Interactions that are inTERmolecular are among more than one molecule of a substance. These are the forces that attract the molecules of a liquid or solid to each other. Intermolecular interactions occur among nonpolar molecules as well as among polar ones, but polar interactions are much stronger than nonpolar ones are.

Polar Molecules:
Polar molecules contain one or more dipoles. They fall into two categories: polar protic, and polar aprotic. Polar protic molecules have protons (hydrogens) attached to electronegative atoms, namely F, O, or N. Polar aprotic molecules still contain dipoles, but there are no protons attached to F, O, or N. It is essential that you understand the distinction between protic and aprotic polar molecules in order to properly understand the organic chemistry substitution mechanisms.

Polar Aprotic Molecules:
Polar aprotic molecules have intermolecular interactions known as dipole-dipole interactions. The positive end of the dipole on one molecule is attracted to the negative end of the dipole on another molecule. This is an electrostatic interaction, and it is governed by Coulomb’s Law. Coulomb’s Law tells us that either a larger charge separation (leading to a larger molecular dipole) or a smaller distance between the molecules will strengthen the dipole-dipole interactions between them. There are many examples of polar aprotic molecules, including ketones, aldehydes, esters, tertiary amines and amides, HCl, HBr, HI, cyanides, nitro groups, and DMSO (dimethyl sulfoxide). Polar aprotic solvents destabilize nucleophiles, making them suitable solvents for Sn2 reactions in organic chemistry.

Polar Protic Molecules:
Polar protic molecules have intermolecular interactions known as hydrogen bonds. In spite of their name, note that hydrogen bonds are not true bonds. Rather, hydrogen bonds are an especially strong type of dipole-dipole interaction. The positive end of a dipole on one molecule (the H) is sometimes so strongly attracted to the negative end of another molecule’s dipole (the F, O, or N) that the H will dissociate from the first molecule and bind to the second one. You are already familiar with this process for water, which autoionizes to form H3O+ and OH-:

2 (H2O) <-> (OH)- + (H3O)+

Water is the most common example of a polar protic molecule, but some other examples include alcohols, carboxylic acids, primary and secondary amines and amides, and hydrogen fluoride. Polar protic solvents stabilize carbocations, making them suitable solvents for Sn1 reactions in organic chemistry.

Nonpolar Molecules:
Molecules that do not contain dipoles are called nonpolar molecules. They are held together by intermolecular interactions that are called London dispersion forces or van der Waals interactions. These interactions are due to temporary, induced dipoles. Where do these dipoles come from? To answer this, you must realize that the electrons in a bond or in an atom are not stationary. Rather, these electrons are constantly in motion. Sometimes, by sheer chance, they are not equally distributed on the atom or in the bond, and this inequality of charge forms a temporary dipole. It is temporary because soon thereafter, as the electrons continue to move, they distribute more equally again, and so the dipole goes away. But in the meantime, that short-lived dipole has affected its neighbors, and it has caused similar dipoles to form in them as well. This is why these dipoles are induced. So you can imagine a bunch of molecules, held together by small dipoles that are constantly forming and unforming. But there are always some dipoles present at any given time, and that is what holds the liquid or solid together.

Note that all substances contain dispersion forces, but these forces are only greatly significant for nonpolar elements and molecules that lack dipole-dipole interactions. In polar compounds, dispersion forces can be neglected.

Question: Is it ok to calculate the emf just as if the cell were galvanic, and then simply switch to a negative value?

Yes, that will work as long as you write one half-reaction as a reduction and the other as an oxidation. Many times both half-reactions will be written as reductions, so check where the electrons are in the equations. If they are on the left side of the equation, it’s a reduction half-reaction, and if they are on the right side of the equation, it’s an oxidation half-reaction. You must always have one of each in any redox reaction, because if something is getting reduced than something else must be getting oxidized.

Question: for standard cell potential: which formula is correct: Eo = Ereduction + Eoxidation? Or Eo(cell) = Eo(cathode) - Eo(anode)? How can this problem be solved?

Consider the following electrode potentials:

A) -0.89 V
B) +0.55 V
C) +1.57 V
D) + 1.91 V

My advice to you is to not use either version of the formula, as it will only confuse you. (Guess I didn’t need to tell you that, right? ) Instead, what you should do is attack any electrochemistry problem by first considering the following:

What is the form in which the two half-cell reactions are written? Generally, but not always, you will see both half-reactions written as reduction potentials. You can determine this because you will see that the electrons are being added on the left side of the equation. Remember, when you add electrons, it’s a reduction. Conversely, if you lose electrons, it’s an oxidation, and you will see the electrons on the right side of the equation.

If both half-reactions are written as reductions, you must turn one of the them backward into an oxidation half-reaction. (If something is being reduced, something else better be getting oxidized because those electrons have to come from somewhere.) The tough part is deciding which half-reaction to turn backward. If you have a galvanic cell (most common case), you will want to turn the half-reaction with the smaller reaction potential backward. (Don’t forget that a number like -1.2 would be smaller than one like -0.3). You do this because galvanic cells will always have a positive Ecell (they have to, because they are spontaneous, and the equation that relates G with Ecell has a minus sign in it: G=-nFEcell). If the cell is electrolytic, then it is not spontaneous, Ecell should be negative (making G positive) and you will turn the half-reaction with the larger reaction potential backward instead. Remember, in either case, when you turn the half-reaction backward, you change the sign of that half-reaction’s reaction potential.

Ok, so now let’s consider your specific example. The first half-reaction (the Cu one) is a reduction half-reaction. Again, I know that because the electrons are being added on the left side; Cu is gaining electrons. The second one, in contrast, is an oxidation half-reaction. The electrons are on the right side; O is losing electrons. Good, so in this case, nothing needs to be turned around, because we already have one reduction and one oxidation. You can see that this cell is an electrolytic one as written. (Add the two reaction potentials together, and you are going to get a negative number for Ecell) Looking at the third equation that combines both half-reactions, you can see that it is written in the same fashion as the two half-reactions above it: Cu is getting reduced, and O is getting oxidized. Thus, again, you have an electrolytic cell, nothing needs to be turned backward because one partner is oxidizing and one is reducing, and you merely need to add the two reaction potentials together to get your -0.89.

Really, there isn’t any need to do any math on this problem. As soon as you recognize that you have an electrolytic cell, there is only one possible right answer for this question, because the others are all positive for Ecell.

Remember that you do not have to multiply the values of the reaction potentials by the number of moles of electrons. The reason why is that reaction potential is an intrinsic property; it is independent of the amount of material you have. That is, one gram of Cu has the same reaction potential as 1 kg of Cu does. (Temperature and density are other examples of intrinsic properties.) The properties where you do have to multiply by the number of moles are extrinsic properties; those do depend on the amount of material. For example, when you use Hess’s law, it matters greatly how much material you have because the more compound you have, the more heat it will give off. (Energy and volume are some other examples of extrinsic properties.)

Calculating logs without a calculator is not as difficult as you might expect it to be. First, you should be familiar with the following log rules:

log (xy) = log x + log y
log (x/y) = log x - log y
log (x)^y = y(log x)

You can use these rules to derive the Henderson-Hasselbalch equation from the Ka expression for an acid:

HA -> A- + H+

Ka = [A-][H+]/[HA]

Taking the -log of both sides gives us:

-log Ka = -log {[A-][H+]/[HA]}

The left side is the definition of pKa:

pKa = -log {[A-][H+]/[HA]}

The right side can be re-written using the log rules:

pKa = -log [A-] - log [H+] - (-log [HA])

The second right side term is the definition of pH:

pKa = -log [A-] + pH + log [HA]

Solving for pH:

pH = pKa + log [A-] - log [HA]

Using the log rules in reverse gives us the equation as it is normally written:

pH = pKa + log [A-]/[HA]

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You should also know the common base 10 logs. For example:

log 1/100 = -2
log 1/10 = -1
log 1 = 0
log 10 = 1
log 100 = 2
and one more that is very helpful for pH estimations: -log 0.5 = 0.3

So let’s say that I’m working on a pH problem. I find out that my [H+] is equal to 3.6 x 10^-5. What is the pH? We don’t have a calculator on any of the pre-health tests, so we need to be clever about this. I don’t know what the -log of 3.6 x 10^-5 is. But here’s what I do know:

-log (1 x 10^-5) = 5
-log (10 x 10^-5 = 1 x 10^-4) = 4

So, the -log (3.6 x 10^-5) must be some number between 4 and 5. That is usually sufficient to solve the problem. Your answer choices might be 3.4, 4.4, 5.4, and 6.4, and you will of course pick 4.4.

If necessary, we can do even better than this. Say two of our answer choices are 4.2 and 4.5. How do we pick between them? (Note that this situation should not come up on your test, but just in case it does, you’ll know what to do!)

Here’s where knowing that -log 0.5 = 0.3 comes in useful. We know that -log (5 x 10^-5) must be 4.3, because we already memorized that -log (0.5) = -log (5 x 10^-1) = 0.3. Since we want the -log (3.6 x 10^-5), it must be a number that is close to 4.3, but a bit higher (because 3.6 is between 1 and 5, then the pH must be between 4.3 and 5). So we can eliminate 4.2, and 4.5 must be the correct answer. (The actual pH value is 4.44.)