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Assignment #6 Intro to REDOX Reactions 1. State whether the following changes are oxidation or reduction. Reactions Classification oxidation or reduction a) b) c) d) e) f) g) h) Fe2+ 2O22N3X5Sn4+ Cu Y1Br2 Fe3+ O2 N2 X7Sn2+ Cu2+ Y32Br- Oxidation Oxidation Oxidation Reduction Reduction Oxidation Reduction Reduction 2. State whether the following changes are oxidation or reduction and write the electrons in the equation on the appropriate side. Reactions Classification oxidation or reduction a) b) c) d) e) f) g) h) i) j) k) l) Reduction Reduction Reduction Oxidation Oxidation Oxidation Oxidation Reduction Oxidation Reduction Reduction Oxidation (place your electrons on the appropriate side) X2 + 2e 2XMn4+ + 2e Mn2+ P4 + 12e 4P3Z5 Z2- + 3e 8S2 S8 + 16e + Cu Cu2+ + 1e Al Al3+ + 3e 3S8 + 48e 24S2Q2 Q2+ + 4e 5P4 + 60e 20P3J4+ + 6e J26N3 3N2 + 18e 3. Write the two half-reactions for each of the following reactions and come up with the net ionic equation. a) Ba(s) + 2HCl(aq) BaCl2(aq) + H2(g) Ox ½ rxn: Ba(s) Ba2+(aq) + 2e Red ½ rxn: 2H+(aq) + 2e H2(g) Net equation: Ba(s) + 2H+(aq) Ba2+(aq) + H2(g) b) 2Al(s) + Zn3(PO4)2(aq) 2AlPO4(aq) + 3Zn(s) Ox ½ rxn: 2(Al(s) Al3+(aq) + 3e) Red ½ rxn: 3(Zn2+(aq)+ 2e Zn(s)) Net equation: 2Al(s) + 3Zn2+(aq) 2Al3+(aq) + 3Zn(s) 4. Write the two half-reactions for each of the following net ionic equations. a) 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) Ox ½ rxn: 2(Al(s) Al3+(aq) + 3e) Red ½ rxn: 3(2H+(aq) + 2e H2(g)) b) Cl2(g) + 2Br-(aq) 2Cl-(aq) + Br2(g) Ox ½ rxn: 2Br-(aq) Br2(g) + 2e Red ½ rxn: Cl2(g) + 2e 2Cl-(aq) c) 3Zn(s) + 2Fe3+(aq) 3Zn2+(aq) + 2Fe(s) Ox ½ rxn: 3(Zn(s) Zn2+(aq) + 2e) Red ½ rxn: 2(Fe3+(aq)+ 3e Fe(s)) d) 2Mn7+(aq) + 10Cl-(aq) 2Mn2+(aq) + 5Cl2(g) Ox ½ rxn: 5(2Cl-(aq) Cl2(g) + 2e) Red ½ rxn: 2(Mn7+(aq)+ 5e Mn2+(aq)) 5. For each of the reactions below indicate the following in the spaces provided: The substance being reduced using the letter R The substance being oxidized using the letter O The reducing agent using the letters RA The oxidizing agent using the letters OA a) Br2(aq) + R / OA Sn2+(aq) O / RA 2Br-(aq) ______ + Sn4+(aq) ______ b) Al3+(aq) R / OA 3Fe2+(aq) O / RA Al(s) + ______ 3Fe3+(aq) ______ c) 2Ag(s) + O / RA Cu2+(aq) R / OA 2Ag+(aq) ______ + d) 2Al(s) + O / RA 6H+(aq) R / OA 2Al3+(aq) ______ + 3H2(g) ______ e) Br2(aq) + R / OA 2I2-(aq) O / RA 2Br-(aq) ______ + I2(l) ______ f) Pb2+(aq) R / OA + 2K(s) O / RA 2K+(aq) + ______ Pb(s) ______ g) Ni2+(aq) R / OA + Ca(s) O / RA Ni(s) + ______ Ca2+(aq) ______ + Cu(s) ______ Assignment #7 Writing Complex REDOX Reactions 1. Write out the half-reaction for the oxidation of manganese(IV) oxide to form the permanganate ion in an acidic solution. MnO2(s) + 2 H2O(l) → MnO4– (aq) + 4 H+(aq) + 3e– 2. Write out the half-reaction for the oxidation of nitrite ions to nitrate ions under basic conditions. NO2–(aq) + 2 OH–(aq) → NO3– (aq) + H2O(l) + 2e– 3. Write out the half-reaction for the reduction of the IO3–(aq) ion to iodine in an acidic solution. 2 IO3–(aq) + 12 H+(aq) + 10e–→ I2 (s) + 6 H2O(l) 4. Write out the half-reaction for the oxidation of nitrogen monoxide to form the nitrate ion in an acidic solution. NO(g) + 2 H2O(l) → NO3– (aq) + 4 H+(aq) + 3e– 5. Write out the half-reaction for the reduction of the nitrite ion to nitrogen monoxide gas under basic conditions. NO2– (aq) + H2O(l) + 1e–→ NO(g) + 2 OH–(aq) 6. Write out the half-reaction of hydrogen peroxide to water in an acidic solution. H2O2(aq) + 2 H+(aq) + 2e–→ 2 H2O(l) 7. Write out the half-reaction of lead(IV) oxide being reduced to lead(II) ions in an acidic solution. PbO2(s) + 4 H+(aq) + 2e–→ Pb2+(aq) + 2 H2O(l) 8. Write out the half-reaction for the reduction of the VO2+(aq) ion to the VO2+(aq) ion in an acidic solution. VO2+(aq) + 2 H+(aq) + 1e–→ VO2+ (aq) + H2O(l) 9. Write out the half-reaction for the reduction of the permanganate ion to manganese(IV) oxide in a basic solution. MnO4– (aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq) 10. Write out the half-reaction for the reduction of the IO4–(aq) ion to the IO3–(aq) ion in an acidic solution. IO4–(aq) + 2 H+(aq) + 2e–→ IO3– (aq) + H2O(l) 11. Write out the half-reaction of oxalate ion (C2O42–(aq)) to carbon monoxide gas under basic conditions. C2O42–(aq) + 2 H2O(l) + 2e– → 2 CO(g) + 4 OH–(aq) 12. Write out a balanced REDOX reaction for the reaction between permanganate and iron(II) ions in an acidic solution. MnO4–(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq) Red ½ rxn: MnO4–(aq) + 8 H+(aq) + 5e– → Mn2+ (aq) + 4 H2O(l) Oxid ½ rxn: (Fe2+(aq) → Fe3+(aq) + 1e–) x5 Net ionic: MnO4–(aq) + 8 H+(aq) + 5 Fe2+(aq) → Mn2+ (aq) + 5 Fe3+(aq) + 4 H2O(l) 13. Cyanide ions are oxidized by permanganate ions in a basic solution. CN–(aq) + MnO4–(aq) → CNO–(aq) + MnO2(s) Red ½ rxn: (MnO4–(aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq))x2 Oxid ½ rxn: (CN–(aq) + 2 OH–(aq) → CNO–(aq) + H2O(l) + 2e–) x3 Net ionic: 2 MnO4–(aq) + H2O(l) + 3 CN–(aq) → 2 MnO2(s) + 3 CNO–(aq) + 2 OH–(aq) 14. Silver is oxidized by permanganate ions in an acidic solution. Ag(s) + MnO4–(aq) → Ag+(aq) + Mn2+(aq) Red ½ rxn: MnO4–(aq) + 8 H+(aq) + 5e– → Mn2+(aq) + 4 H2O(l) Oxid ½ rxn: (Ag(s) → Ag+(aq) + 1e–) x5 Net ionic: MnO4–(aq) + 5 Ag(s) + 8 H+(aq) → 5 Ag+ (aq) + Mn2+(aq) + 4 H2O(l) 15. Balance the following reaction under acidic conditions. Hg(l) + NO3–(aq) + Cl–(aq) → HgCl4–(aq) + NO2(g) Red ½ rxn: (NO3–(aq) + 2 H+(aq) + 1e– → NO2(g) + H2O(l))x3 Oxid ½ rxn: Hg(l) + 4 Cl–(aq) → HgCl4–(aq) + 3e– Net ionic: Hg(l) + 4 Cl–(aq) + 3 NO3–(aq) + 6 H+(aq) → HgCl4–(aq) + 3 NO2(g) + 3 H2O(l) 16. Balance the REDOX reaction between dichromate ion and iodide ion to form chromium(III) ion and solid iodine, which occurs in acidic solutions. I–(aq) + Cr2O72–(aq) → Cr3+(aq) + I2(s) Red ½ rxn: Cr2O72–(aq) + 14 H+(aq) + 6e– → 2 Cr3+(s) + 7 H2O(l) Oxid ½ rxn: (2 I–(aq) → I2(s) + 2e–) x3 Net ionic: Cr2O72–(aq) + 14 H+(aq) + 6 I–(aq) → 3 I2(s) + 2 Cr3+(aq) + 7 H2O(l) 17. Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in REDOX titrations. It reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Write a balanced equation that represents this process. C2O42–(aq) + MnO4–(aq) → CO32–(aq) + MnO2(s) Red ½ rxn: (MnO4–(aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq))x2 Oxid ½ rxn: (C2O42–(aq) + 4 OH–(aq) → 2 CO32–(aq) + 2 H2O(l) + 2e–)x3 Net ionic: 2 MnO4–(aq) + 4 OH–(aq) + 3 C2O42–(aq) → 2 MnO2(s) + 6 CO32–(aq) + 2 H2O(l) 18. Balance the following reaction under basic conditions. BrO3-(aq) + I-(aq) → I2(l) + Br-(aq) Red ½ rxn: BrO3–(aq) + 3 H2O(l) + 6e– → Br–(aq) + 6 OH–(aq) Oxid ½ rxn: (2 I–(aq) → I2(s) + 2e–) x3 Net ionic: 19. BrO3-(aq) + 6 I-(aq) + 3 H2O(l) 3 I2(l) + Br-(aq) + 6 OH-(aq) Copper(II) oxide reacts with ammonia to produce copper solid and nitrogen gas. Write out a balanced REDOX reaction under acidic conditions. CuO(s) + NH3(aq) → Cu(s) + N2(g) Red ½ rxn: (CuO(s) + 2 H+(aq) + 2e– → Cu(aq) + H2O(l))x3 Oxid ½ rxn: 2 NH3(aq) → N2(s) + 6 H+(aq) + 6e– Net ionic: 20. 3 CuO(s) + 2 NH3(aq) 3 Cu(s) + N2(g) + 3 H2O(l) Write out the balanced REDOX reaction for the reaction of perchlorate ion with chromium metal to produce chromite ion and chlorate ion in basic conditions. ClO4-(aq) + Cr(s) → CrO2-(aq) + ClO3-(aq) Red ½ rxn: (ClO4-(aq) + H2O(l) + 2e– → ClO3-(aq) + 2 OH–(aq))x3 Oxid ½ rxn: (Cr(s) + 4 OH–(aq) → CrO2-(aq) + 2 H2O(l) + 3e–)x2 Net ionic: 3 ClO4-(aq) + 2 Cr(s) + 2 OH-(aq) 2 CrO2-(aq) + 3 ClO3-(aq) + H2O(l) Assignment #8 Building REDOX Tables 1. Create a table of REDOX from the following data collected from an experiment. Be2+(aq) Cd2+(aq) Ra2+(aq) V2+(aq) ― √ ― √ ― ― ― ― √ √ ― √ ― √ ― ― Be(s) Cd(s) Ra(s) V(s) Cd2+(aq) + 2e → Cd(s) V2+(aq) + 2e → V(s) Be2+(aq) + 2e → Be(s) Ra2+(aq) + 2e → Ra(s) 2. Create a table of REDOX from the following data collected from an experiment. Zn2+(aq) V2+(aq) Cd2+(aq) Pb2+(aq) Zn(s) ― ― √ √ V(s) √ ― √ √ Cd(s) ― ― ― √ Pb(s) ― ― ― ― Pb2+(aq) + 2e → Pb(s) Cd2+(aq) + 2e → Cd(s) Zn2+(aq) + 2e → Zn(s) V2+(aq) + 2e → V(s) 3. Combine your answers from question 1 and 2 along with activity 13.0 to create a larger REDOX table. Ag+(aq) + 1e → Ag(s) Cu2+(aq) + 2e → Cu(s) Pb2+(aq) + 2e → Pb(s) Cd2+(aq) + 2e → Cd(s) Zn2+(aq) + 2e → Zn(s) V2+(aq) + 2e → V(s) Be2+(aq) + 2e → Be(s) Ra2+(aq) + 2e → Ra(s) 4. Using the evidence from the following REDOX reactions, set up a table of relative strengths of oxidizing and reducing agents. Co(s) + Pd2+(aq) → Co2+(aq) + Pd(s) Pd(s) + Pt2+(aq) → Pd2+(aq) + Pt(s) Mg(s) + Co2+(aq) → Mg2+(aq) + Co(s) Pt2+(aq) + 2e → Pt(s) Pd2+(aq) + 2e → Pd(s) Co2+(aq) + 2e → Co(s) Mg2+(aq) + 2e → Mg(s) 5. The following equations are interpretations of the evidence from the reactions of four metals with various solutions. Make a table of REDOX half-reactions. Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g) Hg(l) + 2 H+(aq) → no evidence Be(s) + Cd2+(aq) → Be2+(aq) + Cd(s) Be(s) + Ca2+(aq) → no evidence Hg2+(aq) + 2e → Hg(l) 2 H+(aq) + 2e → H2(g) Cd2+(aq) + 2e → Cd(s) Be2+(aq) + 2e → Be(s) Ca2+(aq) + 2e → Ca(s) 6. In an experiment four metals were placed into test tubes containing various ion solutions. Their resulting behavior is communicated by the equations below. Create a REDOX halfreaction table. 2 Ce(s) + 3 Ni2+(aq) → 2 Ce3+(aq) + 3 Ni(s) Pt(s) + 2 H+(aq) → no evidence 3 Sr(s) + 2 Ce3+(aq) → 3 Sr2+(aq) + 2 Ce(s) Ni(s) + 2 H+(aq) → Ni2+(aq) + H2(g) Pt2+(aq) + 2e → Pt(s) 2 H+(aq) + 2e → H2(g) Ni2+(aq) + 2e → Ni(s) Ce3+(aq) + 3e → Ce(s) Sr2+(aq) + 2e → Sr(s) 7. An analytical chemist reacts an unknown metal X with a copper(II) sulfate solution, plating out copper metal. Metal X does not react with a zinc nitrate solution. What is the REDOX table for these reactions? Cu2+(aq) + 2e → Cu(s) X2+(aq) + 2e → X(s) Zn2+(aq) + 2e → Zn(s) 8. What groups of metals are eliminated as a possible identity of metal X? Metal X can not be group 1, 2, or 3 Assignment #9 Predicting REDOX Reactions For the following assignment you are to write the net ionic equations for the chemical combinations as well you are to state whether or not the reactions are spontaneous or non-spontaneous. 1. Iron is used in an environment containing aqueous magnesium chloride. An example of this is when one is working under ocean waters. Entities Mg2+ ClFe SRA H2O SOA SRA: Fe(s) Fe2+(aq) + 2e SOA: 2H2O(l) + 2e H2(g) + 2OH-(aq) net: Fe(s) + 2H2O(l) Fe2+(aq) + H2(g) + 2OH-(aq) RA above OA on the list non-spontaneous 2. A lab technician stores an aqueous solution of iron(III) chloride in a nickel plated container. Entities Fe3+ SOA ClNi SRA H2O SRA: Ni(s) Ni2+(aq) + 2e SOA: 2(Fe3+(aq) + 1e Fe2+(aq)) net: 2Fe3+(aq) + Ni(s) 2Fe2+(aq) + Ni2+(aq) OA above RA on the list spontaneous 3. A student uses hydrobromic acid to acidify a solution of potassium dichromate for later use in a REDOX titration experiment. Entities H+ Cr2O72- SOA Br- SRA K+ SRA: 3(2Br-(aq) Br2(l) + 2e) SOA: Cr2O72-(aq) + 14H+(aq) + 6e 2Cr3+(aq)+ 7H2O(l) net: Cr2O72-(aq) + 14H+(aq) + 6Br-(aq) 2Cr3+(aq) + 3Br2(l) + 7H2O(l) OA above RA on the list spontaneous 4. A solution of tin(II) chloride is added to a solution of lead(IV) chloride. Entities Sn2+ SOA ClPb4+ H2O SRA SRA: 2H2O(l) O2(g) + 4H+(aq) + 4e SOA: 2(Sn2+(aq) + 2e Sn(s)) net: 2Sn2+(aq) + 2H2O(l) 2Sn(s) + O2(g) + 4H+(aq) RA above OA on the list non-spontaneous 5. A student uses copper electrodes to test the conductivity of a nitric acid solution. Entities Cu SRA H+ NO3- SOA H2O SRA: Cu(s) Cu2+(aq) + 2e SOA: 2NO3-(aq) + 4H+(aq) + 2e N2O4 (g) + 2H2O(l) net: 2NO3-(aq) + 4H+(aq) + Cu(s) Cu2+(aq) + N2O4 (g) + 2H2O(l) OA above RA on the list spontaneous 6. An iron bolt is exposed to air and water. (assume the air to be just oxygen) Entities Fe SRA O2 SOA H2O SRA: 2(Fe(s) Fe2+(aq) + 2e) SOA: O2(g) + 2H2O(l) + 4e 4OH-(aq) net: 2Fe(s) + O2(g) + 2H2O(l) 2Fe2+(aq) + 4OH-(aq) OA above RA on the list spontaneous 7. A solution of acidic potassium dichromate is mixed with an aqueous solution of hydrogen peroxide. Entities H+ Cr2O72- SOA K+ H2O2 SRA SRA: 3(H2O2(l) O2(g) + 2H+(aq) + 2e) SOA: Cr2O72-(aq) + 14H+(aq) + 6e 2Cr3+(aq)+ 7H2O(l) net: Cr2O72-(aq) + 8H+(aq) + 3H2O2(l) 2Cr3+(aq) + 3O2(g) + 7H2O(l) OA above RA on the list spontaneous 8. A chemistry teacher demonstrates the test for bromide ions by bubbling some chlorine gas through a sodium bromide solution. Entities Na+ Br- SRA Cl2 SOA H2O SRA: 2Br- (aq) Br2(l) + 2e SOA: Cl2(g) + 2e 2Cl-(aq) net: 2Br-(aq) + Cl2(g) 2Cl-(aq) + Br2(l) OA above RA on the list spontaneous 9. Aqueous solutions of tin(II) bromide and iron(III) nitrate are mixed. Entities Sn2+ SRA BrFe3+ SOA NO3H2O SRA: Sn2+(aq) Sn4+(aq) + 2e SOA: 2(Fe3+(aq) + 1e Fe2+(aq)) net: 2Fe3+(aq) + Sn2+(aq) 2Fe2+(aq) + Sn4+(aq) OA above RA on the list spontaneous 10. Two students try etching their initials on a copper plate using hydrochloric acid. Entities H+ SOA ClCu SRA H2O SRA: Cu(s) Cu2+(aq) + 2e SOA: 2H+(aq) + 2e H2(g) net: Cu(s) + 2H+(aq) Cu2+(aq) + H2(g) RA above OA on the list non-spontaneous 11. Sodium metal is added to some water in a typical demonstration of the reactivity of alkali metals. Entities Na SRA H2O SOA SRA: 2(Na(s) Na+(aq) + 1e) SOA: 2H2O(l) + 2e H2(g) + 2 OH-(aq) net: 2Na(s) + 2H2O(l) 2Na+(aq) + H2(g) + 2OH-(aq) OA above RA on the list Spontaneous Assignment #10 Assigning Oxidation Numbers 1. Find the oxidation number for the sulfur atoms in the following compounds. 0 +4 +2 +6 +6 +6 +4 S8 SO2 S2O32- H2SO4 MgSO4 SO42- H2SO3 2. Find the oxidation number for all the phosphorous atoms in the following compounds. +5 +3 P2O5 0 PCl3 +3 P4 +5 H3PO3 Na3PO4 +5 +5 PO43- H3PO4 3. Find the oxidation number for all the carbon atoms in the following compounds. +4 +3 +4 +4 +4 +4 -2 CO2 C2O42- CO32- NaHCO3 H2CO3 CaCO3 C2H5OH +2 +3 0 MnO ClNO C6H12O6 -1 +5 +1 +4 NaH S2O62- HXeO SnCl4 +6 +5 4. Find the oxidation number of each of the underlined species. +3 -4 +1 H3BO3 CH4 OBr3+ +3 +2 +5 VO2 + HNO2 XeO4 +4 6- HCO3- +3 +6 -1 -1 -1 NO+ Cr2O72- CaH2 NH3OH+ Na2O2 MnO42- Na2HPO4 Assignment #11 Using Oxidation Numbers 1. State whether the following changes are Oxidation (OX) or Reduction (RED) in nature. a) AsO3- c) H2O2 e) NH4+ 3+ g) Cr i) Au + k) NH3OH 2. AsO2 O2 N2O5 Cr2O72Au(CN)4NO2 Ox -1 ot 0 Ox -3 to +5 Ox +3 to +6 Ox 0 to +3 Ox -1 to +4 MnO2 CO2 CH4 I2 H2S P2O5 Red +7 to +4 Ox -2 to +4 Red +3 to –4 Red +7 to 0 Red +4 to –2 Neither State whether the following changes are Oxidation (OX) or Reduction (RED) in nature and indicate how many electrons must be lost or gained in the process. Ox or Red a) C2O42- c) ClO3 e) Na2HPO4 g) XO54 i) 2P2O5 k) C12H22O11 3. b) MnO4d) C2H3OH f) C2O42h) H4IO6j) H2SO3 l) Na2HPO4 Red +5 to +4 2CO2 ClPO2 H3 X P4 12C Ox Red Red Red Red none e- lost or gianed Lost 2 Ox or Red b) NiO2 Ni(OH)2 d) C2H5OH 2CO2 f) N2O3 2NH3 h) KMnO4 MnO2 j) S8 8SO42- Gained 6 Gained 1 Gained 9 Gained 20 Red Ox Red Red Ox e- lost or gianed Gained 2 Lost 12 Gained 12 Gained 3 Lost 48 No change For each of the following reactions identify the oxidizing agent and the reducing agent by circling each and labeling it OA or RA respectively. a) Sn + NO3- SnO2 + NO RA OA O2RA b) PbCl4 OA + PbCl2 + O2 c) Fe3+ OA + NH3OH+ RA Fe2+ + N2O d) H3PO2 RA + Cr2O72- OA Cr3+ + H3PO4 e) VO2+ OA + Sn2+ RA VO2+ + Sn4+ f) N2O4 OA + H2 RA HOH + N2 g) CrO2RA + S2O8 OA CrO42- + SO42- h) XeO3 OA + IRA Xe I2 + 4. In the balanced equations below, establish the oxidation numbers for all atoms involved, and use them to identify the substance being oxidized (RA) and the substance being reduced (OA) by circling it and labeling it. +1 +5 –2 -2 +1 -2 +1 OA RA +4 –2 +1 -1 +1 -2 a) 2KClO3(s) + C2H5OH(l) 2CO2(g) + 2KCl(s) + 3H2O(l) +1–2+1 -1 - 0 0 -2+1 b) 2HOH(l) + 2I (aq) I2(s) + H2(g) + 2OH-(aq) OA RA +4 –2 0 -3 +1 +1-2+1 0 +7 –2 c) 2NO2(g) + 7H2(g) 2NH3(g) + 4HOH(g) OA RA +5 –2 +1 - d) 7ClO3 (aq) + 2H+(aq) OA/RA Example of disproportionation +1 -2 Cl2(g) + 5ClO4 (aq) + H2O(l) - Assignment #12 Balancing Equations Using Oxidation Numbers For the following reaction equations, you are to use oxidation numbers to identify the OA, RA, and to balance the equation. Remember to show all your work. +3 -2 +5-2 +5 –2 -1 +5 -1 1. AsO33–(aq) + IO3–(aq) AsO43–(aq) + I–(aq) +3 lost 2e tot lost 2e x3 +5 gained 6e tot gained 6e x1 3 AsO33–(aq) + IO3–(aq) 3 AsO43–(aq) + I–(aq) +7 -2 2. +2 MnO4-(aq) +7 gained 5e tot gained 5e +1 2+ + Fe (aq) + H +2 + (aq) Mn 2+ +2 lost 1e tot lost 1e x5 +3 (aq) +1 -2 3+ + Fe +2 (aq) + H2O(l) +3 MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) +3 -2 +5 -2 +5 -2 -1 +5 -1 3. AsO33-(aq) + BrO3-(aq) AsO43-(aq) + Br-(aq) +3 lost 2e tot lost 2e x3 +5 gained 6e tot gained 6e 3 AsO33-(aq) + BrO3-(aq) 3 AsO43-(aq) + Br-(aq) +1+4-2 4. HPO3-(aq) +4 lost 1e tot lost 1e x2 0 +1 –2 + I2(l) + H2O(l) 0 gained 1e tot gained 2e +1 +5 –2 -1 H2PO4-(aq) I-(aq) +5 + +1 + H+(aq) -1 2 HPO3-(aq) + I2(l) + 2 H2O(l) 2 H2PO4-(aq) + 2 I-(aq) + 2 H+(aq) +2 -2 -3+1 0 0 -3 lost 3e tot lost 3e x2 0 0 +1 -2 5. CuO(s) + NH3(aq) Cu(s) + N2(g) + H2O(l) +2 gained 2e tot gained 2e x3 3 CuO(s) + 2 NH3(aq) 3 Cu(s) + N2(g) + 3 H2O(l) +4 –2 6. SeO32-(aq) +4 lost 2e tot lost 2e 0 + Cl2(g) + -2+1 OH-(aq) 0 gained 1e tot gained 2e +6 –2 -1 SeO42-(aq) Cl-(aq) + +6 +1 -2 + H2O(l) -1 SeO32-(aq) + Cl2(g) + 2 OH-(aq) SeO42-(aq) + 2 Cl-(aq) + H2O(l) +4 –2 -1 +1 –2 +3 –2 +5 –2 +3 +5 +1 7. N2O4(aq) + Br-(aq) + H2O(l) NO2-(aq) + BrO3-(aq) + H+(aq) +4 gained 1e tot gained 2e x3 -1 lost 6e tot lost 6e 3 N2O4(aq) + Br-(aq) + 3 H2O(l) 6 NO2-(aq) + BrO3-(aq) + 6 H+(aq) +1 –1 +3 –2 -1 gained 1e tot gained 2e x3 +3 lost 3e tot lost 3e x2 +6 –2 +1 –2 +1 8. H2O2(aq) + CrO2-(aq) CrO42-(aq) + H2O(l) + H+(aq) +6 -2 3 H2O2(aq) + 2 CrO2-(aq) 2 CrO42-(aq) + 2 H2O(l) + 2 H+(aq) +6 –2 +4-2 +6 gained 3e tot gained 3e x2 +4 lost 2e tot lost 2e x3 +1 –2 +3 –2 +6-2 +3 +6 -2+1 9. CrO42-(aq) + SO32-(aq) + H2O(l) CrO2-(aq) + SO42-(aq) + OH-(aq) 2 CrO42-(aq) + 3 SO32-(aq) + H2O(l) 2 CrO2-(aq) + 3 SO42-(aq) + 2 OH-(aq) +5 –2 +1 –2 -1 0 -1 0 -1 -2+1 10. BrO3-(aq) + I-(aq) + H2O(l) I2(l) + Br-(aq) + OH-(aq) +5 gained 6e tot gained 6e -1 lost 1e tot lost 1e x6 BrO3-(aq) + 6 I-(aq) + 3 H2O(l) 3 I2(l) + Br-(aq) + 6 OH-(aq) +5 –2 0 +1 –2 NO3-(aq) 11. Zn(s) + -3+1 + H2O(l) NH4 0 +5 lost 2e gained 8e tot lost 2e tot gained 8e x4 +2 + (aq) + Zn -3 -2+1 2+ + OH-(aq) (aq) +2 4 Zn(s) + NO3-(aq) + 7 H2O(l) NH4+(aq) + 4 Zn2+(aq) + 10 OH-(aq) +7 -2 12. -3+1 MnO4-(aq) +7 gained 2e tot gained 2e x8 +1 + NH3(aq) + H + +4 -2 (aq) MnO2(aq) + -3 lost 8e tot lost 8e x3 +4 +5-2 +1 -2 NO3-(aq) + H2O(l) +5 8 MnO4-(aq) + 3 NH3(aq) + 5 H+(aq) 8 MnO2(aq) + 3 NO3-(aq) + 7 H2O(l) +5-2 13. IO3-(aq) +4-2 +1 –2 0 + SO2(g) + H2O(l) I2(l) + +5 gained 5e tot gained 5e x2 +4 lost 2e tot lost 2e x5 0 +6 –2 +1 SO42-(aq) + H+(aq) +6 2 IO3-(aq) + 5 SO2(g) + 4 H2O(l) I2(l) + 5 SO42-(aq) + 8 H+(aq) +7 –2 14. ClO4-(aq) +7 gained 2e tot gained 2e x3 0 -2+1 + Cr(s) + OH (aq) - 0 lost 3e tot lost 3e x2 +3 –2 +5 –2 CrO2-(aq) ClO3-(aq) +3 + +1 -2 + H2O(l) +5 3 ClO4-(aq) + 2 Cr(s) + 2 OH-(aq) 2 CrO2-(aq) + 3 ClO3-(aq) + H2O(l) +6 -2 -1 +6 gained 3e tot gained 6e -1 lost 1e tot lost 1e x6 +1 +3 0 +3 0 +1 -2 15. Cr2O72-(aq) + Cl-(aq) + H+(aq) Cr3+(aq) + Cl2(g) + H2O(l) Cr2O72-(aq) + 6 Cl-(aq) + 14 H+(aq) 2 Cr3+(aq) + 3 Cl2(g) + 7 H2O(l) +1 +3 -2 +7 -2 +1 +4 –2 +2 +4 +2 +1 -2 16. H2C2O4(aq) + MnO4-(aq) + H+(aq) CO2(g) + Mn2+(aq) + H2O(l) +3 lost 1e tot lost 2e x5 +7 gained 5e tot gained 5e x2 5 H2C2O4(aq) + 2 MnO4-(aq) + 6 H+(aq) 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) +2 –2 -3 +1 0 -3 lost 3e tot lost 3e x2 0 +1 –2 0 17. CuO(s) + NH3(g) N2(g) + H2O(l) + Cu(s) +2 gained 2e tot gained 2e x3 0 3 CuO(s) + 2 NH3(g) N2(g) + 3 H2O(l) + 3 Cu(s) +7 -2 +1 -2 +1 +7 gained 5e tot gained 5e x2 -2 lost 2e tot lost 2e x5 +2 0 +2 0 +1 -2 18. MnO4-(aq) + H2Se(g) + H+(aq) Mn2+(aq) + Se(s) + H2O(l) 2 MnO4-(aq) + 5 H2Se(g) + 6 H+(aq) 2 Mn2+(aq) + 3 Se(s) + 8 H2O(l) +4 –2 0 +1+6-2 +2+6-2 +1 -2 19. PbO2(s) + Pb(s) + H2SO4(aq) PbSO4(s) + H2O(l) +4 gained 2e tot gained 2e 0 lost 2e tot lost 2e +2 PbO2(s) + Pb(s) + 2 H2SO4(aq) 2 PbSO4(s) + 2 H2O(l) 20. In a reaction of acidified potassium dichromate with zinc, 3.75 g of zinc was reacted. What was the concentration of potassium dichromate if 28.3 mL was used? 1 Cr2O72-(aq) + 3 Zn(s) + 14 H+(aq) 2 Cr3+(aq) + 3 Zn2+(aq) + 7 H2O(l) +6 G 3e TG 6e 0 L 2e TL 2e ng = m = 0.057348218 M +3 +2 nr = Coef r * ng = 0.0191160728 Coef g C = n = 0.675 mol/L v 21. In the following reaction below, you are to find the mass of carbon dioxide that is produced when 36.5 mL of a 1.50 mol/L solution of potassium permanganate was used. 2 MnO4-(aq) + 5 C2O42-(aq) + 16 H+(aq) 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) +7 G 5e TG 5e +3 L 1e TL 2e ng = Cv = 0.05475 +2 +4 nr = Coef r * ng = 0.27375 Coef g m = nM = 12.0 g 22. What concentration of phosphorous acid was used in the reaction below if we got 6.35 g of nitrogen monoxide? We used 3.50 L of the acid. 2 NO3-(aq) + 3 H3PO3(aq) + 2 H+(aq) 2 NO(g) + 3 H3PO4(aq) + 1 H2O(l) +5 G 3e TG 3e +3 L 2e TL 2e ng = m = 0.2115961346 M +2 +5 nr = Coef r * ng = 0.3173942019 Coef g C = n = 0.0907mol/L v 23. What mass of silver was reacted if 5.24 g of nitrogen monoxide was produced? 1 NO3-(aq) + 3 Ag(s) + 4 H+(aq) 3 Ag+(aq) + 1 NO(g) + 2 H2O(l) +5 G 3e TG 3e 0 L 1e TL 1e ng = m = 0.1746084638 M +1 +2 nr = Coef r * ng = 0.5238253915 Coef g m = nM = 56.5 g Assignment #13 REDOX Stoichiometry 1. The silver in a 5.90 g chunk of material was removed and then reacted with and acidified solution of potassium dichromate. It was found that 35.9 mL of a 0.200 mol/L solution of potassium dichromate was required for the reaction. Calculate the following: e) The number of moles of dichromate ions used in the reaction. Answer: 0.00718 mol + 2K2Cr2O7(aq) 2K (aq) + Cr2O7 (aq) C = 0.200 mol/L v = 0.0359 L n = Cv = (0.200 mol/L)(0.0359 L) = 0.00718 mol 1 : 1 ratio 0.00718 mol of Cr2O72-(aq) f) The number of grams of silver that reacted. Answer: 4.65 g Cr2O72-(aq) + 14H+(aq) + 6Ag(s) 7H2O(l) + n = 0.00718 mol nr = 6 x ng 2Cr3+(aq) + 6Ag+(aq) m=? M = 107.87 g/mol m = nrM = (0.04308 mol)(107.87 g/mol) 1 = 4.6470396 g 4.65 g g) The percentage of silver that is in the chunk. Answer: 78.8 % Reacted x 100% = (4.65 g)(100%) = 78.813559 % Actual 5.90 g 78.8 % 2. A 0.400 mol/L solution of tin(II) nitrate was titrated with an acidified potassium permanganate solution. Use the following data below to calculate the molar concentration of the potassium permanganate solution. Volume of tin(II) nitrate used is 10.0 mL Volumes of potassium permanganate solution required Trial 1 2 3 Initial buret reading (mL) 8.70 17.30 26.00 Final buret reading (mL) 17.30 26.00 34.80 2MnO4-(aq) + 16H+(aq) v = 0.00870 L C=? ng = Cv = (0.400 mol/L)(0.0100 L) = 0.00400 mol nr = 2/5ng 0.00160 mol + 5Sn2+(aq) 8H2O(l) + 2Mn2+(aq) Answer: 0.184 mol/L + 5Sn4+(aq) C = 0.400 mol/L v = 0.0100 L C = nr = 0.00160 mol = 0.1839081 mol/L 0.184 mol/L v 0.00870 L 3. In a titration experiment all the bromide ions in an acidic solution were oxidized to bromine by a 0.0200 mol/L potassium permanganate solution. The volume of the bromide ion solution used was 25.0 mL. Use the data below to calculate the concentration of the bromide ions in the solution. Volumes of potassium permanganate solution required Trial 1 2 3 Final buret reading (mL) 17.30 23.00 24.80 Initial buret reading (mL) 2.30 7.00 10.80 Answer: 0.0600 mol/L + 2+ 2MnO4 (aq) + 16H (aq) + 10Br (aq) 8H2O(l) + 2Mn (aq) + 5Br2(l) v = 0.0150 L C = 0.0200 mol/L ng = Cv = (0.0200 mol/L)(0.0150 L) = 0.000300 mol nr = 10/2ng 0.00150 mol C=? v = 0.0250 L C = nr = 0.00150 mol 0.0600 mol/L v 0.0250 L 4. In a REDOX titration 12.50 mL of 0.0800 mol/L potassium dichromate was used in an acidic solution to oxidize tin(II) ions to tin(IV) ions. The volume of tin(II) ions that was oxidized was 10.0 mL. Calculate the concentration of the tin(II) ions in the solution. Also what color change would indicate the endpoint of this reaction? Answer: 0.300 mol/L Cr2O72-(aq) + 14H+(aq) + 3Sn2+(aq) 7H2O(l) + 2Cr3+(aq) + 3Sn4+(aq) v = 0.0125 L C = 0.0800 mol/L ng = Cv = (0.0800 mol/L)(0.0125 L) = 0.00100 mol 3 nr = /1ng 0.00300 mol C=? v = 0.0100 L C = nr = 0.00300 mol 0.300 mol/L v 0.0100 L 5. What volume of 0.0500 mol/L KMnO4(aq) is needed to oxidize all the bromide ions in 25.0 mL of an acidic 0.200 mol/L NaBr solution? Also what color change would indicate the endpoint of this reaction? Answer: 0.0200 L 2MnO4-(aq) + 16H+(aq) + 10Br-(aq) 8H2O(l) + 2Mn2+(aq) + 5Br2(l) v=? C = 0.0500 mol/L ng = Cv = (0.200 mol/L)(0.0250 L) = 0.00500 mol 2 nr = /10ng 0.00100 mol C = 0.200 mol/L v = 0.0250 L v = nr = 0.00100 mol 0.0200 L C 0.0500 L 6. The mass of iodine produced by the reaction of 400 mL of 0.200 mol/L magnesium iodide solution with excess chlorine gas is? Answer: 20.3 g 2+ Cl2(g) + 2I (aq) I2(l) + 2Cl (aq) MgI2 Mg + 2I v = 0.400 L m =? C=0.200 CI = 2C =0.400 C = 0.400 mol/L ng = Cv = (0.400 mol/L)(0.400 L) m = nrM = (0.0800 mol)(253.80 g/mol) = 0.160 mol = 20.304 g 20.3 g nr = 1/2ng 0.0800 mol 7. 200 mL of iron(III) sulfate solution is required to react with 50.0 g of zinc. What is the concentration of the iron(III) ions in the solution? Answer: 7.65 mol/L 3+ 2+ 2+ Zn(s) + 2Fe (aq) 2Fe (aq) + Zn (aq) m = 50.0 g v = 0.200 L C=? ng = m = 50.0 g = 0.7647599 mol M 65.38 g/ mol nr = 2/1ng 1.5295197 mol C = nr = 1.5295197 mol = 7.6475987 mol/L 7.65 mol/L v 0.200 L 8. If 20.0 mL of a calcium iodide solution is required to reduce 100 g of bromine, calculate the concentration of the calcium iodide solution. Answer: 31.3 mol/L Br2(l) + 2I (aq) 2Br (aq) + I2(l) m = 100 g v = 0.0200 L C=? ng = m = 100 g = 0.6257822 mol M 159.80 g/ mol nr = 2/1ng 1.2515645 mol C = nr = 1.2515645 mol = 62.578223 mol/L 62.6 mol/L v 0.0200 L CaI2 Ca2+ + 2IC = ½CI 31.3 mol/L CI = 62.6 mol/L 9. If 100 mL of a 0.500 mol/L solution of magnesium bromide is used to reduce 50.0 mL of a solution of acidic potassium permanganate. Calculate the concentration of the permanganate ion in the solution. Answer: 0.400 mol/L + 2+ 2MnO4 (aq) + 16H (aq) + 10Br (aq) 8H2O(l) + 2Mn (aq) + 5Br2(l) v = 0.0500 L C=1.00 mol/L MgBr2 Mg2+ + 2BrC=? ng = Cv = (1.00 mol/L)(0.100 L) = 0.100 mol 1 nr = /5ng 0.0200 mol v= 0.100 L C =0.500 C = nr = 0.0200 mol 0.400 mol/L v 0.0500 L 2CBr =1. 00 10. Say 20.0 g of calcium metal is required to reduce 250 mL of a solution of chromium(III) sulfate. Calculate the concentration of the chromium(III) sulfate solution. Answer: 2.00 mol/L 3+ 2+ 2+ Ca(s) + 2Cr (aq) 2Cr (aq) + Ca (aq) m= 20.0 g v = 0.250 L C= ? ng = m = 20.0 g = 0.499002 mol M 40.08 g/ mol nr = 2/1ng 0.998004 mol C = nr = 0.998004 mol = 3.992016 mol/L 3.99 mol/L v 0.250 L Cr2(SO4)3 2Cr3+ + 3SO42C = ½CI 2.00 mol/L CCr = 3.99 mol/L 11. 50.0 mL of a 0.600 mol/L aluminium chloride solution is required to reduce 30.0 mL of a gold(III) nitrate solution. Calculate the concentration of the gold ions in the solution. Answer: 1.00 mol/L 2Au3+(aq) + 6Cl-(aq) 3Cl2(g) + 2Au(s) AlCl3 Al3+ + 3ClC=? v = 0.0500 L C=0.600 CCl=3C =1.80 v= 0.0300 L C = 1.80 mol/L ng = Cv = (1.80 mol/L)(0.0500 L) C = nr = 0.0300 mol 1.00 mol/L = 0.0900 mol v 0.0300 L 1 nr = /3ng 0.0300 mol