14.1 D

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Jesse_torres2H
Posts: 41
Joined: Thu Jul 27, 2017 3:00 am

14.1 D

Postby Jesse_torres2H » Tue Feb 20, 2018 8:10 pm

In this question you are supposed to balance the redox equation the answer reads, 8H+ + Cr2O7^2- + 3C2H5OH---> 2Cr^3+ + 3C2H4O + 7H2O. why do you need the 3 in front of C2H5OH and C2H4O

Lorie Seuylemezian-2K
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: 14.1 D

Postby Lorie Seuylemezian-2K » Tue Feb 20, 2018 8:50 pm

This is because we multiply the half reactions with a coefficient in order to cancel out the electrons on both sides.

Manasvi Paudel 1A
Posts: 50
Joined: Sat Jul 22, 2017 3:00 am

Re: 14.1 D

Postby Manasvi Paudel 1A » Sat Feb 24, 2018 5:16 pm

Before adding half reactions in order to get a final reaction, you have to make sure that there is the same amount of electrons in each half reaction that are being either gained or lost. In order to do that you may have to multiply a half reaction by a certain number, which was done here to get those coefficients.

Erik Khong 2E
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 14.1 D

Postby Erik Khong 2E » Sun Feb 25, 2018 6:34 pm

You need to balance both half reactions in order for them to have equal amount of electrons being gained and lost, so they can cancel out for the final redox reaction.


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