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Redox

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Chemistry

Picture the scene. It’s a chilly November night and you’re standing outside in the dark. Beside you, a crackling bonfire greedily works its way through old pallets and branches. A few jagged nails stick out of the half-rotten wood, tinged dull orange-brown with flaking rust. Believe it or not, the burning wood and rusting metal have something in common - they’re both examples of redox reactions.

Redox is a term used to describe reactions involving both oxidation and reduction. These reactions involve a movement of electrons, and are characterised by a change in oxidation states.

  • This article is about redox reactions in chemistry.
  • We will begin by explaining what is meant by the terms redox, oxidation, and reduction. We’ll then look at some common oxidising and reducing agents.
  • Next, we’ll take a look at oxidising states and how we determine the oxidation state of a particular species.
  • We will then explain how to write redox equations and half equations and go through some worked examples.
  • Lastly, we will touch on disproportionation reactions.

Redox Reaction Definition

The term redox is short for reduction-oxidation, and is used to describe reactions involving - you guessed it - both oxidation reactions and reduction reactions. Redox reactions are characterised by a movement of electrons and a change in oxidation states.

Let’s start by looking more closely at the definitions of oxidation and reduction.

Oxidation and reduction

The words oxidation and reduction have a few different meanings in chemistry. The first definition looks at them in terms of oxygen. Take a punt - you can probably guess what oxidation means.

Oxidation is the gain of oxygen.

You can think of reduction as the opposite of oxidation.

Reduction is the loss of oxygen.

For example, when copper reacts with oxygen, it forms copper oxide. The copper is oxidised.

Cu(s) + ½ O2(g) → CuO(s)

But reacting hydrogen with copper oxide separates the copper and the oxygen. The copper oxide is reduced.

CuO(s) + H2(g) → Cu(s) + H2O(l)

Did you notice that we added hydrogen to reduce copper oxide? This leads us to the second set of definitions for oxidation and reduction.

Oxidation is the loss of hydrogen, and reduction is the gain of hydrogen.

However, in chemistry, we tend to use a different definition. It refers to the movement of electrons between species in a reaction.

Oxidation is the loss of electrons, and reduction is the gain of electrons.

There’s a handy acronym that will help you remember this definition: OILRIG.

Redox, OILRIG, StudySmarterThe acronym OILRIG. Anna Brewer, StudySmarter Originals

Let’s revisit the example from before. What happens when copper reacts with oxygen? It forms copper oxide, an ionic compound. Copper oxide is made up of copper ions and oxygen ions, Cu2+ and O2- respectively. To form these ions from neutral atoms, we need to move some electrons around.

  • To turn a copper atom into a copper ion, the atom must lose two electrons. Copper is therefore oxidised.
  • To turn an oxygen atom into an oxygen ion, the atom must gain two electrons. Oxygen is therefore reduced.

Because both oxidation and reduction are happening side by side, this is an example of a redox reaction.

In summary, oxidation can mean:

  • Gaining oxygen
  • A loss of hydrogen
  • A loss of electrons

Likewise, reduction can mean:

  • A loss of oxygen
  • Gaining hydrogen
  • Gaining electrons

Oxidising and Reducing Agents in Redox

We know what oxidation and reduction reactions are. Now let’s look at the species that carry out these reactions.

Oxidising agents

Oxidising agents are species that oxidise another atom, ion, or compound. They are reduced in the process.

Oxidising agents take electrons from another species - they oxidise it. They are also called oxidants. Some particularly strong oxidising agents are fluorine and, perhaps unsurprisingly, oxygen.

There are several factors that affect the strength of an oxidising agent. These include electronegativity, electron affinity enthalpy, and oxidation state, which we’ll look at in just a second. This is because oxidising agents take electrons, so anything that increases the attraction between an atom or ion and an incoming electron will increase the agent’s oxidising strength. For example, fluorine is the most electronegative element in the periodic table, which makes it a very strong oxidising agent. Species with a high oxidation state also tend to be good oxidising agents.

Reducing agents

What do you think reducing agents are? You can probably take a good guess.

Reducing agents are species that reduce another atom, ion, or compound. They are oxidised in the process.

Reducing agents donate electrons to another species - they reduce it. They are also called reductants. Many metals, such as lithium, aluminium, and zinc, are good reducing agents, and so is hydrogen gas (if it is in the presence of a nickel catalyst).

Once again, there is a handy acronym that should help you remember the actions of oxidising and reducing agents in terms of electrons: RAD OAT.

Redox, RADOAT, StudySmarterRAD OAT. Anna Brewer, StudySmarter Originals

Take the example of copper and oxygen again. We know that copper is oxidised and oxygen is reduced. Copper loses two electrons, which oxygen gains. This also means that copper acts as a reducing agent and oxygen acts as an oxidising agent.

You should now feel confident about the terms oxidation, reduction, oxidising agent, and reducing agent. Let's move on to our next topic.

Oxidation States in Redox

Now that we know what redox reactions are, we can look at how to work out which species is oxidised and which species is reduced in a reaction. To do this, we use oxidation states.

Oxidation states are numbers assigned to ions that show how many electrons the ion has lost or gained, compared to the element in its uncombined state. A positive oxidation state shows that the element lost electrons, whilst a negative oxidation state shows that it gained electrons. They can also be referred to as oxidation numbers.

Let’s go through an example.

What is the oxidation state of copper in copper oxide, CuO?

To solve this problem, look at the copper ions in copper oxide. They are ions with a charge of 2+. How do you turn an uncombined atom, which is just a neutral atom on its own, into an ion with a charge of 2+? By losing two electrons! To form a copper ion with a charge of 2+, each copper atom has to lose two electrons. This means that these copper ions have an oxidation state of +2.

When talking about charges, we put the number first: 2+. But when talking about oxidation states, we put the positive or negative symbol first: +2.

Assigning oxidation states

It is all well and good knowing what an oxidation state is, but how do you assign them? There are a few rules you can use to find out a species' oxidation state.

  • All uncombined elements have an oxidation state of 0.
    • e.g. Cl2, Zn, and O2 all have oxidation states of 0.
  • The oxidation states of all the atoms or ions in a neutral compound add up to 0.
    • e.g. In the neutral compound NaCl, Na has an oxidation state of +1, and Cl has an oxidation state of -1. These add up to make 0.
  • The sum of the oxidation states in an ion equals the charge of the ion. This works for both monatomic ions made from one atom, and complex ions made from lots of atoms.
    • e.g. Cl- has an oxidation state of -1 and Ca2+ has an oxidation state of +2.
    • e.g. in the negative nitrate ion NO3-, N has an oxidation state of +5 and the three O each have an oxidation state of -2. These add up to make -1, which is the charge of the ion.
  • In an ion or a compound, the more electronegative atom generally has the negative oxidation state.
    • e.g. in F2O the oxidation state of F is -1, and the oxidation state of O is +2.
  • Some elements take certain oxidation states:
    • Group 1 metals always have an oxidation state of +1
    • Group 2 metals always have an oxidation state of +2
    • Al always has an oxidation state of +3
    • H usually has an oxidation state of +1 (except in metal hydrides)
    • F always has an oxidation state of -1
    • Cl usually has an oxidation state of -1 (except in compounds with O or F)
    • O usually has an oxidation state of -2 (except in peroxides and compounds with F)

We can use this knowledge to calculate oxidation states of unknown elements in compounds and ions.

Calculating oxidation states

The sum of all the oxidation states in a neutral compound must add up to zero, and the sum of all the oxidation states in a complex ion must add up to the charge of the ion - we know this from our rules for assigning oxidation states. But how do we work out the oxidation states of the individual elements within the compound or ion? For this, we can apply our knowledge of fixed oxidation states and work out the unknown oxidation states by deduction.

It can help to follow this process:

  1. Look at the charge of the ion or compound, if any. This will help you know what you are aiming for.
  2. Identify any atoms with fixed oxidation states and write these above the atoms.
  3. Deduce the oxidation states of the remaining atoms, making sure the sum of all the oxidation states adds up to the charge of the ion or compound.

Let’s look at some examples.

What is the oxidation state of phosphorous in PCl5?

In PCl5, we know that each chlorine atom has an oxidation state of -1. -This was one of our rules. Because there are 5 chlorine atoms, their combined oxidation state is -5. This compound is neutral, meaning that the overall sum of the oxidation states must equal 0; so, phosphorus must have an oxidation state of +5.

Have a go at this next example.

What is the oxidation state of sulphur in SO42-?

We know that each oxygen atom has an oxidation state of -2. Their combined total is -8. SO42- has a negative charge of 2-, so it must have an overall oxidation state of -2. The oxidation state of sulphur is therefore +6.

Why are there exceptions to the general rules given above? For example, why does hydrogen have an oxidation state of -1 in metal hydrides?

Well, there are two things going on. Firstly, we know that group 1 metals always have an oxidation state of +1. Secondly, we know that hydrogen is more electronegative than all the group 1 metals, so must take a lower oxidation state. Because metal hydrides are neutral compounds, the sum of the oxidation states must equal 0. Therefore, hydrogen has an oxidation state of -1.

Likewise, oxygen normally has an oxidation state of -2. But in F2O, it has an oxidation state of +2. This is because fluorine is more electronegative than oxygen, so each fluorine atom has a lower oxidation state than oxygen of -1. There are two fluorine atoms in this compound, with a combined oxidation state of -2, so oxygen must have an oxidation state of +2.

We can use oxidation states to see whether species have been oxidised or reduced. Going from a positive oxidation state to a more negative oxidation state means that the atom has gained electrons. It has therefore been reduced. On the other hand, going from a negative oxidation state to a more positive oxidation state means that the atom has lost electrons. It has therefore been oxidised.

Roman numerals in redox

Although some elements only form ions with one oxidation state, some elements can form ions with multiple different oxidation states. To avoid confusion when talking about these ions, we show oxidation states using roman numerals.

For example, when talking about the copper ions in CuO with an oxidation state of +2, we would write copper(II).

Redox Equation

A redox equation is a way of representing a redox reaction.

During a redox reaction, two simultaneous processes occur - reduction and oxidation. We can show these processes using one overall equation that ignores any ions that aren’t oxidised or reduced - that is to say, aren't taking part in the reaction. These are known as spectator ions.

Spectator ions are ions that are present in both the reactants and products of a reaction. They are totally unchanged by the reaction - their physical state, oxidation state, and charge don’t change.

Writing redox equations

We'll now focus on how you can write redox equations, using some worked examples. Here’s one such example: the displacement reaction between magnesium and copper sulfate.

Write a redox equation for the reaction between magnesium and copper sulfate. The overall equation is given below:

Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)

First of all, we need to find out the spectator ions in the equation. These don't change oxidation states. They aren't oxidised or reduced, so we don't need to worry about them.

To help you identify the spectator ions, split the ionic salts into their constituent ions:

Mg(s) + Cu2+(aq) + SO42-(aq) → Mg2+(aq) + SO42-(aq) + Cu(s)

The sulfate ion, SO42-, is present on both sides of the equation. It doesn’t change physical state, oxidation state, or charge. This means that it is a spectator ion. To write an overall redox equation, we simply omit this ion. Here's your final answer:

Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

But this overall equation makes it tricky to see the individual oxidation and reduction processes. To see them more clearly, we need to look at half equations. We'll explore them next.

Half Equations

Remember that redox reactions involve both oxidation and reduction. One species is oxidised whilst another is reduced. This means that there is movement of electrons. If we look at our overall redox equation, it isn’t always easy to see this movement. Instead, we can split our redox equation into two half equations that show the oxidation and reduction reactions separately. One half equation represents the process of reduction, while the other half equation represents the process of oxidation.

Writing half equations

To write half equations, we consider each of the ions or atoms involved in the redox equation separately. We add in electrons to show the processes of oxidation and reduction, and might also have to add in water or hydrogen ions to balance the equation.

These steps should help you learn how to write half equations.

  1. Pick an atom or ion involved in the redox reaction and write out the reactants and products involving it.
  2. Balance the elements apart from oxygen and hydrogen. Like all equations, half equations must be balanced - you must have the same number of moles of each element on both sides of the equation.
  3. Add in water molecules to balance the oxygen atoms on both sides of the equation.
  4. Add in hydrogen ions to balance the hydrogen atoms on both sides of the equation.
  5. Add in electrons to balance the charges.

The only three things you can add into half equations, besides more of the reactant or product, are water, hydrogen ions, and electrons. You can't sneak in oxygen gas, for example.

Let’s look at a couple of examples.

Write half equations for the displacement reaction between bromine and iodide ions. The unbalanced equation is given below:

Br2 + I- → I2 + Br-

First of all, let’s pick a reactant. Let’s start with bromine. Bromine reacts to form bromide ions:

Br2 → Br-

This equation isn’t balanced. There are two bromine atoms on the left but only one on the right. To balance it, we need to double the number of bromide ions:

Br2 → 2Br-

There aren’t any oxygen or hydrogen atoms involved in the equation, so we don’t need to add in any water molecules or hydrogen ions to balance it. However, we do need to balance the charges on both sides of the equation. The left-hand side has a charge of +0, whereas the right has a charge of two lots of -1, equalling -2. To balance the charges in the equation, we need to add in a charge of -2 to the left-hand side. This means we need to add in two negative electrons:

Br2 + 2e- → 2Br-

Both sides now have the same charge. The half equation is balanced. But we’re not done. We now need to write a half equation for iodine. In this reaction, iodide ions react to form iodine:

I- → I2

Balancing the equation in terms of iodine atoms gives us the following:

2I- → I2

This time, the charge on the left-hand side of the equation is -2 and the charge on the right side is +0. This means that we need to add two negative electrons to the right-hand side:

2I- → I2 + 2e-

Your two half equations are complete.

Can you tell which species has been oxidised, and which has been reduced? Here, bromine gains electrons and is reduced, whilst iodide ions lose electrons and so are oxidised.

Now try a trickier example.

Write half equations for the reaction between manganate(VII) ions and iron(II) ions to form manganese(II) ions and iron(III) ions. The unbalanced equation is given below:

MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O

Let’s start with iron. We start with Fe2+ ions which change to Fe3+ ions:

Fe2+ → Fe3+

The number of iron atoms is balanced in this equation - we just need to add electrons to balance the charges:

Fe2+ → Fe3+ + e-

Moving on to the manganate, things get a bit more complicated. Here’s our starting equation:MnO4- → Mn2+There is the same number of manganese atoms on each side of the equation, so we don’t need to worry about them. However, there are four oxygen atoms on the left-hand side but none on the right-hand side. Remember that the only substances we can add in are water, hydrogen ions, and electrons. So, to add more oxygen to the right-hand side, we need to include some water molecules. We need four oxygen atoms, so we add in four water molecules:MnO4- → Mn2+ + 4H2OBut now we have eight hydrogen atoms on the right-hand side and none on the left. We therefore need to add in eight hydrogen ions to the left-hand side, to balance it all out:MnO4- + 8H+ → Mn2+ + 4H2O We’re almost done. However, the charges aren’t balanced. There is an overall charge of +7 on the left and +2 on the right. To make the charges equal, we need to add in 5 negative electrons to the left-hand side:MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

You can also use oxidation states to balance half equations. In our second example, iron went from an oxidation state of +2 to +3. This means that it lost one electron. We therefore add one electron to the right-hand side of the equation to show this loss:

Fe2+ → Fe3+ + e-

Well done! We made it - our two half equations are complete. But now we need to combine them into one overall equation.

Combining half equations

What if we want to go from half equations to a redox equation? You’ll notice that in redox equations, we don’t see any electrons. In order to form redox equations, we combine multiples of the two half equations so the electrons cancel out.

In our first example, we formed the following two half equations:

Br2 + 2e- → 2Br-

2I- → I2 + 2e-

Both sides of the equation have just two electrons. We can therefore simply add the two equations together to make one overall redox equation:

Br2 + 2I- + 2e- → I2 + 2Br- + 2e-

Because two electrons feature on both sides of the equation, they cancel out:

Br2 + 2I- + 2e- → I2 + 2Br- + 2e-

Here's the overall redox equation:

Br2 + 2I- → I2 + 2Br-

That example was fairly simple, but our second example from earlier is a little more tricky. We can see that there is only one electron in the first half equation, but five in the second half equation:

Fe2+ → Fe3+ + e-

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

We need to multiply the first equation by 5 so that both equations feature the same number of electrons:

5Fe2+ → 5Fe3+ + 5e-

We can now add the two equations together. The electrons on each side of the equation cancel out:

MnO4- + 8H+ + 5e- + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ + 5e-

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

This is our final answer.

Disproportionation reactions in redox

Let’s now touch on disproportionation reactions. Before, we only looked at equations where a species was either oxidised or reduced. In disproportionation reactions, both processes can occur.

Disproportionation reactions are reactions where the same species is both reduced and oxidised.

We can see whether a species has been reduced, oxidised, or both by looking at its oxidation states. Here’s an example:

We can see the following:

  • Copper goes from an oxidation state of +1 in Cu2O to O in Cu and +2 in CuSO4.
  • To get from Cu2O to Cu, copper must gain an electron. This means that it is reduced.
  • To get from Cu2O to CuSO4, copper must lose an electron. This means that it is oxidised.

Because copper has been both oxidised and reduced, this is a disproportionation reaction. This is just one example of a type of redox reaction.

Let’s look at some more.

Redox Reactions Examples

There are a few common examples of redox reactions. One of these includes electrolysis.

Electrolysis

Either oxidation or reduction takes place at each electrode in electrolysis.

  • At the cathode, positive cations gain electrons to form neutral atoms. They are reduced.
  • At the anode, negative anions lose electrons to form neutral atoms. They are oxidised.

For example, electrolysis of sodium chloride produces chlorine gas at the anode and hydrogen gas at the cathode. Chloride ions are oxidised whilst hydrogen ions are reduced.

2Cl- → Cl2 + 2e-

2H+ + 2e- → H2

Other examples of redox reactions

Further examples of redox reactions in everyday life are rusting, respiration, and combustion.

  • In combustion, the fuel is oxidised and oxygen is reduced.
  • In rusting, the iron is oxidised while oxygen is reduced.
  • In respiration, electron carriers are oxidised while oxygen is reduced.

Redox - Key takeaways

  • Redox is a term used to describe reactions involving both oxidation and reduction. These reactions involve a movement of electrons, and are characterised by a change in oxidation states.
  • Oxidation and reduction have several definitions in chemistry. However, oxidation is generally taken to mean a loss of electrons, whilst reduction means a gain of electrons.
  • Reducing agents are electron donors that reduce another species and are oxidised themselves.
  • Oxidising agents are electron acceptors that oxidise another species and are reduced themselves.
  • Oxidation states are numbers assigned to ions that show how many electrons they have lost or gained, compared to the element in its uncombined state. A positive oxidation state shows that the element has lost electrons, whilst a negative oxidation state shows that it has gained electrons.
  • There are certain rules to follow when assigning oxidation states.
  • A redox equation is an equation showing the reduction and oxidation processes in a reaction, ignoring any spectator ions.
  • Redox equations can be split into half equations which show the oxidation and reduction processes separately. They also show the movement of electrons.
  • Disproportionation reactions are reactions in which the same species is both reduced and oxidised.

Redox

A redox reaction is a reaction involving both oxidation and reduction.

A simple redox reaction is a displacement reaction between two metals, such as adding magnesium to a solution of Fe2+ ions. The magnesium is oxidised and loses electrons, whilst the iron is reduced and gains these electrons. Examples of more complicated redox reactions include respiration, combustion, and rusting.

We can identify redox reactions because they cause the oxidation state(s) of one or more species to change.

In electrolysis, reduction and oxidation both happen simultaneously. This makes it an example of a redox reaction. At the cathode, positive cations gain electrons and so are reduced, whilst at the anode, negative anions lose electrons and so are oxidised.

In redox reactions, one species is oxidised and another species is reduced. The oxidised species loses electrons, while the reduced species gains these electrons.

Final Redox Quiz

Question

What is a redox reaction?

Show answer

Answer

A reaction involving both oxidation and reduction.

Show question

Question

Which of the following are definitions of oxidation?

Show answer

Answer

Gain of oxygen

Show question

Question

Which of the following are definitions of reduction?


Show answer

Answer

Gain of hydrogen

Show question

Question

Which of the following are redox reactions?


Show answer

Answer

Mg + Cu2+ → Mg2+ + Cu

Show question

Question

What does the acronym OIL RIG stand for?


Show answer

Answer

Oxidation is loss, reduction is gain.

Show question

Question

What does the acronym RAD OAT stand for?


Show answer

Answer

 Reducing agents donate, oxidising agents take.

Show question

Question

What are oxidation states?


Show answer

Answer

Numbers assigned to ions that show how many electrons they have lost or gained, compared to the element in its uncombined state.

Show question

Question

What is the oxidation state of iron in iron(II) chloride?


Show answer

Answer

+2

Show question

Question

What is the oxidation state of oxygen in O2?


Show answer

Answer

0

Show question

Question

What is the oxidation state of S in SO42-?


Show answer

Answer

+6

Show question

Question

What is the oxidation state of S in H2S?


Show answer

Answer

-2

Show question

Question

What is a disproportionation reaction?


Show answer

Answer

A reaction where the same species is both reduced and oxidised.

Show question

Question

True or false? Group 1 metals always have an oxidation state of +1.


Show answer

Answer

True

Show question

Question

True or false? Chlorine always has an oxidation state of -1.


Show answer

Answer

False

Show question

Question

What is the oxidation state of Mg2+?


Show answer

Answer

+2

Show question

Question

Write half equations for the following redox reaction:

    

Cl2 + 2Br- → 2Cl- + Br2

Show answer

Answer

Cl2 + 2e- → 2Cl-

2Br- → Br2 + 2e-

Show question

Question

What three species can you add when balancing half equations?


Show answer

Answer

Oxygen

Show question

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