2023 ASRJC-NYJC-SAJC H3 Chemistry Prelim Answer

H3 Section A A collaboration between ANDERSON SERANGOON JUNIOR COLLEGE NANYANG JUNIOR COLLEGE St. ANDREW’S JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 3 CANDIDATE NAME CLASS CHEMISTRY 9813/01 Paper 1 21 September 2023 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet I n s e r t READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer two questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 42 printed pages and 0 blank page and 1 insert.

2 H3 Chemistry 9813/01 Answer all questions in this section. 1 (a) (i) Zn  ZnO Zn + H 2O  ZnO + 2H+ + 2e– Zn + 2OH–  ZnO + H2O + 2e– (ii) [O] Zn + 2OH−  ZnO + H2O + 2e− [R] O2 + 2H2O + 4e−  4OH− Overall: 2Zn + O2  2ZnO (iii) Zn + ½ O2  ZnO E 0cell = +0.40 – (−1.25) = +1.65 V G0 = – 2 × 96500 × 1.65 = −318450 J mol−1 = −318 kJ mol−1 (3 sig fig) (iv) Since air is a mixture, the pressure of oxygen in the atmosphere is less than 1 bar. This will cause the E0(O2/OH−) to be less positive resulting in a less positive E 0cell and hence a smaller magnitude in the Gibbs free energy change. (b) CO2 + OH−  HCO3− This will decrease the concentration of OH− in the electrolyte and makes the E0(ZnO/Zn) more positive which will make the E0cell less positive. Explanation (not required in answer) ZnO + H2O + 2e– ⇌ Zn + 2OH– E0(ZnO/Zn) = −1.25 V When [OH −] decreases, position of the equilibrium will shift to the right and favor the reduction reaction. This will cause E0(ZnO/Zn) to become more positive. Since E 0cell = E 0(O2/OH–) – E 0(ZnO/Zn), a more positive E 0(ZnO/Zn) will cause the E 0cell to become less positive. OR CO2 + OH− + Zn2+  ZnCO3 + H2O The insoluble zinc carbonate formed can clog up the pores of the carbon electrode and can prevent the reduction of oxygen to hydroxide ions. (c) (i) Oxidation state of O in LiO2: −0.5 Oxidation state of O in Li2O2: –1 Oxidation state of O in Li2O: –2 (ii) O2 + Li+ + e−  LiO2 (1) 2LiO2  Li2O2 + O2 (2) LiO2 + Li+ + e−  Li2O2