The Lego® Model of Pharmacology

In understanding drugs and how they work, it can be helpful to think of the body and its atoms as a giant set of Lego® bricks[1]. A child can use Lego® bricks to make buildings, space ships, cars, etc; and the body can use its atoms to build muscles, bones, signaling molecules, etc. Moreover, just as the child can disassemble her building and then use the bricks to make a car, the body can disassemble its muscle and build bone. Of course, the car requires special bricks (e.g. wheels) which aren’t needed to make a building, and bone needs special atoms (e.g. calcium) which aren’t needed to make a muscle: the number of cars the child can make is limited by the number of wheels she has, and the amount of bone the body can make is limited by the amount of calcium available.

This last point is important. It means that if we want to regulate the number of cars that the child makes, we only need to regulate the number of wheels we allow her to use. We might do this because we have too many cars, and don’t want any more, or we might do so because we don’t have enough buildings (or space ships, or bridges) and want to conserve our bricks to make those instead of cars.

In the body, if we want to regulate the amount of bone we make, we can regulate the amount of calcium there is to make it with. We might do this because we have too much bone, and don’t want any more, or we might do so because we don’t have enough of something else, and want to conserve building materials to make, say, muscle.

Pharmacology (the science of drugs) manipulates the body by interfering with the way it uses its atoms. Continuing with the Lego analogy, drugs are the equivalent of another person adding or removing bricks to the buildings, cars, space ships, etc as they’re being built or after they’re finished; or adding or removing bricks from the box of unused bricks.

[1] Lego® is a registered trademark of the LEGO® Group of companies, which does not sponsor, authorise, or endorse this site

Electrochemistry

ELECTROCHEMISTRY seems to be a subject that lends its self to being confusingly taught. Do we add the half-cells together, or do we subtract one from the other? How do we know which direction the reactions run? Is this set-up voltaic, or electrolytic? Or perhaps it’s galvanic? And how many volts are really involved here?

Let's start with a galvanic cell, also known as a voltaic cell.

The simplest way to look at electrochemistry is to realize that the universe is a brutal place: might makes right. When we couple two reactions, (for instance, two half cells), whichever is stronger (has the largest magnitude (absolute value) for Eº) runs the show, which means that it gets to go the direction it wants to go. And like any chemical reaction, it wants to go in the direction that it's spontaneous.

Now, if we consider the equation

∆Gº = -nFEº

we note that it shows us the relationship between Eº and ∆Gº. F represents Faraday’s number, which is about 10,000, and n indicates the number of electrons involved. F and n will always be positive numbers, so a negative ∆Gº value (indicating a spontaneous reaction) corresponds to a positive Eº value. In other words, a positive Eº value indicates a spontaneous electrochemical reaction.

So, putting these items together, since the biggest Eº value goes the direction it wants to (i.e. it is spontaneous) we give it a positive value. If the Eº value that we’re given is positive, then we're good - the half-cell reaction runs as it is shown. If the value we’re given is negative, then the reaction is written backwards.

Of course, this is only one half of our cell. The half-cell with the big Eº is either generating electrons (electrons are a product) or absorbing them (electrons are a reactant). If it's generating them, then the second half-cell MUST absorb them. If the equation given for the second half-cell shows it absorbing electrons, then we can use the Eº value given. If the equation shows the half-cell generating electrons, then we have to run that reaction backwards, which means we keep the magnitude of the Eº value, but reverse its sign.

If the half-cell with the bigger Eº is absorbing electrons, then the second half-cell MUST generate those electrons. If its equation shows it generating electrons, we use its Eº as noted; if its equation shows it absorbing electrons then we have to reverse the sign of the Eº, but keep its magnitude.

RECAP
1. Biggest Eº runs the show - it goes in the spontaneous direction (direction of positive Eº value)
2. The direction of the second half-cell is determined by whether it has to absorb the electrons generated by the first half-cell, or supply electrons to be absorbed by the first half cell. If its equation is written in the correct direction, then we take the Eº value given - otherwise we reverse its sign.
3. The total Eº is the sum of the two Eº values. The Eº value of the first half-cell is positive. The sign of the second half-cell is determined by the direction it must run in order to cater to the electron needs of the first half-cell.

EXAMPLE
Create a galvanic/voltaic cell using the following 2 half cells:
SO4 + 4 H + 2e -> H2SO3 + H2O Eº = + 0.170
ZnS + 2e -> Zn + S Eº = - 1.440

Which half-cell is the anode, and which the cathode?
Which electrode is positive, and which is negative?
How many volts do we get out of the cell?
(Assume 1M concentrations and 25 degrees C)

NOTE: this all describes a galvanic/voltaic cell. Electrolytic cells are exactly the opposite.

The answers
1. The half-cell with ZnS has a larger absolute value (greater magnitude) for its value of Eº, so it runs in its spontaneous direction.
2. As written, this half-cell yields a negative Eº value; its spontaneous direction is therefore the reverse of how it's written, so we actually have:
Zn + S -> ZnS + 2 e Eº = + 1.440 Volts
3. In its spontaneous direction, the first half-cell generates electrons; the second half-cell MUST therefor absorb them (i.e. use them as a reactant)
4. As written, the second half-cell does have electrons as a reactant, so it's written the correct way:
SO4 + 4 H + 2 e -> H2SO4 + H20 Eº = + 0.170
5. Total voltage equals the voltage of half-cell 1 plus the voltage of half-cell 2:
1.440 + 0.170 = 1.610 Volts
6. OIL RIG: Oxidation Is Loss of electrons, Reduction is gain of electrons - electrons are lost at the Zn half-cell, so this is where oxidation occurs. Reduction occurs at the SO4 half-cell.
7. An Ox; Red Cat: The anode is the site of oxidation, while reduction occurs at the Cathode. The Zn half-cell is the site of Oxidation, and thus is the Anode. The SO4 half-cell is the site of reduction, and thus is the Cathode
8. The Zn half-cell is running the show. It generates electrons and sends them away (repels them). Electrons are repelled by a negative charge, so Zn is the negative electrode. SO4 is absorbing (attracting) electrons; it must have a positive charge to do so.