This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
You're on a roll — here are the solutions to the questions: c) i) • A Lewis acid is an electron pair acceptor. • A Lewis base is an electron pair donor. c) ii) A buffer solution is a solution that resists significant changes in pH upon the addition of small amounts of acid or base. One use of a buffer solution is to maintain a stable pH in biological systems, such as blood, which must remain within a narrow pH range for proper physiological function. c) iv) Step 1: Identify the given values and the relevant equation. The solution is a buffer made from a weak acid (butanoic acid, CH_3CH_2CH_2COOH) and its conjugate base (butanoate ion from potassium butanoate, CH_3CH_2CH_2COOK). [Acid] = 0.01 M [Base] = 0.1 M K_a = 1.5 × 10^-5 Use the Henderson-Hasselbalch equation: pH = pK_a + ([Base][Acid]). Step 2: Calculate pK_a. pK_a = -(K_a) pK_a = -(1.5 × 10^-5) pK_a ≈ 4.82 Step 3: Substitute the values into the Henderson-Hasselbalch equation and calculate the pH. pH = 4.82 + ((0.1)/(0.01)) pH = 4.82 + (10) pH = 4.82 + 1 pH = 5.82 d) i) When the cell is connected: A. The species that is oxidized is Sn(s). (Since Sn^2+/Sn has a more negative standard reduction potential, Sn will lose electrons and be oxidized). B. The species that is reduced is MnO_4^-(aq). (Since MnO_4^-/Mn^2+ has a more positive standard reduction potential, MnO_4^- will gain electrons and be reduced). d) ii) A suitable instrument that can be used to determine the emf of the cell is a voltmeter (or potentiometer). d) iii) The composition and function of the part labeled S: • Composition: A salt bridge, typically containing an inert electrolyte such as potassium nitrate (KNO_3) or potassium chloride (KCl) in a gel. • Function: It completes the electrical circuit by allowing the migration of ions between the two half-cells, thereby maintaining electrical neutrality and preventing charge buildup in the solutions. d) iv) Calculate the emf of the cell. Step 1: Identify the standard reduction potentials for the cathode and anode. The reduction half-reaction is MnO_4^- + 8H^+ + 5e^- Mn^2+ + 4H_2O, so E^_cathode = +1.5 V. The oxidation half-reaction is Sn Sn^2+ + 2e^-, so E^_anode = -0.14 V. Step 2: Use the formula for the standard cell potential. E^_cell = E^_cathode - E^_anode Step 3: Substitute the values and calculate. E^_cell = (+1.5 V) - (-0.14 V) E^_cell = 1.5 V + 0.14 V E^_cell = 1.64 V d) v) Platinum is used in the RHS of the cell because it is an inert electrode. It provides a surface for the electron transfer to occur in the MnO_4^-/Mn^2+ redox couple without reacting itself, as there is no solid reactant or product to serve as an electrode. What's next? 📸