) by 7 so that I have 14electrons on the product side. You can only cancel out all the electrons in the overall equation, if you have equal number of electrons on both sides.
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Re:Redox Half Reactions again
« Reply #11 on: May 20, 2005, 02:08:48 PM »
It is not a net ionic reaction! Why did you split ion IO4(-) into some non existing entities like I(+6) and O4(-2)?

Regardless of the fact that such entities doesn't exist, what will be the charge of IO4 if it will be made of I(+6) a (O4)(-2)?

Are you sure it was IO4(-) from the start, not IO3(-)?

He probably did that to see the oxidation numbers that each atom has in the equation.  There is absolutely nothing wrong with doing that.

For the IO4(-) thing, I don't see what's so unusual about it.  We have perchlorates which are formed with the much smaller chlorine atom, so I would think that periodate is actually a bit easier to form.
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Re:Redox Half Reactions again
« Reply #12 on: May 20, 2005, 04:13:19 PM »
Thank you so much for your help, i think i understand what you did, except on the final balancing of the eqs, why can't you combine all of the I2s?

I think geo has just overlooked it. It is 8 I2.
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Re:Redox Half Reactions again
« Reply #13 on: May 20, 2005, 04:36:25 PM »
LOL. 1 + 7 = 8. I overlooked it. Haha.. I recombined them already.
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Re:Redox Half Reactions again
« Reply #14 on: May 20, 2005, 04:43:31 PM »
He probably did that to see the oxidation numbers that each atom has in the equation.  There is absolutely nothing wrong with doing that.

He did it when asked to write down the net ionic equation, and what he wrote was definitely not an net ionic equation :) That attracted my attention.

BTW: I was just observing long discussion between teachers about the ON and whether to use them or not when teaching how to balance redox equations. How do you say it? Jury is still out?

Quote
For the IO4(-) thing, I don't see what's so unusual about it.  We have perchlorates which are formed with the much smaller chlorine atom, so I would think that periodate is actually a bit easier to form.

I have nothing against periodic acid :) Just the reaction between iodide and iodate is very common and often used when teaching redox reactions or redox titrations, periodate seemed a little bit exotic in this context.
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