The electrolysis of aqueous solutions
5 marks · HardA student carries out the electrolysis of concentrated sodium chloride solution using carbon electrodes as shown below.
State the product formed at the cathode.
State the product formed at the anode.
Write the half-equation for the reaction at the cathode.
Write the half-equation for the reaction at the anode.
Explain why hydrogen forms at the cathode instead of sodium.
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(a) hydrogen
(b) chlorine
(c) \ce{2H^+ + 2e^- -> H2}
(d) \ce{2Cl^- -> Cl2 + 2e^-}
(e) Hydrogen ions are discharged more easily because they are less reactive than sodium ions.
Also accept: Accept \,\ce{2H2O + 2e^- -> H2 + 2OH^-} \, for cathode half-equation; Accept answer in words (hydrogen is less reactive than sodium, so is discharged more easily)
Common mistakes: Confusing cathode and anode; Writing full equations instead of half-equations; Incorrect explanation of reactivity
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Marking points
B1 (a) · 1 mark · Hydrogen identified as the product at the cathode.
B2 (b) · 1 mark · Chlorine identified as the product at the anode.
B3 (c) · 1 mark · Correct cathode half-equation: or .
B4 (d) · 1 mark · Correct anode half-equation: .
B5 (e) · 1 mark · Explanation: Hydrogen ions are discharged more readily as hydrogen is less reactive than sodium.
Worked solution
a: The product at the cathode is hydrogen.
b: The product at the anode is chlorine.
or
e: Hydrogen forms at the cathode because hydrogen ions are discharged more easily than sodium ions, as hydrogen is less reactive than sodium.