Electrolysis
7 marks · HardA science teacher asks students to compare what happens at the cathode during the electrolysis of sodium chloride solution and copper(II) sulfate solution, both using inert electrodes.
Explain why the solution around the cathode becomes alkaline during the electrolysis of sodium chloride solution.
Explain why copper is deposited at the cathode instead of hydrogen during the electrolysis of copper(II) sulfate solution using inert electrodes.
In both electrolysis experiments above, water is present in solution. State one difference in the effect of prolonged electrolysis on the composition of the solution in each case, and explain your answer using the relevant ions.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Hydrogen ions (from water) are reduced at the cathode, producing hydroxide ions which remain, making the solution alkaline. · must show Reference to discharge of H+ or reduction of H+, Production/increase of OH- ions, Resulting alkalinity around the cathode
(b) Copper ions are less reactive and more easily reduced than hydrogen ions, so copper is deposited at the cathode. · must show Cu2+ less reactive than H+, Cu2+ reduced in preference to H+, Deposition of copper at cathode
(c) In NaCl solution, sodium hydroxide forms (solution becomes alkaline). In CuSO4, copper ions are removed so the solution becomes less blue and eventually forms dilute sulfuric acid (acidic). · must show NaCl: sodium hydroxide forms/more OH- and Na+ remain, CuSO4: copper ions removed, solution becomes acidic/dilute sulfuric acid, Ionic explanation for each
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Marking points
M1 (a) · 1 mark · States (or implies) H+ ions from water are discharged/reduced at the cathode.
M2 (a) · 1 mark · States or explains that OH- ions are produced/left behind/increase near the cathode.
A1 (a) · 1 mark · Links increase in OH- to solution becoming alkaline at the cathode.
M3 (b) · 1 mark · Mentions that copper is less reactive than hydrogen or that Cu2+ is preferentially reduced.
A2 (b) · 1 mark · States copper is deposited at the cathode because it is reduced.
M4 (c) · 1 mark · Identifies correct change for each solution: NaCl forms sodium hydroxide/alkaline; CuSO4 becomes acidic/dilute sulfuric acid.
A3 (c) · 1 mark · Describes for each: Na+ and OH- remain (NaOH); Cu2+ removed, H+ and SO4^2- remain.
Worked solution
a) During the electrolysis of sodium chloride solution, hydrogen ions from the ionisation of water are attracted to the cathode and are reduced to form hydrogen gas:
or, considering water directly:
This produces hydroxide ions which stay in the solution, making it alkaline around the cathode.
b) In copper(II) sulfate solution, both Cu2+ ions and H+ ions are present at the cathode, but copper is less reactive than hydrogen and is preferentially reduced:
Copper is deposited at the cathode instead of hydrogen because copper(II) ions accept electrons more readily than hydrogen ions.
c) In sodium chloride solution, prolonged electrolysis removes Cl- and H+, leaving Na+ and OH- ions. The solution becomes sodium hydroxide (alkaline).
In copper(II) sulfate solution, prolonged electrolysis removes Cu2+ (deposited as copper) and OH- (at anode), eventually leaving H+ and SO4^2- ions, so the solution becomes dilute sulfuric acid (acidic).