Standard Electrode Potential — Definition & Overview

We've been investigating redox reactions. So far, we've spent a lot of time balancing redox equations. We determined each atom's oxidation number. Then, we specified which were oxidizing and reducing agents. Then we used either the half-reaction or the oxidation change method. Finally, we made everything more complicated. We introduced acidic and alkaline conditions. But why do redox reactions even matter?

Redox reactions form the basis for batteries. Are you reading this on your laptop or phone? Right now, you're taking advantage of redox reactions! A lithium-ion battery powers your device. At one electrode, a lithium-transition metal compound gets reduced. At the other electrode, a carbon-silicon complex gets oxidized. What's the result? An electric current that keeps your screen lit, wi-fi connected, and apps running. What happens when you charge your device? We reverse the reactions! Today, we'll begin studying how we use redox reactions for electricity.

Cell Potentials

What causes electrons to flow from the anode to the cathode? It's the potential energy difference. Do you remember internal energy from thermodynamics? We talked about what energy is and how it's stored. In chemistry, we hold potential energy via electrostatic interactions. Let's consider a zinc electrode. Where does zinc's potential energy come from? The interaction between its nuclei's protons and the sea of electrons. Let's say we connect it to a copper cathode. Why do the electrons spontaneously flow from zinc to copper? Copper's electrons have less potential energy. Remember, copper is a unique element. Do you remember its electron configuration? It has a full 3d-orbital in exchange for a half-filled 4s orbital.

When something changes its potential energy, it exerts a force. The anode's electrons lower their potential by flowing to the cathode. So, the electrons must have exerted a force. In chemical reactions, the force is the breaking of chemical bonds. But, in a battery, the cells are physically separated. The only connection is the wire. So, the force is the repulsion of the liberated electrons and the wire's electrons. We call it the electromotive force (emf) or voltage. We measure voltage in units of joules per coulomb. So, it's energy per charge. Remember joules equal one newton (force) displacing an object by one meter.

So, what's the overall voltage of a voltaic cell? It's the difference between the voltage of the cathode and anode. So, the formula is $E_{cell}= E_{cathode}-E_{anode}$. But, how do we know the voltage of the cathode or anode?

Standard Hydrogen Electrodes

We need a method to find the potential of a cathode or anode by itself. But, we only have the formula $E_{cell}= E_{cathode}-E_{anode}$. A cell only has a voltage if it has both a cathode and anode. So, we need an electrode with a voltage of zero. But how? Chemists are lazy. So, our easiest option is to define a standard. So, chemists created the standard hydrogen electrode (SHE). Then they said that it would have a potential of zero. What was the benefit? We can use it to find the voltages of every other electrode.

How can we use hydrogen as an electrode? Every other electrode we've seen are metals. But hydrogen's a diatomic gas! The experimental setup is challenging. So, we start with a platinum wire. On the end of the wire is platinum foil. We enclose both in a glass tube. The wire connects to the other electrode. At the same time, we immerse the foil in a one-molar acid solution. Finally, we bubble hydrogen gas through the solution. We connect the gas source to an opening in the glass tube. The gas pressure is one atmosphere.

What's the point of the platinum foil? We need to reduce hydrogen ions to hydrogen gas. The anode loses electrons. They transfer along the wire and into the foil. Remember, the foil is in an acidic solution. So, we initiate the reaction $2H^{+} + 2e^{-} \rightarrow H_{2(g)}$ on the foil's surface. We defined this reaction to have a potential of zero. So, we measure the voltage between the anode and the SHE. Based on our definition, the potential formula becomes $E^{o}_{cell}= \cancel{E^{o}_{cathode}}-E^{o}_{anode}$. Notice that we are now using $E^{o}$. It indicates standard conditions.

Standard Reduction Potentials

After a lot of work, we have a method for finding the potential of an electrode. The convention is to list voltages based on the reduction half-reaction. We call the voltages standard reduction potentials. So, our formula becomes $E^{o}_{cell}= E^{o}_{red}(cathode)-E^{o}_{red}(anode)$. We can calculate the cell's voltage from the electrodes' reduction half-reactions. Notice that the negative sign accounts for the oxidation of the anode. So, even though the anode gets oxidized, we still use the potential from the reduction.

Where do we find the standard reduction potentials? Thankfully, analytical chemists have done the work for us! They've calculated the standard reduction potentials for all the common electrodes. In class, your teacher will give you the data. I've made a table with some of them below. We'll use these values in the example problems.

$$\begin{array}{c|c} Potential\,(V) & Reduction\,Half-Reaction \\ \hline 2.87 & F_{2(g)}+ 2e^{-}\rightarrow 2F^{-} \\ 0.96 & NO_{3}^{-}+ 4H^{+} + 3e^{-}\rightarrow NO_{(g)} + 2H_{2}O \\ 0.80 & Ag^{+}+ e^{-}\rightarrow Ag_{(s)} \\ 0.34 & Cu^{2+}+ 2e^{-} \rightarrow Cu_{(s)} \\ 0.00 & 2H^{+} + 2e^{-} \rightarrow H_{2(g)} \\ -0.76 & Zn^{2+}+ 2e^{-} \rightarrow Zn_{(s)}\\ -1.66 & Al^{3+} + 3e^{-} \rightarrow Al_{(s)} \\-2.71 & Na^{+}+ e^{-} \rightarrow Na_{(s)} \\ -3.05 & Li^{+}+ e^{-} \rightarrow Li_{(s)} \end{array}$$

Notice that some of the voltages are positive and some negative. The positive values indicate reduction. So, when we linked the electrode to the SHE, it gained electrons. The negative values signify oxidation. So, the electrode lost electrons when connected to the SHE. Which elements oxidize? The ones with low electron affinities! The two alkali metals ($Li$ and $Na$) have considerable negative reduction potentials. We know that they give up their valance electron for a full octet. Which elements get reduced? The most electronegative elements. They want electrons! For example, the halogens all have positive voltages. The most electronegative element is fluorine. It has an enormous positive voltage!

Example Problems

1. A voltaic cell is made from silver and copper. Calculate the cell's standard voltage.

First, list the two half-reactions.

$Ag^{+}+ e^{-}\rightarrow Ag_{(s)}\, E^{0}_{red}=+0.80$

$Cu^{2+}+ 2e^{-} \rightarrow Cu_{(s)}\, E^{0}_{red}=+0.34$

Now, we need to determine which metal is the cathode and which is the anode. The cathode will have a more positive potential. Why? In a cell, we reduce the cathode. So, whichever metal reduces more efficiently is the cathode. The reduction potential for silver is +0.80. The reduction potential for copper is +0.34. So, silver will be the cathode.

$E^{o}_{cell}= E^{o}_{red}(cathode)-E^{o}_{red}(anode)=0.80V-0.34V=0.46V$

1. What is the standard cell potential in a cell made from zinc and copper?

First, list the two half-reactions.

$Zn^{2+}+ 2e^{-} \rightarrow Zn_{(s)}\, E^{0}_{red}=-0.76$

$Cu^{2+}+ 2e^{-} \rightarrow Cu_{(s)}\, E^{0}_{red}=+0.34$

Now, we need to determine the cathode. The cathode will be copper because it has a more positive $E^{0}_{red}$. So, it reduces more readily than zinc.

$E^{o}_{cell}= E^{o}_{red}(cathode)-E^{o}_{red}(anode)=0.34V-(-0.76)=1.10V$

(Video) 19.4 How to Calculate Standard Cell Potential | General Chemistry

by Chad's Prep

In this video, Chad describes standard electrode potentials. He starts by explaining the reduction potentials table and how to use it. Then, he illustrates three practice problems. After the practice problems, you'll learn how to identify strong oxidizing and reducing agents from the table. Finally, you'll see how the table can be used to determine a reaction's spontaneity. This is an excellent video for understanding standard cell potentials.