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, i