The Action Potential

Action Potential is the term describing the sending of a signal from one end of an axon (the axon hillock) to the other end of the axon (by the synapse). What follows is the basics of this phenomenon.

Remember that in almost every cell, including neurons, the Na/K pump is continuously running in the background. This pump, of course, pumps three Na+ out for every two K+ it pumps into the cell. In doing so, it creates several gradients:*

1. a Na+ gradient, where Na+ is greater on the outside of the cell than on the inside
2. a K+ gradient, where K+ is greater on the inside of the cell than on the outside
3. an electrical gradient, where the outside of the cell has more positive charges (is more positive than) the inside of the cell. Thus, the inside of the cell is negative in comparison to the outside of the cell, and a polarity exists across the cell membrane. This electrical gradient is known as the membrane potential, and it’s normal value (inside perhaps -70 mV with respect to the outside) is referred to as the resting (membrane) voltage, or resting potential.

Because of these gradients, if we were to open a Na+ channel, Na+ would be drawn into the cell for two reasons:

1. there is more Na+ outside of the cell than inside of it
2. the inside of the cell is negative with respect to the outside of the cell, attracting the positively charged Na+ ions

If we were to open a K+ channel, K+ would be torn between two impulses:

1. the K+ gradient, which would push K+ out of the cell
2. the electrical gradient, which would pull the positively charged K+ ions into the relatively negative interior of the cell

An action potential involves channels for both Na+ and K+. Both channels are voltage-gated (which means that they are triggered to open by changes in membrane potential). Na+ channels open quickly, but after being open for a brief time they lock closed. K+ channels don’t open as fast as the Na+ channels do. K+ channels do not lock closed. Threshold potenital is the membrane potential at which the channels are triggered.

If we focus on the cell body of the neuron, and on its dendrites, we find several ion channels in it. These channels open and close in response to incoming signals, and let positive and negative ions enter and leave the cell. If enough positive ions enter the cell, then the voltage across the membrane (the membrane potential) reaches the threshold potential and triggers those Na+ and K+ channels on the axon hillock. These channels then open.

DEPOLARIZATION:
The Na+ channels open first. Na+ pours into the cell, bringing its positive charge with it. This positive charge continues to grow, and eventually spreads as far as the next voltage-gated Na+ channel. When the positive charge at this next Na+ channel grows to a large enough size, it triggers that Na+ channel (remember, the Na+ channels are triggered to open by the membrane potential). Sodium flows in through this Na+ channel, and the positive charge spreads down to a third Na+ channel, triggering it to open. This process continues, opening the Na+ channels one at a time, with each Na+ channel allowing in the Na+ ions that trigger the next Na+ channel to open. This advancing wave of positive charge is the signal that is passed down from one end of the cell to the other. We call this phenomenon depolarization because the initial polarity across the cell membrane is lost as the positively charged Na+ ions enter the cell. (In fact, we actually end up with the cell interior being slightly positive.)

REPOLARIZATION:
If all we had was Na+ channels, we could send a signal, as described above, but then positive charge would have filled the axon, and there would be no way to send a second signal. So, we have to have a way to reset this membrane potential. This resetting is the job of the K+ channels. Recall that the same membrane voltage changes that trigger the Na+ channels also trigger the K+ channels, but that the K+ channels are slower to react. Eventually, however, the K+ channels do react, and do open, and K+ rushes out of the cell. K+’s exit moves positive charge out of the cell, returning the cell membrane voltage to its normal value of slightly negative inside the cell, and frankly overshooting a bit, so we end up a bit too negative. The Na/K pump helps return the cell from this overshoot (hyperpolarization) to normal membrane potential. We have regained our normal, resting membrane voltage, so we call this repolarization.

DEPALARIZATION + REPOLARIZATION = ACTION POTENTIAL:
The wave of depolarization, followed by its wave of repolarization, is known as the action potential.

REFRACTORY PERIODS:
Above, I noted that the Na+ channels, after being open for a brief moment of time, lock closed. Obviously, as long as they are locked closed, they cannot open, and so cannot participate in an action potential. This period during which the Na+ channels are locked closed is known as the absolute refractory period, since no amount of stimulation can cause another action potential to pass down the neuron.

Following the absolute refractory period is the relative refractory period. Remember how the K+ channels cause us to overshoot our target membrane voltage? Until the Na/K pump has returned the membrane to its proper resting voltage, the cell membrane is too negative, and a larger than normal force is required to change the cell membrane voltage enough to trigger an action potential.

* The pump also creates an osmotic gradient, where the number of particles outside the cell is greater than those inside the cell. The osmotic gradient is not used in the action potential, and is further complicated by the fact that running the pump splits ATP (a single osmotic particle) into ADP and Pi (two osmotic particles), which would tend to cancel the osmotic effect of pumping three ions out of the cell for each two ions pumped in. However, the ADP and Pi are generally swiftly recycled back into ATP, etc, etc, and at this point we're getting to be more complicated than we need to be. Focus on the three gradients already mentioned, and you should be fine.

Dalton's Law of Partial Pressures: HW problems

1. Consider Dalton’s Law of Partial Pressures:
Ina. Restate the law (put it into your own words) Answer
Inb. Consider a closed container with 7 atm of gas inside. There are 2 mol of gas X, 5 mol of gas Y, and 7 mol of gas Z. The gasses do not interact with each other, and thus Dalton’s Law of Partial Pressures applies. What are the partial pressures of each of the three gases? Answer
Inc. Consider another closed container. Inside, there are 2 mol gas A. When we add 3 mol of gas B, we see that the total pressure inside the cylinder increases by 6 atm. If gas A does not interact with gas B, what was the original pressure in the cylinder? Hint: Hint 2 Hint 3 Answer


HINTS:
1.c.
Hint 1: Another way of looking at Dalton’s Law of Partial Pressures tells us that as long as the gases in a container don’t interact, their pressures are independent. In other words, as long as the pressures don’t interact, we can apply PV=nRT to them independently.

Hint 2: Since the pressures of A and B are independent, the partial pressure of gas A doesn’t change when we add gas B.

Hint 3: Since the partial pressure of gas A doesn’t change when we add gas B, we can calculate A’s partial pressure after we add gas B, as the pressure will be the same.


ANSWERS:
Question 1
Part a
Dalton tells us that, if we have more than one gas in an enclosed container, and those gases don't react, then the ratio of each gas's partial pressure to the total of the pressure of all of the gases equals the ratio of the moles of each gas present to the total number of moles of gas present.

Part b
gas X: 1 atm; gas Y: 2.5 atm; gas Z: 3.5 atm
Explanation: Initially, we have:

2 mol gas X	= ? atm
+ 5 mol gas Y	= ? atm
+ 7 mol gas Z	= ? atm
= ? mol	= 7 atm

We can easily fill in the total number of moles:

2 mol gas X	= ? atm
+ 5 mol gas Y	= ? atm
+ 7 mol gas Z	= ? atm
= 14 mol	= 7 atm

Now, we can determine the ratio of moles to atmospheres: 14 moles to 7 atm, which reduces to 2 mol to 1 atm, or 1 mol to 0.5 atm. We then apply this ratio to the individual gases:

2 mol gas X	= 1 atm
+ 5 mol gas Y	= 2.5 atm
+ 7 mol gas Z	= 3.5 atm
= 14 mol	= 7 atm

Note that the pressure each gas individually adds up to the total pressure in the container.


Part c
4 atm
Explanation: Dalton tells us that, if gasses don't interact with each other, for a given volume and temperature the ratio of moles to presure is fixed. When we added three moles of gas B, we saw that it generated 6 atm, telling us that the ratio of moles:atm was 3:6, or 1:2. Since there are two moles of gas A, these moles exert 4 atm.


Last edited July 26, 2008

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.

FAQ

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[Note - this entry may be updated as questions come in, so the "posted" notation at the bottom refers to the time when I first created this FAQ, not when I last updated it.]

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