Will a Precipitate Form?

We’re going to solve two different problems on this page. The first is relatively straight-forward, the second is the same sort of problem, but has an added issue that we’ll need to address.

A simple Problem 

 Let’s start by solving this problem:

A solution is made that is 0.0035 M \(Pb(NO_3)_2\) and 0.070 M \(NaCl\). Will a precipitate form? The \(K_{sp} ~of~ PbCl_2\) = \(1.7 \cdot 10^{-5}\).

There are several things we need to think about before we can start the problem:

• Since we are given the \(K_{sp}\) of \(PbCl_2\) then the relevant reaction must be the dissolving of \(PbCl_2\) even if that seems to come from out of nowhere.
• All nitrates are soluble (check your solubility table). That means that we don’t need to care about the \(NO_3^{-1}\) ions, on the \(Pb^{+2}\) ions, which have a concentration of 0.0035 M.
• All sodium compounds are soluble. So, like the nitrates, we can ignore the \(Na^{+1}\) ions and the \([Cl^{-1}]\) = 0.070 M.

Here is our relevant reaction and the current ICE table:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

It is tempting to think that this reaction must run backwards, since no reactant is present, but this is a solubility problem. The reactant is a solid, is not included in the math and therefore is NOT relevant.

What we actually need to do is to figure out what the question actually is. “Will a precipitate form? means “will a solid appear in the container?” For that to happen, the ions in solution need to stick together and form the solid compound. In other words:

“Will a precipitate form?” = “Does the reaction run LEFT?”

That question, of course, is really asking: Is \(K_{sp} \lt Q_{sp}\)?

And that is simple math.

\(K_{sp}\) > or = or \lt \(Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\)

\(1.7 \cdot 10^{-5}\) ? \([0.0035] \cdot [0.070]^2\)

\(1.7 \cdot 10^{-5} \lt 1.72 \cdot 10^{-5}\)

Since \(K \lt Q\), the reaction will run to the left and the answer is YES (a precipitate will form).

A More Complex Problem

Now, let's try a more complex problem:

A solution is made by mixing 100 mL of a 0.0035 M \(Pb(NO_3)_2\) solution and 200.0 mL of a 0.070 M \(NaCl\) solution. Will a precipitate form? The \(K_{sp} of PbCl_2 is 1.7 \cdot 10^{-5}\).

In the last problem, we were given the concentrations of the ions in the solution which had already been made. In this problem we are given volumes and concentrations of the two solutions that are being mixed. Since each compound will end up in a larger volume than it had been before, the concentrations of each must be determined before we can do the check for precipitation.

Here is the work for the lead II nitrate. We’ll do a FULL version of the math and then find the quicker and easier way.

The full version:

\(100 ~mL ~Pb(NO_3)_2 \cdot \frac{1 L}{1000 mL} \cdot \frac{0.0035~moles}{1 L~solution}= 3.5 \cdot 10^{-4} mols~PbCl_2\)

Then, calculating the new molarity of the \(Pb(NO_3)_2\) in the mixed solution with its volume of 300 mL (0.300 L) requires this math:

\(M~=\frac{moles}{L}=\frac{3.5 \cdot 10^{-4} mols~PbCl_2}{0.300~L}= 0.00117 M\)

There is a simpler and faster way to get the same value. We need to notice that the \(Pb(NO_3)_2\) started as 100 mL and ended up in 300 mL. We can, therefore, do the following math:

\(0.0035 \frac{mols}{L}\cdot \frac{100~mL}{300 ~mL} = 0.00117~M\)

or even more simply:

\(0.0035 \frac{mols}{L}\cdot \frac{1}{3} = 0.00117~M\)

The NaCl goes from 200 mL to 300 mL, so we can calculate that molarity:

\(0.070~M~NaCl \cdot \frac{2}{3}= 0.0467~M~NaCl\)

Finally, we can address the actual question, which was “Will a precipitate form?” which we know means: is \(K_{sp} \lt Q_{sp}\)?

So, with our newly determined concentrations, we can solve the problem: 

\(K_{sp} \gt ~or~ = ~or~ \lt Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\) 

\(1.7 \cdot 10^{-5}\) ? \([0.00117] \cdot [0.0467]^2\) 

\(1.7 \cdot 10^{-5} \gt 2.55 \cdot 10^{-6}\)

So, in this case, the answer is NO (a precipitate will not form).

Finding Solublity in g/100 mL

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the solubility of the compound in \(\frac{g}{100~mL}\)?

This problem exists because we don’t have devices that measure moles directly. Therefore, knowing the molar solubility is not all that helpful. In the end, this problem is about turning molar solubility into a usable value.

We start this problem by finding the molar solubility of the compound. Given that we solved this problem on this page, we’ll jump right to the answer for that portion:

Molar solubility = “x” = 0.016 M.

The first thing we need to do is fix the unit moles, by converting it to grams. That’s a relatively simple problem using molar mass:

\(0.016 \frac{moles~PbCl_2}{L} \cdot \frac{278.10~g~PbCl_2}{1~mole~PbCl_2}=4.45 \frac{g}{L}\)

This is now a useful and workable amount. It tells us that we can dissolve, at most, 4.45 g of \(PbCl_2\) in a liter of solution.

For some reason, that is beyond the reasoning of this author, it was decided that solubilities should be recorded in terms of the grams that will dissolve in 100 mL, rather than in 1 L.

Solving this is easy...IF you think. If you try to set up a conversion process, you will get it wrong. I promise.

Here’s the simple logic. 100 mL is \(\frac{1}{10}\) of a liter. Therefore you can just divide the value we had before by 10.

So, the solubility of \(PbCl_2\) is 0.445 \(\frac{g}{100~mL}\).

Finding Molar Solublity

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the molar solubility of the compound?

This problem is the SAME as finding the concentrations in a saturated solution, with the exception that the answer is ALWAYS “x”. Be sure to read all the way to the bottom of this page to make sure that you understand the implications of that statement!

Our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

Simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x=0.016 M

Molar solubility is “x” therefore the answer is 0.016M.

What's interesting here is that it does NOT matter what the formula for the ionic compound is, the answer is always "x". 

A simple ionic compound like \(AgCl\) would simplify to:

\(K_{sp} = x^2\)

while a compound like \(Pb_3(PO_4)_2\) would work out to be:

\(K_{sp} = 108x^5\)

However, in both cases, the left column of the ICE table would be "some - x" and therefore the molar solubility is just "x".

Finding the Concentrations - Common Ion

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\). What is the concentration of chloride ion 

in a saturated solution that already has \([Pb^{+2}]\ = 0.25~M\)?

There are several things that are implied, but not stated, that we need to recognize:

The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).

The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [0.25 + x] \cdot [2x]^2\)

This will be a “little x” problem, so we can assume that 0.25 + x = 0.25. That simplifies the math to:

\(1.7 \cdot 10^{-5} = (0.25)\cdot 4x^2\)

which gives us x=0.0041 M

Since \([Cl^{-1} = 2x\) we know that \([Cl^{-1}]=0.0082~M\).

This is an example of the common ion effect. Simply put, the presence of lead ions in the solution did not allow as much of the lead II chloride to dissolve as would have in pure water.

Another, perhaps more useful, way to think about the common ion effect is to relate it to Le Chatelier’s Principle. For the reaction:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

additional amounts of \(Pb^{+2}\) ions will drive the reaction to the left, reducing the amount of chloride ion in solution.

Finding Concentrations in a Saturated Solution

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the concentration of chloride ion in a saturated solution?

There are several things that are implied, but not stated, that we need to recognize:

• The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).
• The solution is made by adding solid \(PbCl_2\) to pure water. This is implied by the term saturated solution. If the situation was anything OTHER than solid added to water, we would be told.
• The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

Although writing “some” in the midst of an ICE table seems odd, we can get away with it, since the solid is not part of the equilibrium expression.

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~or~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x = 0.016 M

Since \([Cl^{-1}] = 2x\) we know that \([Cl^{-1}]=0.032~M\).

Solubility

Understanding solubility requires us to understand how things (especially ionic compounds) dissolve, reversible reactions, and the mathematics of equilibrium - especially “little x” problems.

Let’s look at solubility with respect to one particular ionic compound - lead II chloride (\(PbCl_2\)). All of the ideas that are developed here, will then be applicable to other compounds. We can write the process of lead II chloride dissolving as a simple reaction:

\(PbCl_{2~(s)} \rightleftharpoons PbCl_{2~(aq)}\)

Although that reaction is correct, it isn’t very useful. It is much more interesting to write it as an ionic equation:

\(PbCl_{2~(s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

Remembering that solids are not included in the equilibrium expression, we can write the following:

\(K=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

Since this equilibrium expression has an “odd” look (nothing on the bottom), we give it a special name and add a subscript to the K:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

where \(K_{sp}\) stands for the Solubility Product.

There are several important ideas embedded in this:

1. \(K_{sp}\) is always for an ionic solid dissolved in water breaking up into its constituent ions
2. Any problem that mentions \(K_{sp}\) does not need to give the reaction, because it is implied.

There is also some vocabulary that flows from this:

• Saturated solutions are solutions for which \(Q_{sp} = K_{sp}\). In other words, a saturated solution is at equilibrium. Remember that this does NOT mean that the process of dissolving has stopped. It means that the rate at which ions are entering solution is matched by the rate at which ions are coming together and dropping out of solution. Therefore the concentrations of the ions will not be changing, nor will the amount of solid present. In addition, no solid must be visibly present. Saturation is based on the concentrations of dissolved ions only.
• Unsaturated solutions are those that can hold more solute than is currently dissolved. That is \(K_{sp} > Q_{sp}\). If solid is present, then the rate of solvation is faster than the rate of precipitation and the concentrations will be rising. If no solid is present, then the solution will remain as it is. 
• Supersaturated solutions are those where “too much” solute is dissolved. That is, solutions where \(K_{sp} < Q_{sp}\). Generally, we would expect such a situation to result in ions in solution coming together and precipitating, thus decreasing the concentrations of the ions and increasing the amount of solid present in the container. In rare occasions, when the crystalline structure of the solid is complex, it is possible for a solution to temporarily remain in a state of supersaturation until random chance allows the beginning of precipitation or until crystalization is seeded.
• Molar Solublity is the amount of a solid that will dissolve in water in units of \(\frac{moles~solid}{liters}\)

The only way to really understand all of this is, of course, to look at some problems. We’ll look at five different problems: 

1. Finding the concentrations of ions in a saturated solution 
2. Finding the concentrations of ions in a saturated solution in which one ion was already present (this is called the common ion effect) 
3. Finding the molar solubility of an ionic compound 
4. Finding the solubility of an ionic compound in \(\frac{g}{100~mL}\) 
5. Determining whether a precipitate will form when solutions are mixed

Finding the Density of a Solid

Density

Remember that density is a measure of the amount of stuff packed into a space. Different liquids and different solutions will have different densities and determining the density of a solution can help identify the solution.

Mathematically, we use the formula \(d=\frac{mass}{volume}\). This means that to find the density of something, we need to know the mass and volume. There are several ways to do this for a solid.

Finding the Density of a Regular Solid

Finding the density of a regular solid is very straightforward. Mass is measured on a balance and then volume is found by measuring the the solid and doing simple geometric math.

Cubes and Rectangular Prisms

Cylinders

Finding the Density of Irregular Solids

An irregular solid is one whose shape is not a simple geometric figure. This could be anything ranging from a slightly irregular block (where the corners aren’t quite square) to a random unsightly chunk.

Again, mass is easy. We simply put the solid on a balance.

Volume is now the challenge. In such a case, we find the volume using Archimedes Principle. The idea that a solid will displace a volume of water equal to its own volume.

If you aren’t familiar with the Archimedes Principle, the VERY short version of the story is that Archimedes was charged by the king to determine if the man who had made his new crown had cheated him by mixing the gold he had been given with another metal. The only way to identify pure gold at the time was to measure the density. However, the crown, being “crown-shaped” did not have an obvious volume. While pondering how to determine the volume of the crown, Archimedes went to the public baths. When he lowered himself into the tub, a large amount of water splashed over the side and Archimedes realized he could measure the volume of the crown by measuring the water displaced. He was so excited that he jumped up and ran through the town (naked) crying Eureka (I’ve found it).

Using the Archimedes Principle is simple. We find a graduated cylinder that has a wider diameter than the object and fill it part way with water. Carefully measuring the volume. Then we place the object into the cylinder so that it is completely submerged. As a result of this, the water will rise up the cylinder. We can then read the new volume. The volume of the object is simple the difference between the two measured volumes:

\(volume~of~object = volume~with~object - volume~without~object\)

So the density can be found this way:

\(density = \frac{mass}{(volume~with~object - volume~without~object)}\)

Finding the Density of a Liquid

Density

Remember that density is a measure of the amount of stuff packed into a space. Different liquids and different solutions will have different densities and determining the density of a solution can help identify the solution.

Mathematically, we use the formula \(d=\frac{mass}{volume}\). This means that to find the density of something, we need to know the mass and volume. There are several ways to do this for a liquid.

Finding Density of a Liquid

Direct Measurement

The simplest way to do this is to measure the mass and volume directly in something like a graduated cylinder.

First you would find the mass of the empty cylinder, then pour in an amount of the liquid and reweigh it. The difference in the two masses will be the mass of the liquid and the volume can be read directly from the cylinder.

The density would be found using this equation: \(d=\frac{(mass~cylinder~and~liquid - mass~cylinder)}{volume~of~liquid}\)

Although this method is quick and easy, it is limited by the imperfections of the measurements (the inherent error in mass measurement and the human error in volume measurement).

Graphically with Repeated Measurement

We can get past some of the limitations of direct measurement by making a series of measurements (mass and volume) and then graphing mass v. volume. Since density is an intrinsic quality (it doesn’t depend on amount), the data will give a straight line. Here is a series of data points graphed with and without the line of best fit. 

You can see that some of the points are above the line of best fit while some are below it. If we had relied on just one of those points, our density value could be quite a bit off, but using the line “smooths” out those inconsistencies.

Because slope is \(\frac{''rise''}{''run''}\) then in this case, slope will be \(\frac{mass}{volume}\) which is, of course, density.

Pycnometer

Another way to find the density of a liquid is to use a device known as a pycnometer. A pycnometer is simply a container with a known and constant volume. The picture to the right is a commercially produced pycnometer.

When we measure density directly, the weakest part of the process is the human error associated with measuring the volume of liquid in a graduated cylinder. Pycnometers eliminate this error. Because pycnometers are calibrated, the volume measurement can be as reliable as the mass measurement on a good balance, thus providing a more reliable density value.

Of course, pycnometers serve no other purpose, which means that unless a lab is making regular density measurements it is unlikely that such a device will be available.

Corrosion

Corrosion is the name of the reaction that occurs between a metal and oxygen. Although, that would also classify it as combustion, because these reactions happen so slowly, over such a long time, it seems odd to lump them together with fires and explosions. But, whether odd or not, that’s what these reactions are. 

Corrosion Reactions

Here is the reaction for iron:

\(2~Fe + O_2 \rightarrow 2~FeO\)

Iron will then undergo another reaction that takes it to the +3 state, but for the purposes of this page, we’ll focus on the first reaction.

The reaction between iron and oxygen is very slow, but there is a catalyst for this reaction - water. Remember that one of the ways that a catalyst can speed up a reaction is to provide an alternate pathway for the reaction which has a lower activation energy.

When water is present, corrosion can happen as a mechanism whose first step is a pair of half reactions. Oxygen in the air, can react with the water and electrons to make hydroxide ions.

\(O_2 + 4~e^{-1} + 2~H_2O \rightarrow 4~OH^{-1}\)

The source for those electrons is the iron, which loses electrons to become \(Fe^{+2}\) ions:

\(Fe \rightarrow 2~e^{-1} + Fe^{+2}\)

The iron II ions and the hydroxide ions can then combine to give iron II hydroxide (\(Fe(OH)_2\)).

Doubling the iron reaction (to match the number of electrons) and then adding the two half-reactions give us:

\(2~Fe + O_2 + 2~H_2O \rightarrow 2~Fe(OH)_2\)

Iron II hydroxide can then decompose into water and iron II oxide:

\(Fe(OH)_2 \rightarrow FeO + H_2O\)

Putting It ALL Together

Let’s put all of those steps together as a mechanism:

\(Step~1~~~~2~Fe + O_2 + 2~H_2O \rightarrow 2~Fe(OH)_2\)
\(Step2~~~~2~Fe(OH)_2 \rightarrow 2~FeO + 2~H_2O\)
\(Overall~~2~Fe + O_2 \rightarrow 2~FeO\)

Which is, of course, the original reaction at the top of the page. In other words, we have accomplished the same reaction, but through a different mechanism using water as the catalyst.

Because the activation energy for these reactions is much lower than it is for the overall reaction occurring in a single step, the catalyzed reaction occurs much faster. This is why cars in the desert areas of the Southwestern U.S. rust much more slowly than do cars in the wet Southeastern U.S. Adding salt to the water makes it a better conductor for charge which reduces the activation energy even more, which is why cars in the Northeastern U.S. (where they salt the roads in the winter) rust even faster.

Protecting From Corrosion

Separation

Once we understand that corrosion is a reaction with oxygen, catalyzed by water, protection from corrosion becomes clear. Let’s think about our cars. We paint our cars and wax them because the pain forms an airtight and watertight seal over the metal. In addition, wax is non-polar, so water beads up and rolls off of the surface, rather than staying in contact with the metal. Both of these help prevent corrosion.

Aluminum

Another way we avoid having things corrode is to change the metal from which they are made. Aluminum is a more reactive metal than iron, but aluminum oxide is extremely insoluble and it forms an airtight and watertight seal over the metal. In other words, aluminum corrodes a TINY amount and then that corrosion prevents further corrosion.

Sacrificial Anode

A last way we have of preventing (or at least slowing) corrosion, is the use of a sacrificial anode. One of the most common examples of this is galvanized steel.

Galvanized Steel is steel (mostly iron) that has been plated with a thin layer of zinc. When oxygen reacts with the water on galvanized steel, there are now two possible reactions that could supply those electrons. Here they are, listed with their voyages:

\(Fe \rightarrow Fe^{+2} + 2~e^{-1}~~~~~~~~~~0.44~V\)

\(Zn \rightarrow Zn^{+2} + 2~e^{-1}~~~~~~~~~~0.76~V\)

Both of these are positive, which means both could happen readily, but nature will favor the easier (more positive) half-reaction. That means that as long as ANY zinc remains, the electrons will be taken from the zinc and the iron will remain untouched. The zinc “sacrifices itself” to save the iron.

Electroplating

Galvanic cells and batteries use chemistry to produce electricity, but the opposite is also possible. We can use electricity to drive chemistry. This process is called electrolysis and one of the reasons we do it is to plate one metal on to another.

Imagine that you had a well crafted spoon made of tin. No matter how good the workmanship, a tin spoon will never be pretty, since tin is a rather drab gray metal. So, how can we make the spoon more attractive? We could plate it with gold. 

To do this, we’ll need to set up an electrolytic cell, like this:

The container is filled with a solution of gold III nitrate. We are only showing the \(Au^{+3}\) ions and not the nitrate, since only the gold ions will be involved in the chemistry. On the right side is a gold wire connected to the positive end of a power source. As electrons are pulled away from the wire, some of the gold atoms become ions and move into the solution:

\(Au \rightarrow Au^{+3} + 3~e^{-1}\)

The left side has our spoon connected to the negative end of the power source. As electrons move into the spoon, the \(Au^{+3}\) ions are attracted to the spoon, take electrons and become neutral gold atoms:

\(Au^{+3} + 3~e^{-1} \rightarrow Au \)

These atoms join the spoon and create a layer of gold (only a few atoms thick) on the surface. As a result, our spoon will appear to be gold (since you can only see the atoms on the surface) even though it is mostly tin.

 

Purifcation of Metals by Electrolysis

Galvanic cells and batteries use chemistry to produce electricity, but the opposite is also possible. We can use electricity to drive chemistry. This process is called electrolysis and one of the reasons we do it is to purify metals from mixed ore samples.

Imagine that you were digging in your backyard and found a sample of gold. You might think that you had struck pay-dirt, but if the gold is mixed with other metals, like zinc, then you might not be able to sell it. What you need to do then, is to purify the sample. One way to do so would be through electrolysis.

We’ll start with the setup below.

In this, we have a container holding a solution of gold III nitrate. We’ve shown only the gold III ions since the nitrate ions will not be involved in the chemistry. On the left side we have a wire of pure gold connected to the negative end of our power source. The right side is our impure gold sample connected to the positive end of the power source.

On the positive side, several things are happening. Zinc atoms at the surface can lose 2 electrons and become \(Zn^{+2}\) ions (\(Zn \rightarrow Zn^{+2} + 2~e^{-1}\)) and gold atoms on the surface can lose three electrons to become \(Au^{+3}\) ions (\(Au \rightarrow Au^{+3} + 3~e^{-1}\)). BOTH of these reactions can and do occur. Which is happening at any given moment is determined in large part by which atoms are at the surface of the metal sample.

The negative side is where the “magic” occurs. Initially, only gold III ions are present in solution, but very quickly, the solution will contain both gold III ions and zinc ions. BOTH of these ions are attracted to the negative wire and both of these ions can take in electrons and become neutral metal atoms. But, nature doesn’t look at these reactions equally. Here are the reactions and their reduction potentials:

\(Au^{+3} + 3~e^{-1} \rightarrow Au~~~~~~~~~~1.50~V\)

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn~~~~~~~~~~-0.76~V\)

We know that the more positive a voltage is, the more “natural” the reaction is, that it the more likely it is to occur. In this case there is no contest. Although both the gold and the zinc will react on the right (determined by physical arrangement of the ions) only the gold will form on the left. Thus, the impure metal on the right will slowly dissolve and the wire on the left will grow through the addition of pure gold.

Electrolysis - Separation of compounds into elements

Galvanic cells and batteries use chemistry to produce electricity, but the opposite is also possible. We can use electricity to drive chemistry. This process is called electrolysis and one of the reasons we do it is to break compounds down into their elements.

Electrolysis of a Pure Ionic Compound

Gold, as you probably know, is mined. This works because gold has a very low activity. As a result, it is found in pure form in nature. Sodium, near the top of the activity series, is never found in nature as a neutral element. We find compounds containing sodium inos (\(Na^{+1}\)), but never pure sodium. That means that in order to obtain pure sodium, we need to force the sodium ions to take back their electrons to become neutral.

\(Na^{+1} + e^{-1} \rightarrow Na\)

Let’s look at how this could be done. We’ll use NaCl (table salt) as the source of the sodium ions. At room temperature, NaCl is an ionic solid with the ions locked in place. To free them up to move, we’ll melt the salt. This requires a temperature above \(801^oC\) (nearly \(1500^oF\)). Once the salt has melted, the ions will be freer to move around.

We’ll then place two electrodes in the container with the molten salt.

We can then hook an external source of electricity (shown here as a dry cell).

The negative end of the dry cell is giving off electrons and the positive end is pulling them in. That means that the electrode hooked to the negative end ends up with a lot of excess electrons and the other electrode ends up positively charged (missing electrons).

Since the ions that make up NaCl are charged, they will be attracted to opposite electrodes as shown

So, what happens when those ions reach the electrodes?

When the \(Na^{+1}\) ions reach the electrode with the excess electrons, they take an electron and produce pure sodium:

\(Na^{+1} + e^{-1} \rightarrow Na\)

At the same time, when the \(Cl^{-1}\) ions reach the positive electrode, they lose an electron and become neutral. Of course, chlorine atoms ever occur alone, so this process happens to two chlorine ions to produce chlorine gas, as seen here:

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}\)

This reaction is, of course, not “natural” in the sense that the salt on your table never randomly falls apart into pure sodium and chlorine gas. That is why we needed the power source. Let’s take a closer look.

If we look up the two reactions that are occurring in our electrolytic cell on the Standard Reduction Potential Table, we find only the sodium reaction occurring as written.

\(Na^{+1} + e^{-1} \rightarrow Na~~~~~~~~~~~-2.71 V\)

The other reaction is written on the table as a reduction:

\( Cl_2 + 2~e^{-1}\rightarrow 2~Cl^{-1}~~~~~~~~~~1.36 V\)

However, since this reaction is occurring in the opposite direction (as an oxidation, rather than a reduction) we need to reverse the reaction, which will change the sign on the potential:

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}~~~~~~~~~~-1.36 V\)

Adding the two reactions together, will give us a potential of \(- 4.07 ~V\). The negative sign here tells us two things. First, that the reaction has to be pushed in order to occur, and second that the “push” required is more than 4.07 volts. In other words, to make this work, the dry cell at the top would need a voltage of at least 4.08 volts.

Electrolysis of an Aqueous Solution

You remember, from reading about solvation, that the ions in an ionic compound (like NaCl) are also free to move around when they are dissolved in water. So, you might wonder why we had to melt the table salt, rather than just dissolving it in water. It turns out, there is a really good reason.

When NaCl is dissolved in water and placed in an electrolytic cell (like the one pictured above), the same reactions are possible:

\(Na^{+1} + e^{-1} \rightarrow Na\)

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}\)

However, because water is present, there are several other reactions that are also possible. On the Standard Reduction Potential Table, there are three reactions we need to consider. Each of these reactions is relevant because they can occur with ONLY water as a reactant.

The first is about ⅓ of the way down the front side of the table:

\(2~H_2O + 2~e^{-1} \rightarrow H_2 + 2~OH^{-1}~~~~~~~~~~-0.83~V\)

The other two reactions are both on the back side of the table. Because water is a product of the reactions (as written on the table) we will need to flip each of them and then to change the sign on their voltages. The reactions, in the form we need them, are:

\(2~H_2O \rightarrow O_2 + 4~H^{+1} + 4~e^{-1}~~~~~~~~~~-1.23~V\)

\(2~H_2O \rightarrow H_2O_2 + 2~H^{+1} + 2~e^{-1}~~~~~~~~~~-1.77~V\)

All of that means that there are 5 possible reactions that can occur in our cell when we apply the external power. Those 5 reactions are:

\(Na^{+1} + e^{-1} \rightarrow Na~~~~~~~~~~-2.71~V\)

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}~~~~~~~~~~-1.36~V\)

\(2~H_2O + 2~e^{-1} \rightarrow H_2 + 2~OH^{-1}~~~~~~~~~~-0.83~V\)

\(2~H_2O \rightarrow O_2 + 4~H^{+1} + 4~e^{-1}~~~~~~~~~~-1.23~V\)

\(2~H_2O \rightarrow H_2O_2 + 2~H^{+1} + 2~e^{-1}~~~~~~~~~~-1.77~V\)

To determine which reactions actually DO occur, we need to do two things.

First, we need to separate the reactions into oxidations and reductions. This is because there MUST be one of each type of reaction for the process to work. That gives us:

\(Na^{+1} + e^{-1} \rightarrow Na~~~~~~~~~~-2.71~V\)

\(2~H_2O + 2~e^{-1} \rightarrow H_2 + 2~OH^{-1}~~~~~~~~~~-0.83~V\)

and

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}~~~~~~~~~~-1.36~V\)

\(2~H_2O \rightarrow O_2 + 4~H^{+1} + 4~e^{-1}~~~~~~~~~~-1.23~V\)

\(2~H_2O \rightarrow H_2O_2 + 2~H^{+1} + 2~e^{-1}~~~~~~~~~~-1.77~V\)

Secondly, we need to think about what those negative voltages mean. Galvanic cells always have a positive voltage. They are reactions that nature favors. Here, all of the reactions go “against nature.” However, nature is lazy. That is, the reaction that occurs will be the one which is the “least unnatural” - the one with the least negative voltage.

So, for the two reductions, the water reaction is less negative (-0.83 v -2.71) and for the three oxidations, the first water reaction is the least negative (-1.23 v -1.36 and -1.77). That means that when we electrolyze aqueous NaCl, we don’t get sodium (Na) and chlorine (\(Cl_2\)). Instead, we get hydrogen and oxygen.

\(2~H_2O + 2~e^{-1} \rightarrow H_2 + 2~OH^{-1}~~~~~~~~~~-0.83~V\)

\(2~H_2O \rightarrow O_2 + 4~H^{+1} + 4~e^{-1}~~~~~~~~~~-1.23~V\)

Fuel Cells

A fuel cell is a galvanic cell in which the reactants are fed in a constant stream. This essentially creates a battery that never runs out (as long as the fuel supply is not interrupted). There are a number of types of fuel cells and this is an area of active research, but to make the idea clear, we’ll look at a particular type of fuel cell - a Hydrogen Cell.

The Hydrogen Fuel Cell

Although the hydrogen fuel cell is the most commonly used fuel cell now, there are a number of ways that these can be built and used. We’ll look at an older type of fuel cell that uses an alkaline (basic) electrolyte solution.

The Structure

The fuel cell consists of a chamber into which the fuel (hydrogen) is pumped, a chamber into which oxygen is pumped, and two porous electrodes that surround a basic solution, like \(NaOH_{(aq)}\).

The Reactions

On the hydrogen side of the cell, the hydrogen loses electrons to become \(H^{+1}\) ions which combine with hydroxide ions that have moved through the electrode. The \(H^{+1}\) ions and the hydroxide then form water. Some of those water molecules will move through the electrode into the middle.

\(H_{2~(g)} + 2~OH^{-1}_{(aq)} \rightarrow 2~H_2O_{(l)} + 2~e^{-1}\)

On the oxygen side, the \(O_2\) takes 4 electrons and reacts with 2 water molecules that have moved through the electrode to make 4 hydroxide ions which will then move back into the middle.

\(O_{2~(g)} + 4~e^{-1} + 2~H_2O_{(l)} \rightarrow 4~OH^{-1}_{(aq)}\)

Remember that when we balance redox reactions, we need to match the number of electrons in the oxidation and reduction half-reactions. In this case, we need to double the hydrogen reaction. That gives us:

\(2~H_{2~(g)} + 4~OH^{-1}_{(aq)} \rightarrow 4~H_2O_{(l)} + 4~e^{-1}\)

\(O_{2~(g)} + 4~e^{-1} + 2~H_2O_{(l)} \rightarrow 4~OH^{-1}_{(aq)}\)

Adding those two together gives:

\(2~H_{2~(g)} + 4~OH^{-1}_{(aq)} + O_{2~(g)} + 4~e^{-1} + 2~H_2O_{(l)} \rightarrow 4~H_2O_{(l)} + 4~e^{-1} + 4~OH^{-1}_{(aq)}\)

then, cancelling out the species that appear on both sides of the equation, we get this:

\(2~H_{2~(g)} + O_{2~(g)} \rightarrow 2~H_2O_{(l)} \)

In simple terms, our cell is burning water, but doing so electrically.

A Note From History

You probably know that the Apollo 11 mission landed two men on the moon. What you may not know is that Apollo 12 also landed on the moon as did Apollo missions 14-17. The question then becomes, what about Apollo 13?

Apollo 13 was intended to land on the moon as the others did. However, about half-way to the moon, an oxygen tank exploded. The moon was no longer a viable option and it was not clear up until the moment the command module landed in the Pacific Ocean that the crew would survive. The mission was called a “successful failure”, in that they did not land on the moon, but all three astronauts aboard made it back to earth alive and well.

What is noteworthy about Apollo 13, from a chemistry perspective, is that the danger to the crew from a lack of oxygen was not biological. They had a separate supply of oxygen for breathing. The oxygen that was lost was there to run the hydrogen fuel cells needed for all of the electrical systems. The crew was forced to shut down the command module and live for days in the much smaller lunar module, which ran on more traditional batteries.

Lead Storage Batteries

Lead storage batteries, first created by Gaston Plante in 1859, were the first rechargeable battery created. Despite being a very old technology, they are still quite common because they are relatively inexpensive and can provide a large current (lots of electrons flowing at the same time). Their most common use today is the battery found in your gasoline burning car. (Electric cars use lithium-ion batteries.)

The Structure

Lead storage batteries take advantage of the fact that lead has three common charges: 0, +2 and +4, with the middle charge being more stable than the higher, +4 charge. Each cell in the battery is composed of two grids, or thin boxes made of conducting material. The first grid is filled with powdered lead, the other with lead oxide and the two are then placed in a bath of aqueous sulfuric acid (\(H_2SO_4\)).

Because sulfuric acid is diprotic and the first acid is strong, the solution consists primarily of \(HSO_4^{-1}\) ions and \(H^{+1}\) ions.

The battery is comprised of 6 of these cells, that is 12 grids alternating between \(Pb\) and \(PbO_2\).

http://www.vehiclemaintenanceandrepairs.com/car-battery-facts-101/

The Reactions (simplified)

In simplest terms, the reactions in a lead storage battery are simply an exchange of electrons between the +4 lead and the neutral lead.

\(Pb^{+4} + 2~e^{-1} \rightarrow Pb^{+2}\)

\(Pb \rightarrow Pb^{+2} + 2~e^{-1}\)

These two half-reactions give 2 V, so the 6 cells together provide 12V.

Recharging (simplified)

Recharging the battery requires us to force the two reactions above to occur backwards:

\(Pb^{+2} \rightarrow Pb^{+4} + 2~e^{-1}\)

\(Pb^{+2} + 2~e^{-1} \rightarrow Pb \)

We do this by applying MORE than 12V in the “wrong” direction. In your car, this is done by the alternator, which is essentially an electric power plant that runs off of the engine.

In fact, the alternator is what provides the electricity to all of the parts of your car (A/C, radio, lights, etc) when the engine is running. Only when the engine is off, do the electrical systems of your car pull electricity from the battery. That’s why your fuel efficiency is lower in the summer when you run the A/C than it is in the winter. The engine has to burn extra gas to make the electricity to run the A/C.

The Problem with Recharging

When we recharge the battery, we force the \(Pb^{+2}\) ions to either take two electrons and become \(Pb\) or to give up two electrons and become \(Pb^{+4}\). However, we have a problem. Lead II ions (\(Pb^{+2}\)) are somewhat soluble in water. That means that these ions could dissolve into the electrolyte solution in the battery. If they did that, they would leave their positions in the grids. Then, when we try to force those reactions to go backwards, those ions will not be there to react.

We needed to find a way to make sure that the ions were still locked in the grids when the alternator kicked in. The solution was to make the \(Pb^{+2}\) ions insoluble. This is done by having sulfate (\(SO_4^{-2}\)) ions present in the solution. Lead II sulfate (\(PbSO_4\)) is quite insoluble, so when the \(Pb^{+2}\) ions are formed, they react with the sulfate ions, make \(PbSO_4\) and stay put, waiting to be “recharged.”

The Reactions - Complete

Once we understood the need to make the \(Pb^{+2}\) ions insoluble, we addressed this need by using sulfuric acid. The resulting (complete) reactions are these:

\(Pb_{(s)} + HSO_{4~(aq)}^{-1} \rightarrow PbSO_{4~(s)} + H^{+1}_{(aq)} + 2~e^{-1}\)

\(PbO_{2~(s)} + HSO_{4~(aq)}^{-1} + 3~H^{+1}_{(aq)} + 2~e^{-1} \rightarrow PbSO_{4~(s)} + 2~H_2O_{(l)}\)

Lithium-Ion Cells and Batteries

Lithium ion cells and batteries are the newest and most common form of rechargeable battery on the market today. Although lithium-ion cells (in AAA, to D) are available for things like a flashlight, Their most common use is for the rechargeable things in our lives for which we never see or touch the batteries. That includes things like cell phones, laptops and electric cars.

The technology of lithium-ion batteries is a huge area of research and, although what follows is only a surface explanation of what is happening, it may already be out of date by the time you read this.

A lithium-ion cell contains a positive end and a negative end which are connected by an electrolyte mixture that allows lithium ions to move from one end to the other through a separator that does not allow the flow of electrons.

  

The positive end contains one of several compounds such as \(LiCoO_2\). Because this is an ionic compound and because lithium ions are very soluble, the connection between the \(CoO_2^{-1}\) and the \(Li^{+1}\) ion is loose and tentative.

The negative end of the cell has a matrix that can hold lithium ions. This matrix is often made of graphite.

An uncharged (completely used up) lithium-ion cell looks like this:

Charging a lithium-ion battery involves connecting the battery to a power source that can create a large negative charge on the graphite end of the battery and a positive charge on the Co end. This is done by moving electrons through the charger from the Co end to the graphite end.

As a result of this, the \(Li^{+1}\) ions are attracted to the graphite end. They move through the separator and lodge in the graphite matrix.

The result of this process is a charged battery. The power is stored in the potential energy found between the \(Li^{+1}\) ions and the \(CoO_2^{-1}\) ions.

When the battery is placed in an open circuit (when your phone is on, for instance) the lithium ions can flow back to the Co side of the battery which causes electrons to move through the external circuit (your phone).

Dry Cells

Dry cells are the things that most of us commonly call batteries. Specifically those tubes with the bump on the end that come in sizes ranging from AAA to D. What follows is a rough description of the structure and function of a dry cell. It is worth noting that companies are always researching and improving these cells, so what is inside your flashlight may be somewhat different from what is shown here. 

Making things even more complicated, there have been a number of different types of dry cells produced and each type is built and functions slightly differently. We will look at two of the common types here.

The Zinc-Carbon Dry Cell

The Reactions

The reactions that occur in a zinc-carbon cell are the following:

\(Zn \rightarrow Zn^{+2} + 2~e^{-1}\)

\(2~MnO_2 + 2~NH_4Cl + H_2O + 2~e^{-1} \rightarrow Mn_2O_3 + 2~NH_4OH + 2~Cl^{-1}\)

The second reaction mixture is a paste with just enough water to allow ions to move, but not enough to make it "wet" or to risk spilling.

In addition, there is a secondary reaction that the zinc undergoes which is explained below.

The Construction

The basic construction of the zinc-carbon dry cell is shown below.

Electrons are pulled in through the bump on the top and are passed down through the graphite electrode where the \(MnO_2\) takes them.

Those electrons are taken (eventually) from the zinc can that comprises the outer shell.

Two Problems

There are two problems with this that need to be solved. The first is that we need to make sure that the electron theft reaction ONLY occurs by passing electrons through our flashlight (or calculator, etc) and not through direct contact. Up until now we have managed that by putting the half-reactions in separate containers. Here, however, the two half-reactions are in the same container.

The second problem is that we have a charge issue. Remember that in our first galvanic cell, we needed a salt bridge to allow the current to continue flowing. The same issue occurs here. As the zinc reacts, \(Zn^{+2}\) ions collect at the outside. This will make it less and less likely that electrons will move away from this positive charge. At the same time, as electrons enter the middle of the cell, the paste becomes negative which will discourage additional electrons from entering.

Solving Those Problems

The solution to both problems is the porous separator.

The porous separator is a “can” of heavy duty paper that fits just inside the router zinc can. The \(MnO_2\) paste is held inside this paper can and prevented from touching the zinc directly. That solves the first problem. The charge problem is solved because the \(Zn^{+2}\) ions move in through the pores in the paper. They do this because they are attracted to the increasing negative charge inside. When they move, they solve both charge problems. The outer can stops being positive and the positive ions balance the increasing negative charge inside.

Once inside, the \(Zn^{+2}\) ions react with chloride ions (\(Cl^{-1}\)) to form \(ZnCl_2\). This is the secondary Zinc reaction mentioned above and it stabilizes the zinc ions inside the paste.

The Alkaline Dry Cell

Alkaline cells use similar chemistry to the classic zinc-carbon cell, but the paste is basic (alkaline) rather than acidic.

The Reactions

The half-reactions that occur in an alkaline cell are these:

\(Zn + 2~OH^{-1} \rightarrow ZnO + H_2O + 2~e^{-1}\)

\(2~MnO_2 + H_2O + 2~e^{-1}\rightarrow Mn_2O_3 + 2~OH^{-1}\)

In this construction, the Zn is in powder form and mixed in a paste that contains hydroxide. Since both reactants are now in paste form, the structure of the cell is somewhat different.

Here is a diagram (from Wikipedia) showing the structure of an alkaline dry cell.

Notice that, even though the zinc is no longer on the outside, it is still electrically connected to the flat bottom of the cell.

Alkaline cells are generally preferred to zinc-carbon cells these days because they have a higher energy density (you can get more life out of them) and a longer shelf life.

Cells and Batteries

Let’s take a moment to think about the word cell. As we have discussed, a galvanic cell is a pair of half-reactions connected in such a way that the movement of electrons created can be used. The important piece of that definition is the “a” . In other words, a cell is a SINGLE pair of half-reactions. 

That means that if we choose to match the strongest and weakest half-reactions, the highest potential chemistry can give us is 5.92 V, by using the two half-reactions here:

\(Li \rightarrow Li^{+1} + e^{-1}\)

\(F_2 + 2~e^{-1} \rightarrow 2~F^{-1}\)

That raises the question, how do we make 9V batteries?

The answer to that question is that a battery is simply more than one cell connected in series.

Here’s what that means. This is a cell. We may call it a battery, but it isn’t. It’s a cell containing a single pair of half-reactions producing 1.5 V. That's why one of the companies that makes lots of these is called \(Duracel^®\).

On the other hand, this is a battery.

Note that there are two cells, stacked so that the positive end of one, touches the negative end of the other. Connecting the cells in this way, adds their potential, so this flashlight runs on 3V (1.5V + 1.5V).

This distinction is is important when you are learning about galvanic cells, electrolytic cells, hydrogen fuel cells and more although, admittedly, most of us just call them all batteries when we're out in the "real world."

Electrochemical Potential

The last piece to understand about galvanic cells is exactly WHY they work and how cells differ from each other. 

Why cells work

To understand how cells really work, we need to remember that nothing gives away electrons. No positive nucleus has ever given away negative electrons.

What we really mean when we say that something gives away an electron is that it doesn’t fight very hard when something else tries to take an electron. In other words, sodium atoms do NOT give away electrons, but they also don’t hold their electrons very strongly, so they are likely to lose any time another element pulls on its electron.

Comparing what we commonly say and the truth side-by-side will give you a pretty clear picture of why we commonly (and intentionally) misspeak.

However, if we are going to understand galvanic cells in a meaningful way we need to consider the truth. For this we’ll consider these two half-reactions:

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn\)

\(Cu^{+2} + 2~e^{-1} \rightarrow Cu\)

These two reactions are nearly identical. Both involve a +2 ion, pulling in two electrons to make a neutral atom.

If these two reactions are placed in the two beakers of a galvanic cell they will be pitted against each other - the copper trying to take electrons from the zinc and the zinc trying to take electrons from the copper. In simple terms, we will have created a chemical tug-of-war between the two elements for electrons.

Since different atoms are different, we can expect that one of these elements will pull more strongly than the other and will “win” the tug-of-war.

In this particular case, the copper pulls more strongly on the electrons and therefore it wins. That means that the zinc loses.

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn\) -- Loser

\(Cu^{+2} + 2~e^{-1} \rightarrow Cu\) -- Winner

More specifically, instead of \(Zn^{+2}\) ions taking in electrons and making Zn atoms, the reaction will occur the other direction - Zn atoms will lose electrons and create \(Zn^{+2}\) ions. So, the two half-reactions that will actually occur are these:

\( Zn \rightarrow Zn^{+2} + 2~e^{-1}\)

\(Cu^{+2} + 2~e^{-1} \rightarrow Cu\)

Determining Who Wins

The question that may have occurred to you is “How do I know which reaction ‘wins’?”

The answer is actually pretty simple. Chemists, a long time ago, ran a massive tug-of-war tournament between LOTS of half-reactions and ranked them. This ranked list is called a Standard Reduction Potential Table.

To fully understand the table, we need to understand the title. We’ll do that in reverse order.

Table - this is simply an organized list. In this case the organization is based on strength (rather than the alphabet or some other feature). That can make it more difficult to find the half-reaction you are looking for, but the utility of having them in order outweighs those considerations.

Potential - This is the electrical measurement commonly called voltage. The physics definition for potential (amount of work needed to move a unit charge from a reference point to a specific point against an electric field - Britannica.com) is not terribly helpful for our understanding, so let’s come up with our own.

Potential can be thought of as the force with which electrons are moved through a wire. That’s not a perfect definition, and it probably causes pain in your physics teacher’s brain, but it will help us to understand that this is all about.

In a galvanic cell, each half-reaction is pulling on electrons. The potential recorded on the table is a measure of how hard the half-reaction is pulling. The reaction with the more positive potential will “win” and the reaction with the less positive potential will lose. The difference between the two numbers is the potential (or voltage) that will be measured in the wire.

Reduction - All half-reactions on the table are written as a reduction, that is all of them are taking in electrons. Just as in a tug-of-war, each is trying to take the electrons from the other. Even the losers are trying, just not as effectively.

Standard - All of the measurements on the table were made under “normal” or standard conditions. That means that all of the measurements were taken at 298 K (\(25^oC\)). In addition, all ions are present at a concentration of 1.00 M, gases are present at 1.00 atm, and elements and compounds are in their standard states.

This matters, because if you have a huge number of positive ions, even relatively weak ions, they will end up pulling harder than a tiny number of other ions, even if they are stronger. In simple terms 10,000 kindergarten students will pull harder than one professional football player.

Recording all of these values at standard conditions makes the comparison fair and meaningful.

So, the Standard Reduction Potential Table is simply a list of half-reactions showing how strongly they pull on electrons listed from weakest to strongest.

Going back to the reactions from the top of the page, now listed with their potentials:

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn~~~~~~~E^{o} = -0.76V\)

\(Cu^{+2} + 2~e^{-1} \rightarrow Cu~~~~~~~E^{o} =+0.34V\)

From this we can determine two things: 1) copper wins, since it has the more positive potential, and 2) the potential on the galvanic cell that these two half-reactions would make is 1.10 V (the difference between the two values).

Let’s try one other example. Let’s match our zinc reaction to another “competitor.”

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn~~~~~~~E^{o} = -0.76V\)

\(Sr^{+2} + 2~e^{-1} \rightarrow Sr~~~~~~~E^{o} = -2.89V\)

In this case, the zinc is the winner with the more positive potential and the potential on the cell would be 2.13V, with the reactions being:

\(Zn^{+2} + 2~e^{-1} \rightarrow Zn\)

\(Sr\rightarrow Sr^{+2} + 2~e^{-1}\)

Galvanic Cell Construction and Notation

Electrodes

Let’s look at one more aspect of construction for galvanic cells - the electrode (wire) that sticks into the beaker. In the cell we discussed here (pictured below), the wires are simply a way to get electrons from one beaker to the other.

Our only requirements in that case are that the wire not corrode, that it not react with the solutions in the beaker and that it conducts well. 

Given those requirements, the ideal choice is platinum. Most diagrams of galvanic cells (that don’t fall into the category described below) will use platinum for the electrodes. Of course, platinum is really expensive, so there are alternatives.

A second choice would be graphite. Graphite is generally non-reactive and conducts well. Of course it is also brittle. In a high school lab, we often use copper, which although it corrodes faster than platinum is a heck of a lot cheaper.

When the electrode matters - a special category of reaction

Let’s look at another cell that uses these two half-reactions:

\(Zn \rightarrow Zn^{+2} + 2~e^{-1}\)

\(Cu^{+2} + 2~e^{-1} \rightarrow Cu\)

Before, when we needed to put all of the parts of each half-reaction into each beaker, we simply added them to the solution like this.

But, now we have a problem. Zn (without a charge) cannot be added to a solution, since it is a solid metal. The same is true for the Cu on the right side of the second half-reaction.

The solution is simple though. We simply make the electrode out of that metal. So the set-up would begin like this

and a galvanic cell made from these reactions would look like this.

So, as a general rule, if a metal (solid, uncharged, and pure) is part of the half-reaction, then make the electrode out of that metal. Otherwise, make it out of Pt (or graphite).

Convention and Galvanic Cell Notation

A word about “convention”

Often in science and mathematics, things are done “by convention.” This means that everyone agrees on an arbitrary choice to make communication easier. For instance, there is no reason that North should be at the top of maps, but virtually all maps are made that way.

When dealing with galvanic cells, the convention is to put the oxidation half-reaction on the left side and the reduction reaction on the right. Hopefully you can see that this is meaningless. If your teacher has a galvanic cell on their front demonstration table, then the reaction that is on your left is on their right. Chemistry doesn’t care.

This convention only really matters when we decide to use a short-cut to communicate about that cell, as discussed below.

Galvanic Cell Notation

• Oxidation is always listed first
• The oxidation and reduction are separated by the symbol || which represents the salt bridge
• Within each half reaction, each species is separated by a comma (,)
• States of matter are separated by a |

So, the galvanic cell at the top would be represented by the notation

\(Pt|Fe^{+2},~Fe^{+3}||MnO_4^{-1},~H^{+1},~Mn^{+2}|Pt\)

and the cell at the bottom would be noted this way:

\(Zn|Zn^{+2}||Cu^{+2}|Cu\)

Now we need to look at Electro-chemical Potential and the effect of switching reactions.