RI Electrochemistry 2 Tutorial - Answers to Disucssion Questions

3 Suggested solutions to discussion questions 6 (a) Amt of Ag deposited = 0.100 / 107.9 = 9.27 x 10–4 mol (b) Ag+(aq) + e– → Ag 96500 C (1 F) of charge is required to deposit 1 mol of Ag Amt of Ag deposited = 9.27 x 10–4 mol Quantity of charge passed = 9.27 x 10–4 x 96500 = 89.43 C Current passed = 89.43 / (30 x 60) = 0.0497 A (c) Cr3+(aq) + 3e− → Cr(s) 3 x 96500 C (3 F) of charge is required to deposit 1 mol of Cr Amt of Cr deposited = 89.43 3 x 96500 = 3.089 x 10–4 mol Mass of Cr deposited = 3.086 x 10–4 x 52.0 = 0.0161 g 7 (a) Al3+ + 3e– ⇌ Al E = –1.66 V Pb2+ + 2e– ⇌ Pb E = –0.13 V Lead has a less negative reduction potential (than aluminium) and so it is relatively easier for its ion to be reduced to the metal. Hence its oxide can be reduced by C. Aluminium has a very negative reduction potential and so its ion is not easily reduced to the metal. Its ore (e.g. Al2O3) cannot be reduced by C and requires electrolysis. Note: Carbon is a cheap raw material (reducing agent). Electrolysis, however, is expensive due to the high electrical power needed to drive the non -spontaneous reaction. As such, reduction by C is preferred. However, for Al3+, reduction by C cannot occur. As such, it has to be extracted from Al2O3(ore) via electrolysis. Since Al2O3 is insoluble in water, it needs to be in the molten state which adds to the total cost of extraction as the melting process requires high electrical power. (b) A species with a less positive E is more easily oxidised. F2 + 2e– ⇌ 2F− E = +2.87 V Cl2 + 2e– ⇌ 2Cl− E = +1.36 V O2 + 4H+ + 4e– ⇌ 2H2O E = +1.23 V E(O2/H2O) is only slightly less positive than E(Cl2/Cl−). A higher [Cl–] causes E(Cl2/Cl–) to decrease, i.e. less positive than +1.36 V. When concentrated aqueous NaCl (brine) is electrolysed, [C l−] is sufficiently high for E(Cl2/Cl–) to be less positive than E(O2/H2O), so that Cl2 is produced at the anode. E(O2/H2O) is significantly less positive than E(F2/F−). Though a higher [F–] causes E(F2/F–) to decrease, E(O2/H2O) will still be less positive than E(F2/F–), even for aqueous solution containing a very high concentration of F−. If concentrated aqueous sodium fluoride were to be electrolysed, H2O will still be oxidised in preference to F–. O2 will be produced at the anode instead of F2. Hence, F2 cannot be produced by the electrolysis of concentrated aqueous sodium fluoride. Comments Most students did not realise that the question actually required them to contrast between chloride and fluoride. Simply rehashing that the E value of F 2 is more positive than that of chlorine does not earn credit. Stating that there is a “concentration effect” without describing it in greater detail also does not earn credit. In addition, some students compared the wrong E values. When we discuss selective discharge at the anode, we should compare the E value of F2 or Cl2 with that of O2/H2O rather than the E value of H2O/H2.

4 8 (a) From the Data Booklet