Asked by anonymous
please please please help me balance these two equations...
1. K2Cr2O7+H2C2O4.2H2O--> K[Cr(H20)2(C2O4)2].2H2O+CO2
2. K2Cr2O7+H2C2O4.2H2O--> K[Cr(H20)2(C2O4)2].3H2O+CO2
1. K2Cr2O7+H2C2O4.2H2O--> K[Cr(H20)2(C2O4)2].2H2O+CO2
2. K2Cr2O7+H2C2O4.2H2O--> K[Cr(H20)2(C2O4)2].3H2O+CO2
Answers
Answered by
DrBob222
There are rules to follow in balancing redox equations. Here are the rules.
http://www.chemteam.info/Redox/Balance-HalfReactions-Acid.html
Here are some helpful hints.
#1. Both Cr change from +12 on the left to +6 on the right(FOR TWO Cr ATOMS on the right).
C changes from +6 on the left for both C atoms to +8 on the right for TWO C ATOMS of C in CO2. Note also that the C2O4 that is complexed with the Cr on the right does not undergo oxidation.
http://www.chemteam.info/Redox/Balance-HalfReactions-Acid.html
Here are some helpful hints.
#1. Both Cr change from +12 on the left to +6 on the right(FOR TWO Cr ATOMS on the right).
C changes from +6 on the left for both C atoms to +8 on the right for TWO C ATOMS of C in CO2. Note also that the C2O4 that is complexed with the Cr on the right does not undergo oxidation.
Answered by
Kurt
Common we need an example, its complexed. It doesnt change from 12 -6 it goes from 6-3 and carbon goes from 3-4. If the half reactions have to = the entire reaction then how is broken down. This is why students get irritated with posts like this. They know you wont be able to figure it out on your own.
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