Lecture Notes from CHM 1341
26 June 1996

Advanced Stoichiometry

I presented an example of an ugly stoichiometric challenge designed to point out how valuable Julius Caesar's cis-Alpine Gallic strategy is to all aspects of the examined Life: Caesar said, "Divisa et impera" (divide and conquer). In the instance below, we divide a redox reaction into its half reactions to make balancing the whole an easy matter of balancing the parts!

Suppose we attempt to balance the oxidation of sulfite (ion) to sulfate (ion) by the nitrate ion which itself is (must be) reduced to nitric oxide in acid solution.

NO3-(aq) + SO32-(aq)  --/-->  NO(aq) + SO42-(aq)  1st bal O

NO3-(aq) + SO32-(aq)  --/-->  NO(aq) + SO42-(aq) + H2O(l)  now H

NO3-(aq) + SO32-(aq) + 2 H+(aq)  --/-->  NO(aq) + SO42-(aq) + H2O(l)

OK, so we should've used hydronium ion there on the left and added another water on the right accordingly. It still would not be balanced even though all the atoms balance! What's wrong? The charges don't balance! We've got an overall 1 negative change on the left but 2 on the right. Oops. How do we fix it?

Caesar to the rescue. Since this is (fortuitously) a redox reaction, let's balance the half reactions separately.

Oxidation:

SO32-(aq)  --/-->  SO42-(aq)   1st bal O

SO32-(aq) + H2O(l)  --/-->  SO42-(aq)  next H

SO32-(aq) + H2O(l)  --/-->  SO42-(aq) + 2 H+(aq)  finally e-

SO32-(aq) + H2O(l)  ----->  SO42-(aq) + 2 H+(aq) + 2 e-

which correctly concludes that being oxidized involves giving up electrons, two in this case for each sulfite ion oxidized. Note that this was simpler to balance because we have the freedom of fixing "broken" charge balance by flinging around electrons. Neat, eh?

Reduction:

NO3-(aq)  --/-->  NO(aq)   1st balance O

NO3-(aq)  --/-->  NO(aq) + 2 H2O(l)   next H

NO3-(aq) + 4 H+(aq)  --/-->  NO(aq) + 2 H2O(l)  finally e

NO3-(aq) + 4 H+(aq) + 3 e-  ----->  NO(aq) + 2 H2O(l)

which produces 3 electrons for each nitrate reduced. AHA...that's where the charge mismatch lies. So we need to rationalize the donation and acceptance of electrons by finding the least common denominator (6 electrons) and scaling each half reaction accordingly (3 oxidations for every 2 reductions, as written):

3 SO32-(aq) + 3 H2O(l)  ----->  3 SO42-(aq) + 6 H+(aq) + 6 e-

2 NO3-(aq) + 8 H+(aq) + 6 e-  ----->  2 NO(aq) + 4 H2O(l)

which we sum (just like algebraic equations) and cancel like terms from each side (just like algebraic equations) to reveal:

2 NO3-(aq) + 3 SO32-(aq) + 2 H+(aq)  ----->  2 NO(aq) + 3 SO42-(aq) + H2O(l)

which would've been agony trying to guess!

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

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