General Info

Okay, so for those of you who happen to be reading this (but aren’t part of my assignment), this is a “technologically mediated learning tool/resource”. It has been designed/written for University of Wollongong, Chem101 students for an education assignment. I am studying to become a high school science teacher.

Girls, you are welcome to leave comments or email me with any questions or suggestions you may have. Start from the bottom (periodic table) and work your way up (redox).

Please, feel free to offer suggestions for improvement because chances are I’ll keep adjusting it once I have some spare time.

Hope its useful and interesting!

Redox Reactions

Having covered oxidation numbers, we can now look at redox reactions (remember that reduction and oxidation always occur together).

When we deal with reduction and oxidation we usually use half equations. We usually do this when we set up an electrochemical cell. This is when the sheet that is usually on the back of the Periodic Table comes in very handy. It is a list of reduction potentials. This is the list we use to determine what is reduced and oxidised in reactions.

Handy hint: if you ever get confused as to which way it goes, just remember that fluorine likes gaining an electron sooooooo much more than potassium.

The higher the number (or the biggest positive number) is the species that is going to be reduced.

For the half equations, remember to write the electrons on the right side of the equation.

Fe²⁺ + 2e^– → Fe

This is the reduction half equation for iron (II). Oxidation half equations have the electrons on the other side (all you have to do is reverse the equation given on the sheet).

If you have 2 half equations and need to add them together, you need to make sure that the electrons cancel each other out.

We’ll work through a copper and zinc example.

Cu²⁺ + 2e^– → Cu                            E= 0.16V

Zn²⁺ + 2e^– → Zn                            E= -0.76V

Clearly, the copper half equation is going to be reduced. This means we now flip the zinc half equation and because we reverse it, we also change the reduction potential sign.

Zn → Zn²⁺ + 2e^–                            E = 0.76V

Its nice and easy because the electrons already equal each other but sometime you might need to times one of the equations by something to make them equal (think lowest common multiple stuff from maths).

We add them together now and get our final equation and the voltage generated.

Zn + Cu²⁺ → Zn²⁺+ Cu              E = 0.92V

Easy as, right?

So they are the basics of redox reactions. The redox potential isn’t really that important unless you’re trying to calculate the voltage running through an electrochemical cell.

We’ll try something a little bit harder now. These aren’t so much constructing the redox equations, as balancing them out.

Mn 2+  +  BiO3 –  ===>  MnO4 –  +  Bi 3+

First, we split them up. Our two main species/elements Mn and Bi.

Mn²⁺  →   MnO₄^–

So, now we balance the Mn, which is fine. Next, the oxygen gets balanced. There are 4 oxygen atoms on one side, there needs to be 4 on the other side. This is done by adding water to the other side.

4H₂O + Mn²⁺  →   MnO₄^–

We balance the hydrogen now (quick note – this is only for acidic solutions, basic solutions have a slightly different process). Add hydrogen to the other side, so that it is balanced.

4H₂O + Mn²⁺  →   MnO₄^– + 8H⁺

So, as long as all the species present are balanced, we get to balance the charge now. Each side always need to be equal.

We have a +2 charge on one side and an overall charge of +7 on the other side. So we need 5 electrons to balance it out completely. After you’ve done that, you should have a very pretty equation that looks like this:

4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–

Have a go at the other half equation with bismuth.

You should end up with something that looks like this:

6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O

Now, we add both equations together. No, wait! The electrons need to be balanced first. To do that we have to get the lowest common multiple for the electrons. Two electrons and five electrons; the lowest common multiple is 10.

2x (4H₂O + Mn²⁺ →    MnO₄^– + 8H⁺ + 5e^–)                                  8H₂O + 2Mn²⁺ →    2MnO₄^– + 16H⁺ + 10e^–

5x (6H⁺ + BiO₃^– + 2e^–  →  Bi³⁺ + 3H₂O)                                        30H⁺ + 5BiO₃^– + 10e^–  →  5Bi³⁺ + 15H₂O

 

Cancel out any species that appear on both sides of the equation (like the electrons) and voila!

2Mn²⁺ 14H⁺ + 5BiO₃^–   →  5Bi³⁺ + 7H₂O + 2MnO₄^–

Our equation is now balanced. Well, I’m 95% sure that its balanced.

For basic solutions, its exactly the same with just one extra step. After you’ve added hydrogen but before you add any electrons, add an equal number of hydroxide ions to both sides. The hydrogen and hydroxide (on one side) form water.

That’s all there is to it. Simple, huh? Okay, maybe not at first but after practise (see the website below), it gets pretty easy.

http://wc.pima.edu/~skolchens/C152OL/Ch21/REDOX.htm

http://kurtniedenzu.cmswiki.wikispaces.net/file/view/Redox_Equations_WorkSheet.pdf   – I recommend this one

Click to access redox.pdf

Click to access redox_chem220.pdf

Enjoy!

Reduction and Oxidation

Moving in a slightly different direction, we’re going to start reduction and oxidation. They tend to go together in reactions and so are known as redox reactions in general.

A few quick definitions:

Reduction: a gain in electrons, a addition of hydrogen and/or removal of oxygen.

Oxidation: a loss of electrons, a removal of hydrogen and/or addition of oxygen.

Handy hint to remember which one is which – OIL RIG

Oxidation is Loss, Reduction is Gain. Its how I still remember them even after 5 years.

These next two terms are a little bit harder because they seem to be backwards to commonsense.

Reductant: The species being oxidised, or the species that causes reduction.

Oxidant: The species being reduced, or the species that causes oxidation.

In working with redox reactions, we have to understand the concept of oxidation numbers. We assign oxidation numbers to the various elements in a chemical equation. Some of them change, some of them don’t. My favourites tend to be things like hydrogen and oxygen because their oxidation numbers don’t change (+1 and -2 respectively).

The way we assign oxidation numbers is very similar to the way we balance equations, both sides have to equal the same thing.

Oh, and before I completely forget to mention it, an atom by itself has an oxidation number of zero.

I’ll work through an example or two and then its your turn *grins evilly*.

Lets start with aluminium hydroxide. Write down the chemical formula (we’ll assume it has a valency of 3).

Al(OH)₃

We know that oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1. The overall molecules charge is zero. This is where we throw in a bit of algebra.

Al + (-2×3) + (1×3) = 0

Al + -6 +3 = 0

Al – 3 = 0

Al = +3

As we can see, aluminium has an oxidation number of +3.

If we have the compound SO (just for fun, name it for me), what is the oxidation number of sulfur?

Remember that oxygen has an oxidation number of -2.

S + (-2) = 0

S = +2

Sulfur has an oxidation number of +2.

Now that we know how to assign oxidation numbers to different elements, we can tell if a redox reaction occurs. If the oxidation number changes, it is a redox reaction. If it doesn’t change, it isn’t.

N.B: A heads up, occasionally, very occasionally, the oxidation numbers of our standards (oxygen, hydrogen and some others) do change but not often.

Here are some worksheets to have  go at to practise assigning oxidation numbers but there are plenty more out there on the net.

Click to access oxidationNumbersWkst.pdf

Click to access Oxidation%20Numbers%20Worksheet.pdf

Moles and Theoretical Yields

Checklist time:

• Write the chemical formulas for compounds
• Write chemical equations for different chemical reactions
• Calculate moles and molarity

Knowing all these things, we can now calculate theoretical yields and limiting reagents. Well, to be honest, you kinda have to calculate both if you want to know one.

Down to business. This gets a little bit more tricky than what we’ve been through so far. It isn’t anymore difficult, the process is just a bit longer.

Let’s work through an example so that it makes a teensy bit of a smidge more sense.

First things first, write down the chemical reaction. What are you waiting for? Oh right, you haven’t been given it yet. Okay, lets see, hmm…I’ve got it! Its a fun reaction that Year 7 absolutely adore! We’re going to add sulfuric acid to zinc metal. Go ahead and write it down. Done?

H₂SO₄ + Zn →  ZnSO₄ + H₂

We’ll add an excess of sulfuric acid, just to make sure that all the zinc gets used up *claps hands over mouth*, whoops, I’ve just given you the answer to my next question about the limiting reagent. I’ll go through it anyways, just to be thorough.

Limiting reagent. In simple terms – the limiting reagent is just what we have the least of, it means that it limits the production of the products to a certain amount. Since we have added an excess of acid, it means that the zinc has to be our limiting reagent, its not like its self-renewing or anything (and now that just reminds me of that Tim Tam ad, mmm…chocolate). Having an excess actually makes it easier because we can now ignore the acid.

Alright, we’ve done our scientist thing and added the acid to a 5 gram strip of zinc metal. It has made pretty bubbles but now we want to know how much zinc sulfate we made.

First step?

That’s right, figure out how much zinc we had at the start (remembering that any chemical type calculations deal with moles).

Moles of zinc =  5 (g)/65.38 (g/mol)

Moles of zinc = 0.0764 mol

So, do you remember when I said that the stoichiometry of a chemical equation was important later? Well, its later now.

Basically, stoichiometry is a ratio. This example is nice and simple because it is 1:1. What that means is that 1 mole of sulfuric acid reacts with 1 mol of zinc to form 1 mol of zinc (II) sulfate and 1 mol of hydrogen gas. Except, that we don’t have 1 mol, we have 0.0764 moles. Which is fine.

To simplify, for a 1:1 ratio, its just moles of zinc = moles of zinc sulfate.

With that in mind, we can say that the moles of zinc sulfate formed = o.0764 moles.

And now all that’s left is to convert moles back into grams. We re-arrange our equation (moles=mass/molar mass) to calculate mass.

mass = moles x molar mass                                                     but remember to use the molar mass of zinc sulfate now.

mass = 0.0764 (mol) x 161.45 (g/mol)

mass = 12.33 g

Cools! We made nearly 13 grams of zinc sulfate. Well, theoretically we did. A theoretical yield is just how much you calculate that you should make. In labs, you usually calculate a % yield, which is just however much you actually made, divided by the theoretical amount.

So, if we only made 11.51 grams, our percentage yield is about 93%. Not bad!!

Just a quick note about stoichiometry. 1:1 ratios are easy-peasy. Throw in some different numbers and sometimes, its a bit hard to wrap your heads around. I won’t go through this next one completely but hopefully, you guys can.

We’ll add hydrochoric acid to lead (II) nitrate (not that they let us play with lead all that much anymore. Sad face).

2HCl + Pb(NO₃)₂ → PbCl₂ +2HNO₃

Assuming the equation is balanced (which it is but you should always check, you never know when we teachers will pull a fast one), we can see that we need twice as many moles of hydrochloric acid than lead (II) nitrate for it to react completely. So if I gave you a problem that said we had 3 moles of lead (II) nitrate and 5 moles of hydrochloric acid, which one would you say is the limiting reagent?

Yep, you got it, hydrochloric acid. So we’d only make 5 moles of lead (II) chloride at the very most.

Make sense?

Chemical Reactions and Moles

We can tick off writing chemical equations now, or at least we should be able to do that. That being the case, let’s move on to calculating how much of something we actually have and how much we can make.

Chemists have a few magic numbers but one of our favourites is Avogadro’s number (it’s one of our favourites because it is always given to us in exams). Avogadro was a smart cookie that decided that we needed a new measurement for our fun-fun-fun chemicals. He reckoned that we should measure elements and compounds in moles. One mole of a substance is equal to 6.022 x 10²³ atoms/molecules. This is actually how we get our atomic weights. The weights given on the Periodic Table is actually the weight of 1 mole of the element, so we call it the molar weight or molar mass. Some even refer to it as the molecular weight/mass, but any of those terms are acceptable.

Very important: molar mass is measured in grams per mole (g/mol or gmol^-1).

Okay now, calculating moles is actually pretty simple. All you do is divide the mass of the compound (in grams) by the compound’s molar mass.

Moles =  mass (g)/molar mass (g/mol)

Just quickly, we’ll go through an example. Let’s pretend that we have, say, 7.9 grams of magnesium bromide. First things first, write down the chemical formula for magnesium bromide. 3, 2, 1…pens down please. *Drum roll* Magnesium bromide is MgBr_2. 

Next, go calculate the molar mass of magnesium bromide. All you have to do is to add the molar weights together.

Molar mass = (24.305 +79.904 + 79.904) g/mol

Molar mass =  184.113 g/mol

We now have the mass and the molar mass, which means we can now calculate the moles.

Moles = 7.9 (g)/184.113 (g/mol)

Moles = 0.0429 mol              (I know, I know, my significant figures are horrible, but we’ll worry about that later)

So, we can now calculate moles. We can do much the same to calculate the molarity of a solution. We measure molarity is moles per litre (mol/L), meaning we divide the moles by the volume of the solution.

Molarity (M) = moles (mol)/volume (L).

Again, a quick example. Lets pretend that we dissolve the magnesium bromide into 500mL of water. What is the molarity of the solution?

Molarity (M) = 0.0429 (mol)/0.5 (L)

Molarity (M) = 0.858 mol/L

And that’s pretty much how it works in the basics. Check out the next post for theoretical yields of reactions.

Balancing Equations

Moving on to the really awesomely fun stuff now. We can name ionic and molecular compounds now and we know how to write their chemical formulas. Now that we know that, we can move on to writing chemical equations! Yay!

Alrighty, writing a (balanced) chemical equation is just a matter of applying compound names and valencies. We have to make sure that it is balanced. What that means is that, both sides of the equation have the same amount of stuff, it doesn’t need to be in the same order but they have to have the same amount. Think of sharing stuff with a sibling. You’re the reactants and your little brother is the products: why should he get more than you?!

Writing a balanced chemical equation is also very, very, very important (I can’t stress how important) for later, when we start to calculate moles and theoretical yields but we’ll worry about that later, yeah?

We’ll start off with one of my favourite reactions. It won’t blow up (a shame, I know) but it is pretty simple. Let’s add sodium hydroxide and hydrochloric acid together. This is what is known as a neutralisation reaction but it isn’t that important to know that at the moment.

So, before you skip ahead to the answer, would you do something for me? Write down the chemical formula for sodium hydroxide.

Now write down the chemical formula for hydrochloric acid.

Have you done that? If you have, you may continue on and if not, go back and do it. It’s either that or go do the dishes, your choice.

Your equation, for now, should look a little something like this:

NaOH + HCl

We won’t worry about states of matter for now but you should remember to always, and I mean ALWAYS, include states of matter. I have taken marks off students because they haven’t.

So, the trick to writing the next part of the equation is knowing how reactants will rearrange themselves. When you start out, one of the easiest ways is to split these compounds up into their ions.

Na⁺ + OH^– + H⁺ + Cl^–

We now switch the cations and ions. Before, sodium was paired with hydroxide but now, we pair it with chlor-(wait for it) -ide (remember, we have an ion), which gives us sodium chloride. Go on, write that down. This leaves 2 hydrogens and an oxygen. Now, what compound could possibly have 2 hydrogens and an oxygen? Oh right, water.

If you wanted to write it in words, it would be something along the lines of:

sodium hydroxide and hydrochloric acid go to sodium chloride and water

Bit of a mouthful and it makes for sore hands. Chemical equations are much easier.

NaOH + HCl  → NaCl + H₂O

Pretty simple, huh? The  sides balance out, each one has just 1 sodium, 1 chloride, 2 hydrogens and 1 oxygen.

As we can see, this chemical reactions has a 1:1stoichiometry. This means that one molecule of sodium hydroxide reacts with one molecule of acid. This becomes  very important later.

http://misterguch.brinkster.net/pra_equationworksheets.html

Click to access balancing_chem_equations_WS1.pdf

http://www.wfu.edu/~ylwong/balanceeq/balanceq.html

These are some websites that could be useful for practise. Have a go and feel free to ask me or your teachers any questions you might have.

We’ll try one that’s just a little bit harder. Hmm, now what are some of my other favourite compounds? I know, we’ll keep sodium hydroxide (write it down) but we’ll use phosphoric acid instead! (phosphoric acid isn’t used nearly as much as it should be, its a pretty awesome acid).

So, if you don’t already know the molecular formula for phosphoric acid, you should learn it. It’s formula is H₃PO_4.

So, go ahead and write out the left hand side of the equation.

NaOH + H₃PO_{4  →}

This one is a bit trickier than the last one. The phosphate ion has a charge of -3. Balance the chemical formula for sodium phosphate first. The other product is water, and I’d be very worried if you didn’t already know that formula.

Sodium phosphate should look something like this: Na₃PO₄. Remember the crossover method?

Our equation should look something like this now:

NaOH + H₃PO₄ →   Na₃PO₄ + H₂O

All that’s left now is to balance the whole equation (stop groaning, the really hard part is already over). This one isn’t all that hard to balance overall. Start with the sodium; make sure that both sides are balanced for sodium. Go on, do it. Now check for hydrogen and oxygen. If you’ve done that, you can now scroll down to see the final answer.

3NaOH + H₃PO₄   →   Na₃PO₄ + H₂O

Congratulations, you can now balance chemical equations!

Name that Compound!

Okay, so as we know, naming chemicals is a horible, boring, time consuming task. I mean who looks at a substance like this and knows immediately that it just so happens to be 5-bromo-5-iodohexan-2,3-diol. Or Steve, as my friend ended up calling it. Personally, it strikes me more as a Bruce, but each to their own I suppose.

The first step to being able to name these overly-complicated, but not completely useless compounds, is to be able to recognise the individual elemental symbols and their names.

Now, I’m not a fan of rote-learning at all but sometimes that’s just what it takes to master it.

The easy part for most simple compounds is the cation, or positive ion, if you prefer. These include things like sodium (Na+), magnesium (Mg+) or even iron (III) (Fe3+). The nams of cations tend to stay the same as their normal name.

Anions, or negative ions, are a little bit harder. Their normal names change. As an example, for a simple monatomic, like chlorine, the suffix changes to -ide. Same for fluorine and iodine. When we say chlorine, we mean the element chlorine, symbol Cl.  It hasn’t lost or gained any electrons. When it gains an electron, it changes its name (what a lot of paperwork) to chloride.

A word of warning: if you mix them up more than once, I will hunt you down and I will hurt you and make sure you never mix them up again. I am a teacher, do not doubt me.

Just remember, cations first and the anion’s name changes slightly.

When chemists talk about ions, there’s usually some mention of this thing, valency. “What is this?” you ask. Well, short answer: it just means the charge on the ion. Sodium has a valency of 1, chlorine has a valency of -1. Most metals have multiple valencies, just look at iron (I), iron (II) and iron (III). So, when we have ions with different valencies, we have to make sure they balance out so that the overall molecule’s charge is 0. The easiest way to do this is using the “crossover” method.

Ignore the signs, they’re just there to remind you that they are charged.

So, write the charge on top of each ion. Cross them over and then write the number as a subscript. For aluminium oxide to be a molecule, the charges have to equal zero overall. So, since aluminium has a valency of +3, we need 1.5 oxygen ions to cancel that out. The chemical equation for aluminium oxide looks something like Al2O3.

So, those are the basic rules for naming compounds ionic compounds (they are made up of ions). The rules for naming molecular compounds are pretty similar, but we use prefixes like mono-, di-, tri- etc… For these, its whatever happens to be written first.

For example, N2O4 (please imagine that the numbers are subscripts) is called dinitrogen tetraoxide.

Here are some websites that have some good stuff to help you practise:

• http://www.zerobio.com/drag_gr9/chem.htm
• http://reekoscience.com/Games/ChemFormula/default.aspx

Back to the Basics

So, let’s start from the very beginning.

“In the beginning, there was nothing. Then suddenly there was something.”

Okay, the Big Bang is probably a little bit too far back. Let’s just go back to the fundamentals of chemistry for now, yeah?

One of the big things in chemistry, that makes our lives (and calculations and related problems) soooooo much easier is… *drumroll please*… the ever wonderful, the amazing, the always-provided-in-tests PERIODIC TABLE. This is where we will start.

Way back in the day, when elements were being discovered left, right and centre, scientists were playing around with the elements, like they were playing solitaire with a pack of cards. One very bright spark, by the name of Dimitri Ivanovich Mendeleev, developed the Periodic Table that we know so well today. The elements were listed horizontally by their atomic weight and when chemical characteristics began to repeat, a new row was started. Mendeleev’s Periodic Table even predicted the existence of elements that hadn’t been discovered yet!

These are two websites that have “interactive” periodic tables. More information is provided for each element (if so desired).

http://www.ptable.com/

http://www.webelements.com/

In terms of chemistry, the best part of the periodic table is that it displays the chemical symbols for the elements (e.g. Na for sodium, W for tungsten) and the atomic weight (e.g. S – 32.07gmol-1 or C – 12.01gmol-1).