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Author Topic: Redox reaction  (Read 801 times)

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sun725

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Redox reaction
« on: January 05, 2018, 01:20:51 PM »

I have the following redox reaction:

Cr2O72-,14H+ /2Cr3+, 7H2O

And

ClO4-, 2H+ / ClO3-, H2O

I have balanced the half reactions:

Cr2O72-+14H++6e-  :rarrow:2Cr3++7H2O

And

ClO4-+2H+ +2e-  :rarrow:ClO3-+H2O

So I have to add the reactions together, and I tried that, however it didn't matched the answer on the assignment.

They get:
Cr2O72-+8H++3ClO3- :rarrow: 2Cr3+ +4H2O+3ClO4-

I thought they were balanced in the first place when I balanced the half reaction. The number of oxygen doesn't add up on both sides? And why the 4 H2O and 3 infront of both the ClO3^- and ClO4^-
This is really confusing me. I hope someone could clarify this.

Thank you.
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Borek

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Re: Redox reaction
« Reply #1 on: January 05, 2018, 10:01:33 PM »

Hard to tell what went wrong not seeing your final answer. All three equations you listed so far look OK.
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mjc123

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Re: Redox reaction
« Reply #2 on: January 07, 2018, 11:08:20 PM »

Quote
I thought they were balanced in the first place when I balanced the half reaction.
Not so. You have to match the half-reactions to each other. You can see from the half-equations that the dichromate reduction is a 6-electron reduction, while the reduction of perchlorate is a 2-electron reduction (or oxidation, when it goes the other way). So to balance them, you need to have the same number of electrons produced by the oxidation and consumed by the reduction.
The transfer of 6 electrons will reduce 1 dichromate and oxidise 3 chlorates. So the overall reaction is half-reaction 1 - 3 x half-reaction 2, which gives you your final balanced equation.
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