Homework 14.1

Moderators: Chem_Mod, Chem_Admin

Angela_Kim_1N
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

Homework 14.1

Postby Angela_Kim_1N » Thu Feb 09, 2017 10:20 pm



Identify the elements undergoing changes in oxidation state and indicate the initial and final oxidation numbers for these elements.
The solution manual states that Cr is reduced from 6+ to 3+ and C is oxidized from 2- to 1-. What part of the equation are we supposed to look at to figure this out?

Chem_Mod
Posts: 19565
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 885 times

Re: Homework 14.1

Postby Chem_Mod » Thu Feb 09, 2017 10:49 pm

Oxygen typically has an oxidation state of -2, and there are 7 oxygen atoms in , which has an overall charge of -2. This means each Cr must have an oxidation state of +6 in order for the overall compound to have a charge of -2 (). For C in , hydrogen typically has an oxidation state of +1 while hydroxide has a charge of -1. Thus, each C must have an oxidation state of -2 for the entire compound to have a net charge of 0. To figure out the oxidation state of C in , you once again use the oxidation states of hydrogen and oxygen to determine that the oxidation state of each C must be -1 for the overall compound to have a charge of 0.

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

Re: Homework 14.1

Postby Zoe Robertson 2H » Sat Feb 11, 2017 8:35 pm

How do you go about getting the answer for part d? It asks to combine the half reactions to produce a balanced redox equation, and the answer is: 8H+Cr2O72-+3C2H5OH-->2Cr3++3C2H4O+7H2O

I understand how to get to the individual half-reactions, I'm just a bit confused about how to combine them. Thanks!

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

Re: Homework 14.1

Postby Zoe Robertson 2H » Sat Feb 11, 2017 9:11 pm

Zoe Robertson 2H wrote:How do you go about getting the answer for part d? It asks to combine the half reactions to produce a balanced redox equation, and the answer is: 8H+Cr2O72-+3C2H5OH-->2Cr3++3C2H4O+7H2O

I understand how to get to the individual half-reactions, I'm just a bit confused about how to combine them. Thanks!



Realized my problem-- I didn't know to balance the equations so the number of electrons is the same for both equations.

But if anyone else needs clarification: take the balanced half-reactions and multiply each by a factor to make the number of electrons the same for both equations. That way, they cancel out. Add everything on the left of the arrow together, and everything on the right together. That's it!


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest