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)