Sometimes a “rule of thumb” can make you feel dumb – but by learning why it (often) helps you can overcome! Atoms bond by sharing elections – But how fair of an electron share is there? And why should you care? Redox stands for oxidation and reduction reactions which involve electrons “swapping allegiance.” When you learn about redox and/or go look it up online they usually teach you with inorganic reactions that makes things easier because you can really picture electrons coming, going, and flowing from one molecule to another, but when it comes to biochemistry and organic reactions, things can seem a lot murkier and it’s easy to get confused (I know I have/still do!). So I want to try explaining things in a different, biochemistry-focused and dog-analogized way that can hopefully be useful
Molecules (things like water, H2O, or methanol CH3OH) are made up of atoms (each of those Cs & Os) and those atoms are held to one another because they’re sharing pairs of electrons which are really tiny little negatively charged things that whizz around the central part of the atom, the nucleus, which is home to positively charged tiny but not quite as tiny things called protons and some other non-charged things called neutrons.
I like to think of atoms as dog walkers (the protons) trying to keep control over a bunch of dogs (electrons). Especially the super-hyper dogs that are out on the edge (valence electrons). Those hyper dogs are the most likely to get distracted – they can get attracted to nearby dogs getting walked by other owners and drag their walker and the rest of the dogs with it to form a new covalent bond (atoms held together by a shared pair of electrons) or, if they have enough strength, break off. How much strength they’ll need depends on how much their walker is pulling them back.
Their “pulling power” comes from the positive pull of protons and how well that pull can be felt by the electrons. Atoms “come” with a certain number of protons that’s fixed for that element (e.g. oxygen (O) has 8 and always has 8 whereas hydrogen (H) has 1 and will alway have 1 or else it wouldn’t be hydrogen. But the # of electrons can change. And they can get transferred from one molecule to another and we call this electron-moving reduction (gaining electrons) & oxidation (losing electrons). There can’t be one without the other, so we call the combo REDOX reactions.
A way to remember which gains & which loses is OIL RIG: Oxidation is Loss (of electrons) and Reduction is Gain (of electrons). If you have whole electrons coming and going like in batteries where electrodes flow from one side to another this mnemonic does the trick. But when it’s not really whole electrons getting transferred, it’s just electron density shifting, recognizing what’s going on can be tricky… And sometimes it can be hard to see how the concepts you learn in gen chem apply – they really do! They’re just often focusing on and emphasizing different points.
For example, in gen chem you’re usually taught the “shell model” – it’s often convenient to think about electrons as being housed in “shells” like layers of an onion, where the outermost shell is the “valence shell”, These shell models are helpful when trying to predict reactions because atoms want their valence shell to be full – so atoms can share electrons with one another, give up electrons, or take on electrons in order to try to get there to the “ideal” #s of electrons they’d like to have there (often 8, hence the “octet rule”) more here: http://bit.ly/2Aajn2S
This “shell model” helps you see that sodium (Na) which has a single electron in its outer shell, will happily give it up to expose a full “inner shell” (which becomes its new outer shell) and chlorine will happily take that electron to go from 7 valence electrons to that glorious eight. Leaving you with Na+ & Cl- which, having opposite charges, stick around each other until a better offer comes in. These “bonds” in which one dog breaks leash with the original walker all together are called “ionic bonds” and they’re really just strong charge-based attractions.
But atoms can also share custody of their electron “dogs” – for example, 2 chlorine molecules (each with those 7 valence electrons) can pair up – pairing their unpaired electrons in a shared pair so each gets to “own” an octet. So the shell model’s helpful for predicting what’ll react, but the sharing isn’t always fair, and the shell model doesn’t tell you about how fair of a share is there.
These simplified drawings make it look like electrons are fixed in one place. But they’re really whizzing around. And in fact you can’t ever know exactly where one will be. Instead, we can talk about “orbitals” which are basically where electrons are most likely to be (kinda like if you walked around dropping crumbs and then took a picture of where the crumbs are – the places you were the most will be the crumbiest) The molecular version of this is areas of high electron density.
Basically, electronegative atoms can get electrons to hang out near them more often, so the electron density gets pulled towards them (you’re more likely to find electron “crumbs” there. in a pure covalent bond there’s even joint custody but in a polar covalent bond one of the owners is better able to woo the dogs, so even though they share custody, it’s not an even split (the electron spends more time closer to the more electronegative one). And in an ionic bond, one atom gets “full custody.” As we’ll look at more later, we can use oxidation numbers to say – if all the bonds were ionic, who would get to keep the dogs?
When we assign oxidation numbers we pretend that all the bonds – even the real covalent ones are actually ionic ones and then we determine what charge each of the atoms would have – which depends on how many dogs it gets to keep and how many positive protons it has to neutralize the negative charge those dogs bring. Unlike the real “formal charges” that come from actually losing or gaining electrons (like in the Na+ & Cl- case), these oxidation state numbers are *Hypothetical* charges.
Oxidation numbers can be confusing – but they’re really there to help! Because they allow us to imagine what’s going on at the subatomic level at the larger (but still really tiny) molecular level. But sometimes the “helpful things” can be confusing as hell – as I know all too well! And one of the things that I had a really hard time grasping was the whole “oxidation is gaining oxygen and reduction is gaining hydrogen” rule of thumb.
Often in biochemistry and organic chemistry, it’s easier to talk in forms of atoms than electrons. So, one of the things that’s helpful is to remember that oxidation often involves adding oxygen or losing hydrogen (H). And reduction often involves adding hydrogen (H) or losing oxygen. This can be helpful, but it can also be confusing.
Because there’s another common place we see the letter H coming and going – acid/base reactions. But in that case there’s a plus sign (+) after it. Don’t ignore that plus! It tells you that you’re dealing with a PROTON NOT a “whole, normal” hydrogen atom. acid/base reactions involve PROTONation & dePROTONation. dePROTONation refers to the loss of a PROTON, which is H+. An H atom short an electron. And H only had 1 electron to begin with, so H+ really is just a proton (and a neutron, but that’s neutral and not involved here)
The Bronsted-Lowry definition of an acid is something that gives a proton (H+) & a base is something that takes a proton (H+).
The hydrogen referred to in the redox reactions is neutral hydrogen (proton AND electron) NOT H+ (proton alone). The reason the “gaining H is reduction” mnemonic often rings true is that getting an H is basically getting a free electron because it’s really easy to convince it to spend time with you. Kinda like 2 dog walkers and the H dog walker’s dog goes to sniff some other dog and the other owner woos the dog, pulling it closer, while at the same time chatting with the walker so it sticks around.
BUT in an acid/base reaction, if a base picks up a proton, it isn’t gaining an electron, just a proton (a walker but no dog). So getting an H+ is NOT reduction.
Oxidation is a dog leaving or shifting loyalty to another dog, deprotonation is like a walker leaving without a dog.
Reduction is doggy “adoption” – a dog coming ownerless or coming with an owner that lets its dog spend most of the time with you.
Put more conventionally: Losing a proton is acting as an acid. Losing a hydrogen or gaining oxygen is often being oxidized. And conversely, Gaining a proton is acting as a base. Gaining a hydrogen or losing oxygen is often being reduced.
Why H & O in this molecular tic-tac-toe? Oxygen’s kinda like a “dog whisperer” – it’s really good at recruiting dogs & convincing dogs to stay loyal to it. Hydrogen, on the other hand has a pretty hard time keeping the dog it has.
Dog whisper power (ELECTRONEGATIVITY) comes from having a strong whisper (lots of protons) (towards the right of the periodic table) & having the furthest dogs able to hear you (valence electrons nearer the nucleus)(towards top of the periodic table).
So when an oxygen dog walker meets a hydrogen dog walker, and the hydrogen dog starts sniffing one of the oxygen dogs, the oxygen dog walker convinces it to stay. And it “brainwashes” the dog into switching its loyalty from the H to the O. So now the O basically “owns it” (what’s really going on is that the electrons are whizzing around & if you were to take a snapshot, the electrons from pair of shared electrons formed is more likely to be closer to the oxygen).
This “ownership” is reflected in the oxidation state, which basically says, if you were to split up the walkers, who would the dogs stay with? And how does this compare to how many the walker has in its neutral form.
For example, oxygen has 8 protons, so to be neutral it needs 8 electrons. If you were to break an O-H bond, O would have the 6 it started with, and the one from the H, so 7. 6-7 = -1 so the O has an oxidation state of -1 when it’s bound to an H and something else. You could actually remove the H (deprotonate) and that oxidation state wouldn’t change because the electron had already “shifted loyalty”
But now consider an O-O bond like we saw with hydrogen peroxide (H2O2). The Os have equal dog whispering powers, so neither can convince the other’s dog to switch loyalty. So if you were to split the bond evenly , neither “gains” any electrons – they’re left with what they started with – 6 and that’s what they have in neutral form, so you didn’t change oxidation states, but the state you stay with is 0.
You could split that way, and this can happen in the thermal decomposition of hydrogen peroxide (H2O2) we looked at. But that leaves you with free radicals (single lone electrons), which are unstable and go looking for another electron to pair up with.
So then, if those were to find something to join to pick up a partner electron, with something less pully, such as from an H (H NOT H+ which has nothing of use here) it WOULD be gaining an electron – so it WOULD be getting reduced and it’s oxidation state would go to -1.
Alternatively, when something is deprotonated (which is more likely in basic conditions, when there are lots of bases around to take them and the field isn’t already crowded with protons) its oxidation state doesn’t change. Because electrons aren’t moving, just protons, and the electron left behind had already “switched its loyalty”
Because O is so much more electronegative than H, it had “stolen” the H’s electron before the proton even left. When the proton leaves though, it does leave a mark in the form of a lower charge. You aren’t losing an electron, but you ARE losing a proton and the +1 charge it gave.
So the oxygen loses some of its help controlling its electrons. So it can go looking for help. It can pick up another proton, and that will neutralize the charge but it won’t reduce it – you’re not gaining any electrons if you pick up something that doesn’t have any. But if, instead of picking up a proton (acting as a base) it attacks something that’s as or more electronegative as it is? That atom’s going to make it share. so it will be oxidized.
If you take something like an alcohol (something-OH) that O can get oxidized by swapping that H (with its electron) for something more electronegative (and its electron). You still have the same number of electrons, but now they’re not shared evenly – you can’t convince the dog to be loyal to you so if you were to separate the walkers, the dog would stay with the other walker. You’d have 1 less electron, so you’ve been OXIDIZED. And whatever you swap the H (with its electron) onto gets reduced. The key is to find someone to swap with. Things that want to get reduced are oxidants.
Oxidation doesn’t always involve oxygen! But it also doesn’t always involve whole electrons. So how can we track redox at the molecular scale “whenever”? We can’t “microchip” the “dogs” but we can assign and track oxidation numbers.
When thinking about which atoms “own” which electrons, I find the dog analogy useful, but at the bigger molecule scale of things it can be helpful to think in terms of poker chips, which also come in handy in biochemistry because, when it comes to energy, cells operate kinda like a casino – instead of paying out cash, reactions transfer “poker chips” that can later get “turned in” for the real money. The “poker chips” of chemical reactions are electrons – they can get passed from one atom to another (with the giver called the oxidant and the receiver called the reductant). And then those transferred electrons eventually get “cashed out” through a process called oxidative phosphorylation that uses them to power a pump to drive a proton gradient and create ATP, which is like cellular cash. More on ATP here: http://bit.ly/2CHvGWN
A common place to see redox in play is catabolism, the branch of metabolism where you break down big things into little things either for energy, or to make new big things (this making is the anabolism part of metabolism – think of anabolic steroids making big muscles (and big problems when abused).
You can use oxidation numbers to kinda keep a running tally of how many “poker chips” the molecule has. A good fuel comes in which lots but gradually they get transferred to other molecules and then finally “cashed in” for the big payoff in oxidative phosphorylation.
Some common fuels our cells use are carbohydrates like the sugar glucose, proteins, and lipids (fats, oils, and waxes). They come in with different #s of poker chips that they gradually lose until they have no more to lose (they’re at their most oxidized – which, for a carbon, means being attached to 2 oxygens (CO2). In that state, each of the Os forms a double bond with the C, so it owns 4 of the C’s electrons. And C only had 4 – so O owns them all. And if they were to divorce, the O would get to keep all the dogs.
Formally, we can write the oxidation state of the C to be 4-0 = +4
And the oxidation state of the Os to each be 6-(4 from double bond + 2 lone pairs) = 6-8 = -2
That may seem “formal” but those are NOT “formal charges” – they’re not even “real charges” – they’re hypothetical – no divorce has actually happened we’re just saying what the charges *would* be
“Real charges” are called formal charges – whereas the oxidation # is hypothetical. (i.e. oxidation # says – if the walkers were to divorce who would the dog want to stay with whereas the formal charge says these dogs really are divorced and this is how many dogs it has compared to walkers)
To help distinguish between the 2, oxidation states are often given in Roman numerals. So we’d write the oxidation state of that C to be IV, and the O’s would be -II)
Since oxidation states aren’t real charges, the sum of the hypothetical charges should equal the formal charge (the real one). if our molecule doesn’t have a real charge, its “fake charges” should add up to 0. If you add up the oxidation states of all of the atoms in a molecule you should get the actual charge of the molecule. Let’s check: 4 + 2(-2) = 4-4 = 0. Yay!
Not all players come with the same amount of chips – the more reduced the player is the more electrons it has to lose (and energy we have to gain). Take a lipid (fat, oil, or wax) and compare it to a sugar. Even if they have the same number of C’s, most of the C’s in the lipid have lower oxidation #s because they’re linked to H’s which are really weak dog whisperers so if you force them to divorce the dogs stay with the C’s and this would give the C’s a negative charge.
So most of the Cs in a lipid have an oxidation # of -2 or -3. This is like having a lot of poker chips to cash in. What if you look at a sugar like glucose? Here a lot of the C’s have already “cashed in” some of their chips. You see a lot of alcohol (-OH) groups in place of just H’s. And when C’s and O’s share ) doesn’t share fair – so when they get divorced O keeps the dog. So the C’s have higher oxidation numbers (from -1 to +1 depending on what else they’re attached to)