Faraday’s Laws

Faraday's Laws connect the relationship between the electric charge passed through an electrochemical cell and the quantities of substances involved in the reaction. Faraday’s law can be applied to calculate any amounts of reactants or products in redox half reactions that gain or lose electrons.

Use this page to revise the following concepts within Faraday's Laws:

Faraday’s first law

Faraday's first law states the mass of a substance deposited at an electrode is directly proportional to the amount of charge passed through it.

two half cells, one with Zn2+ in the electrolyte and Zn strip as electrode, the other with Cu2+ in the electrolyte and Cu as electrode

Applying Faraday’s first law to the galvanic cell above, the mass of copper deposited at cathode is proportional to the charge passing through the cathode.

Charge (\(Q\)) can be calculated from:

\[Q = It\]

Where:

  • \(Q\) is the electric charge (coulombs, \(C\))
  • \(I\) is the current (amps, \(A\))
  • \(t\) is the time (seconds, \(s\))

Therefore, according to Faraday's first law, if the current is constant, the mass of copper deposited is proportional to time.

Faraday’s second law

Faraday's second law states that to produce 1 mole of metal, 1, 2, 3 or another whole number of moles of electrons must be consumed. This suggests that, with the same amount of charge (or electrons) passing through the galvanic cell, the mole of the metal deposited has a relationship with the charge of the metal. The number of electrons passing through the electrode can be determined from the stoichiometric ratio using the reduction half-equation. For example, for \(\ce{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s), n(e^{-}): n(Cu) = 2:1}\).

Faraday’s constant helps to quantifies \(\ce{n(e^{-})}\) and the charge \(\ce{Q=n(e^{-})×F}\)

  • \(\ce{n(e^{-})}\)is the number of moles of electrons
  • \(F\) is the Faraday constant (\(96500 \text{Cmol}^{-1}\))

The quantitative calculations involved in the electrochemical reactions can be summarised below:

Worked Example

A fuel cell uses \(300 \text{g}\) of methanol (\(\ce{CH3OH}\)) as its fuel and is performed under acidic electrolyte conditions. If the cell operates at a current of \(2.00 A\), how long, in hours, will it take for the methanol to be completely consumed?

Solution

The balanced half equation: \(\ce{CH_3OH(l) + H_2O(l) \rightarrow CO_2(g) + 6H^{+}(aq) + 6e^{−}}\)

\(\ce{M(CH_3OH)=32.0 \text{g mol}^{-1}}\), hence \(\ce{n(CH_3OH) = \frac{300}{32.0} = 9.38 \text{mol}}\)

\(\ce{n(CH_3OH) : n(e^{-}) = 1:6}\), hence \(\ce{n(e-) = 6 \times 9.38 = 56.3 \text{mol}}\)

\(Q = \ce{n(e^{-})} F = 56.3 \times 96500 = 5.43 \times 106 \text{ C}\)

Because \(Q = It\), hence \(t = \dfrac{Q}{I} = \dfrac{5.43 \times 10^6}{2} = 2.71 x 10^{6} s = 754 \text{h}\)