Standard electrode potentials

Why a cell has a voltage

Think of each electrode as pushing electrons into the wire. Zinc pushes harder than copper does. The voltmeter reads the difference in the two pushes.

This causes a problem. We can only measure the difference between two electrodes. We can never measure the push of one electrode alone. So we do what we do with height: we pick a zero. Maps measure height above sea level. Electrochemists measure potential against the standard hydrogen electrode.

The standard hydrogen electrode

The standard hydrogen electrode, SHE, is a piece of platinum foil covered with a fine black powder of platinum. It dips into hydrochloric acid with $\text{H}^+$ at $1\ \text{mol/L}$. Hydrogen gas at $1\ \text{bar}$ bubbles over the foil. The reaction at the foil is:

• $2\text{H}^+(aq) + 2e^- \rightleftharpoons \text{H}_2(g)$

By agreement, the potential of the SHE is exactly $0.00\ \text{V}$ at every temperature.

Standard conditions

A potential depends on the concentrations, the pressures and the temperature. So we fix them. A standard electrode potential, written $E^\circ$ and read "E standard", is measured with:

• ions at $1\ \text{mol/L}$,
• gases at $1\ \text{bar}$,
• pure solids and liquids, and
• a temperature of $298\ \text{K}$.

Measuring a standard electrode potential

To find $E^\circ$ of a couple, join its half-cell to the SHE and read the voltmeter. Two things are noted: the reading and the direction of the electrons.

• Zinc. The voltmeter reads $0.76\ \text{V}$. Electrons flow from the zinc to the SHE, so zinc is the negative terminal. Zinc gives electrons more readily than hydrogen does. So $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$.
• Copper. The voltmeter reads $0.34\ \text{V}$. Electrons flow from the SHE to the copper, so copper is the positive terminal. Copper takes electrons more readily than hydrogen does. So $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$.

By convention, every table lists the reduction half-reaction: oxidised form plus electrons gives the reduced form, $\text{M}^{n+} + ne^- \rightarrow \text{M}$. Here are some values:

• $\text{Mg}^{2+}/\text{Mg}$: $-2.37\ \text{V}$
• $\text{Zn}^{2+}/\text{Zn}$: $-0.76\ \text{V}$
• $\text{Fe}^{2+}/\text{Fe}$: $-0.44\ \text{V}$
• $\text{H}^+/\text{H}_2$: $0.00\ \text{V}$
• $\text{Cu}^{2+}/\text{Cu}$: $+0.34\ \text{V}$
• $\text{Ag}^+/\text{Ag}$: $+0.80\ \text{V}$

A negative $E^\circ$ means the couple gives up electrons more readily than hydrogen does. A positive $E^\circ$ means it takes electrons more readily.

The cell potential

The cell potential is the difference between the two electrodes. We always take the cathode minus the anode, with both written as reduction potentials:

For the Daniell cell, copper is the cathode and zinc is the anode:

• $E^\circ_{cell} = 0.34 - (-0.76) = 1.10\ \text{V}$

This agrees with the voltmeter. Now try a silver and copper cell. Silver has the higher $E^\circ$, so it is the cathode:

• $E^\circ_{cell} = 0.80 - 0.34 = 0.46\ \text{V}$

So the more positive couple is the cathode and the more negative couple is the anode. If your answer comes out negative, you have swapped the electrodes. The cell would run the other way.

Two rules for working with $E^\circ$

• Reverse a half-reaction and $E^\circ$ changes sign. For $\text{Zn}(s) \rightarrow \text{Zn}^{2+} + 2e^-$ the value is $+0.76\ \text{V}$.
• Multiply a half-reaction and $E^\circ$ stays the same. The equation $2\text{Zn}^{2+} + 4e^- \rightarrow 2\text{Zn}$ still has $-0.76\ \text{V}$. The potential tells you the push per electron, not how many electrons there are.

Try the pair 2 and 5 first. It is the Daniell cell. Then try putting copper as the anode and zinc as the cathode.

Practise with new numbers

Try it in Python

The program keeps the potentials in a table and works out the cell potential of every pair.

Common mistakes

• Taking anode minus cathode. Always cathode minus anode.
• Doubling $E^\circ$ when you double the equation. It does not change.
• Forgetting to reverse the sign for an oxidation. The table gives reductions only.
• Using the values for other concentrations. A standard potential holds only at $1\ \text{mol/L}$, $1\ \text{bar}$ and $298\ \text{K}$.

Remember these

• The SHE: platinum, $\text{H}^+$ at $1\ \text{mol/L}$, $\text{H}_2$ at $1\ \text{bar}$. Its $E^\circ$ is $0.00\ \text{V}$.
• $E^\circ$ is a reduction potential, measured against the SHE at $298\ \text{K}$.
• $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.
• Reverse the equation: change the sign. Multiply the equation: no change.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is $E^\circ$ of the SHE, and why?

2. One couple has $E^\circ = +0.80\ \text{V}$ and another has $E^\circ = -0.44\ \text{V}$. Which is the cathode, and what is the cell potential?

3. $E^\circ$ of $\text{Cu}^{2+}/\text{Cu}$ is $+0.34\ \text{V}$. What is $E^\circ$ for $2\text{Cu}^{2+} + 4e^- \rightarrow 2\text{Cu}$?

4. You work out $E^\circ_{cell} = -0.50\ \text{V}$. What does that tell you?