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Answers to Redox Reactions III

Answers to Redox Reactions III. MCQ D D B D C C. 7. C 8. D 9. D 10. A 11. D 12. D. Section B, Question 1 i) -3 ii) 0 iii) +2 iv) +4 v) +5 vi) -3 i) 0 ii) +2 iii) +3 iv) +6 v) +6 i) 0 ii) +4 iii) +6 iv) +6 +6 vi) –2 (Fe 2+ S 2- )

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Answers to Redox Reactions III

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  1. Answers to Redox Reactions III • MCQ • D • D • B • D • C • C 7. C 8. D 9. D 10. A 11. D 12. D

  2. Section B, Question 1 • i) -3 ii) 0 • iii) +2 iv) +4 • v) +5 vi) -3 • i) 0 ii) +2 • iii) +3 iv) +6 • v) +6 • i) 0 ii) +4 • iii) +6 iv) +6 • +6 vi) –2 (Fe2+S2-) • vii) +2 viii) +4

  3. Question 2 Read question, you must underline !! Always underline the reactant, NOT product ! a) Zn (s) + 2H+ (aq)  Zn2+ (aq) + H2 (g) The oxidation no. of zinc increases from 0 (in Zn) to +2 (in Zn2+). b) 2I- (aq) + Cl2 2Cl- (aq) + I2 (s)   The oxidation no. of iodine increases from -1 (in I-) to 0 (in I2). [Cannot write “from -1 (in 2I-) to … 2 is NOT part of formula!!, write as -1 (in I-)

  4. Question 2 • 2Na (s) + 2C2H5OH (l) 2C2H5O -Na+ (aq) + H2 (g) • The oxidation no. of sodium increases from 0 (in Na) to +1 (in C2H5O-Na+) • d) 2F2 (g) + 2H2O 4HF (aq) + O2 (g) • The oxidation no. of oxygen increases from -2 (in H2O) to 0 (in O2). • e) 2SO2 + O2 2SO3 • The oxidation no. of sulfur increases from +4 (in SO2) to +6 (in SO3) • CANNOT write as “+4(in 2SO2) to +6 (in 2SO3)!! • DO NOT include the 2 as the formula!!

  5. Question 2 f)FeO4- + MnO2 + 4H+ Fe3+ + MnO4- 2H2O The oxidation no. of manganese increases from +4 (in MnO2) to +7 (in MnO4-)

  6. Question 3 • 2Fe(s) + 3Cl2 (g)  2FeCl3 (s) • (Recall all halogens are diatomic, hence formula of chlorine gas is Cl2 and NOT Cl) • b) Substance oxidised is iron. (NOT iron (III) atom!!) • Each iron atom loses three electrons to form a iron(III) ion. (Fe  Fe3+ + 3e-) • Aqueous potassium iodide • (To carry out reduction, obviously you must use a reducing agent, hence use KI, a powerful reducing agent which gives a distinctive colour change.) • Fe3+ (aq) + e-  Fe2+ (aq) (Read question which says half equation that illustrates reduction) • (In ‘O’ level, half eqn also known as ion-electron eqn or simply ionic eqn)

  7. Question 4 • Oxidation is the loss of electrons from a substance during a chemical change. • To about 2 cm3 of aqueous bleach, add equal volume of aqueous potassium iodide. The aqueous potassium iodide turns from colourless to brown, showing that the iodide ions have been oxidised by the bleach to form iodine . • (CANNOT write “The solution turns brown” • Don’t know which solution you are referring to, always state the name of solution; must always state initial colour!)

  8. Question 4 • Yes. • Chlorine is oxidised as the oxidation no. of chlorine increases from 0 (in Cl2 )to +1 (in ClO-). • Chlorine is also reduced as the oxidation no. of chlorine decreases from 0 (in Cl2) to -1 (in Cl-). • (To justify whether a reaction is redox, both oxidation and reduction in terms of change in oxidation no. must be clearly explained.) • (This is an example of disproportionation, whereby a substance is both oxidised and reduced at the same time.)

  9. Question 5 • The iodine solution turns from brown to colourless. • (Notice from the eqn that the products are all sodium compounds which are colourless) • No. of mol of iodine = 25/1000 x 0.100 = 0.00250 • 1 mol of I2 reacts with 2 mol of Na2S2O3. • Hence 0.00250 mol of I2 react with • 0.00250x2 = 0.00500 mol of Na2S2O3. • Answers to calculations must be left to 3 sf !!!!

  10. The following are strictly required for calculations (more so for SPA assessment in term 3): • Proper statements • E.g. No. of mol of I2 = … • Concentration of I2 = … • Ratio • E.g 1 mol of HCl reacts with 1 mol of NaOH • instead of 0.100 mol of HCl react with 0.100 mol of NaOH • 3 sf. • E.g. 0.100 mol/dm3, 0.0100 mol/dm3, 5.00 g/dm3 • Marks will be deducted in coming tests for not writing statements/ratio, incomplete working and wrong sf!!

  11. Question 5 d)20.0 cm3 of solution contain 0.00500 mol of Na2S2O3. Hence concentration of Na2S2O3 = 0.00500 20/1000 = 0.250 mol/dm3 e) Concentration of Na2S2O3 in g/dm3 = concentration in mol/dm3 x molar mass = 0.250 x (2x 23 + 2x32 + 3x16) = 39.5 g/dm3 (3 sf) f) Iodine. The oxidation no. of iodine decreases from 0 (in I2) to -1 (in NaI).

  12. Question 6 • Recall no. of e- must be the same for both equations before you add them up. • 2nd half equation x 5: • 5Fe2+  5Fe3+ + 5e- • Overall equation: • MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+ (aq) + 4H2O (l)+ 5Fe3+ (aq)

  13. Question 6 • Acidified potassium manganate (VII) • Acidified potassium manganate (VII) oxidises iron (II) ion to iron (III) ion as the oxidation no. of iron increases from +2 (in Fe2+) to +3 (in Fe3+). • OR The oxidation no. of manganese decreases from +7 (in MnO4-) to +2 (in Mn2+), thus potassium manganate (VII) is being reduced. Since oxidising agent undergoes reduction, potassium manganate (VII) is an oxidising agent. • Manganese (Note question asks for ELEMENT) • The acidified potassium manganate (VII) turns from purple to colourless. (Write full statement here!)

  14. Question 7 • i) No. of mol of Ti2+ = 40.0/1000 x 0.0200 = 0.000800 • ii) No. of mol of Fe3+ = 32.0/1000 x 0.0500 = 0.00160 • iii) 0.000800 mol of Ti2+ react with 0.00160 mol of Fe3+ • 1.00 mol of Ti2+ reacts with 2.00 mol of Fe3+ • Fe3+ + e- Fe2+ • 1 mol of Fe3+ gains 1 mol of e-, hence 2 mol of Fe3+gain 2 mol of e-. • This means that 1 mol of Ti2+ loses 2 mol of e-to form Ti4+ • Oxidation no. of Ti = +4 • Ti2+ Ti4++2e- • Overall equation: 1st half eqn x2 + 2nd half eqn • Ti2+ (aq) + 2Fe3+ (aq)  2Fe2+ (aq) + Ti4+ (aq)

  15. Question 7 (Method 2) • Let the charge on the titanium ion after reaction be Tin+ • From a) iii) we know that 1 mol of Ti2+ reacts with 2 mol of Fe3+. Hence the eqn would be: • Ti2+ + 2Fe3+ 2Fe2+ + Tin+ • Balancing charges on both sides of eqn, we get • 2 + 2(+3) = 4 + n • n = +4 • Hence the higher oxidation state of titanium is +4.

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