What is voltage in electrolysis?

Is that the Q which is measured in coulombs. All I know is Volts = J/C

Can too much voltage be a systematic error in electrolysis- and if so why?

Remember that galvanic cells put out a certain voltage based on the Ecell value. For an electrolytic cell, instead the voltage is the amount of energy required to be put into the system for it to run. The two are mirror images of each. For this reason, do you think it might be a systematic error if you add too much voltage?

Ah right. I did not know that the d shell begins to fill at energy level 3.

So then, why is the electron configuration of K = K:[Ar]4s^1

And not equal to K:[Ar]3d^1?

The reason for this is because the energy of the 4s sub-shell is lower than the energy of the 3d sub-shell.

Remember how a little bit back I mentioned every new year in chemistry, you're told something isn't exactly true? We're back at it again.

The idea that the 3rd shell is lower in energy than the 4th shell is true for all elements up to 20 - and this is why you were taught that back then. As I said, the model you were taught is useful, but it's not ENTIRELY correct. This breaks down once you get into sub-shells, and you learn that the 4th shell starts to have electrons in it BEFORE the 3rd shell is full.

And the reason for this is because the energy level of a shell isn't constant - it actually depends on the sub-shell in question (and when you hit university, you'll learn that even that isn't constant, and that the energy level of each sub-shell changes for some atoms depending on what might be bonding to it. Ooft!). As it turns out, the 3d sub-shell - despite being in the 3rd shell - is HIGHER in energy than the 4s sub-shell - despite being in the 4th shell. That's where that arrow diagram above comes into play - it tells you what the order of the energy levels roughly are: 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s... and you get the idea.

So, when it comes to potassium putting electrons in its energy levels, it has 19 electrons and asks: what's my lowest energy unfilled sub-shell? Well, 1s, of course, so now its electron configuration and electron count is:

1s

^{2}, 17

And if you keep asking this question, you'll fill up the sub-shells like so:

1s

^{2} 2s

^{2}, 15

1s

^{2} 2s

^{2} 2p

^{6}, 9

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2}, 7

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6}, 1

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6} 4s

^{1}, 0

Which is long and tedious, so surely there's a better way? And that way is you can combine the information of that arrow-filling diagram (which tells you the order of the energy levels) and the periodic table (see below for one which tells you which block corresponds to which sub-shell):

So, we know that K is in the s-block and the 4th period, which corresponds to the 4s block. This means that every sub-shell below 4s will be filled:

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6}Since it is in the first group of the s-block, then that means there's only 1 leftover electron to add to the 4s sub-shell, giving us an electronic configuration of:

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6} 4s

^{1}If I were to do this for tungsten (W - group 6, period 6), I'd note that it's in the third row of the d-block, which corresponds to sub-shell 3(because third row)+2(because the d-block starts at energy level 3)=5d. So now I need to fill up all the energy levels below the 5d sub-shell - using the diagram above, that would be the following:

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6} 4s

^{2} 3d

^{10} 4p

^{6} 5s

^{2} 4d

^{10} 5p

^{6} 6s

^{2} 4f

^{14}Next, tungsten is the 4th element in the d-block, so it has 4 electrons in its valence sub-shell, so the final configuration is:

1s

^{2} 2s

^{2} 2p

^{6} 3s

^{2} 3p

^{6} 4s

^{2} 3d

^{10} 4p

^{6} 5s

^{2} 4d

^{10} 5p

^{6} 6s

^{2} 4f

^{14} 5d

^{4}And if you want to double-check that's correct, if you add up all the electrons, you should get 74 (the atomic number of W): 2+2+6+2+6+2+10+6+2+10+6+2+14+4=74

But also, it's worth noting that the study design only requires you to know how to do this up to 36 - so feel free to test with monstrously big ones, like tungsten, but know that you'll never have to go past Krypton. This also technically means all you REALLY need to know is that the 3d electrons are higher in energy than 4s, but not than 4d, and that's why you fill up 4s before you fill up 4d.