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Topic: How do you balance this redox reaction?  (Read 650 times)

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Offline The_fin

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How do you balance this redox reaction?
« on: February 05, 2021, 01:37:21 PM »
Hello
How do you balance this redox reaction:
Cr7N66H96C42O24+ MnO4- :rarrow:  Cr2O72-+ CO2+ NO333-+ Mn2+
My teacher told me to set all oxidation numbers in Cr7N66H96C42O24 to 0.
I have tried to write the half reactions down and this is what i've got:

I think my mistakes are in H and O.
Thank you in advance

Offline AWK

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Re: How do you balance this redox reaction?
« Reply #1 on: February 05, 2021, 01:54:02 PM »
Remove 24 molecules of water and oxidize each element of Cr7C42H48N66 with MnO4- in acidic conditions.
Then scale all the oxidation reactions to the correct number of atoms (e.g. Cr to 7 atoms, etc.) and add the water removed previously.
« Last Edit: February 05, 2021, 02:24:14 PM by AWK »
AWK

Offline The_fin

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Re: How do you balance this redox reaction?
« Reply #2 on: February 05, 2021, 02:26:43 PM »
But have I calculated the electrons for H and O correct?

Offline AWK

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Re: How do you balance this redox reaction?
« Reply #3 on: February 05, 2021, 02:33:37 PM »
H96O24 = 24H2O + 48H
AWK

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