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Topic: Solving redox reactions  (Read 2976 times)

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Offline Lukas1121

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Solving redox reactions
« on: February 10, 2018, 05:31:41 AM »
I have a couple of redox reactions i just can't wrap my head around.

SO2 + Br2 :rarrow: HSO4- + Br-

S on the left side has oxidation number +4 and on the right side +6. Br has the oxidation number 0 but reduces to -1. However, if i increase SO2 to get the desired O4 required on the right side i'm still left with an Extra S. Is it possible to equalize on the left side with H3O+ or is there some other way?

Cr2O72- + H2:rarrow: Cr3+ + S8

In this reaktion i found Cr's oxidation number to be +6 and S to be -2 on the left side and Cr to be +3 and S to be 0 on the right side. However i need at least 8 H2S molecules to satisfy the right side and therefore need to multiply sulphur with 8 and 3 and Cr with 8 and 2. So 24 H2S and 16Cr2. However, on the left side i get a totalt oxidation number of 144, while's on the right side i only get 96. Should i just add more electrons on the left side or have i miscalculated somewhere?

Offline Lukas1121

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Re: Solving redox reactions
« Reply #1 on: February 10, 2018, 05:45:54 AM »
Forgot to mention that all reactions are in an acidic environment

Offline Borek

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Re: Solving redox reactions
« Reply #2 on: February 10, 2018, 06:25:09 AM »
If it is in acidic environment you are free to use both H2O and H+ for balancing hydrogen and oxygen.
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Offline Lukas1121

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Re: Solving redox reactions
« Reply #3 on: February 10, 2018, 06:55:21 AM »
If it is in acidic environment you are free to use both H2O and H+ for balancing hydrogen and oxygen.

I realize this, but before i can equalize with H+ and H2O i need to make sure the oxidation number on each side is the same.

Offline chenbeier

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Re: Solving redox reactions
« Reply #4 on: February 10, 2018, 07:01:36 AM »
Not really necessary.

You have to get the redox pair: here its SO2 and HSO4- and Br2/Br-.
As already mentioned use H+ and H2O to balance the Oxygens on both sides.
The same procedure for Dichromate and chromium-III.

Offline Lukas1121

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Re: Solving redox reactions
« Reply #5 on: February 10, 2018, 07:07:57 AM »
Not really necessary.

You have to get the redox pair: here its SO2 and HSO4- and Br2/Br-.
As already mentioned use H+ and H2O to balance the Oxygens on both sides.
The same procedure for Dichromate and chromium-III.

Okay, but where do i get the addition dioxygen needed for the HSO4?

Offline chenbeier

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Re: Solving redox reactions
« Reply #6 on: February 10, 2018, 02:10:22 PM »
You take water and create H+ for the oxidation and do balance with electrons. Can you develop it now?

 

Offline Hacktacular

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Re: Solving redox reactions
« Reply #7 on: February 23, 2018, 11:15:46 AM »
G'Day Lukas,

Just in case you were still looking for a bit of an explanation...  About halfway down is the process for acidic solutions.

https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions

Basically you split it into your half equation, balance the atoms, throw in water where you need it to balance your oxygen, balance that in turn with hydrogen ions and chuck a few electrons in to get the charges equal.  The website explains it better than I just did to be honest.

As for oxidation numbers...  They should just come out in the wash, I wouldn't worry about them until you've put your two half reactions together.  96 and 144 seem pretty high, usually it'll be like a +/- 10 at the most from what I've done.

Hope it helps.

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