1. Study of the relationships between electricity and chemical reactions

2. Oxidation States  REVIEW: How do we determine the oxidation state? What is the oxidation state of the boldfaced element: ○ P 2 O 5 ○ NaH ○ Cr 2 O 7 -2 ○ SnBr 4

3. Redox Reactions  How do we determine if a reaction is a redox reaction?  How would we define the following vocabulary words: Oxidizing agent (oxidant): agent aquiring electrons from another substance (reduced) Reducing agent (reductant): agent giving up electrons (oxidized)

4. Identifying Oxidizing and Reducing Agents  Cd (s) + NiO 2 (s) + 2 H 2 O (l)  Cd(OH) 2 (s) + Ni(OH) 2 (s)  2 H 2 O (l) + Al (s) + MnO 4 - (aq)  Al(OH) 4 - (aq) + MnO 2 (s)

5. Balancing Redox Reactions  How is balancing a redox reaction different than balancing a normal equation?  How do we split a redox reaction?  Why would using half reactions make the process easier?

6. Balancing equations: acidic solutions  In balancing reactions, we use “skeleton” ionic equation. Predict why this is called the skeleton ionic equation.  Step-by-step procedure: STEP 1: BALANCE EQUATION ○ Balance elements other than H and O ○ Balance O atoms by adding H 2 O as needed ○ Balance H atoms by adding H + as needed ○ Finally balance charge by adding e - as needed

7. Balancing equations: acidic solution cont’ STEP 2: Multiply half-reactions by integers as needed to make the number of electrons lost in the oxidation half- reaction equal the number of electrons gained in the reduction half-reaction STEP 3: Add half reactions (simplify if possible by canceling species found on both sides) STEP 4: Check to make sure atoms and charges are balanced

8. ಅಭ್ಯಾಸ (Kannada)  Given the skeleton equation: MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g)

9. April 10 th, 2013  DO NOW: Predict how balancing a redox reaction in a basic solution is different than in an acidic solution.

10. Basic Solutions  For Basic solutions: Follow same process except for after balancing H +, add OH - to both reactants and products side to “neutralize” the H + ion.

11. 행하다 (Korean)  See Worksheet

13. Voltaic Cells  Energy is released when using a voltaic cell, how is it being used?  How can we define voltaic cell?  Draw a diagram of a voltaic cell in your notebook.

14. Anodes and Cathodes!  How do we differentiate between an anode and a cathode?  Predict how the voltaic cell conducts electricity.  Why is using a salt bridge necessary?  Predict which direction the anions flow.

15. Describing a voltaic cell  Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 I - (aq)  2 Cr 3+ (aq) + 3 I 2 (s) + 7 H 2 O (l) Is a spontaneous reaction in a voltaic cell. A solution containing K 2 Cr 2 O 7 and a solution containing H 2 SO 4 are poured into 1 beaker and a solution containing KI is poured into a separate beaker. 2 pieces of metal that won’t react with either are suspended into both solutions. Indicate the following: ○ Reaction at anode ○ Reaction at cathode ○ Direction of Electron flow ○ Direction of ion flow ○ Signs of electrodes

16. Electricity Flow  Describe the flow of water in a waterfall. Why?  Why do electrons flow spontaneously from an anode to a cathode?  How do we measure energy between the two?

17. Potential Energy  Potential Energy difference between the two electrodes is called cell potential. Also known as: Electromotive force (EMF) ○ Pushes e - through circuit Assume 25°C  standard conditions Standard Cell Potential: E° cell Depends on particular anode and cathode half cells  How might we measure E° cell

18. Reference Cell  Why is it necessary to have a reference half reaction? Allows for measure of other cell potentials directly.  Reference cell: SHE (standard hydrogen electrode)

19. Calculating E° cell  How do we calculate E° cell from the SHE?  IE: oxidation of zinc and reduction of hydrogen ion  How would we represent this standard reduction potential?

20. Reactions  How do we determine how likely a reaction is to occur?  Given that electrons must flow spontaneously, what assumption can we make about the values of the E° red of both the anode and the cathode?

21. April 11 th, 2013  Do Now: Thermodyanmics Quiz. ○ Have out Calculator and Reference Sheet.

22. Spontaneity of Voltaic Cells  How can we use our equation for standard cell potentials as a generalized equation?  Based on this equation and your knowledge of thermodynamics, predict how we determine if the reaction is spotaneous.  NOTE: E = EMF @ nonstandard conditions E°= EMF @ standard conditions  Practice: Is this reaction spontaneous: Cu (s) + 2 H + (aq)  Cu 2+ (aq) + H 2 (g)

23. Reactivity of Metals  How are the standard reduction potentials related to the activity series of metals?  Which is a stronger reducing agent: silver or nickel? Why?

24. Gibbs Free Energy (it never leaves you alone)  How can we related Gibbs free energy to EMF?  Predict the sign of the Gibbs free energy if you have a positive EMF. Why is this to be expected?  Rewrite the equation for a reaction in “standard conditions”  How can we relate E° to K?

25. April 12 th, 2013 DO NOW:  Using standard reduction potentials, calculate standard free energy change and k @ 298K for: 4 Ag (s) + O 2 (g) + 4H + (aq)  4 Ag + (aq) + 2 H 2 O (l)  Suppose the reaction is halved, calculate the change in gibbs free energy at standard conditions, EMF at standard conditions, and k.

26. Non standard Conditions  As a cell is discharged (released all possible electricity) reactants are consumed and products are generated the concentration changes until EMF drops to 0 (dead cell) We want to answer: How is EMF generated under non-standard conditions

27. Concentration is a changin!  Walther Nernst (1864-1941) Established many theoretical foundations of electrochemistry  Effect of concentration on cell EMF can be obtained from effect of concentration on Gibbs Free Energy.  Which equation would we want to use to relate effect of concentration on Gibbs Free Energy. Why?

28. Practice… again.  Calculate the EMF @ 298K generated by the voltaic cell in which the reaction is Cr 2 O 7 2- (aq) + 14H + (aq) + 6 I - (aq)  2 Cr 3+ (aq) + 3 I 2 (s) + 7 H 2 O (l) [Cr 2 O 7 2- ] = 2.0 M [H + ] = 1.0 M [I - ]= 1.0 M [Cr 3+ ] = 1.0 x 10 -5 M

29. One More  If the potential of a Zn-H + cell is at 0.45 V at 25°C when [Zn 2+ ] = 1.0 M and P H 2 = 1.0 atm, what is the H+ concentration?

30. Concentration cells  Voltaic cells constructed by the same species in both half-cells with different concentrations

31. Ni 2+ - Ni reaction  Write the two half reactions and the overall reaction.  How can we calculate the EMF of this cell?  Determine the EMF.  If E° is 0, what produces the driving force?  **NOTE: when concentrations become the same: Q = 1, E = 0

32. pH meters  Generating a potential by a concentration difference  Example: A voltaic cell with 2 hydrogen electrodes: Electrode 1: P H 2 = 1.0 atm and unknown concentration of H+ ions. Electrode 2 is a standard Hydrogen electrode. At 298K, E° = 0.211 V, and the electrical current is observed to flow from electrode 2 to electrode 1. What is the pH of the unknown solution?

33. Electrolysis (Electrolytic cells)  Non-spontaneous redox reactions driven by outside electrical energy IE: decomposition of molten NaCl to Na and Cl2  2 electrodes immersed in either molten salt of solution Process still similar to voltaic cells Battery acts as electron pump

34. Electroplating  How is electroplating different than general electrolysis? Uses electrolysis to deposit a thin layer of metal on another metal to improve beauty or resistance to corrosion Uses active electrodes instead of inert

35. Quantitative Electrolysis  How many moles of electrons are needed to create the following from their ions? Sodium Copper Aluminum  Amount of substance reduced or oxidized in electrolytic cell is proportional to number of electrons passed through cell  Quantity of charge is measured in Coulombs Charge on 1 mol = 96485 C. ○ Coulomb = quantity of charge passing a point in a circuit in 1 s when the current is 1 A (ampere) Coulombs = amperes x seconds

36. Practissimo!  Calculate the number of grams of aluminum produced in 1.00 hour by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.