How do I calculate the chance of an atom (ion) losing an electron?

[Shelnutt2]

So I've read through https://www.physicsforums.com/showthread.php?t=32409" thread on valence electrons but it doesn't really answer my question. I think I am in the right forum, I was debating between here and chemistry but it deals more with atoms. What I want to know is this:

How do I figure out (/calculate) the chance of an atom (ion) losing an electron? Is it based on how many electrons in the valance shell? As it's shell is closer to being filled, or closer to being empty how does it effect the ability to loose an electron?

What I am trying to figure out is how many amps are possible in a flowing fluid.

What I did was I found the total number of valance electrons contained in all of the negative ions in regular tap water. I found out the ions, then I looked up the model, to find out how many valence electrons. From there I divided the weight of the valance electrons by the weight of the entire atom. I then found out what % of weight was valence electrons. Knowing that, I think just multiplied out the % by the mg/l in tap water. I then had several smaller mg/l number which is about equal to the total weight of valence elections per mg/l. From that I divided by the weight of an electron, finding the number of valence electrons per liter. Multiplied by the charge of an electron to get charge / liter. Then I multiplied by my flow in liters / second and I got charge / second or amps. I know the formula for amps is I = q * n * v * A, but v * A = flow, so I can keep that.

Bicarbonate = 113 mg /l *.000134 => .0151 mg/l
Sulfate = 25 mg/l *.000137 => .00342 mg / l
Chloride = 21 mg / l *.000124 => .00260 mg/l

Total valence electrons are 0.0212 mg /l . That is .0212 * 10^-6 kg / l.
Mass of electron is 9.1093818 * 10^-31 . Divide .0212*10^-6 kg/l / 9.1093818 *10^-31 kg = 2.318 * 10^22 electrons/liter. Charge of an electrons is 1.60217646 * 10^-19. q * n = (1.60217646 * 10^-19) * (2.318 * 10^22) = 3,714.63 coulombs/liter

I = q * n * flow = 3,714.63 coulombs/liter* .667 liters/second = 2,477.6 amps.

I have no clue if this is right, or not. Can anyone give me some pointers?

 

[alxm]

How do I figure out (/calculate) the chance of an atom (ion) losing an electron?

Well, electrons don't just pack their bags and leave, as it were. The electron needs some place to go.
Or to put it more concretely: Imagine a single atom in a vacuum (at low temperature). The chance that an electron will leave, that it'll spontaneously ionize, is more or less zero. The rule you learn is that an atom with a noble-gas electron structure has lower energy, but you need to remember that this is given that it's got other atoms to interact with.

Is it based on how many electrons in the valance shell? As it's shell is closer to being filled, or closer to being empty how does it effect the ability to loose an electron?

The explanation you get (or I got) in 8th grade Chemistry is that atoms 'strive' to have noble-gas structure, so that elements on the left side of the periodic table will 'want' to lose electrons, and those on the left will 'want' to gain them. As for why that is, I gave a more detailed answer to this in https://www.physicsforums.com/showthread.php?t=293273".

What I am trying to figure out is how many amps are possible in a flowing fluid.

That depends on the voltage of course. And if the voltage is low, the resistance will be limited by the amount of free ions in the liquid that can act as charge carriers. But at higher voltages, the electric potential will ionize the liquid. And if you put in a very big amount of electricity, you're going to be dealing with a plasma (ionized gas).

In general, charge conduction is limited by the number of ions/degree of ionization, and the ease at which they can move through solution, which depends on the properties of the ions, the viscosity, etc. Conductivity is not easily calculated, but various models for ideal cases (strong and weak electrolytes) exist. See a textbook on electrochemistry or physical chemistry.
(Picking up the one I happen to have closest to me at the moment, Atkins, "Physical Chemistry" sec 24.7 "The conductivity of electrolytic solutions")

 

[charng]

Calculating the chance of an atom (ion) losing an electron is a complex process that depends on various factors, such as the energy levels and electron configurations of the atom, the surrounding environment, and any external influences. It is not as simple as just looking at the number of valence electrons.

In order to accurately calculate the probability of an atom losing an electron, you would need to consider the ionization energy of the atom, which is the amount of energy required to remove an electron from the atom. This value can vary depending on the specific atom and its electron configuration.

Additionally, the stability of the atom and the strength of its bonds can also play a role in the likelihood of electron loss. Atoms with a stable electron configuration are less likely to lose an electron, while those with a less stable configuration may be more prone to losing an electron.

In your calculation, you have considered the number of valence electrons in various ions found in tap water, but this does not necessarily reflect the probability of those atoms losing an electron. It is also important to note that the charge of an electron is not constant and can vary depending on the environment.

Overall, predicting the chance of an atom losing an electron is a complex process that requires a thorough understanding of the atom and its surroundings. Your calculation may provide an estimate, but it is not a reliable way to determine the probability of electron loss. It would be best to consult with a chemistry or physics expert for a more accurate and comprehensive calculation.