Practice problems at the bottom!

The electrochemistry unit at CU begins with redox reactions. Make sure you understand the basics of this before continuing through this article.

A voltaic cell, a.k.a. a galvanic cell, requires several parts. In official terms, you’ll need a cathode, an anode, and a conductive substance (usually wire) and salt bridge connecting the two. Less formally, you’ll need: 

  • 2 pieces of solid metal

  • 2 solutions containing the cationic form of the solid metal that is suspended in the solution

  • A wire

  • A salt-bridge

  • A voltmeter to measure the amount of electricity flowing (optional)

When we put all these together, we end up with a schematic that looks like this:

Here we have two pieces of solid metal (dark blue and dark green boxes), two solutions containing cations of the same metal (light blue and light green), a salt bridge (orange), and a wire with a voltmeter. For this example, let’s say blue is Copper and green is Zinc, and the salt bridge will contain NaCl. In the light blue solution, we will have copper sulfate (\(CuSO_4\)) and the light green zinc sulfate (\(ZnSO_4\)). As these are both soluble, they will dissociate into \(Cu^{2+}\) and \(SO_4^{2-}\) as well as \(Zn^{2+}\) and \(SO_4^{2-}\), respectively. Updating our schematic, we have:

Before we get into why, let’s talk about what is happening conceptually. Once we connect the circuit, electrons will start flowing. The solid zinc is going to begin losing electrons, which will flow up the wire and left, through the voltmeter, and into the solid copper. As the piece of copper begins gaining electrons (buzzword: gaining electrons means what?), it will become more negative, thus attracting the positive copper ions. Each ion will gain two electrons, and we’ll be left with solid copper. If we were to mass the solid copper before this process begins, and at intervals throughout, we would find that it would gradually increase. This gained mass is literally just more solid copper forming on the initial piece, the source being the copper ions in solution. If we look at the other side, as electrons leave the solid zinc, the atoms will become cations. Specifically, because the zinc cation has a +2 charge, every 2 electrons that leave Zn will form 1 zinc cation, which will “drain” into the solution (losing electrons means….). As this process continues over time, we will see four things:

  1. The copper metal gaining mass

  2. The zinc metal losing mass

  3. The copper sulfate solution loses copper ions

  4. The zinc sulfate solution gains zinc ions

Hopefully you recognized the redox reaction occurring with solid and ionic metals gaining and losing electrons. On one side oxidation is taking place, and the other side is being reduced. Which is which? Remembering OIL RIG (or Leo the Lion, see redox reactions for more info), we know that the oxidized side loses electrons. In our example, this is the zinc side, which means that the copper is being reduced and gaining electrons. From this information, we can determine which side is the cathode and which is the anode. The cathode is the side that is reduced, and the anode is the side that is oxidized. You’ll have to memorize these terms. In the past I’ve heard people use vowels versus consonants (A goes with O, C goes with R). Usually I just yell “AAAYYOOOHH” in my head and this does the job.  Typically, the anode is drawn on the left with electrons flowing to the right, however professors will often switch it up to make things more challenging. Either way, this is where the shorthand notation comes from. Starting with the solid anode:

Solid anode | cationic anode || cationic cathode | solid cathode

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

Officially, we should include the state and molar concentrations of the solutions. For example: 

Zn(s) | \(Zn^{2+}\) (1.0M) | | \(Cu^{2+}\) (0.5M) | Cu(s)

Note: these concentrations are arbitrary in this example as I didn’t declare in the schematic what the concentration was. On your test, this will most likely be included as you’ll need them for some other types of questions. 

At this point we’ve covered all components except the salt bridge. This is an addition that allows the reaction to continue longer. For example, if we didn’t have the salt bridge, the zinc sulfate solution would continue gaining zinc cations. The more ions we get, the more cations we have in a confined space. These will begin to repel each other, causing the reaction to lose steam until it can’t force anymore solid zinc into solution. The same opposite is true for the copper side. Copper cations will continue decreasing, eventually creating a much greater concentration of sulfate anions than copper cations. If we didn’t have the salt bridge, this reaction would slow down, slow down some more then eventually stop altogether. However, by introducing a salt bridge, we can counterbalance the buildup of charges with the opposite ion from the salt. As positive zinc ions increase, negative chloride ions will flow into the solution, maintaining a neutral solution. At the same time, positive sodium ions will flow into the copper sulfate solution, replacing the now missing positive copper ions as the solution cation. 

Now you may be wondering how we know that the electrons will flow from the zinc solid to the copper solid and not the other way around. This is determined using the table of standard reduction potentials and the half reactions of the redox reaction taking place. Looking at this table we can see that \(\mathrm{Cu}^{2+}+2 e^- \rightarrow \mathrm{Cu}(\mathrm{s})\) has a value of +0.34, and \(\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s})\) has a value of -0.76. This means that it requires 0.34 or -0.76 volts, respectively, to reduce the cation to a metal. For the redox reaction to occur, we need one to be reduced and one to be oxidized. To find the oxidized reaction, we just reverse the reduction reaction, and flip the sign of the standard reduction potential, now it becomes the standard oxidation potential. For the reaction to be spontaneous, thus NOT require an outside source of power, our total cell potential must be positive. Using this information, we can determine that zinc must be oxidized and copper reduced. Flipping the zinc equation so Zn(s) is oxidized, and switching the sign of the potential to +0.76, we can then add the equations and cell potentials:

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) \(E^o_{cell} = 0.34\)

\(\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}+2e^{-}\) \(E^o_{cell} = 0.76\)

\(\mathrm{Cu}^{2+}+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}\) \(E^o_{cell} = 1.10 \)

We should balance this equation if needed. When doing this, make sure the number of electrons on either side of the equation are equivalent.  By changing the stoichiometry of the equation, the cell potential does not change. In other words, if we need to multiply all parts of the reduction half-reaction by 2, we will not multiply the \(E^o_{cell}\) by 2, it just stays as is. 

This is a pretty straight forward process for standard conditions, that is the solutions are at 1.0M, pressure is at 1atm, etc. To calculate the non-standard conditions, we must use a slightly more complicated equation: 

$$\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\mathrm{o}}_\mathrm{cell}-(\mathrm{RT} / \mathrm{nF}) \ln \mathrm{Q}$$

Where \(E^o_{cell}\) is the potential at standard conditions determined using the standard reduction potential table, R is the ideal gas constant (8.31 J/molK), T is temperature in Kelvin, n is the moles of electrons transferring, F is Faraday’s constant (96,485C/mol e-), and Q is the reaction quotient. If the temperature is 25C, the equation can be simplified further to:

$$\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\mathrm{o}}_{cell}-(0.0257 / \mathrm{n}) \ln \mathrm{Q}$$

$$\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\mathrm{o}}_{cell}-(0.0592 / \mathrm{n}) \log \mathrm{Q}$$

I prefer to think of this style of equation as the new, non-standard \(E_{cell}\) is equal to the \(E_{cell}\) at standard conditions, plus an adjustment for the non-standard condition. The reaction quotient, Q, is calculated using the following equation:

For the reaction, \(a A+b B \rightarrow c C+d D\),

$$Q=\frac{[C]^{c}[D]^{d}}{[A]^{a} [B ]^{b}}$$

See below for a step by step example problem. 

Could you force the reaction to go the other way and make the non-spontaneous reaction occur? Yes, by adding a power source to the circuit, you can transform your voltaic cell into an electrolytic cell, and the power source will force the electrons to flow the opposite direction. The same process occurs, however now the circuit will “flow” in the direction yielding a negative cell potential. Spontaneity is a characteristic of a voltaic cell.

You’ll see questions stemming from any point in this process and asking about any other point. However, by going through each step, you can double check that you didn’t make any careless mistakes trying to go quickly. Here are some question types you’ll run into and suggestions for how to approach them. 

Conceptual questions

These will usually be in the “which of the following is true” or “which isn’t true” format with possible answers being common misconceptions or half true statements. Make sure when you’re answering these you read the whole answer, and I would strongly suggest you draw out what the answers are saying. If it asks about a balanced equation, write out the equation. If it’s asking which way the electrons are flowing, complete the diagram and draw the arrow the electrons are flowing. These questions are intentionally not intuitive and it’s very easy to get tripped up, but most can be solved going through the steps above and determining each piece of the puzzle.

Equation questions

These questions will either have you simply calculating the cell potential at standard conditions, or trickier questions will change some conditions such as temperature or concentrations. For example, a question may say: 

What is the \(E_{cell}\) for a voltaic cell made with 0.3M copper and 1.2M silver?

For starters, let’s write down the equation we’ll need. Since there is no change to temperature, we can use the simplified version:

$$\mathrm{E}_{\text { cell }}=\mathrm{E^o}_{\text { cell }}-(0.0592 / \mathrm{n}) \log \mathrm{Q}$$

Now we need \(E^o_{cell}\).

According to the table of standard reduction potentials, we have: 

$$1 e^{-}+A g^{+}(a q) \rightarrow A g(s) E^{\circ}_{cell}=0.80$$

$$\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \quad E^o _{cell}=0.34$$

Which one gets oxidized? In order to maintain a spontaneous reaction, we must flip the copper half-reaction.

$$\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}-E^o _{cell}=-0.34$$

And adding up our cell potentials we get +0.46. If we instead flipped the silver reaction, our cell potential would be -0.46, non-spontaneous. 

Updating our equation: 

$$\mathrm{E}_{\mathrm{cell}}=0.46-(0.0592 / \mathrm{n}) \log \mathrm{Q}$$

Next, we should find out what n is. Let’s combine our half reactions: 

$$1 e^{-}+A g^{+}(a q) \rightarrow A g(s)$$

$$\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}$$

In reaction two we have 2 electrons per copper, and reaction one we have 1. We must double the first reaction to balance the equation:

$$2 e^{-}+2 \operatorname{Ag}^{+}(a q)+C u(s) \rightarrow 2 \operatorname{Ag}(s)+C u^{2+}+2 e^-$$

We have 2 electrons transferring, so n = 2. Simplifying our equation fully, we can see the electrons will cancel out and we’re left with:

$$2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Cu}^{2+}$$

And our \(E_{cell}\) equation looks like:

$$E_{\text { cell }}=0.46-(0.0592 / 2) \log Q=0.46-(0.0296) \log Q$$

Finally, we need to calculate Q. To do this, all we do is plug the concentrations into either the top or bottom, and raise this value to its coefficient. One thing to remember is that on aqueous and gaseous components are included in the Q equation:

\(aA = 2Ag^+\)

\(bB\) is not included, copper is solid

\(cC\) is not included, silver is a solid

\(dD = 1Cu^{2+}\)

\(Q=\frac{[0.3 M]}{[1.2 M]^{2}}\)

\(Q = 0.2083\)

Filling this into the rest of the equation:

$$\mathrm{E}_{\text { cell }}=0.46-(0.0296) \log (0.2083)$$

$$\mathrm{E}_{\text { cell }}=0.46-(0.0296)(-0.681)$$

$$\mathrm{E}_{\text { cell }}=0.46-(0.0296)(-0.681)$$

$$\mathrm{E}_{\text { cell }}=0.46-(-0.0201)$$

$$\mathrm{E}_{\text { cell }}=0.46+0.0201$$

$$\mathrm{E}_{\text { cell }}=0.48$$

And there’s your answer. 

So what does this answer mean? This is telling us that at these concentrations, the reaction is “more spontaneous.” Extrapolating from this concept, we could potentially find ourselves in a situation where certain conditions may push a cell potential negative, causing it to be non-spontaneous. 

Concept Question: At what cell potential will there be no reaction, spontaneous or non-spontaneous? 

Answer: When \(\mathrm{E}_{\text { cell }}\) is positive, the reaction is spontaneous. When \(\mathrm{E}_{\text { cell }}\) is negative, the reaction is non-spontaneous. When \(\mathrm{E}_{\text { cell }}\) is zero, there is no reaction, as there is no force one way or the other.