SASS Chem Electrolysis AfL Ans

Page 1 ST ANDREW’S SECONDARY SCHOOL NAME TEACHING GROUP CHEMISTRY (G3) 6092 14.1 Electrolytic Cells ELECTROLYSIS Assessment for Learning 1 [B] At anode (ions attracted: Cl– & OH–) • [Cl–] > [OH–]: concentration effect applies. • Thus, Cl– ions are selectively discharged over OH– ions, and lose e– to form Cl2 gas. • Oxidation: 2 Cl– (aq) ® Cl2 (g) + 2 e– At cathode (ions attracted: M2+ & H+) • M forms M2+; valency = 2, based on MCl2. • Position of M is lower than H in the reactivity series: between Cu & Ag. • Thus, M2+ ions are selectively discharged over H+ ions, and gain e– to form solid M. • Reduction: M2+ (aq) + 2 e– ® M (s) 2 [A] • Position of Cu is lower than H in the reactivity series. • Thus, Cu2+ ions are selectively discharged over H+ ions, and gain e– to form Cu metal at cathode. • This led to the increase in mass of the cathode. 3 [A] • Negative electrode (cathode) attracts all 3 cations in the electrolyte (Cu2+, H+, Mg2+). • Position of Cu is the lowest amongst them in the reactivity series. • Thus, Cu2+ ions are selectively discharged over H+ and Mg2+ ions, and gain e– to form Cu metal at cathode. 4 None of the options is the correct answer... Option A: Ba2+ ions gain electrons at the anode cathode. Option B: Ba2+ ions lose gain electrons at the cathode. Option C: Cl– ions gain lose electrons at the anode. Option D: Cl– ions lose electrons at the cathode anode. 5 [C] • Position of Cu is lower than H in the reactivity series. • Thus, Cu2+ ions are selectively discharged over H+ ions, and gain e– to form Cu metal at cathode. • [Cu2+] continues to decrease during the electrolysis, causing the blue colour of the solution to fade. Option A: A pink-brown solid is deposited at the anode cathode. Option B: Bubbles form at the negative positive electrode. (O2 evolved: OH– are discharged) Option D: The negative electrode becomes smaller bigger. (size increases due to Cu deposited) SOLUTIONS

Page 2 6 [B] At anode (ions attracted: Cl– & OH–) • [Cl–] > [OH–]: concentration effect applies. • Thus, Cl– ions are selectively discharged over OH– ions, and lose e– to form Cl2 gas. • Cl2 formed dissolves back into solution: Cl2 + H2O ® HCl + HOCl (OCl– bleaches litmus). At cathode (ions attracted: Na+ & H+) • Position of Na is higher than H in the reactivity series. • Thus, H+ ions are selectively discharged over Na+ ions, and gain e– to form H2 gas. • [H+] decreases as the reaction progresses: [H+] < [OH–]. Litmus turns blue colour. 7 [A] • X forms at negative electrode ⇒ X is a cation ⇒ X is metallic / H+ (molten compound) ⇒ X is a metal. • Y forms at positive electrode ⇒ Y is an anion ⇒ Y is a non-metal. 8 [B] • Bulb lights = Circuit is closed = Electrolyte used. • Gas produced at each electrode: Cathode = H2 ; Anode = O2 / Cl2. 9 [C] • Oxidation occurs at anode. With a diatomic molecule formed at the anode, the electrodes used cannot be active, ie. inert. • X– ions: X has a valency of 1, X is a halogen (Group VII). • To selectively discharge X2,electrolyte used must be concentrated. 10 [B] • Reduction occurs at the cathode; Oxidation occurs at the anode. • Note the use of inert electrodes. Eqn 2: (At anode) As inert electrodes are used, ions in the electrolyte are involved in electrolysis. Hence, OH– ions (in the electrolyte) undergo oxidation. Eqn 3: (At cathode) As Cu is lower than H in the reactivity series, Cu2+ ions are selectively discharged. Hence, Cu2+ ions (not H+) undergo reduction.

Page 1 ST ANDREW’S SECONDARY SCHOOL NAME TEACHING GROUP CHEMISTRY (G3) 6092 14.2 Simple Cells ELECTROLYSIS Assessment for Learning 1 [D] • The further apart two metals are in the reactivity series, the greater the voltage produced. 2 [B] • The further apart two metals are in the reactivity series, the greater the voltage produced. • Using a porous membrane allows certain ions to pass through, usually H+. This ensures electrical conduction within the electrolyte (ie. closed circuit) while preventing both solutions used from mixing. 3 [B] Statement 1 is correct: Mg/Ag is further apart compared to Mg/Cu. Statement 2 is correct: Both positive and negative ions are required for electrical conduction. Statement 3 is wrong: ‘Amount of e–’ varies with the valency of the metal. Consider Na & Al. Na is more reactive than Al. Each mole of Na loses 1 mole of e–, whereas each mole of Al loses 3 moles of e–. 4 [D] • Zn/Cu: As Zn is more reactive than Cu, it undergoes oxidation and loses e– to form Zn2+ ions. Option A: Electrons move through the electrolyte external circuit. Option B: Mass of the zinc electrode increases decreases. Option C: The electrolyte changes its colour to blue. No change in the colour of the electrolyte. 5 [D] • Electrolyte must contain mobile ions serving as mobile charge carriers that conduct electricity in a complete circuit. Option A: Electrodes of the same metal used. (With both electrodes having the same electric potential energy (same metal), no electricity is being produced as there is no e– flow between them (ie. pd = 0 V).) Option B: Solid NaCl used. (As the ions in the crystal lattice are still held in their fixed positions, they are not mobile and cannot conduct electricity.) Option C: Liquid C2H5OH (ethanol) used. (As ethanol molecules are covalently-bonded, they are electrically neutral. Hence, they will not be attracted to the electrodes and cannot conduct electricity.) SOLUTIONS

Page 2 6 [B] • This electrical circuit may be considered as having the cell with Zn/Cu electrodes as a battery (simple cell) connected to the cell with electrodes 1 and 2 (electrolytic cell). • Zn is more reactive than Cu: Zn loses electrons to electrode 1: electrode 1 identified as the negatively-charged cathode of an electrolytic cell. • Since electrode 1 is the cathode, electrode 2 will be the positively-charged anode of an electrolytic cell. At electrode 1 (e– rich): Reduction occurs. Cu2+ ions are selectively discharged instead of H+ ions. [Cathode] ⇒ Cu2+ + 2 e– ® Cu At electrode 2: Oxidation occurs. To complete the circuit, e– are lost to the Cu electrode. [Anode] OH– ions are selectively discharged instead of SO42– ions. ⇒ 4 OH– ® O2 + 2 H2O + 4 e– 7 [C] • e– travels from the more reactive metal electrode to the less reactive metal electrode. • Metal X undergoes oxidation to form ions by losing its e–. Option A: Copper is the anode cathode. Option B: Mass of the copper electrode decreases remains the same. Option D: Metal X is below above copper in the reactivity series. Page 3 6 An electrical circuit was set up using two simple cells as shown. Both electrodes 1 and 2 are made of graphite. Which row shows the products formed at electrodes 1 and 2? Electrode 1 Electrode 2 A Cu metal H2 gas B Cu metal O2 gas C H2 gas Cu metal D O2 gas Cu metal 7 The diagram shows a simple cell made using copper and metal X as electrodes. The direction of electron flow in the cell is indicated with arrows. Which statement about this simple cell is correct? A Copper is the anode. B Mass of the copper electrode decreases. C Mass of the metal X electrode decreases. D Metal X is below copper in the reactivity series. A D inert electrode copper dilute sulfuric acid aqueous copper(II) sulfate electrode 2zinc electrode 1 6 9701/51/M/J/14© UCLES 2014 2 An experiment was set up to investigate how the cell potential of a cell containing a metal, M, in contact with an aqueous solution of its ions, Mn+(aq) (where n = 1, 2 or 3), changed as Mn+(aq) was diluted. Since a standard hydrogen half-cell was not available, a standard half-cell consisting of silver in contact with a 1 mol dm–3 solution of silver ions was used to connect to the half-cell with M in contact with