# Difference between revisions of "Chemistry Lab/Electrochemistry"

Electrochemistry was a topic for the event Chemistry Lab in 2011 and 2012.

## Basic Information

A redox reaction, or an oxidation/reduction reaction, occurs when one reactant is oxidized, or loses electrons, and one reactant is reduced, or gains electrons. A simple way to tell the difference is OIL RIG (Oxidation Is Losing; Reducing Is Gaining) or LEO says GER (Lose Electrons - Oxidize; Gain Electrons - Reduce). The oxidizing agent is reduced, and the reducing agent is oxidized.

A half-reaction is exactly what it sounds like - half a reaction. It focuses exclusively on one portion of the reaction, either oxidation or reduction.

The oxidation number of an atom in a molecule refers to the electrons being shared. If, in a covalent bond, it is basically "giving" an electron to the other atom, then that counts as a +1 on it's oxidation number. If, in a covalent bond, it is "taking" an electron from the other atom, then that counts as a -1 on it's oxidation number. Something that is important to realize is that it is not actually giving or taking an electron, that only occurs in ionic bonds, but that it or the other atom is getting the electron more often.

When something is oxidized its oxidation number increases, and when something is reduced it's oxidation number decreases.

The octet rule states that all atoms (excluding H and He) are trying to get an octet (8) electrons. H is trying to get either 2 or 0 electrons, and He already has 2.

## Figuring Out Oxidation Numbers

When figuring out the oxidation number of an atom in a molecule, there are two simple rules to follow.

1. The most electronegative atom gets the electron.
2. The oxidation numbers add up to the charge of the molecule. If you have a charge put it on the atoms which have not reached an octet of electrons. Change their oxidation number accordingly.

Take the example of ethanol, $CH_3CH_2OH$.

Number the C atoms. There is one on the left that will be called C1. There is another one in the middle that will be called C2.

C1 has 3 bonds with H atoms. In each bond C1 is the more electronegative. So each of the 3 H atoms have an oxidation number of +1. So far C1's oxidation number is -3. C1 has another bond that is with C2. They each have the same electronegativity, so there is no change in their oxidation numbers. Thus, C1 has an oxidation number of -3.

C2 so far has an oxidation number of 0 from the C1-C2 bond. It is also bonded to 2 H atoms, giving it an oxidation number of -2 and giving each of the H atoms an oxidation number of +1. Lastly, it has a bond with the O atom. O is more electronegative, so C2's oxidation number is -1.

O so far has an oxidation number of -1 from the C-O bond. It is also bonded with an H atom. Since it is more electronegative, it gets the electron. Thus, the H atom has an oxidation number of +1, and O has an oxidation number of -2.

Last but not least, add up all the oxidation numbers. All 6 of the H atoms have an oxidation number of +1. C1 has an oxidation number of -3. C2 has an oxidation number of -1. O has an oxidation number of -2.

6(+1) -3 -1 -2 = 0

This is the charge of ethanol, so the procedure has been completed.

## Balancing Oxidation/Reduction Reactions

Redox reactions follow a simple set of steps to solve.

A simple example of separating a reaction into half-reactions, balancing mass and charge, and combining the reactions.
1. Split it into 2 half-reactions. Usually, the basis of a redox reaction will be given with two compounds reacting to form two more compounds. Aside from oxygen and hydrogen, each reactant will have a corresponding element with a product. These two compounds will form one half-reaction; the other two form the other. Balance the coefficients of key compounds if necessary, i.e. Balancing Cr with Cr2O7. Don't worry about balancing extra hydrogens or oxygens yet.
2. Balance all non-hydrogen or oxygen elements. You may have balanced the key components of the half-reaction, but sometimes you'll have a pesky oxygen or hydrogen messing things up. To get this to balance, you must add ions to the other side to balance the reaction out. This process differs for acidic vs. basic solutions. Do NOT try to balance one equation with the other, this step ONLY comes at the end.
3. If it's in an acidic medium:
1. Balance oxygen by adding H2O to the appropriate side. Add a coefficient to this H2O if necessary.
2. Balance hydrogen by adding H+ to the appropriate side. This can be on either side, depending on how many H2O's you added. Add a coefficient to this if necessary.
3. Balance the charge by adding electrons. Now you've balanced the equation in terms of elements, but the charge may not be balanced yet. To balance this, add electrons to the side with a higher charge until the total charge of each half of the half-reaction is the same.
4. If it's a basic medium
1. Balance oxygen as above.
2. Balance hydrogen. Instead of balancing this with H+, you need to balance it with OH-. This means that you may get extra oxygens. If this happens, add another H2O on the other side and continue adding OH- until it balances. There is also another method that is detailed below.
3. Balance the charge as above.
5. Now, we can add the reactions together to come up with our final reaction. Multiply each reaction by an integer so that there are the same number of electrons on each side (i.e. they cancel out). This means that the electrons of one half-reaction should be on the OPPOSITE side of the electrons in the other half-reaction. If this is not the case, go back and check your work. More than likely, there's a mistake in there somewhere.
6. Combine the half-reactions and cancel. The elctrons should cancel out completely, and H2O's and H+'s may cancel somewhat. If you have time, it's usually a good idea to make sure that the equation is balanced by elements and by charge.
• Second method for balancing redox reactions in basic solution: If it's in a basic medium, add OH- to each side of the final equation until all H+ is gone; then, cancel again. Remember that OH- + H+ ---> H2O in this step.

## Activity Series

An activity series.

One task participants may be asked to complete in this event is to construct an activity series based on what ions react with others. This activity series dictates what elements oxidize more easily than others.

One of the most common ways that you will have to make an activity series in Chem Lab is through performing single replacement reactions by putting metal strips in a metal solution and seeing if there is a reaction. Each team will have a set of solutions and metals, and will have to perform each possible combination of metal to solution. A table can be formed recording which combinations result in reactions. The metal that reacts the most is the one that oxidizes the most easily, while the metal that does not react at all is the one that reduces the most easily. Once complete, the activity series should look similar to the one at right. If not, you likely made a mistake and should recheck your work.

## Electrochemical Cells

Eelectrochemical cells results in an exchange of electrons in a redox reaction. There are two main types of electrochemical cells.

### Voltaic Cells

A voltaic, or galvanic, cell is composed of two metals connected by a salt bridge. It uses the electron exchange to generate the current. It consists of two half-cells, each of which contains a metal solution with that metal submerged in it.

The cell in which oxidation occurs is called the anode, and the cell in which reduction occurs is called the cathode. You can remember this by knowing that reduction has a "c" in it, and cathode starts with a "c". Electrons flow from the anode to the cathode.

Another, more sure fire way of telling where oxidation and reduction occurs is as so: the cathode is positive (you can remember this by "cats are positive", even if they're not). Thus, electrons flow to the cathode, meaning that reduction is occurring at the cathode. This means that oxidation occurs at the anode.

A porous barrier allows ions to flow from the anode compartment to the cathode compartment and vice versa, balancing the charge. The electrode compartments are called half-cells.

Salt bridges may be used as an alternative to porous barriers.

### Electrolytic Cells

Electrolytic cells use a current to decompose chemical compounds.

One principle use of electrolytic cells is to electroplate objects such as nails and silverware.

Coulombs is a measure of charge. Current is a measure of the flow of electrons. The SI unit of current is the ampere, expressed as C/s or coulombs per second.

Coulombs = amperes $\times$ seconds

Given the current run through an electrolytic cell and the time it is run, you can calculate the number of coulombs. There are 1.602E-19 coulombs in an electron. From the amount of coulombs you may calculate the number of electrons used to reduce.

Say that you are electroplating copper onto a plate.

$Cu^{2+} + 2e^- \to Cu$

Given the number of electrons used to reduce the copper ions, you may calculate the amount of Cu electroplated onto the plate. This is applicable to any electroplating situation.

## Electron Potential

There are two different analogies for understanding electron potential or voltage.

One is water. Electron potential corresponds to the water pressure. The higher the pressure, the stronger the stream that flows. Electron potential does not correspond to the strength of the stream, since different sized pipes with the same water pressure will have different strength streams.

The second analogy is height. Higher electron potential corresponds to higher height. From higher height you can drop, while doing work, to lower height.

Something to note is that electron potential is not absolute, it is with respect to. Standing on a 10 ft. high cliff and dropping a ball is the same as standing on the edge of a 10 ft. deep pit and dropping a ball (ignoring changes in gravity: analogies are not perfect). It is the same way with electron potential. You must define a zero before you can say what the electron potential is. Because of this it is quite possible to have negative electron potential.

## Electromotive Force (emf)

The emf of a cell, measured in volts, is the potential difference between the cathode and the anode of a cell. It tells you how much potential there is to do work. Electromotive means "causing electron motion".

### Measuring emf

It is quite easy to measure emf. Take a voltmeter and touch the probes to the cathode and anode. The voltmeter will tell you what the voltage difference, or emf, is.

### Standard Reduction (Half-Cell) Potentials

Reduction Potentials tell you how much something "wants" to reduce. For example, $Cu^{2+}$ (with a reduction reaction of $Cu^{2+} + 2e^- \to Cu$) has a higher reduction potential then $Fe^{2+}$ (with a reduction reaction of $Fe^{2+} + 2e^- \to Fe$). This means that $Cu^{2+}$ "wants" to reduce more then $Fe^{2+}$.

Like all potentials, reduction potentials are not absolute and have to be with respect to something.

Standard Reduction Potentials, denoted $E_{red}^{\circ}$ are the reduction potential with respect to a reference reaction. This reference reaction is the reduction of $H^+$.

$2H^+ + 2e^- \to H_2$

An electrode designed to produce this half reaction is called a standard hydrogen electrode (SHE) or the normal hydrogen electrode (NHE). A SHE consists of a piece of platinum foil covered with finely divided platinum. The electrode is encased in a glass tube so that hydrogen gas at STP can bubble over the platinum. (STP refers to 273 Kelvin and 1 atm) The solution contains $H^+$.

Here is a table of standard reduction potentials:

Standard Reduction Potentials in Water at 25 $^{\circ}$C
Reduction Half-Reaction Potential(V)
$F_2 + 2e^- \to 2F^-$ +2.87
$MnO_4^- + 8H^+ + 5e^- \to Mn^2+ + 4H_2O$ +1.51
$Cl_2 + 2e^- \to 2Cl^-$ +1.36
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$ +1.33
$O_2 + 4H^+ + 4e^- \to 2H_2O$ +1.23
$Br_2 + 2e^- \to 2Br^-$ +1.06
$NO_3^- + 4H^+ + 3e^- \to NO + 2H_2O$ +0.96
$Ag^+ + e^- \to Ag$ +0.80
$Fe^{3+} + e^- \to Fe^{2+}$ +0.77
$O_2 + 2H^+ + 2e^- \to H_2O_2$ +0.68
$MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$ +0.59
$I_2 + 2e^- \to 2I^-$ +0.54
$O_2 + 2H_2O + 4e^- \to 4OH^-$ +0.40
$Cu^{2+} + 2e^- \to Cu$ +0.34
$2H^+ + 2e^- \to H_2$ 0[defined]
$Ni^{2+} + 2e^- \to Ni$ -0.28
$Fe^{2+} + 2e^- \to Fe$ -0.44
$Zn^{2+} + 2e^- \to Zn$ -0.76
$2H_2O + 2e^- \to H_2 + 2OH^-$ -0.83
$Al^{3+} + 3e^- \to Al$ -1.66
$Na^+ + e^- \to Na$ -2.71
$Li^+ + e^- \to Li$ -3.05

Table from "The Central Science".

### Calculating emf

To calculate the emf of a cell, simply take the standard reduction potential of the cathode and subtract the standard reduction potential of the anode from it. Because electron potential measures potential energy per charge, the stoichiometric coefficients in the half-reactions do not affect the value of the standard reduction potential or the emf of the cell.