Ok, please demonstrate for us how you would go about balancing the following with half reactions like you said.
Who is us?
What is giving you trouble? Can you not identify the two half reactions? You ignore the h+ and the water and then write down the two half reactions using what is left. one half reaction involves Mn, the other has Cr, C, and N.
Sure the problem tedious, but aren't you using a computer?
Balance every element except H,O
2Cr7N66H96C42O24 = 7Cr2O7-2 + 84CO2 + 132NO3-
MnO4 = Mn+2
Now balance the O using H20.
Then balance the H using H+
2Cr7N66H96C42O24 + 565 H2O = 7Cr2O7-2 + 84CO2 + 132NO3- + 1322 H+
MnO4- + 8H+ = Mn+2 + 4H2O
The rest requires a bunch of tedious arithmetic and nothing more. Do you really need me to do that stuff?
Then add e- as needed to each half reaction so each half reaction has balanced charges. If electrons are on the same side in the two half reactions, then it wasn't an ox-redox. Quit.
2Cr7N66H96C42O24 + 565 H2O = 7Cr2O7-2 + 84CO2 + 132NO3- + 1322 H+ + 1176e-
MnO4- + 8H+ +5e- = Mn+2 + 4H2O
Multiplty each equation by whatever required so that each half reaction involves the same number of electrons. (5 and 1176 respectively) then add two half reactions, remove species that are on two sides of equation. This should clear out all electrons. Clear fractions as needed. Reduce if possible. Done
I believe the answer in your display is correct, but I did not do a complete check.
Edited by NoNukes, : No reason given.
Edited by NoNukes, : No reason given.
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