RI Electrochemistry 1 Tutorial - Answers to Discussion Questions

7 ANSWERS TO PRACTICE QUESTIONS 4 Cell (1) (a) (i) Cr 2O72–(aq) + 14H+(aq) + 6e ⇌ 2Cr3+(aq) + 7H2O(l) E = +1.33 V ... (1) Cu 2+(aq) + 2e ⇌ Cu(s) E = +0.34 V ... (2) Reduction: Cr 2O72– (aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7H2O(l) Cathode: Pt electrode in Cr2O72–(aq) / Cr3+(aq) half–cell Oxidation: Cu(s)  Cu2+(aq) + 2e Anode: Copper electrode in Cu2+(aq) / Cu(s) half–cell Overall: Cr2O72–(aq) + 14H+(aq) + 3Cu(s)  2Cr3+(aq) + 7H2O(l) + 3Cu2+(aq) Note: Students are to note that both the half and full equations are written with  instead of ⇌ seen in the Data Booklet. (ii) Ecell = Ecathode – Eanode = +1.33 – (+0.34) = +0.99 V (iii) Cell (2) (a) (i) H 2O2(aq) + 2H+(aq) + 2e ⇌ 2H2O(l) E = +1.77 V O2(g) + 2H+(aq) + 2e ⇌ H2O2(aq) E = +0.68 V Reduction: H 2O2(aq) + 2H+(aq) + 2e  2H2O(l) Cathode: Pt electrode in H2O2(aq) / H2O(l) half–cell Oxidation: H 2O2(aq)  O2(g) + 2H+(aq) + 2e Anode: Pt electrode in O2(g) / H2O2(aq) half–cell Overall: 2H2O2(aq)  2H2O(l) + O2(g) Note: Students are to note that both the half and full equations are written with  instead of ⇌ seen in the Data Booklet. (iii) Ecell = Ecathode – Eanode = +1.77 – (+0.68) = +1.09 V V T = 298 K Pt Cu salt bridge [Cu2+(aq)] = 1 mol dm–3 [Cr2O72– (aq)] = [Cr3+(aq)] = [H+(aq)] = 1 mol dm–3 e–

8 (iv) (b) (i) Cu2+ precipitates out as CuCO3 and so [Cu2+(aq)] decreases. This shifts the position of equilibrium of (2) to the left so that E(Cu2+/Cu) becomes less positive. Hence, Ecell becomes more positive (i.e. Ecell > Ecell). (ii) Br2 + 2e– ⇌ 2Br– E = +1.07 V Br– reduces Cr2O72 to Cr3+. Note: Ecell = +1.33 – (+1.07) = +0.26 V > 0 (feasible) This decreases [Cr 2O72(aq)] but increases [Cr 3+(aq)] so that the position of equilibrium of (1) shifts to the left and so E(Cr2O72/Cr3+) becomes less positive. Hence Ecell becomes less positive (i.e. Ecell < Ecell). (c) To decrease Ecell, either (1) decrease E(H2O2/H2O) or (2) increase E(O2/H2O2). Change (1): E(H2O2/H2O) can be decreased by using H 2O2 or H + of a lower concentration (i.e. < 1 mol dm–3) the H2O2/H2O half–cell. Change (2): E(O2/H2O2) can be increased by using O 2 of a higher pressure (i.e. > 1 bar), H+ of a higher concentration (i.e. > 1 mol dm–3) or H2O2 of a lower concentration (i.e. < 1 mol dm–3) in the O2 / H2O2 half–cell. 5 From Data Booklet, E(H+/H2) = 0.00 V; E(Ag+/Ag) = +0.80 V In the reaction 2H+(aq) + Cd(s)  Cd2+(aq) + H2(g), Cd is oxidised. Thus Ecell = Ecathode – Eanode = E(H+/H2) – E(Cd2+/Cd) = 0.00 – E(Cd2+/Cd) = +0.40 V  E(Cd2+/Cd) = –0.40 V In the reaction Pd2+(aq) + 2Ag(s)  2Ag+(aq) + Pd(s), Pd2+ is reduced. Thus Ecell = Ecathode – Eanode = E(Pd2+/Pd) – E(Ag+/Ag) = E(Pd2+/Pd) – 0.80 = +0.19 V  E(Pd2+/Pd) = +0.99 V Therefore, potential of Pd relative to Cd is +0.99 – (–0.40) = +1.39 V V T = 298 K Pt [H2O2(aq)] = [H+(aq)] = 1 mol dm–3 Pt O2 (1 bar) [H2O2(aq)] = [H+(aq)] = 1 mol dm–3 salt bridge e–

9 6 (a) (i) O2 + 4H+ + 4e  2H2O Note: This half-equation can be found in the Data Booklet. (ii) Reduction: O2 + 4H+ + 4e  2H2O (  3) Oxidation: CH 3OH + H2O  CO2 + 6H+ + 6e ( 2) Overall: 2CH 3OH + 3O2  4H2O + 2CO2 (iii) From Data Booklet, E(O2/H2O) = +1.23 V Ecell = Ecathode – Eanode Thus, +1.18 = +1.23  E(CO2/CH3OH) E(CO2/CH3OH) = +1.23  (+1.18) = +0.05 V (iv) Any one of the following:  Methanol has a more efficient storage (by volume or by weight since it is a liquid) compared to compressed hydrogen gas.  Hydrogen gas is more volatile and explosive, thus requires high pressure storage system but not methanol (less explosive). (b) (i) CH3CH2OH + 3O2  2CO2 + 3H2O (ii) G = H  TS G = (1367  103)  (298)(140) = 1.33 106 J mol1 = 1330 kJ mol1 (iii) Ecell = = +1.15 V 7 (a) (i) A saturated KC l solution is used to keep the concentration C l − constant in the reference electrode. Examiner Comments  This question was poorly done. The hint was given in the question regarding the nature of a reference electrode (fixed composition and constant potential). (ii) The E value is more positive than +0.241 V. At standard conditions, the KC l concentration is 1.00 mol dm 3, which is lower than the saturated KC l solution. Therefore, the equilibrium position of the half-equation lies more to the right, resulting in a more positive E value. Note: Concentration of a saturated solution of KCl is 4.55 mol dm–3 at 20 ºC. Examiner Comments  Students who lost marks here were careless in their reading of the question. The E value (+0.241 V) given is not at standard conditions.  When answering questions, students should be clear in referring to E or E. E is used to represent the potential at any conditions that are not standard. (b) (i) Ag+(aq) + e ⇌ Ag(s) E = +0.80 V Since the Ag+/Ag electrode has the more positive E, it is the cathode. The S.C.E. is the anode.  6- 1 -1 (1 . 3 3 × 1 0 J m o l ) 12 (96500 C mol )

10 (ii) Ecell = +0.80 – 0.241 = +0.559 V Examiner Comments  This question was generally well done.  Students are reminded to express thei r answers in 3 s.f. and with units. (iii) At very low concentrations of Ag+, small increases in [Ag+] result in an exponential/ large increase in Ecell as shown by the steep gradient of the graph. Examiner Comments  Some students described the shape of t he graph in terms of rate, e.g. increased rapidly, increased at a greater rate, etc. This is not accepted as there is no time scale on the graph.  A number of students discussed the gen tle gradient at relatively higher [Ag +]. This is not accepted as it doesn’t answer the question. 8 (a) Ecell = +0.80 – (+0.34) = +0.46 V (spontaneous) (b) Ecell = +1.07 – (+0.77) = +0.30 V (spontaneous) (c) Ecell = –1.66 – (+1.36) = –3.02 V (not spontaneous) (d) Ecell = –0.43 – (–0.83) = +0.40 V (spontaneous) (e) Ecell = +1.52 – (+0.68) = +0.84 V (spontaneous) (f) They are both oxidising agents hence they cannot react with each other. (g) Zn2+ + 2e ⇌ Zn E = –0.76 V ....... (1) Cr2O72 + 14H+ + 6e ⇌ 2Cr3+ + 7H2O E = +1.33 V ....... (2) Cr3+ + e ⇌ Cr2+ E  = –0.41 V ....... (3) Cr2+ + 2e ⇌ Cr E  = –0.91 V ....... (4) Cr3+ + 3e ⇌ Cr E  = –0.74 V ....... (5) Combining half reactions (1) and (2), Ecell = +1.33 – (–0.76) = +2.09 V (spontaneous) Combining half reactions (1) and (3), Ecell = –0.41 – (–0.76) = +0.35 V (spontaneous) Combining half reactions (1) and (4), Ecell = –0.91 – (–0.76) = –0.15 V (not spontaneous) Thus, Zn will reduce Cr2O72 to Cr3+ and further reduce Cr3+ to Cr2+. 9 (a) MnO4–(aq) + 8H+(aq) + 5e ⇌ Mn2+(aq) + 4H2O(l) E = +1.52 V Cl2(g) + 2e ⇌ 2Cl( a q ) E = +1.36 V Cr2O72(aq) + 14H+(aq) + 6e ⇌ 2Cr3+(aq) + 7H2O(l) E = +1.33 V Under standard conditions, MnO4 can oxidise Cl to Cl2, i.e. Ecell = +1.52 – (+1.36) = +0.16 V > 0 In the titration determination of iron(II) ions, the amount of Fe2+ ions present is determined by the amount of oxidant it reacts with. If MnO4 were used, some MnO4 will be consumed by Cl instead of reacting solely with Fe2+. This would result in a larger amount of MnO 4– used than required i.e. the results would be inaccurate.

11 On the other hand, under standard conditions, Cr2O72 cannot oxidise Cl to Cl2, i.e. Ecell = +1.33 – (+1.36) = –0.03 V < 0 Hence all Cr2O72 will be solely used to react with Fe2+ ions and thus K2Cr2O7(aq) can be used for the titration determination of iron(II) ions in the presence of dilute HCl(aq