Chapter 14 Question 1

Moderators: Chem_Mod, Chem_Admin

Angelica Nava-1E
Posts: 14
Joined: Wed Sep 21, 2016 2:56 pm

Chapter 14 Question 1

Postby Angelica Nava-1E » Mon Feb 06, 2017 6:37 pm

The question is to identify the elements undergoing changes in the oxidation state and indicate the initial and final oxidation state.

The equation is H+(aq) + Cr2O7^2-(aq) + C2H5OH(aq) ----> Cr^3+(aq) + C2H4O(aq) + H20(l)

The answer is that Cr is reduced from 6+ to 3+ and C is oxidized from 2- to 1-. But from what I could see Cr goes from 4- to 3+ and C dosent change charge. What am I doing wrong?

Maggie Bui 1H
Posts: 35
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

Re: Chapter 14 Question 1

Postby Maggie Bui 1H » Mon Feb 06, 2017 9:08 pm

I'm not sure how you're calculating the charge, but first, it's good to start with any transition metals if you see any because they typically change charge in redox equations.

You can find the oxidation number of chromium (Cr) by looking at oxygen. Oxygen's common oxidation state is 2-. There's 7 of them, so that'd be an overall charge of 14- if we were looking at oxygen alone, but the ion's (Cr2O7) charge as a whole is 2-. Take that into account and you get 14- - (2-) = 12-. This accounts for the fact that the ion has an overall charge of 2-. Now, you still have a charge of 12- to balance. Since there are 2 chromium ions, the deficit in charge balance should be split equally between them, so you'd balance 12- with 12+. 12+/2 is 6+, so each chromium has a charge of 6+. You see on the other side that Cr is 3+, so you know that Cr goes from 6+ to 3+, a gain in electrons, so it's being reduced.

If you repeat this process for the other reactants and products, you'd see that carbon changes charge. Keep in mind the common oxidation states: hydrogen is commonly 1+ and oxygen is 2-.

Jonathan Sarquiz 3F
Posts: 32
Joined: Fri Jul 15, 2016 3:00 am

Re: Chapter 14 Question 1

Postby Jonathan Sarquiz 3F » Mon Feb 06, 2017 9:16 pm

Let's break it down.

H: as both H+ and H2O, the hydrogen has an oxidation state of +1.

Cr: in Cr2O7-2, each oxygen has a -2 charge so (-2)(7) = -14; with a overall -2 charge, we are left with two Cr atoms to split the +12 charge. Thus Cr as a reactant has an oxidation state of +6. As a product, Cr has an oxidation state of +3. Thus chromium is being reduced.

C: in C2H5OH, each O has a -2 oxidation state and each hydrogen has a +1 oxidation state. (-2 + 6x1 = 4) The two C atoms split the 4 charge, thus each C has an oxidation state of -2. In the product C2H4O, oxygen has a -2 oxidation state and each hydrogen has a +1 oxidation state. (-2 + 4x1 = 2) The two C atoms split the 2 charge, thus each C has an oxidation state of -1. The carbon is being oxidized.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 2 guests