1 Electrochemistry Chapter 19

2 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. Electron Transfer reactions

3 2Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) Electrochemical processes are oxidation- reduction 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 2Mg (s) + O 2 (g) 2Mg O (s)

4 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. 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. Important Concepts

5 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, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1.

6 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. 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. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = -2H = +1 3x(-2) ? = -1 C = +4 Find the oxidation number of carbon in HCO 3 - ? 4.4

7 Redox Reaction Example A redox reaction: ON: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) oxidation reduction Cu(s) is oxidized, therefore it is a reducing agent. Ag + is reduced, therefore it is an oxidizing agent

8 We can envision breaking up the full redox reaction into two half reactions: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) oxidation reduction Cu(s) Cu 2+ (aq) + 2e - Ag + (aq) + e - Ag(s) “half-reactions”

9 Note that the half reactions are combined to make a full reaction: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) oxidation 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.

10 Identify the species being oxidized and reduced in the following (unbalanced) reactions: ClO I - I 2 + Cl oxidation reduction -1 0 NO Sb Sb 4 O 6 + NO oxidation reduction

11 Balancing Redox Equations 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr Reduction: Fe 2+ Fe Balance the atoms other than O and H in each half- reaction. Cr 2 O Cr 3+

12 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O Cr H 2 O 14H + + Cr 2 O Cr H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe e - 6e H + + Cr 2 O Cr H 2 O 6.If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe e - 6e H + + Cr 2 O Cr H 2 O

13 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. 6e H + + Cr 2 O Cr H 2 O 6Fe 2+ 6Fe e - Oxidation: Reduction: 14H + + Cr 2 O Fe 2+ 6Fe Cr H 2 O 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 H 2 O.

14 Step 1: Identify and write down the unbalanced half reactions. CuS(s) Cu 2+ (aq) + SO 4 2- (aq) oxidation NO 3 - (aq) NO(g) reduction Balance the following reaction: CuS(s) + NO 3 - (aq)  Cu 2+ (aq) + SO 4 2- (aq) + NO(g) CuS(s) + NO 3 - (aq) Cu 2+ (aq) + SO 4 2- (aq) + NO(g) oxidation reduction -2 +6

15 Step 2: Balance atoms and charges in each half reaction. Use H 2 O to balance O, and H + to balance H (assume acidic media). Use e - to balance charge. CuS(s) Cu 2+ (aq) + SO 4 2- (aq) NO 3 - (aq) NO(g) 4H 2 O ++ 8H + + 8e - + 2H 2 O4H + +3e - +

16 Step 3: Multiply each half reaction by an integer such that electrons cancel. CuS(s) Cu 2+ (aq) + SO 4 2- (aq) NO 3 - (aq) NO(g) 4H 2 O ++ 8H + + 8e - + 2H 2 O4H + +3e - + x 3 x 8 Step 4: Add the two half reactions and cancel like species. 3CuS + 8 NO H + 8NO + 3Cu SO H 2 O

17 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 NO H + 8NO + 3Cu SO H 2 O + 8OH - 3CuS + 8 NO H 2 O 8NO + 3Cu SO H 2 O + 8OH - 4 The final equation in basic medium is 3CuS + 8 NO H 2 O  8NO + 3Cu SO OH -

18 Balance the following redox reaction occurring in basic media: BH4- H2BO3-BH4- H2BO oxidation ClO 3 - Cl reduction BH ClO 3 - H 2 BO Cl -

19 BH 4 - H 2 BO 3 - ClO 3 - Cl - 3H 2 O ++ 8H + + 8e - + 3H 2 O6H + +6e - + x 3 x BH ClO Cl - + 3H 2 BO H 2 O Here, the net balanced equation contains no H + or OH -, and thus it can be used as is for acidic or basic solutions.

20 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.

21 Salt bridge/porous disk: allows for ion migration such that the solutions will remain neutral. Ionic Conduction: Salt Bridge

22 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 = = 29 And so on.