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Topic: H2CrO4 + Fe(NH4)2(SO4)2  (Read 5157 times)

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Offline kriggy

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H2CrO4 + Fe(NH4)2(SO4)2
« on: May 04, 2013, 01:57:56 PM »
Hi. I have a homework about calculation of the amount of Cr in steel. I know how to do it etc but I cant figure out the stechiometry and products.
"Cr in steel was oxidized ot H2CrO4 and after removing the oxidant, Fe(NH4)2(SO4)2 was added. Unreacted amout of Fe2+ was titrated by KMnO4..."
How does H2CrO4 react with Fe(NH4)2(SO4)2? The only thing comming to my mind is:
H2CrO4 + Fe(NH4)2(SO4)2 -> FeCrO4 + 2 (NH4)HSO4.
Thanks a lot
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Offline Borek

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #1 on: May 04, 2013, 03:55:27 PM »
Hint: CrO42- is a relatively strong oxidizer.
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Offline kriggy

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #2 on: May 05, 2013, 04:25:11 AM »
Thats true.. so something like this?:

2 H2CrO4 + 6 Fe(NH4)2(SO4)2  = Cr2(SO4)3 + 3 Fe2(SO4)3 + 6 NH4OH + 2H2O
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Offline Borek

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #3 on: May 05, 2013, 05:43:07 AM »
You got the right idea, although I would go with just

CrO42- + 4Fe2+ + 8H+ :rarrow: Cr2+ + 4Fe3+ + 4H2O

CrO42- + 3Fe2+ + 8H+ -> Cr3+ + 3Fe3+ + 4H2O

as it reflects the stoichiometry and doesn't muddy the water with all unnecessary spectators.
« Last Edit: May 05, 2013, 02:47:16 PM by Borek »
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Offline kriggy

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #4 on: May 05, 2013, 07:28:33 AM »
And the amonia just doesnt react?
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Offline Borek

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #5 on: May 05, 2013, 08:13:32 AM »
Reaction needs low pH, ammonia starts and ends as NH4+.
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Offline kriggy

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #6 on: May 05, 2013, 02:22:09 PM »
 Just one more thing.. Why it doesnt stop at Cr3+? I mean.. Cr6+ is strong oxidizing agent but Cr2+ is strong reducing agent which is oxidized by air to Cr3+.
Thanks for answers, you helped me alot.
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Offline Borek

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Re: H2CrO4 + Fe(NH4)2(SO4)2
« Reply #7 on: May 05, 2013, 02:46:42 PM »
Sorry, my mistake. Corrected.

Note that technically at these low pHs you don't have chromate, but dichromate, so it should be Cr2O72- on the left. I put chromate in the equation to keep it close to the problem as worded.
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