7 Electrochemistry

7.1 Redox Reactions

7.5 Stranger Things / Electrodes

**7.8 Counting Coulombs, Moles, and Joules**

7.9 Batteries

7.10 Battery Facts

7.42 Learning Outcomes

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Electricity allows us to calculate the amount of charge passed (\(q\), in coulombs, C) by multiplying the electric current (\(I\), in amps, A) by the time (\(t\), in seconds, s). The formula for this is simply...

\[q = I\cdot t \]

The reason this works is because and amp (A) is actually a coulomb per second (C/s). So as long as your time (\(t\)) is in seconds, multiplying current times time will give you coulombs (C).

A faraday is the amount of charge contained by exactly one mole of electrons. That amount of charge is 96485 C. So the faraday is basically a conversion factor to switch from number of moles of electrons to number of coulombs and vice versa.

So to get the total number of coulombs passed for a balanced chemical reaction (aka: a "mole of rxn") you just multiply the number of moles of electrons (\(n\)) by the faraday constant, \(F\).

\[q = n\cdot F\]

Now if we set each of the above equations equal to each other and do some quick algebra we get the following helpful equation allowing us to convert current, time, moles of electrons, and the faraday into moles of the reaction:

\({\displaystyle I\cdot t\over \displaystyle n\cdot F}= {\rm moles\;of\;reaction}\) *

Once again, an amp of current \(I\) is the rate of charge passing in coulombs/second. Because of this, you should just cross out amps when you see it and replace it with C/s. The term "moles of reaction" relates directly to your overall balanced redox reaction or even half-reaction if your only looking at that.

**Electroplating**: When we electroplate a metal, we are just running a "mole of reaction" as much as we want until we plate the amount of metal that we want. Pretty much all metals electroplate by the reduction reaction of the metal in some monatomic ionized state:

M^{n+}(aq) + ne^{–} → M(s)

Notice how you use *n* moles of electrons to deposit one mole of metal in the +*n* oxidation state. You'll need to know how to use that formula (*) and calculate not just the amount of metal, but also the current and/or time if asked that way.

**Not just for electoplating**... realize that ANY redox reaction can be quantified via that formula (*). If you are generating a gas, that equation will tell you how many moles of gas are made. Then you can put that number into the ideal gas law and you've got the volume, or the pressure. That equation is just another piece of the pie called reaction stoichiometry - it brings electricity into the game.

Sometimes we just want to know the total amount of energy (sometimes called electrical work) that is being produced by a given redox reaction under a given set of conditions. Luckily, the rules and formulas of electricity have us covered there as well. Just looking at unit factors (conversion factors) in the electric world you find out that a joule (energy) can also be expressed as a volt-coulomb (V C). To be specific: 1 J = 1 V C. Now think about it, our redox reactions have a potential in volts, and the number of coulombs is just the moles of electrons, \(n\), times the faraday, \(F\). Put it all together and we get:

electrical energy (or work) = \(n\cdot F\cdot E\)

If you have a *standard* cell, you would use \(E^\circ\) in place of the \(E\), if *non-standard*, then use the Nernst equation to calculate \(E\)) and then use that potential. Also note that the units out of this equation is joules, or just plain J. It is often a few hundred thousand joules which means you will often need to convert J into kJ to get a more "reasonable" number.

Your local electric company or utility charges you for energy. They do not count in joules though - that would be a freakin' huge number of joules. They bill you in kilowatt-hours (kW·hr or KWH). It is actually a nice energy unit in that you can estimate energy usage by multiplying the power in kilowatts by the time of use in hours. A watt is a power unit and is the same as a joule per second (J/s). So if you want to know how many joules is in a kilowatt-hour you multiply by 1000 first to change kW into just W. Next multiply by 3600 which is the number of seconds in an hour. So that is 3.6 million joules for one kW·hr. So my actual bill for electricity in August of 2020 (yeah, it's really crazy hot in August and the AC runs a lot) was for 1883 KHW (which happened to be about $210 - ouch!). So how many joules is that? 6,778,800,000 J!! Maybe scientific notation would help, that is 6.78 × 10^{9} J or about 6.8 billion joules of energy. I think I prefer the 1883 KWH version.

Let's imagine a big bowl of electrons. The bowl is holding exactly one mole of electrons. Those electrons each have a tiny amount of negative charge, but because there is an entire mole, that charge adds up to 96485 coulombs (the faraday, \(F\)). If we just divide the total charge by the counting amount of electrons which is Avogadro's number, \(N_{\rm A} = 6.022\times 10^{23}\) we get the following:

\[{F\over N_{\rm A}}=q_{\rm e}\]

Where \(q_{\rm e}\) is the amount of charge on one single electron:

\[q_{\rm e}={96485\over 6.022\times 10^{23}}= 1.602\times 10^{-19} \;{\rm C}\]

And... we also know that a proton has the same amount of charge as an electron, but just opposite in sign. Therefore, a proton has the same amount of charge which is \(1.602\times 10^{-19}\) C, \(q_{\rm p}\). The other term often used for this tiny amount of charge is **elementary charge** and has the symbol \(\rm e\).

Well, thanks to resent revisions in the definition of a kilogram and the mole, we now have *exact* numbers for elementary charge and the faraday constant. There is no uncertainty now, the following complete values are infinitely precise by definition.

**elementary charge** = e = 1.602176634 × 10^{–19} C

**faraday constant** = *F* = 96485.33212 C/mol

That is a lot of digits... be glad that we are happy to just use a rounded version of each of those numbers.