Archive – Coulombs notes – Link to pdf
Coulomb’s Law Notes
Almost everything we talk about in chemistry is related to electrons. The number of electrons and how they are arrange give atoms, molecules, and ions their chemical reactivities. What keeps atoms, ions, and molecules together is related to attraction of the nucleus (protons) to these very small negatively charged particle (electrons). Thus how these electrons are being attracted is a vital concept in chemistry. The force of electrostatic attraction between positive charges (protons) and negative charges (electrons) is measured in coulombs (C) and is described as the coulombic force of attraction. Electronegativity is a limited term used to quantify this attraction and is not universal for the description of the forces that attract proton and electrons. It is almost always better to describe the attraction forces between protons and electrons throughout all of chemistry using the concept of coulombs.
The coulomb (symbol: C) is the International System of Units (SI) unit of electric charge. It is the charge (symbol: Q or q) transported by a constant current of one ampere in one second:
It is equivalent to the charge of approximately 6.242×1018 (1.036×10−5 mol) protons, and −1 C is equivalent to the charge of approximately 6.242×1018 electrons. Notice that the fundamental charge of proton is equivalent to an electron but just opposite in sign.
The coulomb is in honor of Charles-Augustin de Coulomb (1736 – 1806) was a French physicist who determined the relationship between the size of electrostatic forces and distance between them (inverse square law) and the same relationship in magnetic forces (electrostatic and magnetic Forces were not yet unified).
*Connections- Coulombs Law! – F = K (constant) Q1 x Q2
The Force (F) of attraction between 2 point points charges of opposite charge (Q1 & Q2) is directly proportional to the size of the charges and inversely related to the square of the distance between these charges.
-IF the size of the opposing point charge increases then the Force of Attraction will be greater proportionally. We will see this in insoluble salts (ionic compounds) that cannot dissolve in a couple of weeks.
Example: NaCl (s) → Na+ (aq) + Cl– (aq)
sodium chloride can break into its ions in water because the coulombic force between positive and negative ions in the crystal of NaCl are weak compared to the ions attracted to water. We say they are soluble.
MgO (s) → does not break apart into its ions in water
because the Mg+2 and O-2 ions have HIGHER point charges THUS the Force (F) between the ions is TOO LARGE for water to break apart thus it is insoluble and remains as solid in water.
– IF the distance increases between the 2 point charges then the F decreases (inverse square relation). We will look at this inverse square in our atomic structure discussion later in the course. But in terms of the outermost negative electrons ( Q1) and their Proximity (d) to the positive nucleus (Q2) the farther away the electrons the lower the coulombic attractive force that these electrons feel, thus these elements generally lose electrons (oxidize). We see this in the sodium metal demo in water.
Oxidation : Na → Na+1 + 1 e– + Energy
Notice the lowered Force is amplified because of the squaring of the larger distance that metals have between their nucleus and outermost (valance electrons). Increasing the size of the denominator will always Increase the entire value of the formula.
IF the distance decreases between the 2 point charges then the F increases . In terms of the outermost negative electrons (Q1) and the Proximity (d) to the positive nucleus (Q2) the CLOSER the electrons are to the nucleus, the greater the coulombic attractive force that these electrons feel, thus these elements generally gain electrons (reduce). We see this every time you see a combustion reaction where oxygen grabs electrons from metals causing them to corrode or change into a new compound or in combustion when oxygen grabs electrons from carbon (bigger nonmetal)
2Fe + O2 → 2FeO + Energy
Pure Iron RUST
C6H10O5 + 6O2 → 6CO2 + 5H2O + Energy
Reduction half reaction for both: O + 2e– → O-2
In both cases Oxygen became reduced or pulled the electrons because of its higher coulombic attraction due to electrons Closer to the nucleus (smaller d) in part due to the atomic radius being smaller. In the case of the combustion reaction of the cotton ball oxygen has both smaller d and greater (Q2) due to having a greater nuclear charge (more protons) than carbon. That is how Oxygen (a nonmetal) can pull an electron from another nonmetals (carbon)!!!!!!!
What is the oxidation state of C in glucose? see bottom of page
What is the oxidation state of C in CO2? see bottom of page
Elemental Sodium Demo: Its all about Coulombs Law!!!
Na is highly reactive because of it large radius, thus a low coulombic attraction to its outer most valence electron. It easily loses its electrons to the hydrogen in the water. The water breaks in to OH– to H (with the electron from sodium). The H (with the electron from sodium) bonds with itself to make H2. This H2 reacts and explodes with oxygen in the air and with the heat released from sodium getting a stable electron configuration (when losing an electron).
1s22s22p63s1 → 1s22s22p6
Na → Na+1 + 1 e– + Energy
So the heat from the above reaction is used to start the next reaction. It is actually the explosion of the hydrogen with oxygen that resulted in the video.
2H2 + O2 = 2H2O + Energy
In Lab 1 – Constructing a voltaic Cell – How did you determine the which electrode (the conducting object where the redox reaction actually occurs) is the Anode (place of oxidation = Anode = An Ox) and which electrode was the Cathode (place of reduction = Cathode = Red Cat)?
Glucose oxidation state: C = O
(it is actually a little bit more complicated in the 3d structure but its averages about zero.)
CO2 oxidation state = +2