Faraday's Laws
Use this page to revise the following concepts within Faraday's Laws:
In electrolysis, the amount of electric current passing through the cell and the time the cell operates both impact the amount of chemicals produced. The relationship between these quantities is explained through Faraday’s Laws. Chemists use these relationships to control how much of a chemical is produced via electrolysis, or how thick a layer of metal is when it is electroplated on an object.
Electroplating
One application of electrolysis is electroplating, where a thin layer of a metal is plated onto the surface of an object. A common example are ‘tin cans’ that are used to store food. The cans are actually made of steel, which is made from iron. This metal is reasonably reactive and can corrode over time. To prevent this, the can is electroplated with tin, which is less reactive, and protects the can.
To undergo electroplating the object
is immersed in an electrolyte bath containing tin ions (\(Sn^{2+})\). The can is placed at the cathode, where \(Sn^{2+}\) is reduced to Sn on the surface of the can, coating it, according to the reduction half equation:
\[Sn6^{2+}(aq) + 2e- \rightarrow Sn(s)\]
The anode is pure tin, which oxidises into the solution, according to the oxidation half equation:
\[Sn(s) \rightarrow Sn^{2+}(aq) + 2e^-\]
Faraday’s First Law
When an electrolytic cell operates, the mass of the metal converted to ions at the anode, the mass of metal deposited at the cathode and the electric charge that passes through the cell are all directly proportional.
NoteThe charge passed through the cell is proportional to the mass deposited at the cathode. This is known as Faraday’s first law. It is represented as: \[m \propto Q\] |
The charge, \(Q\), is measured in Coulombs, \(C\). It is equal to the current, \(I\) (in Amperes, \(A\)), multiplied by the time, \(t\) (in seconds), that the current flows through the cell, according to:
\[Q = It\]
Faraday’s Second Law
In an electroplating cell, when the same amount of electric charge is passed through an electrolyte, the amount of metal deposited at the cathode depends on the molar mass of the substance and the number of electrons required for its reduction.
The same amount of charge may cause different masses of different metals to be plated at the cathode. If converted to mole amounts, there is a relationship between the charge supplied, the charge on the metal ion and the mole amount of metal produced.
NoteTo produce a whole mole of metal, a whole number mole of electrons is required |
The charge from a mole of electrons is one Faraday \((F)\) or 96 500 \(C\), so the charge passed through a cell can be calculated as:
\[Q = n(e^-) F\]
\[Q = n(e^-)\text{ mol} \times 96 500 \text{Cmol}^{-1}\]
Calculations using Faraday’s Laws
The relationships, \(Q = It\) and \(Q = n(e^-)F\) can be used to calculate quantities involved in electrolytic cells, such as the mass produced or the time taken for a mass to be deposited at the cathode.
Worked Example
Calculating the mass of a product
What is the mass of silver produced in an electrolytic cell if the cell operates at 20\(A\) for 15 minutes?
Calculate the charge passing through the cell using \(Q = It\)
\[\begin{align}Q &= 20 x 15 x 60 \\ Q &= 18000 C \end{align}\]
Calculate the mole of electrons this represents
\[\begin{align} Q &= n(e^-)F \\ n(e) &= \frac{Q}{F} \\ n(e) &= \frac{18000}{96500} \\ n(e) &= 0.187 mol \end{align}\]
Calculate the mole of silver using the mole ratio
\[Ag^+ + e^- \rightarrow Ag\]
\[\begin{align}n(Ag) &= n(e^-) \\ n(Ag) &= 0.187 mol \end{align}\]
Calculate the mass
\[\begin{align} m(Ag) &= nM \\ m(Ag) &= 0.187 \times 107.86 \\ m(Ag) &= 20.11 g\end{align}\]
Worked Example
Calculating the time to produce a chemical
How long would it take to plate 35.0g of copper at the cathode of an electroplating cell, operating at 5A?
Calculate the mole of copper
\[\begin{align}n(Cu) &= \frac{m}{M} \\ n(Cu) &= \frac{35.0}{63.5} \\ n(Cu) &= 0.55 mol \end{align}\]
Calculate the mole of electrons required
Using \(Cu^{2+} + 2e^1 \rightarrow Cu\)
\[\begin{align}n(e) &= 2\times (Cu) \\ n(e) &= 1.10 mol\end{align}\]
Calculate the charge on the electrons
\[\begin{align}Q &= n(e)F \\ Q &= 1.10 \times 96 500 \\ Q &= 106378 C\end{align}\]
Calculate the time
\[\begin{align}Q &= It \\ t &= \frac{Q}{It} \\ t &= \frac{106378}{5} \\ t &= 21275s\end{align}\]
In minutes
\[t = \frac{21275}{60} = 355 \text{min}\]
In hours
\[t = \frac{354}{60} = 5.91h\]
