Cell notation and cell emf

The number on the battery

A torch cell has 1.5 V printed on it. A car battery has 12 V. What does that number mean?

It is the emf of the cell. The letters stand for electromotive force, but the emf is not a force. It is the potential difference between the two electrodes when the cell delivers no current. We write it as $E_{cell}$, and when every ion is at $1\ \text{mol/L}$ it is the standard value $E^\circ_{cell}$.

You met half-cells, the Daniell cell and $E^\circ$ in Class 11. Now we use them to write and read any cell, and in the next two lessons to find the energy and the equilibrium constant that sit behind the emf.

Emf and the voltage under load

Why must we say "no current"? When a current runs, the cell itself pushes the ions and the electrons through its own solutions and electrodes. This internal path has a resistance $r$. A part of the emf is spent there, so the voltage across the terminals is smaller:

The graph is a straight line. At zero current the voltage equals the emf. The more current the cell gives, the lower the voltage. This is why the emf is measured with a voltmeter of very high resistance, or with a potentiometer, which draws no current at the moment of balance.

Notice that the emf stays the same in the sim. Only the voltage across the terminals changes. A battery that lights a torch dimly may still show its full emf on a high-resistance voltmeter. The cell is worn out because its internal resistance has grown, not because the emf has gone.

Practise with new numbers

Writing a cell in one line

Chemists write the whole cell on one line. The rules are:

• The anode, where oxidation happens, goes on the left. The cathode, where reduction happens, goes on the right.
• A single bar $\mid$ is a boundary between two phases, such as a solid and a solution, or a gas and a solution.
• A double bar $\parallel$ is the salt bridge.
• Species in the same phase are separated by a comma: $\text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq)$.
• The state of each species is written, (s), (l), (g) or (aq). A concentration or a gas pressure may follow in brackets.
• When no solid metal takes part, an inert electrode is written at the end. Platinum is the usual choice. It carries electrons but does not react.

Here is a cell that uses platinum at both ends.

Read it from left to right, one phase at a time: platinum, the iron solution, the bridge, the chloride solution, chlorine gas, platinum. The left platinum dips into a solution with both iron ions. The right platinum has chlorine gas bubbling over it.

From the line to the cell reaction

The left side is the oxidation and the right side is the reduction. Follow four steps.

1. Write the half-reaction at the left electrode as an oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$.
2. Write the half-reaction at the right electrode as a reduction: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$.
3. Multiply so that the electrons cancel. Here the oxidation is doubled.
4. Add the two. The overall reaction is $2\text{Fe}^{2+} + \text{Cl}_2 \rightarrow 2\text{Fe}^{3+} + 2\text{Cl}^-$, and $n = 2$.

The number $n$ is the number of moles of electrons that pass for the equation as written. It matters in the next lesson.

Now the emf. We take $E^\circ$ of each couple from the table of reduction potentials. The iron couple $\text{Fe}^{3+}/\text{Fe}^{2+}$ has $E^\circ = 0.77\ \text{V}$ and the chlorine couple $\text{Cl}_2/\text{Cl}^-$ has $E^\circ = 1.36\ \text{V}$.

The right electrode is the cathode, so this is the same as cathode minus anode. The voltmeter in the picture reads $0.59\ \text{V}$.

From a reaction to a cell

Sometimes you are given the reaction and asked for the cell. Work backwards.

Take $\text{Cd}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cd}^{2+}(aq) + 2\text{Ag}(s)$. Cadmium loses electrons, so it is oxidised and goes on the left. Silver ions gain electrons, so they go on the right.

With $E^\circ(\text{Cd}^{2+}/\text{Cd}) = -0.40\ \text{V}$ and $E^\circ(\text{Ag}^+/\text{Ag}) = 0.80\ \text{V}$:

Cadmium has the lower $E^\circ$, so it sits on the left. We will use this cell again in the next two lessons.

What the sign of the emf says

The emf is worked out for the cell as written. A positive value means the reaction runs from left to right, as the line says. A negative value means it runs the other way.

Try the cell $\text{Pt}(s) \mid \text{Br}_2(l) \mid \text{Br}^-(aq) \parallel \text{Fe}^{3+}(aq), \text{Fe}^{2+}(aq) \mid \text{Pt}(s)$. The right electrode is the iron couple at $0.77\ \text{V}$ and the left is bromine at $1.09\ \text{V}$.

• $E^\circ_{cell} = 0.77\ \text{V} - 1.09\ \text{V} = -0.32\ \text{V}$

The cell as written does not run. Iron(III) cannot oxidise bromide ions. If you swap the two sides, the emf becomes $+0.32\ \text{V}$ and the cell works, with bromine oxidising iron(II).

Three facts about the emf

• The emf does not depend on the size of the electrodes or the volume of the solutions. A big cell and a small cell with the same chemicals have the same emf. This is why we call it an intensive property.
• The emf does not change if you multiply the equation. Doubling the equation doubles $n$, not $E^\circ$.
• Changing a concentration changes the emf. The value $E^\circ$ is only for $1\ \text{mol/L}$, $1\ \text{bar}$ and $298\ \text{K}$. Lesson 3 shows by how much.

Practise with new numbers

Try it in Python

The program keeps the potentials in a table, works out the emf of three cells and says which way each one runs.

Common mistakes

• Taking the left electrode minus the right one. The emf is the right-hand potential minus the left-hand one.
• Putting a comma where a bar belongs, or the other way round. A bar separates phases. A comma separates species in the same phase.
• Leaving out the platinum. When no metal takes part, an inert electrode must be written.
• Doubling $E^\circ$ when the equation is doubled. The emf stays the same. The number $n$ doubles.
• Reading the label of a battery as the voltage under load. The emf is the voltage at zero current.

Remember these

• Emf: the potential difference of a cell when no current flows. Under load, $V = E - Ir$.
• Cell line: anode on the left, cathode on the right, a bar for each phase boundary, a double bar for the salt bridge.
• $E^\circ_{cell} = E^\circ_{right} - E^\circ_{left}$. A positive value means the cell runs as written.
• The emf does not depend on the size of the cell or on how the equation is multiplied.

Check yourself

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

1. A cell has an emf of $1.64\ \text{V}$ and an internal resistance of $0.30\ \Omega$. It delivers $2.5\ \text{A}$. What is the voltage across its terminals?

2. Why is a platinum wire used in the electrode for the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple?

3. Write the cell for the reaction $\text{Zn}(s) + \text{Cl}_2(g) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Cl}^-(aq)$.

4. A cell line gives $E^\circ_{cell} = -0.15\ \text{V}$. Does the reaction run as written?

5. The equation of a cell reaction is multiplied by 3. What happens to $E^\circ_{cell}$ and to $n$?