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Answers to Electrolysis III (Discussion)

Answers to Electrolysis III (Discussion). MCQ D A B C B C C A D A. 11. D D D D A D A D B C. Answer to MCQ 18 No. of mol of Cu deposited = 16/64 = 0.250 Cu 2+ + 2e -  Cu 1 mol of Cu is produced by 2 mol of electrons

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Answers to Electrolysis III (Discussion)

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  1. Answers to Electrolysis III (Discussion) • MCQ • D • A • B • C • B • C • C • A • D • A • 11. D • D • D • D • A • D • A • D • B • C

  2. Answer to MCQ 18 No. of mol of Cu deposited = 16/64 = 0.250 Cu2+ + 2e- Cu 1 mol of Cu is produced by 2 mol of electrons 0.250 mol of Cu is produced by 0.500 mol of electrons Let the charge on titanium ion be n. Tin+ + ne-  Ti Since same quantity of electricity deposit 16 g of Cu & 6 g of Ti, hence no. of electron to deposit Ti is also 0.500. No. of mol of Ti = 6/48 = 0.125 0.125 mol of Ti is produced by 0.500 mol of electron. 1 mol of Ti is produced by 0.500/0.125 = 4 mol of electrons. Hence n = 4

  3. Section B, Question 1 • a) Cathode: Ca2+ + 2e- Ca • Anode: 2Br- Br2 + 2e- • Product at cathode: calcium; • Product at anode: bromine gas • (Read from question that electrolyte is molten CaBr2) • Cathode: 2H+ + 2e- H2 • Anode: 4OH- 2H2O + O2 + 4e- • Product at cathode: hydrogen gas • Product at anode: oxygen • (Read from question that electrolyte is dilute NaCl, hence OH- is preferentially discharged. )

  4. Section B, Question 1 c) Cathode: Cu2+ (aq) + 2e- Cu (s) Anode: 4OH- (aq)  2H2O (l) + O2 (g) + 4e- Product at cathode: copper Product at anode: oxygen gas (and water)

  5. Question 1 d) Cathode: Cu2+ (aq) + 2e- Cu (s) Anode: Cu (s)  Cu2+ (aq) + 2e- Product at cathode: copper (Read question, it says “electrolysis of aq. Copper(II) sulfate using copper electrodes, hence at anode copper dissolves.)

  6. Question 2 • *a) i) Steel will be corroded/oxidised by chlorine produced at the anode. Iron in steel will be oxidised by chlorine gas to form iron (III) chloride • (Recall Cl2 is an oxidising agent, 2Fe + 3Cl2 2FeCl3) • ii) graphite (give an inert electrode) • b) 2Cl- Cl2 + 2e- • c) Effervescence of a yellowish-green and pungent gas. • Hydrogen gas • (Read Q, electrolyte is conc NaCl, since Na+ is not discharged in aqueous form, H+ is preferentially discharged to give hydrogen gas).

  7. Section B, Question 3 • A: tin B: mild steel • (Note question asks for materials, thus the answer food can is not accepted for electrode B.) • *b) Reaction at A (anode): Sn  Sn2+ + 2e- • Reaction at B (cathode) : Sn2+ + 2e-  Sn • (Although tin is in group IV, it has 2 oxidation states, +2 and +4, just like lead. You should infer from the diagram, electrolyte being tin(II) sulfate that the oxidation state here is 2 and not 4.)

  8. Question 3 *c) At the anode, each tin atom loses 2 electrons to form a tin(II) ion. At the cathode, each tin(II) ion gains 2 electrons to form a tin atom. Thus for every tin(II) ion that is discharged at the cathode, one tin(II) ion is formed at the anode. Hence concentration of the electrolyte remains the same.

  9. Question 3 • Dilute sulfuric acid (any strong acid) • (Rust is hydrated iron (III) oxide which is a base. Base reacts with acid – sec 3 work) • **e) Electroplating uses less tin and hence it is more economical/prevents wastage. • (Read from question 1st line: Food cans are made of mild steel with a layer of about 10-7 cm thick, and from the part which says “steel used to be plated by dipping it in molten tin forming a layer of tin 0.005 cm thick.)

  10. - + Silver anode Spoon at cathode Aqueous silver nitrate Question 4 a) • Anode: Ag  Ag+ + e- • Cathode: Ag+ + e-  Ag • (MUST label cathode and anode in your answers)

  11. Question 5 a) Cu2+, H+, OH-, SO42- b) Cathode: Cu2+ (aq) + 2e- Cu (s) Anode: 4OH- (aq)  2H2O (l) + O2 (g) + 4e- (Must specify cathode and anode in your answer) *c) No. of mol of Cu in 3.20 g = mass in g/Mr in g = 3.20/64 = 0.0500 1 mol of Cu is formed from 1 mol of CuSO4. Hence, 0.0500 mol of Cu is formed from 0.0500 mol of CuSO4. Volume of CuSO4 solution = 250cm3. Hence concentration = 0.0500  250/1000 = 0.200 mol/dm3(3 sf)

  12. Question 5 • The blue electrolyte would fade and turn colourless. • As copper (II) ions are reduced or gain electrons to form copper, concentration of copper (II) ions in solution would decrease over time. • [Must use the term “concentration” here!]

  13. Question 6 • Impure copper should be the at the anode while pure copper could be at the cathode. • The electrolyte should be changed to aqueous copper (II) sulfate. • b) Cu (s)  Cu2+ (aq) + 2e- (Question asks for state symbols) • % purity of impure copper = 48.0/52.0 x 100 • = 92.3 %

  14. Question 7 *a) In solid state, the ions in copper (II) oxide are held in fixed positions by strong electrostatic forces. The ions are not mobile, hence the solid is not able to conduct electricity. b) Ammonia gas exists as molecules. There are no mobile ions. Also, there are no mobile electrons as the electrons are either used to form covalent bonds or are located in electron shells. Thus ammonia gas does not conduct electricity. Aqueous ammonia consists of ammonium and hydroxide ions. These ions are mobile in aqueous state and thus is able to conduct an electric current.

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