Redox Reactions or Oxidation Reduction Reactions

Redox Reactions  / Oxidation and reduction reactions


These are reactions where electrons are transferred from one species (atom, molecule or ion) to another. We can write 'half' equations to show only what happens to the species losing electrons or a different 'half' equation to show the species gaining electrons.
The whole equation is put together by making sure that the numbers of electrons are balanced in each half equation and adding them together (when the electrons will cancel out)
Oxidation
This is the name given to removal of electrons from a species - the reagent causing the loss of electrons is called the oxidising agent

Example: Mg(s) Mg2+ + 2e In this (half) equation the magnesium atom loses electrons and becomes an ion.

Reduction
This is the gain of electrons - the species donating the electrons is called the reducing agent

Example Fe3+ + 3e Fe(s) In this (half) equation the iron(III) ion gains three electrons to become an atom.

Redox reactions
Obviously the electrons leave one species and go to another. Consequently reduction has to be accompanied by oxidation and vice versa. For this reason reactions involving transfer of electrons are called reduction and oxidation or redox for short

Example 3Mg(s) + 2Fe3+ 2Fe(s)+ 3Mg2+ The electrons from the magnesium are transferred to the iron(III) ions

Summary

Loss of electrons = Oxidation Gain of electrons = Reduction	Mnemonic (memory aid) OIL-RIG
			Oxidation Is Loss Reduction Is Gain

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10.1.2: Calculate the oxidation number of an element in a compound. Oxidation numbers should be shown by a sign (+ or -) and a number, eg +7 for Mn in KMnO4.

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Oxidation number
This is the apparent valency of an atom within a compound. It is usually considered as if the element were bonded ionically to allow the apparent number of electrons gained or lost to be assessed.
The sum of all the oxidation numbers in a compound must add up to 0. By convention, the oxidation number is written as a Roman numeral in the name, eg. iron II sulphate, sulphur VI oxide.
The oxidation number of an uncombined element is always zero (0)
Calculating the oxidation number
There are some elements that virtually always have the same oxidation number and these can be used to calculate the oxidation numbers of the atoms in question.
Hydrogen, for example always has an oxidation number of -1 when bonded to a metal (more electropositive element) and +1 when bonded to a more electronegative element (non-metal). Oxygen is always -2 (except when in the form of the peroxide ion when it has an O-O bond giving it an oxidation number of -1). Group 1 and 2 metals usually have an oxidation number of 1+ and 2+ respectively.

Example - Calculate the oxidation number of sulphur in sulphuric acid H2SO4 Hydrogen = +1 oxidation number Oxygen = -2 oxidation number Therefore: (2 x H) + S + (4 x O) = 0 2 + S -8 = 0 S = 6

Example - Calculate the oxidation number of nitrogen in calcium nitrate Ca(NO3)2 Calcium is in group 2 = +2 oxidation number Oxygen = -2 oxidation number Therefore: (+2) + [(2 x N) + (6 x -2)] = 0 +2 + 2N -12 = 0 2N = 10 N = +5

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10.1.3: State and explain the relationship between oxidation numbers and the names of compounds. Oxidation numbers in names of compounds are represented by Roman numerals, eg iron(II) oxide, iron(III) oxide.

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Names of compounds
Where there is any doubt about the oxidation state of an element within a compound it is stated using Roman numerals immediately after the ambiguous element. For example Iron compounds may be iron in the oxidation state +2 or +3 - it must therefore be stated as iron II or iron III in the compound name.
In the examples above the full systematic name for sulphuric acid is sulphuric(VI) acid and calcium nitrate is calcium nitrate(V)

Example - Name the following compound - FeSO4 Oxidation state of the oxygen = -2 Oxidation state of the sulphur = +6 Therefore oxidation state of the iron = - (+6 - 8) = +2 The name of the compound FeSO4 is iron(II) sulphate

Example - Name the following compound - TiCl4 Oxidation state of the chloride = -1 Therefore oxidation state of the titanium = - (- 4) = +4 The name of the compound TiCl4 is titanium(IV) chloride

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10.1.4: Identify whether an element is oxidised or reduced in simple redox reactions, using oxidation numbers. Appropriate reactions to illustrate this can be found in topics 3 and 11. Possible examples include: iron(II) and (III), manganese(II) and (VII), chromium(III) and (VI), copper(I) and (II), oxides of sulphur and oxyacids, halogens and halide ions.

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Oxidation and reduction
As stated above, for the purposes of oxidation and reduction the oxidation number can be thought of as the apparent ionic charge of an atom within a compound. For example, in sulphuric acid the sulphur is in the VI (6+) oxidation state. For the purposes of redox we can consider that it has an ionic charge of +6 (even though it is clearly covalently bonded). This makes it easier to follow any transfer of electrons.
If the sulphur changes to an oxidation state of IV during a chemical reaction then it has gone from an apparent ionic charge of +6 to a charge of +4, i.e. it has gained two electrons (negative charges). It has therefore been reduced (gain of electrons) in the process.

Examples 2FeCl2 + Cl2 2FeCl3 The iron changes from 2+ to 3+ and is therefore oxidised (removal of electrons) The chlorine gains an electron to go from 0 to -1 and is therefore reduced (addition of electrons) Zn + CuSO4 Cu + ZnSO4 The zinc changes from oxidation state 0 to +2 (removal of electrons) it is oxidised  The copper changes from 2+ to 0 and is oxidised and is therefore reduced (addition of electrons) Cr2O72- + 3SO2 + 2H+ 2Cr3+ + 3SO42- + H2O The chromium changes from +6 to +3 and is therefore reduced (gain of electrons) The sulphur changes from +4 to +6 and therefore loses electrons = oxidation (loss of electrons) 2KI + Br2 2KBr + I2 The iodide ions (oxidation number = -1) change to iodine (oxidation number = 0) : oxidation Bromine (element, oxidation number = 0) changes to bromide ions (oxidation number = -1) : reduction 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O The iron changes from 2+ to 3+ and is oxidised (removal of electrons) The manganese atom changes from +7 to +2 and is therefore reduced (addition of electrons)

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10.1.5: Define the terms oxidising agent and reducing agent.

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Oxidising agents
These are the chemicals that cause the oxidation in a redox reaction. We call the reacting compounds in a reaction the reagents (short form of the words reacting agents).
We consider that the removal of electrons from a species is oxidation and these electrons have to be taken away by another compound or species. This species that attracts the electrons is said to be the oxidising agent i.e. the reagent that causes the oxidation.
Reducing agents
Similarly the reagent that causes reduction in a redox reaction is said to be the reducing agent.
The oxidising agent takes the electron and is itself reduced, the reducing agent loses the electrons and is itself oxidised.

2KI	+ Br2		2KBr	+ I2
Iodide ions get oxidised	Bromine gets reduced
Iodide - reducing agent	Bromine - oxidising agent
Cr2O72-	+ 3SO2 + 2H+		2Cr3+	+ 3SO42- + H2O
Chromium VI gets reduced	Sulphur IV gets oxidised
Chromium VI oxidising agent	Sulphur IV reducing agent

Fundamentals of Acid Base Chemistry

Fundamentals of Acid-Base Chemistry

Definitions of Acidity and Basicity

For more than two-hundred years, chemists have struggled to come up with a way to describe acid-base reactions that is at the same time physically relevant, specific enough to be accurate, and general enough to include everything that should be considered an acid-base relationship.

Svante Arrhenius first defined acids to be proton (H⁺) donors and bases to be hydroxide ion (OH^-) donors in aqueous solution. The Arrhenius model of acids and bases is summarized by the following two reactions:

Figure %: The Arrhenius model of acids and bases, where A = acid and B = base

At the time that Arrhenius proposed these definitions, water was virtually the only solvent used in chemistry, and nearly all known acids and bases contained protons (H⁺) and hydroxyl groups (OH), respectively. His definition was sufficient for the chemistry that was understood then. But progress in chemistry necessitated new definitions: it was discovered that ammonia behaves like a base, and HCl donates protons in non-aqueous solvents. The Bronsted-Lowry model of acids and bases serves that need by describing acids as proton donors and bases as proton acceptors. These definitions remove the role of solvent and allow bases like ammonia and fluoride ion to be classified as bases, so long as they bond to protons. The Bronsted-Lowry model implies that there is a relationship between acids and bases (acids transfer protons to bases) and allows us to define conjugate acids and conjugate bases, as seen in .

Figure %: The Bronsted model of acids and bases

You should note in the figure above that the conjugate acid of the base, BH⁺, acts as an acid in the reverse reaction by donating a proton to the conjugate base, A^-, of the acid HA.
Despite the usefulness of the Bronsted-Lowry definition, there is an even more general definition of acids and bases provided by G. N. Lewis. The Lewis model of acids and bases proposes that an acid is an electron pair acceptor while a base is an electron pair donor. This model of acidity and basicity broadens the characterization of acid-base reactions to include reactions like the following which do not involve any hydrogen transfers. The nitrogen atom in ammonia donates an electron pair to complete the valence octet of boron.

Figure %: Example of a Lewis acid-base reaction

Because we are more interested now in describing terms and processes that involve proton transfers (pH, titration), we will focus on the Bronsted-Lowry definitions of acids and bases. We will leave consideration of the Lewis model of acids and bases for studying reactions in organic chemistry.

Reactions of Acids and Bases with Water

As you may have noted already in the acid-base reactions above, we use arrows in both reaction directions to indicate that these are equilibrium processes. Proportions of reagents and products at equilibrium can be described by an equilibrium constant. The equilibrium constant given in is for the reaction of an acid, HA, with water as shown.

Although water is a reactant in the above reaction and belongs in the equilibrium constant, its value of 55.6 M in aqueous solution is so large in comparison with the change in water concentration at equilibrium that we will assume that the value of [H₂O] is constant. Using that assumption, we will define the acid dissociation constant, K _a, in to be the following:

Figure %: Definition of the acid dissociation constant

From the form of the above equation we can see that stronger acids, those that dissociate to a greater extent, will have larger values of K _a whereas weaker acids will have smaller values of K _a. A practical range for K _a values runs from 10^-12 for very weak acids to 10¹³ for the strongest acids. Knowing this practical range of acidity constants will aid in judging how reasonable your answers are when you calculate values for K _a in problems.
In an analogous way, we define K _b, the base constant in to be the following:

Figure %: Definition of the base dissociation constant

Stronger bases have larger values of K _b while weaker bases have smaller values of K _b. K _b's of typical bases in inorganic chemistry tend to have a range of values between 10^-11 and 10³.
As you may have discovered in our above discussion, water can act as both an acid and as a base. For this reason water is said to be amphiprotic. Water is often incorrectly termed amphoteric. An amphiprotic species like water can either donate or accept a proton. Amphoteric species can both donate and accept hydroxide ions, as water cannot. The following reaction in , called the autoionization of water, has the equilibrium constant K _w defined in the manner of K _a and K _b. The dissociation constant K _w for water is 1 x 10^-14 at room temperature (298 K), and tends to rise with higher temperatures.

Figure %: The autoionization of water

Knowing that K _w = 1 x 10^-14 is useful because the relationship between K _a and K _b for a conjugate acid-base pair is K _a * K _b = K _w. Therefore, you can calculate the K _a of the conjugate acid of a base when given its K _b. This point is proved in below.

Figure %: The autoionization constant of water equals the product of the acid and base dissociation constants of a conjugate acid-base pair.

From the proven expression, we see that strong acids (large K _a's) form weak conjugate bases and weak acids form strong conjugate bases.

Mole Calculations

One of the main problems that beginning chemistry students have is in doing conversions between grams, moles, and molecules (or atoms). Usually, a question will be asked of you in the following form:
How many moles are in 22 grams of copper metal?
If you're confused by this problem, don't worry. Most people are when they start doing this kind of problem. To make life easier for you, I put together a "road map" which tells you exactly what you need to do to convert between atoms (or molecules), grams, and moles.

You should read this picture the same as you would read a subway map. For example, if you want to go from King Street in Alexandria to National Airport on the Blue line of the Metro, you need to first go to the Braddock Road station. The same thing is true here - if a problem tells you to go from atoms to grams, you need to first go through moles before you do anything else. In our example that we're discussing, though, we are making a one-stop trip.
Example: How many moles are in 22 grams of copper metal?
In all problems like this, you need to go through four steps to find a solution.

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The Four Steps to Solving Mole Problems:

Step 1: Figure out how many parts your calculation will have by using the diagram
Looking at the diagram above, we can see that we are going between grams and moles, which is a one-step conversion. Furthermore, we can see that we need to use the atomic mass of copper as our conversion factor.
Step 2: Make a T-chart, and put whatever information the problem gave you in the top left. After that, put the units of whatever you were given in the bottom right of the T, and the units of what you want to find in the top right.
In this case, the problem gave you "22 grams of copper" as the starting information. Because this is what you were given, put "22 grams of copper" in the top left of the T. Since "grams of copper" is the unit of what you were given, put this in the bottom right of the T. Since you want to find out how many moles of copper are going to be made, put "moles of copper" as your unit in the top right. When you've done this, your calculation should look like this:

Step 3: Put the conversion factors into the T-chart in front of the units on the right.
As we saw from the "map", the conversion factor between grams and moles is the atomic mass of copper. Because we measure atomic mass in grams, you need to put the atomic mass in front of the unit "grams of copper". What do you put in front of moles? Whenever you do a calculation of this kind, you need to put "1" in front of moles, like you see here:

Step 4: Cancel out the units from the top left and bottom right, then find the answer by multiplying all the stuff on the top together and dividing it by the stuff on the bottom.
In this case, you'd multiply 22 by one and divide the result by 63.5. Your answer, 0.35 moles of copper:

And that's how you do a one-step problem of this kind!

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Solving Two-Step Mole Calculation Problems:

What happens if we need to solve a problem that requires we not just go from one box in the next in our diagram, but across the entire diagram? Well, it means that we need to do two steps in our calculation. Let's see that "map" again to see what I mean:

If we were asked to convert 22 grams of copper to atoms of copper, we'd have to go from one end of the map to the other. Instead of doing a simple one step calculation, we'd need to do a two-step calculation, with the first step going from grams to moles and the second step going from moles to atoms.
How can we solve this kind of problem? Well, we start off by doing the same thing that we did in our last example: We had to convert grams to moles before, and we can see from the map that we have to convert grams to moles now, too. To refresh your memory, here's the calculation from last time:

In the next step, we do the same thing over again, except that we need to add another T to the T-chart. When you do this, take the units of the thing at the new top left and put them on the bottom right (in this case, moles). Then take the units of what you want (in this case, atoms) and put it in the top right. Finally, put in your conversion factors, which from the chart above is Avogadro's number, or 6.02E23. Since this number refers to the number of atoms in a mole of a substance, we put this in front of "atoms of copper". Again, put the number "1" in front of moles, because we're saying that there are 6.02E23 atoms in ONE mole of an element.
When we add all these terms in, we can cross out the units that cancel out, as shown. To get the answer, multiply all the numbers on the top together and divide by the numbers on the bottom. Your answer should then be set up like this:

And that's how you do mole problems!