Electrolysis

Today I'm going to talk about electrolysis, the final thing that you need to learn about redox in Year 11 Chemistry.

Remember how I said that sometimes reactions might not occur, depending on where the elements sit on the Standard Reduction Potentials chart? (See Redox Equations if you can't remember.) Well, I lied (or at least I think I did...). It is possible to make these reactions occur, but they won't occur spontaneously. That's where electrolysis comes in- an electrical current is passed through a substance to make a reaction occur. For example, you can electrolyse sodium chloride to give you sodium and chlorine, and I'm pretty sure you can also electrolyse water to give hydrogen and oxygen.

To electrolyse something, you need some equipment: mainly a battery, some wires, and inert substances such as graphite. Oh, and you also need the solution that you want to electrolyse. The solution is known as the electrolyte while the inert substances are the electrodes.(You can have electrodes that aren't made of inert substances, but if they aren't inert then they might react in the electrolysis process. Sometimes this comes in handy, like in electroplating which I'll tell you about in a future post, but most other times you'll want to keep the electrodes inert.)

Here's a nice little diagram of how to set up the equipment:

Now, the way I think this works is that the battery provides electrons for oxidation and reduction processes to occur. Electrons flow through the battery from the positive end to the negative end, then out the negative end into the rest of the circuit, then back up the positive end. Therefore, electrons are flowing into the electrode known as the cathode, making the cathode negatively charged, and flowing out of the electrode known as the anode, making the anode positively charged. Because of this, positively- and negatively- charged ions flow to the cathode and the anode, respectively.

How will I remember all this?! you might ask. Well, you could try my Chemistry teacher's trick. The anode is positive because it starts with an A and getting an A+ is a really good thing, while the cathode is negative because it starts with a C and that's just average. Also you might find it interesting (though not really that helpful) that positive ions, also known as cations, flow to the cathode, while negative ions, also known as anions, flow to the anode. There's also another acronym that might help you remember stuff:

AN OIL RIG CAT
At the Anode, Oxidation, which Is Loss takes place while Reduction which Is Gain takes place at the Cathode.

Writing half-equations for electrolysis of molten salts is the same as writing half-equations anywhere else (see Redox Equations for more info). If you're using solutions, however, things get a bit trickier, as a solution indicates that water comes into it as well. In this case, you need to write down all of the possible oxidation reactions and all of the possible reduction reactions. FYI, here are the oxidation and reduction reactions for water:

Oxidation: 2H₂O -> O₂ + 4H⁺ + 4e^-

Reduction: 2H₂O + 2e^- -> H₂ + 2OH^-

(By the way, Wikipedia pointed out to me that if you combine these two equations you get 2H₂O -> H₂ + O_{2 : the electrolysis of water.)}

Then you need to go back to the good ol' Standard Reduction Potentials table and look at those numbers on the side. Compare the standard reduction potentials of the two elements that might possibly be oxidised/ reduced. The element you need is the one with the highest standard reduction potential. (Be careful, though, as some of the reactions might be listed the other way around on your sheet. If this is the case, then you need to change the sign on the standard reduction potential value from positive to negative or vice versa. It may be easier just to look for the highest absolute value.)

(Actually I need to double check that last paragraph- this is the section where I get all confused with what's what. I'm sure it's not that confusing. One day I'll sort it out... though I can get some consolation from the fact that solutions isn't strictly required knowledge in Chemistry 2AB...)

How to Deal with Qualified Drivers

Qualified Drivers- the bane of every Learner Driver. Some qualified drivers are retards, but I get the feeling that many learners who hate the qualified drivers (hereafter referred to as Q-drivers) tend to hate them just because they're there. I've decided to write this guide to combat the average learner's fear (for lack of a better word) of Q-drivers. After all, being able to handle traffic is all part of being a driver- and yes, that means being able to handle the fact that there are other Q-drivers around and some of them are pretty nasty towards L-drivers.

Q-drivers come in all different varieties. There's the nice ones and the indifferent ones, the tailgaters and the show-offs. And yes, there's the smart Q-drivers and the idiotic Q-drivers. Just because you have your driver's licence doesn't mean that you're a smart, safe, courteous driver. Conversely, just because you're still on your Ls doesn't mean that you're a dumb, dangerous, discourteous driver. L-drivers have as much right to be on the road as Q-drivers do, provided that they have L-plates, their learner's permits and a supervisor. I'm going to repeat that again, just to stress my point: you have as much right to be on the road as any Q-driver, so you deserve to be afforded the same amount of respect and courtesy as a Q-driver. Don't let other people's attitudes get in your way of becoming good at driving.

One thing that I found really helped me in becoming used to other Q-drivers is by pretending that I am one- one that follows the road rules, anyway. Since I got into this frame of mind, other benefits compounded on this. I've been beeped at a couple of times by impatient drivers, but because I don't see myself as that L-driver that all of the Q-drivers hate, I just go, "bloody impatient driver" and go on my merry way. Of course, this tactic might not work for you as everyone's mind is different.

If you get a tailgater, you could try some of the tactics that I've gathered from Q-drivers. (However, if you try one of these, you should be sure that the driver behind is definitely a tailgater and that it isn't a case of the car just appearing to be close when you look at it in the rearview mirror.) Here are some tactics:

1. Tap the brake pedal not so much that you start slowing down, but enough so that the brake lights come on.
2. Let go of the accelerate pedal.
3. Use the windscreen spray thingo. If the wind's right the water might blow onto their windscreen. My mum did this on the freeway to a jerk who was tailgating her...
4. If there's an intersection nearby, indicate so that they think that you're turning and will slow down to allow you to turn.
5. If there's a place where they can overtake, slow down to allow them to overtake. However, tailgaters will probably overtake on their own as soon as they have the opportunity to do so.
6. You could always simply ignore the tailgater. Or you could pull faces at them in your rearview mirror. Whatever.

Whatever you do, though, NEVER NEVER NEVER go over the speed limit just to appease a Q-driver. If there's a camera nearby, it's you who'll end up with the fine, not them.

And that's pretty much it for my article on how to handle Q-drivers. If you're still struggling you could try and drive an automatic so that you don't have to worry about Q-drivers and gears at the same time. I'll have to admit that I'm learning on automatic (both my parents have automatics) so I don't have any advice for you if you stall your car in the midst of a whole bunch of Q-drivers. I can, however, offer some consolation: the Q-driver in front of me stalled their car at an intersection. If Q-drivers can stall their cars then really there's nothing that hateful about L-drivers stalling their cars either.

As always, if you have anything else to say on the matter, please leave a comment!

How to Turn Left

Hiya all, today I'm going to post about driving.

When you start learning how to drive, even the simplest little tasks like turning left and right can seem daunting, so I've composed a nice little guide on how to turn left.

First of all, your preparation is pretty important. You should slow down as much as possible before you turn as it is easier to control your car at a lower speed. As my driving instructor says, cars are like men: they can only do one thing at a time. They can slow down, or they can turn, but they can't really do both at the same time, so make sure you do most of your slowing down before you get to the road you want to turn into. 15-25km/h is a good speed for turning at, less if it's raining.

When you turn left, the position of your car is also important. Most suburbian roads widen out at the end for a reason, and that reason is to help you turn left! Make full use of that curvy bit to help you turn. Here's a diagram (I even labelled the car for you, hehe):

The trick is that, to get in this position in the first place, you would already have had to have put some turn on the wheel. That way, when you actually make the turn, you don't have to turn the wheel much more.

Of course there are some roads that won't enable you to do this, but most of the time you can get a nice left-hand turn position going. Even when you're not turning left at the end of the road but down some other random road instead, you'll still find that there are curvy bits where the roads intersect and you can use these to your advantage.

I have another tip for you and that is in regards to steering. If you find that you can't steer sharply enough, try this steering method. To turn left, grab the top of the wheel with your left hand and pull anticlockwise. Push up anticlockwise with your right hand. Repeat the process as needed. To turn right, just reverse the directions.

To sum up- here are the steps that you should be taking when you do a left-hand turn on a normal suburban road:

1. Check your mirrors
2. Start slowing down
3. As you slow down, indicate that you're about to turn left.
4. Get into the left-hand turning position.
5. If you're on a terminating road about to enter a different road, you need to stop and check for traffic in the left lane. If it's all clear, then you can go.

By the way, if visibility is good, as in REALLY good and you can see for quite a while down the road, and you can see that there aren't any cars coming, then you don't need to stop completely at the end of a terminating road unless there's a stop sign. (You must stop before stop signs- even if you're stopped behind a car stopped at a stop sign, once that car moves off, you need to stop AGAIN at the stop sign BEFORE you move off. I learned this the hard way with my driving instructor- at least I didn't learn the hard hard way in a driving test...)

Hopefully that helps you all with turning left... if anyone has any other tips, or if there's anything incorrect in this post, let me know!

Radicals 足 刀 日

I said I was going to post 3 a day, but I kind of didn't stick to that promise. Ah well.

Here are 3 more radicals for you to learn today.

足， ⻊ = foot

This character exists as a standalone character, but it has a slightly different form that appears on the left hand side of some characters, particularly those to do with feet.

足 - foot
跑 - run
踢 - kick (踢足球 - play football/soccer)
跃 - leap, jump
蹦 - leap, jump, spring
跳 - leap, jump, spring, bounce (the list of English translations keeps growing...)
跟 - with, to follow
踩 - step on, trample

刀，刂 = sword, knife

One way of remembering that the two characters are related is that the second one is really just the first one with the top bit removed and the left bit squished a bit closer towards the right bit- wait, that wasn't really all too helpful... There's also a third form that I don't know how to type and can't find one that I can copy or paste (though to be honest I'm not really looking very hard). It looks like the top of the character 色 or 龟.

刀 - sword, knife
剪 - scissors, shears
色 - colour
龟 - turtle
兔 - rabbit
划 - stroke (of a Chinese character), to scratch
切 - cut
免 - avoid, avert, escape (免费 - free of charge)
刚 - just
别 - other, don't

日 - sun, day

This character evolved from a pictograph. I'm sure you can find its origins on a quick Google search.

日 - sun, day
旧 - old
旦 - dawn
晓 - dawn, daybreak
显 - apparent, obvious, show, display, illustrious and influential
明 - bright (昆明 - Kunming, capital of Yunnan Province in China, which is where the Hanyu Qiao Chinese Proficiency Competition is going to be held this year)
时 - time
旭 - rising sun
早 - early, morning
晚 - night
量 - to measure
晴 - sunny

Redox equations

This is possibly the reason why I hate redox so much. The actual equations themselves aren't too bad, it's just working out whether a reaction occurs or not. My brain always gets mixed up reading that stupid reduction potentials chart... argh...

Anyway. The first two types of redox reactions that I'm going to talk about are metal and halogen displacement reactions.

In metal displacement reactions, a metal reacts with a metal ion in such a way that the originally solid metal becomes an ion and the original metal ion becomes a solid metal. Confused yet?

For example, if you have solid potassium reacting with a solution of sodium chloride (i.e. sodium ions and chloride ions), you'd end up with solid sodium and a solution of potassium chloride (i.e. potassium ions and chloride ions). The net ionic equation would look like this:

K + Na⁺-> Na + K⁺

In this reaction, K has been oxidised and Na has been reduced.

Note that if you had solid sodium and a solution of potassium chloride, they would not react. But how can you tell just by looking at the reactants if you're going to have a reaction or not? This is where the Standard Reduction Potentials table at the back of your data sheet comes in.

The closer an element is to the bottom of the table, the more it wants to be oxidised and become a positive ion. For the most part, groups I and II elements (alkali metals and alkaline earth metals) are at the very bottom and directly above them is mainly transition elements (with exceptions such as aluminium and water which randomly snuck in). I'm very curious as to why this is so, but I'm too lazy to look it up myself.

When you have the reactants of a metal displacement reaction, take a look at the Standard Reduction Potentials chart. Which element has the strongest desire to be a positive ion? If it's already a positive ion, no reaction will occur. If it isn't, then a reaction will occur. That's basically how it works.

Halogen displacement reactions are the opposite of metal displacement reactions. Here's an example:

F₂ + 2Br^- -> 2F^- + Br₂

Now, when working out whether a reaction will occur or not, you have to do the opposite of what you did with the metal displacement reactions as this time we're dealing with negative ions, not positive ones.

If the bottom of the table lists the elements that want to be oxidised and become positive ions, then it also follows that the top of the table lists the elements that want to be reduced to negative ions. When working out halogen displacement reactions, work out which element has the strongest desire to be a negative ion. Then think about whether that element is already a negative ion or not. If it's already a negative ion, then no reaction will occur. If not, then a reaction will occur.

(By the way, there are ways to force the reactants to react even when they don't really want to, but I'm not going to go into that for now.)

Now, next up is half equations. In Chemistry 2AB, you only have to deal with the dead easy half equations, but there's also acidic conditions and stuff that you can learn about too.

You can write half equations for any kind of redox reaction, whether it be metal displacement, halogen displacement, or a myriad of other types of reactions that I don't know the names of. As long as an element is being reduced and another is being oxidised, you can write half equations.

Basically, you write separate equations for the reductant and the oxidant. Then you add in electrons to balance the charges.

F₂ + 2e^- -> 2F^- 

2Br^- -> Br_{2 +} 2e^-

Just make sure that the number of Fs or Brs or whatever are balanced on either side of the equation, and then chuck in some electrons to balance the charges as well.

You can also write balanced net ionic equations as well simply by adding together the two half-equations and then cancelling off the electrons on either side. If the two half-equations involve different numbers of electrons, however, care needs to be taken.

If you have different numbers of electrons in the two equations, what you first need to do is multiply each by a certain number so that the number of electrons are the same. For example, if you have 3 electrons in the first equation and 2 in the second, you can multiply the first equation by 2 and the second by 3 so that you have 6 electrons in both equations. Then you can add the two equations together, cancelling out the electrons on both sides. It's just like simultaneous equations in maths really.

Earlier, I mentioned half equations in acidic conditions. Normally, when you have a question asking for these, it's because you have some crazy polyatomic ion like the chromate ion. The steps for working out these are as follows:

1. Balance the number of atoms on each side (except for hydrogen and oxygen- we'll get to them later).

MnO₄^- -> Mn²⁺

2. Add water in order to balance out the number of oxygen atoms.

MnO₄^- -> Mn²⁺ + 4H₂O

3. Add hydrogen ions (acidic conditions, remember?) to balance out the number of hydrogen atoms.

MnO₄^- + 8H⁺ -> Mn²⁺ + 4H₂O

4. Calculate overall charge on each side.

In above equation, the left hand side has an overall charge of +7 while the right hand side has an overall charge of +2.

5. Add electrons to balance out the charges.

MnO₄^- + 8H⁺ + 5e^- -> Mn²⁺ + 4H₂O

So that's pretty much all you need to know and more on redox equations! Next up- electrolysis!

Basics of Redox

Argh I hate redox. I'm not too bad at it, but it annoys me for some reason. Anyway, rant over, it's time for me to try and explain it.

Now, redox is all about the transfer of electrons. When an atom gains electrons and thus has its charge reduced, we say that it has undergone reduction. When an atom loses electrons, we say that it has undergone oxidation. The word "oxidation" comes from the fact that there are lots of cases in which the addition of oxygen makes another atom lose electrons.

When do redox reactions occur? Well, you know very well that the transfer of electrons happens when atoms become ions. (If you didn't already know this, go here to find out more.) But there are other times when electrons are transferred too. It's easier to explain this by talking about oxidation numbers, so that's what I'm going to talk about first.

You can assign each element in a compound an oxidation number. If it's just a single element you're looking at, like Fe or O2 or Na, the oxidation number is 0. Hydrogen usually has an oxidation number of +1, but if it's a metal hydride, then it has an oxidation number of -1. (And, as said before, if it's on its own, like in H2, then it has an oxidation number of 0.) Similarly, oxygen usually has an oxidation number of -2, but there are a few exceptions where this is different (I'll have to check what the exceptions are). In ionically-bonded compounds, the oxidation number of each monoatomic ion (ion with only one atom (ion?), like Na+) is simply the charge on the ion (for example, the oxidation number of Na in NaCl is +1 because Na has a +1 charge). Assign these elements their oxidation numbers first.

Now, in a neutrally-charged compound, all of the oxidation numbers have to add to 0. Similarly, in a charged compound or polyatomic ion, all of the oxidation numbers have to add to the charge on the ion or compound. Simple, isn't it?

Now I'm going to talk about carbonic acid, H2CO3. Let's start off by assigning it some oxidation numbers.

Since this isn't a metal hydride, the oxidation number of H is +1. That was easy, wasn't it?

Now, CO₃<sup>2- has a -2 charge. Therefore, the oxidation numbers of C and O need to add up to -2. Each O has an oxidation number of -2. Three Os have a combined oxidation number of -6. C must have an oxidation number of 4 because 4-6 = -2, the overall charge on the ion.

That wasn't too hard, was it? (If it was, don't feel stupid, just ask me to clarify.) Now for two more examples: carbon dioxide and water.

Water's easy. Each H has an oxidation number of +1 (the two Hs combined have an oxidation number of +2) and O has an oxidation number of -2. The overall charge is 0.

Carbon dioxide isn't much harder. Each O has an oxidation number of -2, and because carbon dioxide is neutral, the oxidation number of C has to balance out the oxidation number of the Os. The oxidation number of C is therefore 4.

Now, the decomposition reaction of carbonic acid to carbon dioxide and water is not a redox reaction. Why? Well, it's because none of the elements has changed oxidation number! If any of the elements had changed oxidation number, then it would be a redox reaction. (Sorry, I thought this decomposition reaction would be a redox reaction as well, but soon realised that it wasn't. One day I'll put a better example up.)

In a redox reaction...

The element that loses electrons, thereby obtaining a higher oxidation number, has been oxidised and is known as the reducing agent or reductant (as it reduces the other element).

The element that gains electrons, thereby having its oxidation number reduced, has been reduced and is known as the oxidising agent or oxidant (as it oxidises the other element as it is being reduced).

Remember that if electrons are being lost then they have to have somewhere to go to, and vice versa (if electrons are being gained the atom needs to gain them from somewhere).</sup>

By the way, there's a nice little mnemonic to help you remember which is the oxidant and which is the reductant. It's OIL RIG. It stands for Oxidation Is Loss of electrons, Reduction Is Gain of electrons.

I'm too tired to do any more for now. At least I've done the first four dot points here (explain oxidation and reduction as electron transfer, calculate oxidation numbers, identify and name oxidants and reductants and identify redox reactions using oxidation numbers). I think I'm good for now. Only 5 dot points left and I'll have gone through pretty much the whole Chemistry 2AB course- unless you count all of the Applied Chemistry stuff, of course.

Naming and Drawing Organic Compounds

Because I'm pretty lazy, I just photocopied my Chemistry holiday homework for this one.

Sorry that it's on a slight angle.

Rightio, let's start with 1a. It's always a good place to start!

First things first- identifying which homologous series the structure belongs to. There's only single bonds here, so it's an alkane and will therefore end with an -ane ending.

Secondly, we work out what the longest carbon chain is. No, it's not 4, it's actually 6. There's three on the top row, and then the third one joins on to the three on the bottom row. (Due to the nature of single bonds, rotations and flips and whatnot don't change the order of bonding- sorry, couldn't think of a clearer term. Anyway, diagrams don't even accurately show where the atoms are in relation to each other.) Therefore, we know that there's a hex- in there somewhere. So far, we've got ourselves "hexane."

But what about that extra CH3 joined on to the third atom? Well, that's called a "alkyl group." Because there's only one C atom, it's "methyl-." (If there were two carbons, it'd be ethyl-. Three would make it propyl-, and so on.)

Now, we also have to assign the alkyl group a number so that we know where to find it on the chain! Counting in one direction, the alkyl group's on the 3rd carbon. Counting in the other direction, it's on the 4th carbon. Always work with the lower numbers. Since 3 is lower than 4, we'll stick with 3. Finally, since there's only one methyl group, we don't need to worry about adding di- or tri- before the "methyl." (You don't need to add mono- either- if there's no prefixes, it's implied that there's only one.) Therefore, the final name is 3-methylhexane. Yay!

We can write this all down as a series of steps:

1. Figure out which homologous series the structure belongs to. This will give the suffix.
2. Count the number of carbons in the longest carbon chain.
3. Figure out the names and locations of all attachments to the chain that aren't hydrogen atoms. (This includes alkyl groups and halogens like bromine and iodine.) If there's more than one of anything, indicate this with a prefix like di- or tri-. Also, when giving locations of groups, make sure to give all the attachments the lowest possible numbers. List all attachments in alphabetical order.

Let's try these steps with 1e.

Homologous series- Alkane, as all single bonds. Hence -ane ending.

No. of carbons- 10, so dec-.

Extra bits and pieces:

1 x Chlorine at 2

1 x Fluorine at 5

3 (tri-) x Methyl at 3, 3 and 9

1 x Propyl at 5

Final name: 2-chloro-5-fluoro-3,3,9-trimethyl-5-propyldecane

I won't give worked examples of the other ones because I have to wake up at 6am tomorrow and I'm not a morning person, but I will give some pointers.

NO2 goes by the prefix nitro- and NH2- goes by the prefix amino-. However, you don't really have to worry about these in 2AB Chemistry.

You also don't need to worry about cycloalkyl groups in 2AB (as seen in 3c). They work pretty much the same way as an alkyl group. The trick is knowing whether you're dealing with a cycloalkane with alkyl groups or an alkane with cycloalkyl groups.

You do need to worry about cycloalkanes with alkyl groups. Labelling these is exactly the same as labelling alkanes with alkyl groups. The only difference is that cycloalkanes obviously don't have ends. The numbering goes clockwise. Just try and make it so that each alkyl group or halogen has the lowest numbering possible. Halogens should get preference for the lowest numbers.

Now for drawing hydrocarbons. I'm going to pick a semi-hard example so that I'll cover pretty much everything and will only have to explain once. The example I'm going to pick is 2i: 2,2-dichloro-6-methyl-4-propylheptane.

Drawing hydrocarbons is easier, in my opinion. Here are my steps:

1. Draw all of the carbons required.

2. Draw bonds between them according to the functional group.

3. Add all of the other bits and pieces on.

4. Draw on all of the hydrogen atoms. Yes, you have to do this step. My Chem teacher told us that this guy from another class offered his brother $5 if he would draw on all of the hydrogen atoms for him. The brother looked at how many he had to draw and said, "Not worth it!"

Obviously, you need to draw all of the bonds on but I kind of ran out of room and couldn't be bothered rearranging stuff.

So that's pretty much all of the basic stuff on naming and drawing organic compounds. I've only really covered alkanes, but alkenes and alkynes aren't too different. The only thing is that you need to put the location of the double and triple bond just before the -ene or -yne suffix (e.g. but-2-ene). If you want me to clarify, just ask!