6K.1 7th edition

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Katie Frei 1L
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

6K.1 7th edition

Postby Katie Frei 1L » Tue Feb 26, 2019 10:44 pm

Could someone explain why for the balanced redox reaction in problem 6K.1) of the 7th edition of the textbook, there ends up being 3 C2H5OH(aq) and 3 C2H4O(aq) as opposed to just the coefficient of 1 for each of these compounds? Thank you!

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

Re: 6K.1 7th edition

Postby Kobe_Wright » Tue Feb 26, 2019 11:06 pm

In order to balance the waters, and hydrogens, and electrons of both half reactions you'll need to multiply by those coefficients.


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