CHEMSHEETS.co.uk 2-July-2016 Chemsheets A2 1076 Page 1
www.CHEMSHEETS.co.uk2-July-2016Chemsheets A2 1076Page 1
TASK 1 – Oxidation states1)Calculate the oxidation state of the stated element in the following species:a) Fe in Fe 0FeCl3 3b) Cl in Cl2 0ClO3 52)FeCl2 2-–ClO 1K2FeO4 6Cl2O7 7[Fe(H2O)6]2 2NaCl -1Calculate the oxidation state of each element in the following:SO2S 4SS0H 2O 2H 1O -1NaHNa 1H -1C 4O -22–SO3S 6O -2CO3H 2SH 1S -2Cr2O3Cr 3O -2NH3N -3H 1CrO3Cr 6O -2NO2N 4O -2KMnO4K 1Mn 7O -2K2MnO4K 1Mn 6O -2Cu2OCu 1O -2CuOCu 2O -2NaCuCl2Na 1Cu 1N2N0NO3Cl–N 5–SO43)O -2O -2Cl -12–S 6O -2Cl -1Give the oxidation state of the transition metal in each of the following species.a) [Co(H2O)6]b) [CrCl6]2 Co 23–c) [Co(NH3)6]Cl2e) [FeO4]2–Cr 3f) [Mn(CN)6]Co 2g) [Ni(CO)4]d) [Co(NH3)5Cl]Cl2 Co 34–h) [Ni(EDTA)]2–Fe 6i)K2[CoCl4]Co 2Mn 2j)K3[AuF6]Au 3Ni 0k) (NH4)2[IrCl6] Ir 4Ni 2l)Na[Mn(CO)5] Mn -1TASK 2 – Writing half equations–a) 2I I2 2eb) Fe2 –3 Fe e2 c) Fe Fe 2e–––d) 2Cl Cl2 2e2-–– e) SO4 2e 4H SO2 2H2Of) VO2 g) SO4 – H2O VO2 e 2H2–– 8e 10H H2S 4H2O–h) H2O2 O2 2e 2H2–– 3 i)Cr2O7 6e 14H 2Cr 7H2Oj)C 2O 4k) Hg2l)2–2 – 2CO2 2e–– 2e 2Hg–2lO3 10e 12H www.CHEMSHEETS.co.uk l2 6H2O2-July-2016Chemsheets A2 1076Page 2
TASK 3 – Combining half equations3 2 2 1)Zn 2 Fe Zn 2Fe2)Cr2O7 6Fe 14H 2Cr 7H2O 6Fe2–2 – 3 3 3)H2SO4 8I 8H H2S 4H2O 4I24)2MnO4 10Cl 16H 2Mn 8H2O 5Cl2–– –2 2 5)2MnO4 5H2O2 6H 2Mn 8H2O 5O26)2IO3 10I 12H 6I2 6H2O–– TASK 4 – Redox reactions or not?Redoxreaction?EquationDisproportionation ntReducingagentFe2O3 3CO 2Fe 3CO2üCOFe2O3Fe2O3COMg 2HCl MgCl2 H2üMgHClHClMgMgO 2HCl MgCl2 H2OûAl2O3 2Fe 2Al Fe2O3üFeAl2O3Al2O3FeüH 2O 2H 2O 2H 2O 2H 2O 2üCl2Cl2Cl2Cl2[Co(H2O)6]2 –2- 4Cl [CoCl4] 6H2OûNa2O H2O 2NaOHû2H2O2 2H2O O2ü2NaBr H2SO4 Na2SO4 2HBrûCl2 2NaOH NaCl NaOCl H2Oü www.CHEMSHEETS.co.uk2-July-2016Chemsheets A2 1076Page 3
TASK 5 – General AS Redox Questions1)–i)ii)iii)iv)v)vi)b)H2SO4 6Br 6H S 4H2O 3Br2––oxidised Br , reduced H2SO4, oxidising agent H2SO4, reducing agent Brc)Cu Cu 2e –HNO3 H e NO2 H2O 2 2HNO3 2H Cu 2NO2 2H2O Cu2)3)–a)2Br Br2 2e–– N2 6H2O 2NO3 10e 12H3 2 – V H2O VO e 2H– H2SO4 6e 6H S 4H2O–– NO3 8e 10H NH4 3H2O–– 2BrO3 10e 12H Br2 6H2O– 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 ZnCl2 H2redoxZn 0 2, H 1 0b) CuO 2 HCl CuCl2 H2Onot redoxc) MnO2 4 HCl MnCl2 Cl2 2 H2O redoxMn 4 2, Cl -1 0d) Cl2 H2O HCl HOClCl 0 -1 and 1 (disproportionation)redoxa) oxidising agent takes away electrons from another speciesreducing agent adds electrons to another species b) oxidised Mg, reduced H , oxidising agent H , reducing agent Mgc) Na2SO4Na2S2O3Na2SO3S8KClO3KClOKHNa2O2Na2O4)a) i)Na 1, S 6,Na 1, S 2,Na 1, S 4,S0K 1, Cl 5,K 1, Cl 1,K 1, H -1Na 1, O -1Na 1, O -2 O -2O -2O -2O -2O -2–2 H 2e H22-– 2-ii) SO4 2e 2H SO3 H2O–iii) H2O2 O2 2e 2H–– iv) 2IO3 10I 12H 6I2 6H2O–v) 2I I2 2eb)5)––– 2IO3 10I 12H 6I2 6H2O–––3 2 a) 2MnO4 10Cl 16H 2Mn 8H2O 5Cl22 2–b) 6MnO4 10Cr 11H2O 6Mn 5Cr2O7 22H www.CHEMSHEETS.co.uk 2-July-2016Chemsheets A2 1076Page 4
TASK 6 – The effect of changing conditions on electrode potential-3 a) 298K, 100kPa H2(g), 1.0 mol dm H (aq)b) by definitionc) i)negative, equilibrium shifts left to side with less gas molecules to reduce pressure of H2ii)no effect, allows for faster rate of transfer of electrons but has no effect on potentialiii)no effect, as concentration of H 1.00 mol dmiv)positive, equilibrium shifts right to lower concentration of Hv)negative, equilibrium shifts left in endothermic direction to reduce temperature -3 TASK 7 – Writing conventional representations of cells2 2 3 Zn(s) Zn (aq) Cu (aq) Cu(s) 2– 3 Pt(s) H2(g) H (aq) Cr2O7 (aq), H (aq), Cr (aq) Pt(s)2 2 Al(s) Al (aq) Pb (aq) Pb(s) 2 Ni(s) Ni (aq) H (aq) H2(g) Pt(s)– 2 Fe(s) Fe (aq) MnO4 (aq), H (aq), Mn (aq) Pt(s)2 Ag(s) Ag (aq) Zn (aq) Zn(s)TASK 8 – Electrode potentials1)Eº –2.71 V2)emf –0.44 – 0.22 –0.66 V3) a) emf –0.13 – –0.76 0.63 V2 b) Eº (Pb /Pb) would become more negative as equilibrium shifts left to increases concentration of Pbtherefore emf would decrease and become less positive2 ions;4) a) emf 0.15 – –0.25 0.40 Vb) emf 0.80 – 0.54 0.26 Vc) emf 1.07 – 1.36 –0.29 V5)1.02 1.36 – ELºELº 1.36 – 1.02 0.34 V6) a) 2.00 ERº – –2.38ERº 2.00 – 2.38 –0.38 Vb) 0.54 –2.38 – ELºELº –2.38 – 0.54 –2.92 Vc) –3.19 ERº – 0.272 ERº –3.19 0.27 –2.92 V2 7) a) Cr(s) Cr (aq) Zn (aq) Zn(s) emf 0.15 V2 3 2 b) Cu(s) Cu (aq) Fe (aq), Fe (aq) Pt(s) emf 0.43 V–– 2 c) Pt(s) Cl (aq) Cl2(g) MnO4 (aq), H (aq), Mn (aq) Pt(s) emf 0.15 V www.CHEMSHEETS.co.uk2-July-2016Chemsheets A2 1076Page 5
TASK 9 – Using the electrochemical series1) a) Ni2 Zn Ni Zn2 2 2 b) E (Ni /Ni) E (Zn /Zn) and therefore Ni2) 2Ag Cu 2Ag Cu 2 ––2 gains electrons from Zn2 E (Ag /Ag) E (Cu /Cu) and therefore Ag gains electrons from Cu3)–2 ––MnO4 will oxidise Cl to form Cl2 as E (MnO4 /Mn ) E (Cl2/ Cl ) and therefore MnO4 gains electrons from Cl2––2–3 –Cr2O7 will not oxidise Cl to form Cl2 as E (Cr2O7 /Cr ) E (Cl2/ Cl ) and therefore Cr2O7–electrons from Cl3 –3 2 –Fe will not oxidise Cl to form Cl2 as E (Fe /Fe ) E (Cl2/ Cl ) and therefore Fe–Cl 3 2––does not gaindoes not gain electrons from–4) a) Pt(s) H2(g) H (aq) Br2(l) Br (aq) Pt(s)emf 1.09 V anode H /H2–Br2 H2 2Br 2H– E (Br2/Br ) E (H /H2) and therefore Br2 gains electrons from H22 3 2 b) Cu(s) Cu (aq) Fe (aq), Fe (aq) Pt(s)emf 0.43 V2 anode Cu /Cu2Fe3 Cu 2Fe3 2 2 Cu2 2 E (Fe /Fe ) E (Cu /Cu) and therefore Fe2 3 gains electrons from Cu–Cu (aq) 2 e Cu(s)5) a) the equilibriumshifts left if concentration of Cu2 -3is 0.500 mol dm and so E becomes less than 0.34 Vb) right hand electrode is anodec) from right hand side to left hand side6) a) Yes: 2 E (H /H2) E (Fe /Fe) and therefore H gains electrons from Fe 2H Fe H2 Fe 2 2 b) No:E (H /H2) E (Cu /Cu) and therefore H cannot gain electrons from Cuc) No:E (Cr2O7 /Cr ) E ( Cl2/ Cl ) and therefore Cr2O7d) Yes:E (MnO4 /Mn ) E (Cl2/ Cl ) and therefore MnO4 gains electrons from Cl2–3 –2 –––– 2–cannot gain electrons from Cl–––2 2MnO4 10Cl 16H 2Mn 8H2O 5Cl2e) Yes:3 2 2 E (V /V ) E (Mg /Mg) and therefore VMg 2V3 www.CHEMSHEETS.co.uk Mg2 2V3 gains electrons from Mg2 2-July-2016Chemsheets A2 1076Page 6
TASK 10 – Commercial cells1) a) emf 1.10 V b) Ag2O 2H Zn 2Ag H2O Zn2 2 c) Zn(s) Zn (aq) Ag2O(s) H (aq) Ag(s)d) chemicals would be used up; goes flat when one or more of reactants runs out2) a) NiO(OH) MH Ni(OH)2 Mb) Ni(OH)2 M NiO(OH) MHc) can reuse rather than throw away3) a) CH3OH 3O2 2CO2 4H2Ob) 1.21 Vc) does not go flat; does not need re-charging www.CHEMSHEETS.co.uk2-July-2016Chemsheets A2 1076Page 7
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 .
www.CHEMSHEETS.co.uk 14-Jul-12 Chemsheets A2 009 7 TASK 3 – pH of strong bases 1) Calculate the pH of the following solutions. a) 0.15 mol dm-3 KOH b) 0.05 mol .
www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1001 Page 3 As the temperature changes, the average energy of the particles increases and so the
www.CHEMSHEETS.co.uk 10-Mar-2016 Chemsheets A2 1014 Page 7 Example 2 Ethanol has the formula C 2 H 5 OH and is used as a fuel (e.g. for cars in Brazil). It burns .
www.CHEMSHEETS.co.uk 10-Jul-12 Chemsheets AS 029 3 CALORIMETRY The enthalpy change for a reaction can be found by measuring the temperature change in a reaction.File Size: 1MB
www.CHEMSHEETS.co.uk 16-July-2016 Chemsheets A2 1078 b) By reduction of nitrile compounds This is the commonest way as it only forms one amine rather than a mixture.
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metal carbonate metal oxide carbon dioxide (on heating) CaCO 3 CaO CO 2 TASK 4 – WRITING BALANCED EQUATIONS 1) Balance the following equations.
Session 10 – Black Holes. Brief Description. Students learn about black holes, the densest objects in the Universe. They learn that the collapsing . core of a star forms a black hole and do an activity that shows how the density of a stellar core increases as the core collapses even though the mass remains the same. They then engage in a kinesthetic activity to model how a black hole affects .
