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Unit 4: Electrochemistry Chemistry 3202 Introduction Electrochemistry involves the study of electron movement or electric current in chemical reactions. Electrochemical reactions involve the transfer of electrons between reactants. Terms oxidation reduction oxidizing agent reducing agent half-reaction oxidation number redox reaction electrochemical cell electrolytic cell electrolysis electroplating fuel cell pyrometallurgy hydrometallurgy CHAPTER 18 Introduction eg. Ag(s) + CuCl(aq) → Cu(s) + AgCl(aq) Which species loses an electron? Which species gains an electron? Which species has no electron change? Oxidation and Reduction OXIDATION occurs when an apple turns brown or steel rusts upon exposure to air (reaction with oxygen) the Loss of Electrons in a reaction is called Oxidation (LEO) a substance is oxidized when it loses electrons Oxidation and Reduction REDUCTION the Gain of Electrons in a reaction is called Reduction (GER) a substance is reduced when it gains electrons LEO GER Oxidation Is Loss Reduction Is Gain Oxidation and Reduction eg. Why is this NOT an example of an electrochemical reaction? CaCl2 + 2 AgNO3 → Ca(NO3)2 + 2 AgCl Oxidation and Reduction Oxidation-Reduction reactions are usually referred to as redox reactions. Oxidation and Reduction eg. Cu(s) metal placed in AgNO3(aq) Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) Cu(s) + 2 Ag+(aq) + 2 NO3-(aq)→ Cu2+(aq) + 2 NO3-(aq)+ 2 Ag(s) Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) NET IONIC EQUATION: Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) loses 2 electrons – Cu is oxidized Ag+ gains 1 electron – Ag+ is reduced NO3-(aq) is not changed by this reaction and is called a spectator ion Cu Oxidation and Reduction eg. Write the NET IONIC EQUATION for the reaction between Cl2(g) and NaI(aq). Identify the species being oxidized and the species being reduced. Cl2(g) + NaI(aq) → NaCl(aq) + I2(s) Oxidizing and Reducing Agents the substance that is oxidized in a redox reaction is called the reducing agent (RA). the substance that is reduced in a redox reaction is called the oxidizing agent (OA). Oxidizing and Reducing Agents eg. Identify the OA and the RA in the following redox reactions Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s) Half Reactions A redox reaction can be separated into two parts called half-reactions. The oxidation half-reaction shows the loss of electrons. The reduction half-reaction shows the gain of electrons. Half Reactions eg. Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) oxidation: Cu(s) →Cu2+(aq) + 2e- reduction: 2 Ag+(aq) + 2e- → 2 Ag(s) Half Reactions Write the half-reactions for: Sn4+(aq) + Pb2+(aq) → Sn2+(aq) + Pb4+(aq) F2(g) + 2 Br -(aq) → 2 F-(aq) + Br2(s) Write the overall equation for: Al3+ + 3e- →Al(s) Mg(s) → Mg2+(aq) + 2e- Oxidation and Reduction p. 715 #’s 1-4 p. 716 #’s 5-7 Oxidation Numbers The oxidation number is the actual or hypothetical charge on an element or ion using an assigned set of rules. species Clion Mg2+ ion Mg atom C in CH4 oxidation # -1 +2 0 -4 Determining Oxidation Numbers (See p. 724) 1. Pure elements → oxidation # = 0 eg. 1. Cl2(g), Mg(s), P4(s), Fe(s) Simple ions → oxidation # = ion charge Cl- → oxidation # = -1 Mg 2+ → oxidation # = +2 eg. Determining Oxidation Numbers 3. Hydrogen has an oxidation number of +1 except in metal hydrides. eg. H is +1 in NH3, H2O, and HCl H is -1 in NaH and CaH2 4. Oxygen has an oxidation number of -2 except in peroxides such as H2O2 and in OF2. Determining Oxidation Numbers 5. If H or O are not in a covalent compound, the atom with the highest electronegativity has the oxidation number that matches its charge. eg. NF3 vs. NI3 Determining Oxidation Numbers The sum of all oxidation numbers in a compound is ZERO. eg. CH4 H → C must be ??? Determining Oxidation Numbers The sum of all oxidation numbers in a polyatomic ion equals the ion charge. CO32O → C + 3(-2) = -2 C = +4 eg. Page 726 - #’s 9 – 12 Extra Practice (oxidation #’s) Determine the oxidation number of: a) S in SO2 d) Cr in Cr2O72b) Cl in HClO4 e) I in MgI2 c) S in SO422. What is the oxidation number of nitrogen in each of the following? N 2O NO NO2 NH3 N 2H 4 NaNO3 N2 NH4Cl 1. Identifying Redox Reactions Redox reactions can be identified by looking at the oxidation numbers of the species involved. Changes in oxidation numbers indicate a reaction is a redox reaction. Identifying Redox Reactions An increase in oxidation number indicates oxidation (ie. the reducing agent) A decrease in oxidation number indicates a reduction (ie. the oxidizing agent) Identifying Redox Reactions eg. Use oxidation numbers to show why this is a redox reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) -4 0 -2 -1 Identifying Redox Reactions eg.Using oxidation numbers, explain why this is NOT a redox reaction: CaCO3(s) → CaO(s) + CO2(g) Page 728 - #’s 13 – 15 Handout Electro #1 Extra Practice (identifying redox) 1. Determine whether the following are redox reactions. For the redox reactions, identify the OA and the RA. a) Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq) b) Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) c) Br2(l) + 2 NaI(aq) → 2 NaBr(aq) + I2(s) Extra Practice (identifying redox) d) 2 NaCl(l) → 2 Na(l) + Cl2(g) e) HCl(aq) + NaOH(aq) → HOH(l) + NaCl(aq) f) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) g) 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) Balancing Redox Equations Redox equations are written by first balancing the ½ reactions and then combining the balanced ½ reactions. Three ways to balance redox equations: reactions in acidic solution reactions in basic solution standard reduction potentials table Reactions in Acidic Solution (p. 732) 1. 2. 3. 4. 5. Write an unbalanced half reaction Balance everything but O and H Balance O by adding water Balance H by adding H+ Balance the charges with é Reactions in Acidic Solution (p. 732) balance each ½ reaction using the 5 steps multiply the ½ reactions to cancel electrons combine the ½ reactions CHARGE AND # OF ATOMS MUST BE BALANCED Reactions in Acidic Solution eg. Balance each ½ reaction in acidic solution. a) MnO4-(aq) → Mn2+(aq) b) HNO2(aq) → N2O(g) Page 732 #’s 19, 20 MnO4-(aq) → Mn2+(aq) MnO4-(aq) → Mn2+(aq) MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) HNO2(aq) → N2O(g) 2 HNO2(aq) → N2O(g) 2 HNO2(aq) → N2O(g) + 3 H2O(l) 4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l) 4 e- + 4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l) (Reduction: OA is HNO2(aq) ) Reactions in Acidic Solution eg.Balance the following redox equation in acidic solution. IO3-(aq) + NO2(g) → NO3-(aq) + I2(s) Reactions in Acidic Solution Solution: write and balance each half reaction IO3-(aq) → I2(s) NO2(g) → NO3-(aq) Reactions in Acidic Solution IO3-(aq) → I2(s) 2 IO3-(aq) → I2(s) 2 IO3-(aq) → I2(s) + 6 H2O(l) 12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l) 10 e- + 12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l) (Reduction: OA is IO3- ) Reactions in Acidic Solution NO2(g) → NO3-(aq) NO2(g) → NO3-(aq) H2O(l) + NO2(g) → NO3-(aq) H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 e(Oxidation: RA is NO2(g) ) Oxidation: H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 eReduction: 10 e- + 12 H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l) 10 H2O(l) + 10 NO2(g) → 10 NO3-(aq) + 20 H+(aq) + 10 eOverall equation: 4 H2O(l) + 10 NO2(g) + 2 IO3-(aq) → I2(s) + 10 NO3-(aq) + 8 H+(aq) Extra Practice (balance in acidic conditions) CH3OH(l) + MnO4-(aq) → CH2O(l) + Mn2+(aq) x5 CH3OH(l) → CH2O(l) CH3OH(l) → CH2O(l) + 2 H+(aq) + 2 e- x2 MnO4-(aq) → Mn2+(aq) 5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 5 CH3OH(l) → 5 CH2O(l) + 10 H+(aq) + 10 e10 e- + 16 H+(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) __ CH3OH(l) + __ H+(aq) + __MnO4-(aq) → __ CH2O(l) + __Mn2+(aq) + __ H2O(l) p. 739 # 27 Balance the following redox equation in acidic solution. N2O(g) + SO42-(aq) → HNO2(aq) + H2SO3(aq) N2O(g) → HNO2(aq) 3 H2O(l) + N2O(g) → 2 HNO2(aq) + 4 H+(aq) + 4 e- SO42-(aq) → H2SO3(aq) SO42-(aq) + 4 H+(aq) + 2 e- → H2SO3(aq) + H2O(l) Reactions in Basic Solution 1. 2. 3. 4. 5. 6. 7. Write an unbalanced half reaction Balance everything but O and H Balance O and H the same Adjust for basic conditions with OH (=H) Balance the charges with é Cancel H+ and OHBalance charges by adding electrons Reactions in Basic Solution balance each ½ reaction using the 7 rules from p. 733 multiply the ½ reactions to cancel electrons combine the ½ reactions CHARGE AND # OF ATOMS MUST BE BALANCED Reactions in Basic Solution eg. Balance this ½ reaction in basic solution ClO-(aq) → Cl-(aq) Reactions in Basic Solution ClO- → Cl- + H2O 2 H+ + ClO- → Cl- + H2O 2 OH- + 2 H+ + ClO- → Cl- + H2O + 2 OH2 H2O + ClO- → Cl- + H2O + 2 OH- 1 H2O + ClO- → Cl- + 2 OH2e- + H2O + ClO- → Cl- + 2 OH- Reactions in Basic Solution eg. Balance the following redox equation under basic conditions. SO42-(aq)+ NO(g) → SO32-(aq) + NO2-(aq) 2 H+ + SO42- → SO32- + H2O 2 OH- + 2 H+ + SO42- → SO32- + H2O + 2 OH- 2 H2O + SO42- → SO32- + H2O + 2 OHH2O + SO42- → SO32- + 2 OH2 e- + H2O + SO42- → SO32- + 2 OH- H2O + NO → NO2- + 2 H+ 2 OH- + H2O + NO → NO2- + 2 H+ + 2 OH- 2 OH- + H2O + NO → NO2- + 2 H2O 2 OH- + NO → NO2- + H2O 2 OH- + NO → NO2- + H2O + 1 e- Reactions in Basic Solution 2 OH- + NO → NO2- + H2O + 1 eMultiply by 2 to cancel electrons!! 4 OH- + 2 NO → 2 NO2- + 2 H2O + 2 e2 e- + H2O + SO42- → SO32- + 2 OH- 2 OH- + 2 NO + SO42- → SO32- + 2 NO2- + H2O ClO− + CrO2− → CrO42− + Cl2 CrO2− → CrO42− ClO− → Cl2 p. 734 #’s 23, 24 p. 739 #’s 27, 28 Handout Electro #2 Handout Electro #3