Topic: How do you balance this redox reaction? (Read 928 times)

The_fin · « **on:** February 05, 2021, 01:37:21 PM »

Hello
How do you balance this redox reaction:
Cr₇N₆₆H₉₆C₄₂O₂₄+ MnO₄^-

Cr₂O₇^2-+ CO₂+ NO₃₃^3-+ 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

AWK · « **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.

The_fin · « **Reply #2 on:** February 05, 2021, 02:26:43 PM »

But have I calculated the electrons for H and O correct?

AWK · « **Reply #3 on:** February 05, 2021, 02:33:37 PM »

H₉₆O₂₄ = 24H₂O + 48H

Topic: How do you balance this redox reaction? (Read 928 times)

The_fin

How do you balance this redox reaction?

AWK

Re: How do you balance this redox reaction?

The_fin

Re: How do you balance this redox reaction?

AWK

Re: How do you balance this redox reaction?

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