Galvanic Cells

Why Redox Reactions Are So Important

Let’s think about the redox reaction here:

\(MnO_4^{-1} + 8~H^{+1} + 5~Fe^{+2} \rightarrow Mn^{+2} + 4~H_2O + 5~Fe^{+3}\)

Remember that we balanced this by separating it into two half-reactions and balancing those half-reactions separately. Here are the balanced half-reactions before we matched the number of electrons:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

The fact that this reaction can be broken into two separate halves tells us two very important, and related things.

1. Redox reactions can violate the collision theory
2. The manganese reaction could happen with ANY reaction that fed it electrons, not just the iron reaction here, and vice versa.

Let’s examine those two issues, one at a time.

Redox reactions can violate the collision theory

Remember that the collision theory (explained fully here) states that “In order to react, particles must collide with enough energy and with correct orientation.”

We’re going to focus our discussion here on the first part of that statement; “particles must collide”. As explained here, this is because in order to form the bonds necessary in the products, the atoms must be together.

What makes redox reactions different is that NONE of the elements from the first half-reaction are required to form the products in the second half-reaction and vice versa. That means that the need for collision has been eliminated.

In fact the only thing that is transferred from one half reaction to the other is electrons.

The Manganese reaction could happen with ANY reaction

What this tells us is that the two half-reactions that make up an overall redox reaction can be swapped and exchanged in a virtually infinite number of reactions. As long as one half-reaction takes electrons and the other loses them, the pairing will work.

Putting These Ideas Together

The product of these ideas has two different names. It is sometimes called a Galvanic Cell, after Luigi Galvani, who first demonstrated the idea by using redox reactions to cause contraction in a frog’s leg. Alternatively, it is called a Voltaic Cell, named after Allesandro Volta, who demonstrated the same idea in an inorganic setting with a force meter. Neither man fully understood what was happening however.

Let’s use the reaction above to understand how such a cell works.

First, we’ll need to picture the construction of the cell. Although there are several ways that a cell can be constructed, we’ll begin with a class cell construction. This consists of two beakers, each filled with a solution that contains all of the ingredients for one of the half reactions.

Into each beaker, we’ll submerge a wire,and then connect those two wires with a third. This will allow the transfer of electrons from one half-reaction to the other.

Now, if the reactions occur, as shown below, electrons will move from the iron beaker to the manganese beaker.

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

As you know, electrons moving through a wire is electricity. That means that we can use that flow of electrons to do electric work, like lighting a light bulb or running a cell phone.

Of course, there is a hitch. As shown, this cell won’t work. We are missing a crucial piece.

The Problem

Before we discuss what is missing, it is important to understand the problem with our system as shown.

The simplistic way to understand the issue is to remember that in order for electricity to flow, you must have a complete circuit. The cell, as drawn, does not have that. Keep in mind that we can’t just put another wire running between the beakers. Given that the reaction on the left is taking electrons and the reaction on the right is giving them, another wire would just allow more electrons to flow in the same direction. It would NOT complete the circuit.

To really understand what is going on, we need to think about those moving electrons and the charge they carry. As electrons move into the beaker on the left, the charge in that beaker will become more and more negative. As a result, electrons will become less and less likely to continue moving into the beaker. Very quickly in fact, electrons will cease to enter at all. At the same time, as electrons leave the beaker on the right, it will become more and more positive, making it less and less likely that electrons will continue to leave.

The Solution

What is needed then, is a way to balance charge. Specifically, we need a way to keep the beaker on the left from becoming too negative and the beaker on the right from becoming too positive.

We do this by adding a salt bridge. This is a piece of glass tubing, bent into a U-shape. It is then filled with a solution of a salt (an ionic compound) and capped at each end by a porous plug (we can use a piece of cotton for this purpose).

If we use potassium chloride (KCl) for the salt, we know that, in solution, we will find separate ions of potassium (\(K^{+1}\)) and chloride (\(Cl^{-01}\)).

The cotton plugs serve two purposes: 1) they allow us to turn the salt bridge upside down without spilling the solution and 2) they will allow ions to slowly travel in and out of the tube.

Inverting the salt bridge into our two beakers, will allow the cell to operate.

As the electrons move from the right to the left the ions in the salt bridge will migrate into the beakers to maintain neutral charge. For each electron that moves into the left beaker, a potassium ion (\(K^{+1}\)) moves into the beaker maintaining the charge. For each electron that leaves the right beaker, a chloride ion (\(Cl^{_1}\)) moves into the beaker maintaining that charge.

There are, of course, a few more details about the  construction of galvanic cells, the notation we use for Galvanic cells and the amount of power we can get from various cells.