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Electrochemistry Chapter 19 1 Electron Transfer reactions • Electron transfer reactions are oxidation- reduction or redox reactions. • Electron transfer reactions result in the generation of an electric current (electricity) or be caused by imposing an electric current. • Therefore, this field of chemistry is called ELECTROCHEMISTRY. 2 Electrochemical processes are oxidationreduction reactions in which: The energy released by a spontaneous reaction is converted to electricity or Electrical energy is used to cause a nonspontaneous reaction to occur 0 0 2Mg (s) + O2 (g) 2Mg O2 + 4e3 2+ 2- 2Mg O (s) 2Mg2+ + 4e- Oxidation half-reaction (lose e-) 2O2- Reduction half-reaction (gain e-) Important Concepts OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized. 4 Review of Oxidation numbers Numbers associated with atoms and ions (can also be fractions), where: 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 3. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 5 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3Find the oxidation number of carbon in HCO3- ? 6 O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 4.4 Redox Reaction Example A redox reaction: Cu(s) + 2Ag+(aq) ON: 0 +1 oxidation Cu2+(aq) + 2Ag(s) +2 0 reduction Cu(s) is oxidized, therefore it is a reducing agent. Ag+ is reduced, therefore it is an oxidizing agent 7 We can envision breaking up the full redox reaction into two half reactions: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) reduction oxidation Cu(s) Ag+(aq) + e- Cu2+(aq) + 2eAg(s) “half-reactions” 8 Note that the half reactions are combined to make a full reaction: Cu(s) + 2Ag+(aq) oxidation Cu2+(aq) + 2Ag(s) reduction The important thing to remember: electrons are neither created nor destroyed during a redox reaction. They are transferred from the species being oxidized to that being reduced. 9 Identify the species being oxidized and reduced in the following (unbalanced) reactions: reduction +5 ClO3- + I- -1 I2 + Cl- -1 0 oxidation NO3- + Sb +5 10 Sb4O6 + NO 0 oxidation +3 reduction +2 Balancing Redox Equations The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? 1. Write the unbalanced equation for the reaction in ionic form. Fe2+ + Cr2O72Fe3+ + Cr3+ 2. Separate the equation into two half-reactions. +2 +3 Fe2+ Oxidation: +6 Reduction: Cr2O7 Fe3+ +3 2- Cr3+ 3. Balance the atoms other than O and H in each halfreaction. Cr2O722Cr3+ 11 4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O 5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ 6e- + 14H+ + Cr2O72- Fe3+ + 1e2Cr3+ + 7H2O 6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e12 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 7. Add the two half-reactions together and construct the final equation. The number of electrons on both sides must cancel. You should also cancel like species. Oxidation: 6Fe2+ Reduction:6e- + 14H+ + Cr2O7214H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 6e2Cr3+ + 7H2O 6Fe3+ + 2Cr3+ + 7H2O 8. Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. You should combine H+ and OH- to make H2O. 13 Balance the following reaction: CuS(s) + NO3-(aq) g Cu2+(aq) + SO42-(aq) + NO(g) Step 1: Identify and write down the unbalanced half reactions. oxidation -2 CuS(s) + NO3-(aq) Cu2+(aq) + SO42-(aq) + NO(g) +5 14 +6 reduction +2 CuS(s) Cu2+(aq) + SO42-(aq) oxidation NO3-(aq) NO(g) reduction Step 2: Balance atoms and charges in each half reaction. Use H2O to balance O, and H+ to balance H (assume acidic media). Use e- to balance charge. 4H2O + CuS(s) 3e- + 4H+ + NO3-(aq) 15 Cu2+(aq) + SO42-(aq)+ 8H+ + 8eNO(g) + 2H2O Step 3: Multiply each half reaction by an integer such that electrons cancel. 4H2O + CuS(s) Cu2+(aq) + SO42-(aq)+ 8H+ + 8e- x 3 3e- + 4H+ + NO3-(aq) 3CuS + 8 NO3- + 8H+ NO(g) + 2H2O x 8 8NO + 3Cu2+ + 3SO42- + 4H2O Step 4: Add the two half reactions and cancel like species. 16 Step 5. For reactions that occur in basic solution, proceed as above. At the end, add OH- to both sides for every H+ present, combine to yield water on the H+ side. 3CuS + 8 NO3- + 8H+ 8NO + 3Cu2+ + 3SO42- + 4H2O + 8OH- + 8OH- 3CuS + 8 NO3- + 8H2O 4 8NO + 3Cu2+ + 3SO42- + 4H2O + 8OH- The final equation in basic medium is 3CuS + 8 NO3- + 4H2O g 8NO + 3Cu2+ + 3SO42- + 8OH17 Balance the following redox reaction occurring in basic media: BH4- + ClO3BH4-5 ClO3- +5 18 H2BO3- + ClH2BO3- oxidation +3 Cl- -1 reduction 3H2O + BH46e- + 6H+ + ClO3- H2BO3- + 8H+ + 8eCl- + 3H2O x3 x4 3 3 BH4- + 4 ClO3- 4 Cl- + 3H2BO3- + 3H2O Here, the net balanced equation contains no H+ or OH-, and thus it can be used as is for acidic or basic solutions. 19 Galvanic Cells In redox reactions, electrons are transferred from the oxidized species to the reduced species. Imagine separating the two half cells physically, then providing a circuit through which the electrons travel from one half cell to the other. 20 Ionic Conduction: Salt Bridge Salt bridge/porous disk: allows for ion migration such that the solutions will remain neutral. 21 Calculate your score in the second exam: Suppose your score on the exam paper was x: The final score = ([x+2]/42)*40 If the points on your paper were 38, then your score in the exam is: ([38+2]/42)*40 = 38. If your score on the exam paper is 28, your final score is: ([28+2]/42)*40 = 28.57 = 29 And so on. 22