## 14.1 Book HW

704578485
Posts: 62
Joined: Fri Sep 25, 2015 3:00 am

### 14.1 Book HW

Question 14.1 involves the compounds Cr2O7^2- and C2H5OH reacting with an H+ ion to form Cr^3+, C2H4O, and water. After looking at the answer in the solutions manual it says Cr is reduced from 6+ to 3+ which matches my answer. However, it says C is oxidized from -2 to -1. I am having trouble understanding how you get that.

Ani Galfayan 1H
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

### Re: 14.1 Book HW

By looking at each atom individually, we can count the oxidation states and calculate the unknown. Breaking up C2H5OH, we know that each O has an oxidation state of -2 and each H, +1. We want the total to equal 0 so we do the following
2(C) + 6(+1) + (-2) = 0
C = -4/2 = -2
We do the same for C2H4O...
2(C) + 4(+1) + (-2) = 0
C = -2/2 = -1
therefore C is oxidized from 2- to 1-

Rachel Lipman
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

### Re: 14.1 Book HW

That makes complete sense, but say for question 14.3 where S is technically 2- as well as O, how do you determine that in SO42- that S has to have the oxidation number 6+?

Jacob Afable 3J
Posts: 41
Joined: Fri Sep 25, 2015 3:00 am

### Re: 14.1 Book HW

Rachel Lipman wrote:That makes complete sense, but say for question 14.3 where S is technically 2- as well as O, how do you determine that in SO42- that S has to have the oxidation number 6+?

Since the total charge of SO4 is 2-, Oxygen has a charge of 2- but since there are four oxygen atoms present you should multiply the 2- charge by 4 to get a total charge of 8- for Oxygen. In order for both the Sulfur and Oxygen charges to = 2-, the Sulfur needs to have an oxidation number 6+. 6-8=-2. Hope this helps. :)