silvery liquid brown gas D.C. supply − + Carbon (cathode) Carbon (anode) Molten lead(II) bromide Heat Copyright Eagle Beacon Global The diagram shows a smal...
Question 1 Report
The diagram shows a small electrolytic cell being used by a metal-recycling club to separate molten lead(II) bromide. The compound is heated until it is a liquid, then a direct current is passed through it using two carbon electrodes. The student records the current shown in Table 1. A brown gas forms near one electrode and a silvery liquid forms near the other electrode.
Table 1 shows the electrical conditions used.
(a) Name the ion that moves towards the positive electrode. [1]
(b) Give the observations at both electrodes. [2]
(c) Complete the two half-equations for the reaction.
2Br- → __________ + __________
Pb2+ + __________ → Pb [3]
(d) Explain why lead(II) bromide must be molten rather than solid for electrolysis to take place. [2]
(e) Calculate the charge transferred when a current of 1.5 A flows for 12 minutes. [1]
Answer Details
(a) Negative bromide ions move to the positive electrode, so the ion is bromide, Br- [1].
(b) At the positive electrode there is brown bromine gas/vapour [1]. At the negative electrode there is silvery-grey liquid lead [1].
(c) At the positive electrode bromide ions lose electrons: \[\mathrm{2Br^-\rightarrow Br_2+2e^-}\] [2]. At the negative electrode lead ions gain electrons: \[\mathrm{Pb^{2+}+2e^-\rightarrow Pb}\] [1].
(d) Lead(II) bromide must be molten because its ions are then free to move [1]. Moving ions carry charge and complete the circuit [1]. In a solid ionic lattice, ions are fixed in place.
(e) \[12\text{ min}=720\text{ s}\] \[Q=It=1.5\text{ A}\times720\text{ s}=1080\text{ C}\] The stem labels this as [1], but the supplied scheme awards a mark for the time conversion and a mark for the calculation.
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