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Topic: Sodium Chromate ( Redox )  (Read 12790 times)

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Offline delusioned

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Sodium Chromate ( Redox )
« on: March 08, 2008, 12:43:04 AM »
The compound, Na3CrO4 (does this exist? i'm not sure if it does exist or is it a typo-error with my question paper), is an unstable dark green solid. The addition of dilute sulphuric acid to this produces a solution containing chromium (III) ions and dichromate (VI) ions, Cr2O7^2-. When a 1.00g sample of the impure green solid was reacted in this way, the Cr(VI) in the resulting solution reacted with an excess of potassium iodide to liberate 5.0 x 10^-3 moles of iondine, I2.

(a) Calculate the oxidation state of chromium in Na3CrO4, and construct a balance equation for its reaction with dilute acid.

I need some help with this question,
Firstly, does the compound actually exist? What I know is that CrO4 has a oxidation state of -2, doesn't it mean that Sodium Chromate is Na2CrO4 then?
Secondly, any idea what products are formed in the equation?

I need to get my assignment done soon, please help.
Thanks in advance.

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Re: Sodium Chromate ( Redox )
« Reply #1 on: March 08, 2008, 04:38:06 AM »
You are told everything you need. Na3CrO4 is not a typo - and it should be fairly ease for you to calculate charge on CrO4. Na3CrO4 is neutral. What is Na charge?
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Offline delusioned

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Re: Sodium Chromate ( Redox )
« Reply #2 on: March 08, 2008, 05:34:45 AM »
Yes, I have managed to get the oxidation state of Cr.
But still thinking of the balanced equation,
below is what I've thought of thus far, however it seems impossible..

Na3CrO4 + H2SO4 -> Na2Cr2O7 + Cr2(SO4)3 + H2O

any hints regarding this?

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Re: Sodium Chromate ( Redox )
« Reply #3 on: March 08, 2008, 05:52:06 AM »
Try to balance net ionic reaction. Start outlining half reactions - you will have the same reactant, but different products. Such a process is called disproportionation.

To be honest - I have not heard about Na3CrO4, but I know that compounds with Cr(V) do exist.
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Offline delusioned

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Re: Sodium Chromate ( Redox )
« Reply #4 on: March 08, 2008, 06:14:57 AM »
CrO4^2- + 8H+  +3e  ->   Cr^3+  + 4H2O     
2CrO4^2  + 2H+           ->   Cr2O7^2-   + H2O

These? (not to sure if I'm allowed to use posts to pen down my workings)

But where do the sulphate ion come in?
Sorry, but I'm kinda lost..

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Re: Sodium Chromate ( Redox )
« Reply #5 on: March 08, 2008, 07:48:37 AM »
CrO4^2- + 8H+  +3e  ->   Cr^3+  + 4H2O     
2CrO4^2  + 2H+           ->   Cr2O7^2-   + H2O

These?

No! You have to start with CrO43- - that's what is disproportionating, that's the reactant in question, initially present in the solution. Don't stick to the compounds you know - this is a new one.

So more like (unbalanced)

CrO43- + H+ -> Cr3+ + H2O

Think how the second half reaction should look alike now.

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But where do the sulphate ion come in?
Sorry, but I'm kinda lost..

Sulphate is just a spectator, just like Na+. You know what the net ionic reaction is?
« Last Edit: March 08, 2008, 09:00:14 AM by Borek »
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Offline ©h£m1§†®¥

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Re: Sodium Chromate ( Redox )
« Reply #6 on: March 06, 2009, 10:13:14 AM »
CrO43- + 8H+ ---> Cr3+ +4H2O
2CrO43- + 2H+ ---> Cr2O72- + H2O +2e-
Is this correct? And is the oxidation state of Cr in Na3CrO4 +5?


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Re: Sodium Chromate ( Redox )
« Reply #7 on: March 06, 2009, 12:49:50 PM »
CrO43- + 8H+ ---> Cr3+ +4H2O

No, that's not OK. Charges on both sides of reaction must be identical. You are missing electrons.

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And is the oxidation state of Cr in Na3CrO4 +5?

Yes. Assuming it exists ;)
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