2-July-2016 Chemsheets A Page 1

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1 2-July-2016 Chemsheets A Page 1

2 TASK 1 Oxidation states 1) Calculate the oxidation state of the stated element in the following species: a) Fe in Fe 0 FeCl 3 +3 FeCl 2 +2 K 2FeO 4 +6 [Fe(H 2O) 6] b) Cl in Cl 2 0 ClO ClO +1 Cl 2O 7 +7 NaCl -1 2) Calculate the oxidation state of each element in the following: SO 2 S +4 O -2 H 2O 2 H +1 O -1 S S 0 NaH Na +1 H -1 SO 3 S +6 O -2 2 CO 3 C +4 O -2 H 2S H +1 S -2 Cr 2O 3 Cr +3 O -2 NH 3 N -3 H +1 CrO 3 Cr +6 O -2 NO 2 N +4 O -2 KMnO 4 K +1 Mn +7 O -2 N 2 N 0 K 2MnO 4 K +1 Mn +6 O -2 NO 3 N +5 O -2 Cu 2O Cu +1 O -2 Cl Cl -1 CuO Cu +2 O -2 SO 4 2 S +6 O -2 NaCuCl 2 Na +1 Cu +1 Cl -1 3) Give the oxidation state of the transition metal in each of the following species. a) [Co(H 2O) 6] 2+ Co +2 e) [FeO 4] 2 Fe +6 i) K 2[CoCl 4] Co +2 b) [CrCl 6] 3 Cr +3 f) [Mn(CN) 6] 4 Mn +2 j) K 3[AuF 6] Au +3 c) [Co(NH 3) 6]Cl 2 Co +2 g) [Ni(CO) 4] Ni 0 k) (NH 4) 2[IrCl 6] Ir +4 d) [Co(NH 3) 5Cl]Cl 2 Co +3 h) [Ni(EDTA)] 2 Ni +2 l) Na[Mn(CO) 5] Mn -1 TASK 2 Writing half equations a) 2I I 2 + 2e b) Fe 2+ Fe 3+ + e c) Fe Fe e d) 2Cl Cl 2 + 2e e) SO e + 4H + SO 2 + 2H 2O f) VO 2+ + H 2O VO e + 2H + g) SO e + 10H + H 2S + 4H 2O h) H 2O 2 O 2 + 2e + 2H + 2 i) Cr 2O 7 + 6e + 14H + 2Cr H 2O j) C 2O 2 4 2CO 2 + 2e k) Hg e 2Hg l) 2lO e + 12H + l 2 + 6H 2O 2-July-2016 Chemsheets A Page 2

3 TASK 3 Combining half equations 1) Zn + 2 Fe 3+ Zn Fe 2+ 2) 2 Cr 2O 7 + 6Fe H + 2Cr H 2O + 6Fe 3+ 3) H 2SO 4 + 8I + 8H + H 2S + 4H 2O + 4I 2 4) 2MnO Cl + 16H + 2Mn H 2O + 5Cl 2 5) 2MnO 4 + 5H 2O 2 + 6H + 2Mn H 2O + 5O 2 6) 2IO I + 12H + 6I 2 + 6H 2O TASK 4 Redox reactions or not? Equation Redox reaction? Disproportionati on reaction? Species oxidised Species reduced Oxidising agent Reducing agent Fe 2O 3 + 3CO 2Fe + 3CO 2 ü CO Fe 2O 3 Fe 2O 3 CO Mg + 2HCl MgCl 2 + H 2 ü Mg HCl HCl Mg MgO + 2HCl MgCl 2 + H 2O û Al 2O 3 + 2Fe 2Al + Fe 2O 3 ü Fe Al 2O 3 Al 2O 3 Fe [Co(H 2O) 6] Cl [CoCl 4] H 2O û Na 2O + H 2O 2NaOH û 2H 2O 2 2H 2O + O 2 ü ü H 2O 2 H 2O 2 H 2O 2 H 2O 2 2NaBr + H 2SO 4 Na 2SO 4 + 2HBr û Cl 2 + 2NaOH NaCl + NaOCl + H 2O ü ü Cl 2 Cl 2 Cl 2 Cl July-2016 Chemsheets A Page 3

4 TASK 5 General AS Redox Questions 1) a) i) 2Br Br 2 + 2e ii) N 2 + 6H 2O 2NO e + 12H + iii) V 3+ + H 2O VO 2+ + e + 2H + iv) H 2SO 4 + 6e + 6H + S + 4H 2O v) NO 3 + 8e + 10H + + NH 4 + 3H 2O vi) 2BrO e + 12H + Br 2 + 6H 2O b) H 2SO 4 + 6Br + 6H + S + 4H 2O + 3Br 2 oxidised = Br, reduced = H 2SO 4, oxidising agent = H 2SO 4, reducing agent = Br c) Cu Cu e HNO 3 + H + + e NO 2 + H 2O 2HNO 3 + 2H + + Cu 2NO 2 + 2H 2O + Cu 2+ 2) State whether the following three reactions are redox reactions or not. For those that are redox reactions, clearly indicate any changes in oxidation state. a) Zn + 2 HCl ZnCl 2 + H 2 redox Zn 0 +2, H +1 0 b) CuO + 2 HCl CuCl 2 + H 2O not redox c) MnO HCl MnCl 2 + Cl H 2O redox Mn +4 +2, Cl -1 0 d) Cl 2 + H 2O HCl + HOCl redox Cl 0-1 and +1 (disproportionation) 3) a) oxidising agent = takes away electrons from another species reducing agent = adds electrons to another species b) oxidised = Mg, reduced = H +, oxidising agent = H +, reducing agent = Mg c) Na 2SO 4 Na +1, S +6, O -2 Na 2S 2O 3 Na +1, S +2, O -2 Na 2SO 3 Na +1, S +4, O -2 S 8 S 0 KClO 3 K +1, Cl +5, O -2 KClO K +1, Cl +1, O -2 KH K +1, H -1 Na 2O 2 Na +1, O -1 Na 2O Na +1, O -2 4) a) i) 2 H + + 2e H 2 ii) SO e + 2H + SO H 2O iii) H 2O 2 O 2 + 2e + 2H + iv) 2IO I + 12H + 6I 2 + 6H 2O v) 2I I 2 + 2e b) 2IO I + 12H + 6I 2 + 6H 2O 5) a) 2MnO Cl + 16H + 2Mn H 2O + 5Cl 2 b) 6MnO Cr H 2O 6Mn Cr 2O H July-2016 Chemsheets A Page 4

5 TASK 6 The effect of changing conditions on electrode potential a) 298K, 100kPa H 2(g), 1.0 mol dm -3 H + (aq) b) by definition c) i) negative, equilibrium shifts left to side with less gas molecules to reduce pressure of H 2 ii) no effect, allows for faster rate of transfer of electrons but has no effect on potential iii) no effect, as concentration of H + = 1.00 mol dm -3 iv) positive, equilibrium shifts right to lower concentration of H + v) negative, equilibrium shifts left in endothermic direction to reduce temperature TASK 7 Writing conventional representations of cells Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Pt(s) H 2(g) H + (aq) Cr 2O 7 2 (aq), H + (aq), Cr 3+ (aq) Pt(s) Ni(s) Ni 2+ (aq) H + (aq) H 2(g) Pt(s) Al(s) Al 3+ (aq) Pb 2+ (aq) Pb(s) Fe(s) Fe 2+ (aq) MnO 4 (aq), H + (aq), Mn 2+ (aq) Pt(s) Ag(s) Ag + (aq) Zn 2+ (aq) Zn(s) TASK 8 Electrode potentials 1) Eº = 2.71 V 2) emf = = 0.66 V 3) a) emf = = V b) Eº (Pb 2+ /Pb) would become more negative as equilibrium shifts left to increases concentration of Pb 2+ ions; therefore emf would decrease and become less positive 4) a) emf = = V b) emf = = V c) emf = = 0.29 V 5) 1.02 = E Lº E Lº = = 0.34 V 6) a) 2.00 = E Rº 2.38 E Rº = = 0.38 V b) 0.54 = 2.38 E Lº E Lº = = 2.92 V c) 3.19 = E Rº E Rº = = 2.92 V 7) a) Cr(s) Cr 2+ (aq) Zn 2+ (aq) Zn(s) emf = V b) Cu(s) Cu 2+ (aq) Fe 3+ (aq), Fe 2+ (aq) Pt(s) emf = V c) Pt(s) Cl (aq) Cl 2 (g) MnO 4 (aq), H + (aq), Mn 2+ (aq) Pt(s) emf = V 2-July-2016 Chemsheets A Page 5

6 TASK 9 Using the electrochemical series 1) a) Ni 2+ + Zn Ni + Zn 2+ b) E (Ni 2+ /Ni) > E (Zn 2+ /Zn) and therefore Ni 2+ gains electrons from Zn 2) 2Ag + + Cu 2Ag + Cu 2+ E (Ag + /Ag) > E (Cu 2+ /Cu) and therefore Ag + gains electrons from Cu 3) MnO 4 will oxidise Cl to form Cl 2 as E (MnO 4 /Mn 2+ ) > E (Cl 2/ Cl ) and therefore MnO 4 gains electrons from Cl Cr 2O 7 2 will not oxidise Cl to form Cl 2 as E (Cr 2O 7 2 /Cr 3+ ) < E (Cl 2/ Cl ) and therefore Cr 2O 7 2 does not gain electrons from Cl Fe 3+ will not oxidise Cl to form Cl 2 as E (Fe 3+ /Fe 2+ ) < E (Cl 2/ Cl ) and therefore Fe 3+ does not gain electrons from Cl 4) a) Pt(s) H 2(g) H + (aq) Br 2(l) Br (aq) Pt(s) emf = V anode = H + /H 2 Br 2 + H 2 2Br + 2H + E (Br 2/Br ) > E (H + /H 2) and therefore Br 2 gains electrons from H 2 b) Cu(s) Cu 2+ (aq) Fe 3+ (aq), Fe 2+ (aq) Pt(s) emf = V anode = Cu 2+ /Cu 2Fe 3+ + Cu 2Fe 2+ + Cu 2+ E (Fe 3+ /Fe 2+ ) > E (Cu 2+ /Cu) and therefore Fe 3+ gains electrons from Cu 5) a) the equilibrium Cu 2+ (aq) + 2 e Cu(s) shifts left if concentration of Cu 2+ is mol dm -3 and so E becomes less than V b) right hand electrode is anode c) from right hand side to left hand side 6) a) Yes: E (H + /H 2) > E (Fe 2+ /Fe) and therefore H + gains electrons from Fe 2H + + Fe H 2 + Fe 2+ b) No: E (H + /H 2) < E (Cu 2+ /Cu) and therefore H + cannot gain electrons from Cu c) No: E (Cr 2O 2 7 /Cr 3+ ) < E ( Cl 2/ Cl ) and therefore Cr 2O 2 7 cannot gain electrons from Cl d) Yes: E (MnO 4 /Mn 2+ ) > E (Cl 2/ Cl ) and therefore MnO 4 gains electrons from Cl 2MnO Cl + 16H + 2Mn H 2O + 5Cl 2 e) Yes: E (V 3+ /V 2+ ) > E (Mg 2+ /Mg) and therefore V 3+ gains electrons from Mg Mg + 2V 3+ Mg V July-2016 Chemsheets A Page 6

7 TASK 10 Commercial cells 1) a) emf = V b) Ag 2O + 2H + + Zn 2Ag + H 2O + Zn 2+ c) Zn(s) Zn 2+ (aq) Ag 2O(s) H + (aq) Ag(s) d) chemicals would be used up; goes flat when one or more of reactants runs out 2) a) NiO(OH) + MH Ni(OH) 2 + M b) Ni(OH) 2 + M NiO(OH) + MH c) can reuse rather than throw away 3) a) CH 3OH + 3O 2 2CO 2 + 4H 2O b) V c) does not go flat; does not need re-charging 2-July-2016 Chemsheets A Page 7