Redox Reaction question in Practice Midterm  [ENDORSED]

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Redox Reaction question in Practice Midterm

Postby Laura_funes_1j » Fri Feb 10, 2017 10:41 pm

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

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Re: Redox Reaction question in Practice Midterm  [ENDORSED]

Postby Chem_Mod » Fri Feb 10, 2017 11:26 pm

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
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Re: Redox Reaction question in Practice Midterm

Postby Michelle Kam 1F » Fri Feb 10, 2017 11:34 pm

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

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