Faraday's laws of electrolysis

How much?

Lesson 1 told us what forms at each electrode. The next question is how much. A plating shop needs to know how long to leave a tray of spoons in the bath. A factory needs to know how much current to use for a tonne of aluminium.

Michael Faraday answered this in the 1830s, long before electrons were known. His result has two parts, called Faraday's laws of electrolysis.

Charge and current

Current is charge flowing each second, $I = Q/t$. So when a steady current flows, the charge that passes is:

The unit of charge is the coulomb. With the current in amperes and the time in seconds, $Q$ is in coulombs. A common slip is to use minutes. Always change to seconds.

Batteries are rated in ampere-hours. One ampere-hour is $1\ \text{A} \times 3600\ \text{s} = 3600\ \text{C}$.

The faraday

One electron carries a charge of $1.602 \times 10^{-19}\ \text{C}$. One mole of electrons is $6.022 \times 10^{23}$ electrons. So the charge on one mole of electrons is:

This is the Faraday constant, or one faraday. It is the bridge between electricity and moles. If $Q$ coulombs pass, the number of moles of electrons is $Q/F$.

The first law

Faraday's first law: the mass of a substance that forms at an electrode is proportional to the charge that passes through the cell.

So $m = ZQ$, where $Z$ is the electrochemical equivalent. It is the mass that one coulomb deposits. With $Q = It$ this reads $m = ZIt$.

The graph shows the law for three metals. Each line is straight and starts at the origin. Double the charge and the mass doubles. Different metals have different slopes, and that is the second law.

Working it out in four steps

You can use $Z = M/zF$ if you wish, but the safest way is to follow the moles. Take a half-reaction, $\text{M}^{z+} + ze^- \rightarrow \text{M}$. Then:

1. Charge. $Q = It$, in coulombs.
2. Moles of electrons. $\dfrac{Q}{F}$.
3. Moles of metal. Each ion needs $z$ electrons, so the moles of metal are $\dfrac{Q}{zF}$.
4. Mass. Multiply by the molar mass: $m = \dfrac{Q}{zF} \times M$.

Worked example. A current of $1.55\ \text{A}$ flows for $40$ minutes through copper sulphate solution. How much copper deposits at the cathode?

• $Q = 1.55 \times 40 \times 60 = 3720\ \text{C}$.
• Moles of electrons: $\dfrac{3720}{96500} = 0.0385\ \text{mol}$.
• Each $\text{Cu}^{2+}$ takes 2 electrons, so $0.0193\ \text{mol}$ of copper forms.
• Mass: $0.0193 \times 63.5\ \text{g/mol} = 1.22\ \text{g}$.

Check the size. A little over one gram from a small current in under an hour is reasonable.

If the time is the unknown, rearrange. We want $t$, so $t = \dfrac{mzF}{MI}$. If the current is steady, a longer time gives a bigger mass in proportion.

The second law

Join several cells in a line, one after another. The same current flows through all of them for the same time, so the same charge passes through each.

Faraday's second law: when the same charge passes through different electrolytes, the masses of the substances formed are in the ratio of their equivalent masses.

The equivalent mass is the molar mass divided by $z$, the number of electrons for each ion. It is the mass produced by one mole of electrons.

• Silver: $\dfrac{107.9}{1} = 107.9\ \text{g}$ for each mole of electrons.
• Copper: $\dfrac{63.5}{2} = 31.75\ \text{g}$.
• Aluminium: $\dfrac{27.0}{3} = 9\ \text{g}$.

So in a series of cells the masses of silver, copper and aluminium are in the ratio $107.9 : 31.75 : 9$. Suppose you know the molar mass of a metal and the mass that a known charge deposits. Then you can work out $z$, which is how the charges on ions were first found.

Gases

Gases are measured as volumes. At $273\ \text{K}$ and $1\ \text{atm}$, one mole of any gas takes up $22.4\ \text{L}$.

For the electrolysis of water, count the electrons for each gas:

• Hydrogen: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$. One mole of $\text{H}_2$ needs 2 mol of electrons, so each mole of electrons gives $\tfrac{1}{2}\ \text{mol}$, which is $11.2\ \text{L}$.
• Oxygen: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$. One mole of $\text{O}_2$ needs 4 mol of electrons, so each mole of electrons gives $\tfrac{1}{4}\ \text{mol}$, which is $5.6\ \text{L}$.

So the volumes at the cathode and the anode are in the ratio $2 : 1$. This matches the formula $\text{H}_2\text{O}$ and is easy to check in a school laboratory.

When the current does other jobs

Sometimes only part of the current makes the product you want. The rest may make hydrogen, or heat. The current efficiency is the share of the charge that does the wanted job:

Here $m_{actual}$ is the mass really deposited and $m_{full}$ is the mass that the whole charge would give.

In every calculation above we assumed that all the current went to one reaction.

Try the same charge with metals 1 and 4. Why does aluminium give so little mass compared with silver? Count the electrons that each ion needs, and look at the molar masses.

Practise with new numbers

Try it in Python

The program passes the same charge through four cells in series and works out the mass of metal in each.

Common mistakes

• Using minutes in $Q = It$. The time must be in seconds.
• Dividing by one electron for every metal. Copper needs 2, aluminium 3, silver 1.
• Using the molar mass without dividing by $z$. The equivalent mass is $M/z$.
• Saying that heavier atoms always give more mass. The ion charge matters too: aluminium has a small $M/z$.
• Forgetting that two moles of electrons are needed for each mole of $\text{H}_2$.

Remember these

• $Q = It$. $F = 96500\ \text{C/mol}$ is the charge on one mole of electrons. One ampere-hour is $3600\ \text{C}$.
• First law: $m = ZQ$. In moles: $m = \dfrac{Q}{zF}M$.
• Second law: the same charge gives masses in the ratio of $M/z$.
• Per mole of electrons: $\tfrac{1}{2}\ \text{mol}$ of $\text{H}_2$ and $\tfrac{1}{4}\ \text{mol}$ of $\text{O}_2$.
• Current efficiency is the actual mass over the expected mass, times 100 per cent.

Check yourself

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

1. How many coulombs are needed to deposit $0.68\ \text{mol}$ of copper from a $\text{Cu}^{2+}$ solution?

2. The same current flows for the same time through silver nitrate and through copper sulphate. Which cell gives the larger number of moles of metal?

3. A student finds the charge for a deposit using a time in minutes and gets an answer 60 times too small. What was the mistake?

4. The current is doubled and the time is halved. What happens to the mass deposited?