Galvanic Cells

From our phones and computer to the battery in our car electrochemistry powers most of the world around us. While a multitude of different types of electrochemical cells exist the most foundational of them all is the galvanic cell.

Electrochem Basics

Before we jump into the specifics of galvanic cells let’s look at the electrochemistry basics that govern all of the cells we will explore in this section.

First and foremost, electrochemical reactions are just separated REDOX reactions with a wire in between. This makes the direct transfer of electrons from one reactant to the other impossible since they aren’t in contact. As a result, transferred electrons must flow through the wire. While this might not seem that important, it is. Since it provides us a way to harness the current generated by the REDOX reaction and power phones, watches, etc.

As with any REDOX reaction, it has one reactant that is oxidized and another that is reduced. In electrochemical cells, the same is true and occurs at electrodes, pieces of metal that act as the sites of oxidation and reduction. Each electrode has a special name depending on the process that occurs. When the electrode is the site of reduction we call it the cathode and when it is the site of oxidation we call it the anode.

You can remember this using the mnemonic red cat an ox which stands for reduction is cathode anode is oxidation.

Lastly, we will discuss the electromotive force (EMF) of these cells or how much voltage they can produce. This is what we are used to dealing with when discussing the voltage of a circuit and they are synonymous concepts. We just discuss voltage using a different lens in physics.

The Galvanic Cell

Most of the batteries on supermarket and convenience store shelves are galvanic (AKA voltaic) cells. These cells produce energy by harnessing spontaneous REDOX reactions. As such, they have negative ∆G values, meaning they release energy to the environment, and positive EMF values, meaning they produce a voltage.

To see exactly how this works let’s take a look at a galvanic cell and apply our basic electrochemical terminology to describe what is going on.

In the galvanic cell above two electrodes, one zinc (Zn) and one copper (Cu) are connected together with a wire and a voltmeter. Looking at the Zn electrode we can see that electrons are leaving this side and traveling to the Cu electrode. Therefore, the Zn electrode is the anode as it is the site of oxidation or electron loss and the Cu electrode is the cathode since it is the site of reduction or electron gain. Regardless of what cell we talk about electrons will always travel from the anode to the cathode. Since the electrons are attracted to the cathode it is labeled as the positive terminal since opposites attract and vice versa for the anode.

Here the voltmeter also shows us the EMF of the cell at +1.21 V or it produces 1.21 J of energy for every coulomb of charge that it moves. Since current and electron flow are opposite one another current will flow from the cathode to the anode. Additionally, since the electrons are attracted to the cathode side it means that the Cu electrode is positively charged.

As the reaction progresses the Zn metal electrode will erode as it is converted into Zn2+ which is soluble while Cu2+ will be deposited on the metal electrode as it is converted into solid copper.

As this occurs another interesting phenomenon happens the anode solution becomes more and more positive as Zn2+ dissolves into the solution while the cathode solution becomes more and more negative as Cu2+ leaves the solution and deposits on the copper electrode.

Over time this will cause the reaction to stop since the electrons will become more and more attracted to the positive charge of the anode. In order to combat this tendency and keep the reaction going a salt bridge donates anions and cations to each side in order to neutralize the building charge.

In summary, a galvanic cell is a spontaneous electrochemical cell with a negative ∆G and a positive V. In the cell oxidation occurs at the anode and reduction at the cathode with electrons flowing from the anode to the cathode. Since current flow and electron flow are opposites current will flow from the cathode to the anode.

In addition to representing these cells in pictures, chemists use shorthand notation or a cell diagram to display electrochemical cells.

For our Zn and Cu cell the notation is as follows:

Zn (s) | Zn2+ (0.5M) || Cu2+ (0.5 M)| Cu (s)

In the notation | indicates the boundary between the solid electrode and its soluble ionic component. While || indicates the separation between the two beakers. By convention, the anode is always listed first and the cathode second which you can remember since it follows the order of the alphabet. Lastly, the reactions are always listed from products to reactants on both the anode and cathode side.

Reduction Potentials in the Galvanic Cell

All of these notations are great, but why did the electrons travel from zinc to copper in the first place? In short reduction potentials (ERed) or how badly an element wants to be reduced. The higher the reduction potential the more the elements wants to be reduced.

Looking at both zinc’s and copper’s ERed it becomes clear why the electrons travel from zinc to copper. Here copper has a larger reduction potential and as a result, gains electrons from zinc who is more than happy to give up its electrons.

Alternatively, we could flip the sign of an element’s reduction potential around and look at their oxidation potential (Eox) or how much an element wants to be oxidized. Therefore as an element’s reduction potential goes up its oxidation potential goes down and vice versa. In this light zinc has a much higher oxidation potential and as a result, freely gives up any electrons it might have to copper who really wants them.

In a biological context, this is why electrons in the electron transport chain transfer from NADH and FADH2 all the way to the terminal electron acceptor oxygen. Oxygen has the highest reduction potential and therefore wants the electrons the most.

Calculating Ecell

We can use these reduction and oxidation potentials to predict how much voltage will be generated by the cell as a whole. While we will go over the exact equation to use I find that it is easier to remember the equation if we understand how it is setup.

To begin we will think of reduction potentials as forces where a positive sign indicates a pulling force and a negative sign indicates a pushing force. Therefore elements with negative reduction potentials such as zinc push electrons away from themselves while copper with its positive reduction potential pulls electrons towards itself.

If we draw these out as vectors we can see that both arrows point towards copper and sum to 1.21 V which is the EMF or Ecell of the reaction. What happens if we have two elements with positive signs or two with negative.

The same thing except here the vectors will be opposing one another with the larger vector ultimately winning.

This leads us to our equation:

Since the cathode is the site of reduction and the anode is the site of oxidation we can transform this equation to apply to electrochemical cells as follows:

\[ E_{cell}=E_{cathode}-E{anode}

To see this equation in action let’s use the electron transport chain as an example and calculate the total overall voltage produced by the electron transport chain. Since the starting point of the electron transport chain is NADH or FADH2 we will pull the voltage value for either from the table below.

ETC ComponentEcell (V)
NAD+ → NADH -0.32
CoQ → CoQH2+0.045
Cytochrome C (Fe3+) → Cytochrome C (Fe2+)+0.25
O2 → H2O+0.81

Only NADH is presented so we will start off with a -0.32 V reduction potential from there we will need to look at the terminal electron acceptor, oxygen. Oxygen has a reduction potential equal to +0.81 V.

Now we need to determine which one will be reduced and which one will be oxidized so we can plug them into our equation. Since oxygen has a higher reduction potential it will be reduced and NADH will be oxidized. Now we can plug both numbers into our equation and solve for the overall voltage produced by the electron transport chain.

\[E_{cell}=E_{red}-E_{ox} /to E_{cell}= 0.81V \minus (-0.32V) \to 1.13V

So from the first step to the last, the electron transport chain generated 1.13V throughout the entire chain. We still have more to discuss when it comes to galvanic cells but we will look at that later when we discuss thermodynamics and equilibria of electrochemical cells. For now let’s move onto electrolytic and concentration cells.