Redox Reaction question in Practice Midterm [ENDORSED]

[Laura_funes_1j]

Midterm Winter 2016
The following reaction is used in the acidic solution in the Breathalyzer test to determine the level of alcohol in the blood:

H+(aq) + (Cr2O7)-2(aq) + C2H5OH(aq) ---> (Cr)3+(aq) + C2H4O(aq) + H2O(l)

I need help figuring out how to get a balanced redox equation

[Chem_Mod]

First you must identify which substances are being oxidized and which are being reduced.

Then write each half-reaction.

Balancing the half-reactions follows the procedure starting on pg. 563 of the textbook since this is in acidic media.

It would be best if you could describe more specifically where you are stuck.

[Michelle Kam 1F]

First, we need to find which is oxidized and which is reduced.
We can calculate the charge for Cr in Cr2O7²⁺ by knowing that Oxygen usually has a charge of 2-.
charge of Cr + 7(-2) = -2
Cr has a charge of 6+.
The reactant Cr has a charge of 6+ and the product Cr has a charge 3+.

In similar way, you can find the charge of Carbon on each side.

Cr goes from a charge of 6+ to 3+
C goes from a charge of 2- to 1-.

We can conclude that Carbon is oxidized and Cr is reduced.

C2H5OH --> C2H4O + 2H+ + 2e-
We add two H+ to make up for the two more H+ on the reactant side.
then we also add 2 electrons to balance the charge of 2+ (coming from 2 H+ atoms) on the product side.

In similar ways, you can get the reduction reaction of
6e- + Cr2O7 2- + 14H+ --> 2Cr3+ + 7H2O

Before we smash the two equations together, we multiply our Carbon oxidation reaction by 3 to match the number of electrons on the two equations.

The final balanced equation is...
Cr2O72- + 3C2H5OH + 8H+ --> 3C2H4O + 2Cr3+ + 7H2O

[Giselle_zamora_1L]

How do you know that the changes in carbon, that's where I'm stuck. Is it something you just know/memorize?

[Hasmik Dis 2F]

How do you know that the changes in carbon, that's where I'm stuck. Is it something you just know/memorize?

The changes in carbon are determined by the oxidation numbers. So for example, we have to see how C's oxidation number changes in C2H5OH --> C2H4O.
This is determined by the other elements in the compound.
O has an oxidation of -2 (that's just a fact) and H is +1. In C2H5OH, the H's equal to +6 (since each H is +1). So now we have a total charge of +4 (6-2). However, we need to make sure our charges balance to 0, since C2H5OH is a neutral compound. We need to balance, and the way to do that is to make C = -2 and since we have 2 C's, it makes it -4. The same process goes for the other compound.