IChO 2003 · Theory problem 9

Charge needed to electrolyse water

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The complete electrolysis of 1 mol of water requires the following amount of electric charge (F is the Faraday constant):

(a) F ( )

(b) (4/3) F ( )

(c) (3/2) F ( )

(d) 2 F ( )

(e) 3 F ( )

Official English text of the problem.

Answers

  • 9: (d) \(2F\). Mark the box for (d); leave the other boxes blank.

Solution

The overall electrolysis reaction is

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)}. \]

The molecular species are water (O), hydrogen ([H][H]), and oxygen (O=O).

Using acidic half-equations, with hydrogen ions \(\ce{H+}\) ([H+]):

  • At the cathode, reduction occurs: \[ \ce{4H+ + 4e- -> 2H2}. \]

  • At the anode, oxidation occurs: \[ \ce{2H2O -> O2 + 4H+ + 4e-}. \]

Adding these gives the overall reaction above. Thus, electrolysing \(2\ \mathrm{mol}\) of water requires \(4\ \mathrm{mol}\) of electrons to pass through the circuit. For \(1\ \mathrm{mol}\) of water,

\[ n(\ce{e-}) = 1\ \mathrm{mol}\times\frac{4}{2} = 2\ \mathrm{mol}. \]

By Faraday’s law,

\[ Q=n(\ce{e-})F=(2\ \mathrm{mol})F. \]

This is two faradays of charge, written \(2F\) in the options, so (d) is correct. The electrode charges are not added: the same charge passes through both electrodes.

The answers were checked against the official answers published by the organisers.