9.7: Galvanic (Voltaic) and Electrolytic Cells
Throughout this unit we’ve centered most of our discussion around the topics of entropy and Gibbs Free Energy. However, there’s another aspect of energy we haven’t discussed and that is electricity! Electrochemistry is the study of redox reactions and how we can use redox reactions to produce electrical energy. During this unit we’ll take a look at cell potentials, which are a measure of the voltage released during redox reactions and connect that voltage to spontaneity and thus equilibrium. We’ll also look at how electrical power can be used to make non-spontaneous redox reactions occur.
Review of Redox Reactions
Before jumping into the thermodynamics of electrochemistry, let’s look back to unit 4
and review what exactly a redox reaction is. A redox reaction
, also known as an oxidation-reduction reaction
, is a reaction in which electrons are transferred from a reducing agent
to an oxidizing agent
. The reducing agent is oxidized
(loses electrons) and the oxidizing agent is reduced
(gains electrons). A good acronym to use to remember this is OIL RIG
(oxidation is loss, reduction is gain). The language of reducing and oxidizing agents won’t be used on the exam, but it’s helpful to know.
An example of a redox reaction is: 2AgNO3 + Cu → Cu(NO3)2 + 2Ag. We see that copper begins at an oxidation number of 0 and ends with an oxidation number of +2. Silver starts with an oxidation number of +1 and ends with an oxidation number of 0. Therefore, copper was oxidized and silver was reduced. Electrons were transferred from copper to silver.
Image From SciencePhoto
We can write the above reaction in terms of an oxidation and a reduction half-reaction that add together to form the overall reaction:
Copper is oxidized to form Cu2+: Cu → Cu2+ + 2e-
Ag+ is reduced to form Ag: 2Ag+ + 2e- → 2Ag
Note that we multiplied the second reaction by 2 to balance the electrons.
When redox reactions occur, the electrons experience an electromotive force. This is essentially the force that pushes electrons from a reducing agent to an oxidizing agent. The stronger this force the more spontaneous the reaction. Electromotive force is measured in volts. We’ll be using these measurements to calculate cell potential.
Let’s take a look at an example of how to calculate the voltage of a redox reaction by using what we call reduction potentials
. Standard reduction potentials are the voltage of a reduction. Note that voltages can
be negative. This just means that it takes
energy to make that reaction happen. Reduction potentials are calculated constants that will be given to you. The table that we’ll be referencing can be found here
Consider the redox reaction: Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s).
First we can split the reaction into two half reactions:
Zn → Zn2+ + 2e- (oxidation)
Pb2+ + 2e- → Pb (reduction)
Looking at our chart, we see that the reduction potential for Pb2+ is -0.13V and for Zn2+ the reduction potential is -0.76V. However, our half-reaction for Zn2+ is an oxidation. Flipping a reduction potential reaction negates the potential, meaning the potential for Zn → Zn2+ + 2e- is +0.76V. Finally, we can add these two numbers together to find that the potential for the overall reaction is -0.13 + 0.76 = 0.63V.
NOTE: Multiplying a half-reaction does not change the potential!
To find the E value for any given redox reaction, you can use the equation E = Ered - Eox. In unit 9.8 we’ll look at calculating Ecell using cathodes and anodes of galvanic cells which will give us a new form of this equation, but for now think about reductions and oxidations. The equation you’ll see on the formula sheet and the one you’ll learn is . You may wonder then why we added instead of subtracted in the last example, but that’s because we negated the reduction potential of Zn2+, meaning we did actually subtract. We could have also done: (-0.13V) - (-0.76V) = +0.63V.
Now that we’ve refreshed redox reactions and learned a bit about reduction potentials, let’s look at what happens when the oxidizing agent and reducing agent are separated and connected through a wire. Remember that the oxidizing agent gains electrons and the reducing agent loses electrons. Therefore electrons will travel through the wire from the reducing agent to the oxidizing agent. The force at which these electrons travel is known as cell potential notated as Ecell.
The part of the cell where oxidation occurs is known as the anode and reduction occurs at the cathode. Electrons travel from the anode to the cathode. A good way of remembering this is the phrase “An Ox, Red Cat” which tells you that the anode is where oxidation occurs and the cathode is where reduction occurs.
These cells are known as galvanic or voltaic cells. Let’s take a look at one for the reaction Cu + 2Ag+ → 2Ag + Cu2+:
Image From BCCampus
In this image we see two half-reactions occurring. One of these is the reduction of Ag+ into silver metal and the other is oxidation of Cu into Cu2+. Therefore we have a copper anode and a silver cathode. When a wire connects these two electrodes in solutions of their respective ions (Cu2+ and Ag+), electrons can travel from the copper to the Ag+ and create Ag. In the end we’ll see the copper anode shrink and the silver cathode grow because we are losing Cu and forming Cu2+ and losing Ag+ to form Ag. We can add a voltmeter to the wire to see the electromotive force of this reaction which we find to be +0.46V.
We also see a salt bridge connecting the two solutions. This is to ensure that ions continue flowing because eventually, the reaction would stop. The purpose of the salt bridge is to act as a source of spectator ions that can migrate into each of the half cells to preserve neutrality. Salt bridges are filled with inert ions that will not interact with the reaction. In this case we used NaNO3. Typically it is a highly soluble salt.
As a practice exercise, see if you can write out the half-reactions and find the cell potential using the equation we looked at earlier. You should get the same number (Ecell = 0.46V). As another exercise, try to logically understand why we can write that Ecell = Ecathode - Eanode using the fact that E = Ered - Eox. This equation for Ecell is the one that is most commonly used.
For a galvanic cell, we assume that the Ecell is greater than 0. This means that the reaction occurs spontaneously. However, if we want to drive a non-spontaneous redox reaction, we can use voltage to make it happen. These cells are electrolytic cells. Let’s look at an example. Suppose we want to rip apart a solution of Na+ and Cl- ions to form the original Na and Cl2 that made it up. This is a nonspontaneous redox reaction: 2NaCl → 2Na + Cl2. We can use a battery to use electromotive force to rip the electron off of Cl- and add it to Na+.
Let’s first show mathematically that this reaction is nonspontaneous. Using half reactions we can calculate Ecell for this redox reaction:
2Na+ + 2e- → 2Na (E = -2.71V)
2Cl- → Cl2 + 2e- (E = -1.36V)
The anode is where oxidation occurs so Eanode = -1.36. The cathode is where reduction occurs so Ecathode = -2.71. We know that Ecell = Ecathode - Eanode = -2.71 - (-1.36) = -1.35V. Because Ecell for this redox reaction is negative, it will not occur spontaneously. Therefore, we have to use a battery (or some other source of electricity) to actively push the electrons from the anode to the cathode.
Image From TropicSU
Because we have molten NaCl (the ions themselves), we use inert electrodes to collect the products, in this case Cl2 gas and Na metal. As the electrons are pushed by the voltage of the battery, the Cl- in solution is oxidized into Cl2 and Na+ ions are reduced into Na meaning gas bubbles will accumulate on the anode and sodium metal will accumulate on the cathode. Note that the voltage of the battery must be greater than or equal to the voltage of the overall redox reaction. In this example, we must use at a minimum a 1.35V battery.