1. Chapter 18 Electrochemistry

2. Redox Reaction Elements change oxidation number  e.g., single displacement, and combustion, some synthesis and decomposition Oxidation--oxidation number increases Reduction--oxidation number decreases  Both must occur in a reaction--two half reactions oxidizing agent is reactant molecule that causes oxidation  contains element reduced reducing agent is reactant molecule that causes reduction  contains the element oxidized

3. Rules for Assigning Oxidation States 1. Free elements have an oxidation state = 0 2. Monatomic ions have an oxidation state equal to their charge. 3. The sum of the oxidation states of all the atoms in a compound is 0. 4. The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion. 5. The oxidation number of fluorine is always -1 in compounds with other elements.

4. Rules for Assigning Oxidation States 6. Chlorine, bromine and iodine always have oxidation numbers of -1 except when bonded to O or F. 7. The oxidation number of oxygen is almost always -2; the oxidation number of hydrogen is almost always +1. Exceptions: --When oxygen is in the form of a peroxide (O 2 2- ), the oxidation number is -1. --When hydrogen forms a binary compound with a metal, the oxidation number is -1 and the compound is called a hydride.

5. Oxidation and Reduction Oxidation occurs when an atom’s oxidation state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction Reducing agentOxidizing agent

6. Identify the element that is oxidized and the element that is reduced in each of the following reactions. What is the oxidizing and the reducing agent in each reaction? 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

7. Common Oxidizing Agents

8. Common Reducing Agents

9. Balancing Redox Reactions 1. Assign oxidation numbers --determine element oxidized and element reduced 2. Separate the reaction into oxidation and reduction half- reactions. 3. Balance half-reactions by mass a. First balance elements other than H and O b. Balance O using H 2 O c. Balance H using H + 4. Balance each half-reaction by charge by adding electrons to the reactants side of the reduction and the product side of the oxidation. 5. Multiply half-reactions by integers to make # electrons the same in both half-reactions 6.Add half-reactions and cancel the electrons to produce a balanced equation. 7.For reactions that occur in acidic solutions, skip to step 9. 8.For reactions that occur in basic solutions, add the same # of OH - as H + to both sides of the equation. 9. Check that reaction is balanced for mass and charge.

10. Practice - Balance the Equation H 2 O 2 + KI + H 2 SO 4  K 2 SO 4 + I 2 + H 2 O

11. +1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2 oxidation reduction ox:2 I -1  I 2 + 2e -1 red:H 2 O 2 + 2e -1 + 2 H +  2 H 2 O tot2 I -1 + H 2 O 2 + 2 H +  I 2 + 2 H 2 O 1 H 2 O 2 + 2 KI + H 2 SO 4  K 2 SO 4 + 1 I 2 + 2 H 2 O

12. Electric Current Flowing Directly Between Atoms

13. Redox Reactions & Current Redox reactions involve the transfer of electrons from one substance to another. Therefore, redox reactions have the potential to generate an electric current. In order to harness the energy produced by moving electrons, we need to separate the half reactions.

14. Voltaic (or Galvanic) Cell

15. Electrochemical Cells Electrochemistry -- the study of redox reactions that produce or require an electric current. The conversion between chemical energy and electrical energy is carried out in an electrochemical cell Spontaneous redox reactions take place in a voltaic cell (galvanic cell). Non-spontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy.

16. Electrodes Anode electrode where oxidation occurs Cathode electrode where reduction occurs

17. Voltaic (Galvanic) Cell the salt bridge is required to complete the circuit and maintain charge balance

18. Current and Voltage Current is the number of electrons that flow through the system per second unit = Ampere 1 A of current = 1 Coulomb/second 1 A = 6.242 x 10 18 electrons/second Electrode surface area dictates the number of electrons that can flow Potential difference is the difference in potential energy between the reactants and products (between electrodes) unit = Volt 1 V of force = 1 J (of energy)/Coulomb (of charge) The voltage that drive electrons through the external circuit Amount of force pushing the electrons through the wire is called the electromotive force, emf

19. Cell Potential The difference in potential energy between the electrodes in a voltaic cell is called the cell potential The cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode The cell potential under standard conditions is called the standard emf, E° cell 25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

20. Cell Notation Shorthand description of Voltaic cell electrode | electrolyte || electrolyte | electrode Oxidation half-cell on left, reduction half-cell on the right | = phase barrier if multiple electrolytes in same phase, a comma is used rather than | often use an inert electrode || = salt bridge

21. Fe(s) | Fe 2+ (aq) || MnO 4  (aq), Mn 2+ (aq), H + (aq) | Pt(s)

22. Standard Reduction Potential The cell potential cannot be measured for a half-reaction We need to compare them to an arbitrary standard. We select as a standard half-reaction the reduction of H + to H 2 (or the oxidation of H 2 to H + ) standard hydrogen electrode, SHE Standard conditions [H + ]=1M, 25˚C Potential difference (E˚)= 0V

24. Half-Cell Potentials SHE reduction potential is defined to be exactly 0 V Reduction potentials are tabulated for many half- reactions Change sign to get oxidation potential E° cell = E° oxidation + E° reduction E° oxidation =  E° reduction When adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half- reactions to balance the equation

27. Calculating E˚ for a cell reaction Calculate E˚ for the reaction: Cu 2+ (aq) + Fe 2+ (aq)  Fe 2+ (aq) + Cu (s) 1) Break up the redox reaction into half-reactions 2) Find the reduction potential (E˚)for each half-reaction 3) Change the sign on E˚ for the oxidation half-reaction 4) Add E˚ (anode) + E˚ (cathode) 5) If the result is negative, the reaction occurs in the opposite direction.