(a)(i) List three factors which affect the discharge of ions during electrolysis (ii) Describe in outline, the purification of copper by electrolysis. (b) t...

(a)(i) Three factors that affect the discharge of ions during electrolysis

• The position of the ion in the electrochemical (discharge) series; the ion lower in the series is discharged in preference.
• The concentration of the ion in the electrolyte; a high concentration can favour the discharge of an ion that would otherwise not be discharged.
• The nature of the electrode used (whether it is inert, e.g. platinum/carbon, or attackable, e.g. copper).

(a)(ii) Purification of copper by electrolysis

• A thick block of impure copper is made the anode and a thin strip of pure copper is made the cathode.
• The electrolyte is acidified copper(II) tetraoxosulphate(VI) solution, \( CuSO_4 \).
• When current flows, the impure anode dissolves: \( Cu \rightarrow Cu^{2+} + 2e^- \).
• The copper ions migrate to the cathode and are deposited there as pure copper: \( Cu^{2+} + 2e^- \rightarrow Cu \).
• Soluble impurities such as iron and zinc remain in solution, while insoluble impurities such as silver and gold fall below the anode as anode mud.

(b)(i) Completed table

The quantity of electricity is \( Q = I \times t \), with \( I = 0.20\ A \) in each case.

Sample calculation: \( Q = 0.20 \times 900 = 180\ C \), and so on for the other rows.

(b)(ii) Graph of mass of M deposited against quantity of electricity passed

The points lie on a straight line passing through the origin, showing that the mass deposited is directly proportional to the quantity of electricity passed.

(b)(iii) Mass of M deposited by the same current for 20 minutes

Time \( t = 20 \times 60 = 1200\ s \), so the quantity of electricity is \[ Q = I \times t = 0.20 \times 1200 = 240\ C. \]Reading up from \( Q = 240\ C \) on the horizontal axis to the line of best fit, and across to the vertical axis, the mass deposited is 0.08 g. (Check: the gradient is \( \dfrac{0.24}{720} = 3.33\times10^{-4}\ \text{g C}^{-1} \), so \( 3.33\times10^{-4} \times 240 = 0.08\ g \).)

(b)(iv) Law verified

Because the graph is a straight line through the origin, the mass deposited is proportional to the quantity of electricity passed. The experiment therefore verifies Faraday's first law of electrolysis.