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, Ag+ + e− ⇌ Ag E = +0.80 V Cu2+ + 2e− ⇌ Cu E = +0.34 V Ni2+ + 2e− ⇌ Ni E = −0.25 V A metal with a more negative (or less positive) E value is more readily oxidised. At the anode, Ni is preferentially oxidised to Ni2+ and then Cu is also oxidised (as Ni is a minor impurity) to Cu2+ and these ions enter the electrolyte solution. Silver with the most positive E value is not oxidised but is collected below the anode as ‘anode sludge’. A metal ion with a more positive (or less negative) E value is more readily reduced. At the cathode, since Cu2+ has a more positive E value than Ni2+, Cu2+ is reduced to Cu and deposits on the cathode as copper metal. Ni2+ remains in the electrolyte solution and is not reduced at the cathode. (b)(i) Cu2+ + 2e− → Cu 2 x 96500 C (2 F) of charge is required to deposit 1 mol of Cu Quantity of charge passed = 2.00 x 23.0 x 60 = 2760 C nCu = nCu2+ = 2760 2 x 96500 = 0.01430 mol Mass of Cu deposited = nCu x Ar of Cu = 0.01430 x 63.5 = 0.908 g Increase in mass of cathode is expected to be 0.908 g. (b)(ii) nNi2+ = nNi(C4H7N2O2)2 = 𝑚𝑎𝑠𝑠 𝑀 = 0.492 58.7+2(12.0×4+1.0×7+14.0×2+16.0×2) = 1.704 x 10−3 mol Mass of Ni oxidised = 1.704 x 10−3 x 58.7 = 0.100 g Mass of Ag = mass of anode sludge = 0.0500 g Mass of Cu removed = 0.950 − 0.100 − 0.0500 = 0.800 g 9 (a) Cathode: 2H2O + 2e– → H2 + 2OH– Anode: 2Br– → Br2 + 2e– Comments • Students need to indicate whether a half equation is a cathode or anode reaction. • Na+ will not be reduced at the cathode as E(Na+/Na) is less positive than E(H2O/H2). • When writing equations for anode or cathode reactions, you must use a ‘⎯→’ sign, NOT ‘⇌’ sign. • The anode is where oxidation occurs, so electrons must appear on the product side of the equation. The opposite is true at the cathode. (b) S2O32– + 4Br2 + 5H2O → 2SO42– + 10H+ + 8Br– Amt of S2O32– = 35.60 × 0.5001000 = 0.0178 mol 2SO42– + 10H+ + 8e– ⇌ S2O32– +5H2O 8Br– ⇌ 4Br2 + 8e– Number of moles of electrons = 8 x 0.0178 = 0.142 mol Comments • A number of students are unable to write a correct balanced redox equation as they failed to begin by writing 2 half equations. • It must be clear in your answer how the mole ratio between S 2O32- and electrons was derived.

5 (c) Since both cells are arranged in series, Crn+ + ne– → Cr(s) Amt of Cr = 3.68 52.0 = 0.0708 mol n = 0.142 / 0.0708 = 2.01 ≈ 2 (whole number) Comments • n has to be expressed as a whole number. • n = amt of electrons / amt of Cr, not the other way round. 10 Number of Cu atoms in 0.1 m length = 0.1 3.0 x 10–12 = 3.333 x 1010 Number of Cu atoms in 0.1 m x 0.1 m electrode = (3.33 x 1010)2 = 1.111 x 1021 Number of Cu atoms if electrode was coated with a total depth of 2000 atoms (1000 atoms on each side) = 2000 x 1.111 x 1021 = 2.222 x 1024 Amount of Cu atoms if electrode was coated with a total of 2000 atoms = 2.222 x 1024 6.02 x 1023 = 3.691 mol Cu2+(aq) + 2e– → Cu(s) Amt of electrons transferred = 2(3.691) = 7.382 mol Since Q = It = neF (4.0)(t) = 7.382(9.65 x 104) t = 178110 s = 178110 3600 h = 49.5 h