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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: The_fin on February 05, 2021, 01:37:21 PM

Title: How do you balance this redox reaction?
Post by: The_fin on February 05, 2021, 01:37:21 PM

Hello
How do you balance this redox reaction:
Cr₇N₆₆H₉₆C₄₂O₂₄+ MnO₄^- :rarrow:  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:
(https://i.imgur.com/CCsewXx.jpg)
I think my mistakes are in H and O.
Thank you in advance

Title: Re: How do you balance this redox reaction?
Post by: AWK 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.

Title: Re: How do you balance this redox reaction?
Post by: The_fin on February 05, 2021, 02:26:43 PM

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

Title: Re: How do you balance this redox reaction?
Post by: AWK on February 05, 2021, 02:33:37 PM

H₉₆O₂₄ = 24H₂O + 48H