(a) Give the products of electrolysis of dilute copper (II) tetraoxosulphate(VI) solution using the following materials as electrodes: (i) carbon rod; (ii) ...
Question 1 Report
(a) Give the products of electrolysis of dilute copper (II) tetraoxosulphate(VI) solution using the following materials as electrodes:
(i) carbon rod;
(ii) copper rods
(b) For each of the process in (a) above;
(i) write the anodic half reaction;
(ii) state how electrolysis affects the pH of electrolyte
Answer Details
(a) Products of electrolysis of dilute copper(II) tetraoxosulphate(VI) solution
- (i) Carbon (inert) electrodes: copper is deposited at the cathode; oxygen gas is evolved at the anode. The blue colour fades and the solution becomes acidic (tetraoxosulphate(VI) acid remains).
- (ii) Copper electrodes: copper is deposited at the cathode; the copper anode dissolves (loses mass) so no gas is evolved. The concentration and colour of the solution stay essentially the same.
(b) For each process
(i) Anodic half reaction
- Carbon anode: \[2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-\]
- Copper anode: \[Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-\]
(ii) Effect on pH
- With carbon electrodes, oxygen is released and H+ ions accumulate in solution, so the pH decreases (the electrolyte becomes more acidic).
- With copper electrodes, copper dissolves at the anode at the same rate as it deposits at the cathode, so the Cu2+ concentration is unchanged and the pH remains essentially constant.
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