1. Balancing Redox Reactions: Acid Conditions Balance the following redox equation: Cr 2 O 7 2- (aq) + HNO 2 (aq) --> Cr 3+ (aq) + NO 3 - (aq) (acidic) Oxidation Reaction: Cr 2 O 7 2- (aq) ----> Cr 3+ (aq) Cr 2 O 7 2- (aq) ----> 2Cr 3+ + 7H 2 O Balance Cr and O Cr 2 O 7 2- (aq) + 14H + ----> 2Cr 3+ + 7H 2 O Balance H Cr 2 O 7 2- (aq) + 14H + + 6e - ----> 2Cr 3+ + 7H 2 O Balance charge Reduction Reaction: HNO 2 (aq) ---> NO 3 - (aq) HNO 2 (aq) + H 2 O ---> NO 3 - (aq) HNO 2 (aq) + H 2 O ---> NO 3 - (aq) + 3H + HNO 2 (aq) + H 2 O ---> NO 3 - (aq) + 3H + + 2e -

2. Balancing Redox Reactions: Acid Conditions Balance the following redox equation: Cr 2 O 7 2- (aq) + HNO 2 (aq) --> Cr 3+ (aq) + NO 3 - (aq) (acidic) Step 3: Multiply the half reaction with the lowest number of electrons by the stoichiometric coefficient to make the number of electrons in each half reaction the same. Cr 2 O 7 2- (aq) + 14H + + 6e - ----> 2Cr 3+ + 7H 2 O HNO 2 (aq) + H 2 O ---> NO 3 - (aq) + 3H + + 2e - Multiply by 3 Cr 2 O 7 2- (aq) + 14H + + 6e - + 3HNO 2 (aq) + 3H 2 O ---> 2Cr 3+ + 7H 2 O + 3NO 3 - (aq) + 9H + + 6e -

3. Balancing Redox Reactions: Acid Conditions Cr 2 O 7 2- (aq) + 14H + + 6e - + 3HNO 2 (aq) + 3H 2 O ---> 2Cr 3+ + 7H 2 O + 3NO 3 - (aq) + 9H + + 6e - Step 4: Cancel the compounds found on both side of the arrow Cr 2 O 7 2- (aq) + 5H + + 3HNO 2 (aq) --> 2Cr 3+ + 4H 2 O + 3NO 3 - (aq) Step 5: Check work. Done!

4. Balancing Redox Reactions: Acid Conditions MnO 4 - (aq) + Br - (aq) ---> MnO 2 (aq) + BrO 3 - (aq) Step 1: Half reactions Mn: +7 --> +4Gain 3e -, Reduction Br: -1 --> +5Lose 6e -, Oxidation Step 2: Balance half reactions Oxidation: Br - (aq) --> BrO 3 - 3H 2 O + Br - (aq) --> BrO 3 - 3H 2 O + Br - (aq) --> BrO 3 - + 6H + 3H 2 O + Br - (aq) --> BrO 3 - + 6H + + 6e - Reduction: MnO 4 - --> MnO 2 MnO 4 - --> MnO 2 + 2H 2 O MnO 4 - + 4H + --> MnO 2 + 2H 2 O MnO 4 - + 4H + + 3e - --> MnO 2 + 2H 2 O

5. Balancing Redox Reactions: Acid Conditions MnO 4 - (aq) + Br - (aq) ---> MnO 2 (aq) + BrO 3 - (aq) Step 3: We have 6 electrons transferred in the oxidation reaction and 3 electrons in the reduction reactions. Multiply the reduction reaction by 2 and combine with the oxidation reaction. 2MnO 4 - + Br - + 3H 2 O + 8H + + 6e - --> BrO 3 - + 6H + + 6e - + 2MnO 2 + 4H 2 O Step 4: Cancel molecules found on both sides of the reaction arrow. 2MnO 4 - + Br - + 2H + --> BrO 3 - + 2MnO 2 + H 2 O Step 5: Done!

6. Balancing Redox Reactions: Basic Conditions Cr(OH) 3 (s) + ClO 3 - (aq) ---> CrO 4 2- (aq) + Cl - (aq) Step 1: Same as acidic conditions. Identify the 2 half reactions Cr: +3 --> +6Lose 3 electrons, Oxidation Cl: +5 --> -1Gain 6 electrons, Reduction Step 2: Balance the half reactions Oxidation reaction: Cr(OH) 3 (s) ---> CrO 4 2- Cr(OH) 3 (s) + H 2 O ---> CrO 4 2- Cr(OH) 3 (s) + H 2 O ---> CrO 4 2- + 5H + Cr(OH) 3 (s) + H 2 O ---> CrO 4 2- + 5H + + 3e -

7. Balancing Redox Reactions: Basic Conditions Cr(OH) 3 (s) + ClO 3 - (aq) ---> CrO 4 2- (aq) + Cl - (aq) Step 2 (cont’d): Balance the half reactions Reduction reaction: ClO 3 - (aq) ---> Cl - ClO 3 - (aq) ---> Cl - + 3H 2 O ClO 3 - (aq) + 6H + ---> Cl - + 3H 2 O ClO 3 - (aq) + 6H + + 6e - ---> Cl - + 3H 2 O Step 3: Multiply the oxidation reaction by 2 to get the same # of electrons and combine the equations 2Cr(OH) 3 (s) + 2H 2 O + ClO 3 - (aq) + 6H + + 6e - --> Cl - + 3H 2 O + 2CrO 4 2- + 10H + + 6e -

8. Balancing Redox Reactions: Basic Conditions 2Cr(OH) 3 (s) + 2H 2 O + ClO 3 - (aq) + 6H + + 6e - --> Cl - + 3H 2 O + 2CrO 4 2- + 10H + + 6e - Step 4: Remove things on both sides of the reaction arrow 2Cr(OH) 3 (s) + ClO 3 - (aq) --> Cl - + H 2 O + 2CrO 4 2- + 4H + Step 5: Now for the tricky part. Count the number of protons and add the same number of hydroxide ions to BOTH sides. 2Cr(OH) 3 (s) + ClO 3 - (aq) + 4 OH - --> Cl - + H 2 O + 2CrO 4 2- + 4H + + 4 OH - OR 2Cr(OH) 3 (s) + ClO 3 - (aq) + 4 OH - --> Cl - + 5H 2 O + 2CrO 4 2-

9. 12.3: The Structure of Galvanic Cells A Galvanic Cell is a device in which electric current is produced by a spontaneous chemical reaction A battery is a series of galvanic cells connected in series so that the voltage of the battery is the sum of the voltages of the individual cells Voltage is the ability to push electric current through a circuit

10. 12.3: The Structure of Galvanic Cells Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s) Electrons from the zinc are transferred to the copper If we put a wire between the two conductive electrodes (one of which will be Zn), we could capture the flow of electrons The solution between the two electrodes is an ELECTROLYTE (perhaps copper nitrate?)

11. 12.3: The Structure of Galvanic Cells Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s) The ANODE is the electrode where oxidation takes place The CATHODE is the electrode where reduction takes place Electrons are PUSHED from the ANODE to the CATHODE Galvanic cells have the cathode marked with a ‘+’ sign and the anode with a ‘-’ sign Electrons come from the anode, hence the ‘-’ sign Electrons are going to the cathode, so they are drawn to the ‘+’ sign

12. 12.3: The Daniel Cell Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s) By separating the electrodes from each other and the electrolyte solution, the current has to flow through the wire The anode is pure zinc surrounded by zinc sulfate (and the clay pot) The Cathode is pure copper The electrolyte solution is copper (II) sulfate

13. Galvanic Cells We can use other materials at the electrode Take your car battery for example. We can dip a platinum wire into a solution of sulfuric acid to complete the reduction half reaction 2H + + 2e - --> H 2 This arrangement of an electrode is called a half cell. The Zn electrode is a Daniel cell is another example of a half cell

14. 12.4: Cell Potential and Gibbs Free Energy We describe the Cell Potential, E, of a cell as the ability to force electrons through a circuit Cells with high cell potentials generate high voltages and vice versa for cells with low potentials On you consume all of the ions or metals in the electrodes, the battery is dead The redox reaction is at equilibrium

15. 12.4: Measuring Cell Potential The SI Unit of cell potential is the Volt A volt is defined as the potential necessary for one Coulomb (a unit of charge times time) falling through a 1V potential to generate 1J of energy 1VC = 1J This means that cell potential is an example of non- expansion work We need to relate  G to E

16. Cell Potential and Gibbs Free Energy At constant T & P, we know that the free energy is equal to the maximum non-expansion work that can be done by a system  G = w e The work done when a number of moles, n, of electrons travels through a potential difference, E, is their charge times the cell potential w e = - neN A En = # of moles e = 1.602x10 -19 C N A = 6.022 x10 23 e - /mole But, we can simplify eN A to Faraday’s constant, F, which is equal to 9.6485x10 4 C/mole e - w e = -nFE Combine the 2 equations and you get  G = -nFE

17. Cell Potential and Gibbs Free Energy  G = -nFE If the cell potential is positive,  G is negative and the cell will form products in the redox reaction spontaneously If the cell potential is negative,  G is positive and the cell will form the reactants of the redox reaction spontaneously We will cell the ideal or maximum potential of a cell the Electromotive Force, emf of the cell Real life cells always produce a cell potential that is smaller than the emf

18. Example The emf of the Daniel cell for certain concentrations of copper and zinc ions is 1.04V. What is the  G under these conditions?

19. Example The reaction taking place in the solver cell used in cameras and watches is: Ag 2 O (s) + Zn (s) --> 2Ag (s) + ZnO (s) The emf when the battery is new is 1.6V. What is the  G of the cell?

20. Standard Reaction Free Energy  G = -nFE ° E° is the standard EMF of the cell All gases are 1bar All solutes are 1M All liquids and solids are pure The free energy,  G°, will change based up the molar amounts of materials used, but E° doesn’t The ability to push/pull electrons in independent of how many materials there are Make sense? (There’s the ‘n’ term in the  G° equation)

21. 12.5: Cell Notation 1.The cell voltage is affected by the liquids/electrolytes mixing 2.In order to harness the maximum potential of a cell, we separate the electrodes/solutions and use a salt bridge to complete the circuit 3.A salt bridge is a gel of concentrated aqueous ions that allows the circuit to complete Allows the ions to maintain a balance of charge between the cells without actually touching The electrolyte in the salt bridge doesn’t participate in the redox reaction

22. 12.5: Cell Notation 1.We show the presence of a salt bridge in the notation of a reaction by a double line: || For example: In the Daniel cell previously described, we describe it as: Zn (s) | Zn 2+ (aq) | Cu 2+ (aq) | Cu (s) This doesn’t show how the electrodes are separated though. The salt bridge is between the 2 half cells or electrodes, so we write it as: Zn (s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu (s) Typically, the anode (-) is on the left and the cathode (+) is on the right Oxidation occurs on the left and reduction occurs on the right

23. Example Write the cell reaction occurring for the cell: Pt (s) | H 2 (g) | HCl (aq) | HgCl 2 (s) | Hg (s)

24. Example Write the cell reaction occurring for the cell: Pt (s) | H 2 (g) | H + (aq) || Co 3+ (aq) | Co 2+ (aq)

25. The sign of the emf is the same as the sign of the right hand electrode in the cell diagram What does this mean? If the emf is negative, then the reverse reaction is spontaneous

26. We can build a galvanic cell out of nearly any combination of electrodes Each electrode has a Standard Potential, E°, which is a measure of the electron pulling power of the electrode Because of this, standard potentials are always written as reduction reactions 12.6: Standard Potentials

27. As we go down the table, you’ll notice that the E° values become negative. What does this mean? If E° is negative, what is  G°? The reverse reaction is spontaneous

28. 12.5: Building a Cell with Standard Potentials The standard potential of a cell is equal to the difference of the standard potentials of the 2 electrodes. E° (cell) = E° (right electrode) - E° (left electrode) Or E° = E° R - E° L Or E° = E° reduction - E° oxidation If E° is greater than zero, then the cell works as written

29. 12.5: Building a Cell with Standard Potentials The more positive the potential, the greater the electron pulling power of the reduction half reaction This means the redox couple is a strongly oxidizing pair The more negative the potential, the greater the electron donating power of the oxidation half reaction This means the redox couple is a strongly reducing pair

30. Example The standard potential of the Ag+/Ag electrode is 0.80V and the standard EMF of the cell Pt (s) | I 2 (s) | I - (aq) || Ag + (aq) | Ag (s) at the same temperature. What is the standard potential of the I 2 /I - electrode?

31. Non-traditional Standard Potentials In general, the most negative E° values are on the left and the most positive are on the right side of the periodic table. Does this make sense given our understanding of periodic trends? Especially, Z eff and electronegativity

32. The Electrochemical Series The more negative the standard potential, the greater its reducing strength Only redox couples with negative standard potential values can reduce hydrogen

33. The Electrochemical Series The species on the left of each equation are potential oxidizing agents The species on the right are potential reducing agents The relationship trend is called the Electrochemical Series F 2 is a strong oxidizing agent Li + is an extremely poor oxidizing agent Lithium metal is a strong reducing agent

34. Predicting the feasibility of a possible redox reaction Standard electrode potentials (redox potentials) are one way of measuring how easily a substance loses electrons. In particular, they give a measure of relative positions of equilibrium in reactions such as: Zn 2+ + 2e - Zn (s)E°= -0.76V Cu 2+ + 2e - Cu (s)E°= +0.34V The more negative the E° value, the further the position of equilibrium lies to the left. Remember that this is always relative to the hydrogen equilibrium - and not in absolute terms. The negative sign of the zinc E° value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E° value shows that it releases electrons less readily than hydrogen. Whenever you link two of these equilibria together (either via a bit of wire, or by allowing one of the substances to give electrons directly to another one in a test tube) electrons flow from one equilibrium to the other. That upsets the equilibria, and Le Chatelier's Principle applies. The positions of equilibrium move - and keep on moving if the electrons continue to be transferred. Using the Electrochemical Series

35. Predicting the feasibility of a possible redox reaction The two equilibria essentially turn into two one-way reactions: The equilibrium with the more negative (or less positive) E° value will move to the left. The equilibrium with the more positive (or less negative) E° value will move to the right. Using the Electrochemical Series

36. Will magnesium react with dilute sulfuric acid? Mg 2+ + 2e - Mg (s)E°= -2.37V 2H + + 2e - H 2 (s)E°= 0.0V Using the Electrochemical Series

37. Will copper react with dilute sulfuric acid? Cu 2+ + 2e - Cu (s)E°= +2.37V 2H + + 2e - H 2 (s)E°= 0.0V Using the Electrochemical Series