1. Redox Reactions & Electrochemical Cells

2.  Type of reaction based on transfer of electrons between species  Used to be based on gain or loss of O  Now: gain or loss of electrons  Oxidation: loss of electrons  Reduction: gain of electrons  Numerous reactions/applications  Single displacement  Combustion and other organic reactions  Rust and corrosion  Batteries Oxidation Reduction

3. Oxidation Numbers Oxidation numbers are numbers assigned to atoms that reflect the net charge an atom would have if the electrons in the chemical bonds involving that atom were assigned to the more electronegative element

4.  Rules!  Oxidation Number Rules  The oxidation number for an atom in its elemental form is always zero.  The oxidation number of a monoatomic ion = charge of the monatomic ion.  Examples:  Oxidation number of S -2 is -2.  Oxidation number of Al +3 is +3.  The oxidation number of all Group 1A metals = +1 (unless elemental).  The oxidation number of all Group 2A metals = +2 (unless elemental).  Hydrogen has two possible oxidation numbers:  +1 when bonded to a nonmetal  -1 when bonded to a metal  Oxygen has two possible oxidation numbers:  -1 in peroxides (O 2 - 2 )  -2 in all other compounds...most common  Fluorine always has an oxidation number of -1.  The sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.  The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the polyatomic ion.

5.  NO 2 O: 2 x (-2) = -4 N: must be +4 to make sum = 0  K 2 CrO 4 Separate into ions: K +1 and CrO 4 -2 K : +1 For CrO 4 -2 : O: 4 x (-2) = -8 Cr: ion must add up to -2, so Cr = +6 Examples

6.  To determine change in oxidation states, compare oxidation numbers from one side of an equation to another  CaCO 3 → CaO + CO 2 CaCO 3 -Ca: +2 -CO 3 -2 : O is -2, so C must be +4 CaO -Ca+2 and O-2 CO 2 - O 2: O is 2 x (-2) = -4, so C is +4 No changes in oxidation numbers, so not a redox reaction!!! Has an element been reduced or oxidized?

7.  Fe + CuSO 4 → FeSO 4 + Cu  Fe: 0Fe: +2 Lost electrons! Oxidized!  Cu: +2Cu: 0 Gained electrons! Reduced!  S: +6S: +6 Same oxidation state  O: -2O: -2 Same oxidation state Example #2

8.  All oxidation reactions involve a reduction occurring at the same time. That is why they are called redox: one cannot occur without the other When one substance is oxidized, it is in turn reducing another substance. Therefore it is called the “reducing agent”. When a substance is reduced, another substance is oxidized so the reduced substance is known as the “oxidizing agent’  Fe+Cu 2+ → Fe 2+ + Cu Fe is oxidized so is the reducing agent Cu is reduced so is the oxidizing agent

10. Zn + 2HCl → Zn 2+ + H 2 +2Cl -  In this reaction Zn + 2H + → Zn 2+ + H 2 (Cl does not change oxidation state  Zn is oxidized to Zn 2+ (loses 2 electrons)  reducing agent (Reduces H + )  H + is reduced to H 2 (gains 2 electrons)  oxidizing agent (Oxidizes Zn) Another Example:

11. Strongest Oxidizing AgentWeakest Reducing Agent Ba 2+ (aq) Ba (s) Ca 2+ (aq) Ca (s) Mg 2+ (aq) Mg (s) Al 3+ (aq) Al (s) Zn 2+ (aq) Zn (s) Cr 3+ (aq) Cr (s) Fe 2+ (aq) Fe (s) Cd 2+ (aq) Cd (s) Tl + (aq) Tl (s) Co 2+ (aq) Co (s) Ni 2+ (aq) Ni (s) Sn 2+ (aq) Sn (s) Cu 2+ (aq) Cu (s) Hg 2+ (aq) Hg (s) Ag 2+ (aq) Ag (s) Pt 2+ (aq) Pt (s) Au 1+ (aq) Au (s) Weakest Oxidizing AgentStrongest Reducing Agent

12.  Half reactions are a way to separate the substance that is oxidizing from the substance that is being reduced. It is useful when balancing an oxidation reduction reaction  For the reaction in the previous slide (Zn + 2HCl → Zn 2+ + H 2 +2Cl - ) the half reactions would be:  Zn →Zn 2+ + 2e- (oxidation)  2H + + 2e- →H 2 (reduction: notice that the equation is balanced) Half reactions

13.  Another example:

14. Balancing more complex reactions using the half reaction method When balancing a more complex reaction it is useful to first split the reaction into two half reactions, balance the half reactions and then put them back together to get a net equation. The following steps will be useful: 1. Write the half reaction for the oxidation and reduction reactions 2. Balance elements other than H & O 3. Balance O with H 2 O 4. Balance H with H + 5. Balance charge by adding electrons as needed 6. Multiply each half reaction if necessary to make the number of electrons in each half reaction equal 7. Add the half reactions & cancel out like terms where possible

15. Balancing Redox Reactions (continued) Cr 2 O 7 2- + C 2 H 6 O → Cr 3+ + C 2 H 4 O What is oxidized? Reduced? Write the half reactions for each Oxidized: C 2 H 6 O → C 2 H 4 O Reduced: Cr 2 O 7 2- → Cr 3+ Balance all atoms but H & O C 2 H 6 O → C 2 H 4 O Cr 2 O 7 2- → 2Cr 3+ Continue next slide…..

16. Balance O using H 2 O C 2 H 6 O → C 2 H 4 O Cr 2 O 7 2- → 2 Cr 3+ + 7 H 2 O Balance H using H + (this is assuming an acid solution..we will look at basic solutions next) C 2 H 6 O → C 2 H 4 O + 2 H+ 14 H+ + Cr 2 O 7 2- → 2 Cr 3+ + 7 H 2 O Continued on next slide….

17. Balance charge by adding the appropriate number of electrons to either side (pay attention to which is oxidation and which is reduction!) The net charge does not have to equal zero, they just have to equal each other. This is done by adding electrons. To determine the number of electrons required, find the net charge of each side the equation. C 2 H 6 O → C 2 H 4 O + 2 H+ 0 +2 14 H+ + Cr 2 O 7 2- → 2 Cr 3+ + 7 H 2 O +12 +6 2e- must be added to the first equation on product side and 6e- to the reactant side of the second equation (Check! Does this make sense???)

18. Almost Done! Multiply each reaction (if necessary) to make the number of electrons lost equal to the number of electrons gained 6 e- + 14 H+ + Cr 2 O 7 2- → 2 Cr 3+ + 7 H 2 O 3 x (C 2 H 6 O → C 2 H 4 O + 2 H+ + 2 e) The final step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which are on both sides of the arrow. 6 e- + 14 H+ + Cr 2 O 7 2- + 3 C 2 H 6 O → 2 Cr 3+ + 7 H 2 O + 3 C 2 H 4 O + 6 H+ + 6 e- The equation can be further simplified by subtracting out 6 e- and 6 H+ ions from both sides of the equation to give the final equation. 8 H+ + Cr 2 O 7 2- + 3 C 2 H 6 O → 2 Cr 3+ + 7 H 2 O + 3 C 2 H 4 O  Note: the equation is completely balanced in terms of having an equal number of atoms as well as charges.

19. Balancing Redox Reactions : another example! Balance following reaction: CuS (s) + NO 3 - (aq) → Cu 2+ (aq) + SO 4 2- (aq) + NO (g) 1. Split into 2 Half-Reactions (Which is oxidized? Reduced?) CuS → Cu 2+ + SO 4 2- NO 3 -1 → NO Balance all atoms but H &O Already balanced!

20. Balance O using H 2 O CuS + 4H 2 O → Cu 2+ + SO 4 2- NO 3 -1 → NO + 2H 2 O Balance H using H+ CuS + 4H 2 O → Cu 2+ + SO 4 2- + 8H+ NO 3 -1 + 4H+ → NO + 2H 2 O Balance Charge CuS + 4H 2 O → Cu 2+ + SO 4 2- + 8H+ + 8e- NO 3 -1 + 4H+ + 3e- → NO + 2H 2 O

21. Multiply to get charges equal for both equations 3 x (CuS + 4H 2 O → Cu 2+ + SO 4 2- + 8H+ + 8e-) 8 x (NO 3 -1 + 4H + + 3e- → NO + 2H 2 O) Add and cancel like terms 3CuS + 12H 2 O + 8NO 3 -1 + 32H + + 24e- → 3Cu 2+ + 3SO 4 2- + 24H + + 8NO +16H 2 O + 24e- 8H + 4H 2 O Final equation: 3CuS (s) + 8 NO 3 -1 (aq) + 8H + (aq) → 3Cu 2+ (aq) + 3 SO 4 2- (aq) + 8NO (g) +4H 2 O (l)

22. Check mass balance and charge balance in equation  Left  3 x Cu  3 x S  8 x N  24 x O  8 x H (8 x 1 -) + (8 x H +) = 0  Right  3 x Cu  3 x S  8 x N  24 x O  8 x H (3 x 2 + )+(3 x 2 - ) = 0

23. What if the solution was basic? In these examples we have assumed the solution was acidic - we added H + to balance the equation. To make a basic solution, we will add enough OH- ions (to both sides of the equation) to neutralize the H + added in the reaction. Balance the equation in acid, then add enough OH - to turn the H + into H 2 O. You must add OH - to both sides, so that one side then ends up with excess OH -, and the solution is now a base

24. Look at our previously balanced equation: 3CuS (s) + 8 NO 3 -1 (aq) + 8H + (aq) → 3Cu 2+ (aq) + 3 SO 4 2- (aq) + 8NO (g) +4H 2 O (l) If we add 8OH- to both sides, this is what happens: 3CuS (s) + 8 NO 3 -1 (aq) + 8H + (aq) →3Cu 2+ (aq) +3 SO 4 2- (aq) + 8NO (g) +4H 2 O (l) + 8OH - 8OH - 8H 2 O 4 Final equation: 3CuS (s) + 8 NO 3 -1 (aq) + 4H 2 O aq) →3Cu 2+ (aq) +3 SO 4 2- (aq) + 8NO (g) + 8OH - Notice! After adding OH -, we have 8 waters on the left and 4 on the right. If we simplify, the waters on the right cancel and there are 4 waters remaining on the left. The right side now has an excess of 8 OH -, so the solution is a base

25. Balancing Redox Equations Practice Balance in acid: H 2 C 2 O 4 + MnO 4 -1 → Mn 2+ + CO 2 Balance in base: CN - + MnO 4 -1 → CNO -1 + MnO 2

26. Redox Reactions in Electrochemistry Redox reactions may be used to produce electricity through chemical work. These reactions are what we use to make batteries Two Types of Electrochemical Cells: 1. Galvanic 2. Electrolytic Galvanic Cell - Converts chemical potential energy into an electrical potential to perform work Electrolytic Cell- Uses electrical energy to force a chemical reaction to happen that would not otherwise occur

27. What happens when a strip of zinc is put into a solution of copper sulfate? CuSO 4 + Zn → Cu + ZnSO 4 The half reactions would be: Cu 2+ + 2 e − → Cu Zn → Zn 2+ + 2 e − Galvanic (Voltaic) Cells

28. Zinc is added to a blue solution of copper(II) sulfate The blue color disappears…the zinc metal reacts, and solid copper metal precipitates on the zinc strip The zinc is oxidized and the copper ions are reduced

29. The galvanic cell If the half reactions in a redox reaction are separated into two “cells” and connected by a wire, the reaction will still occur. This is a simple form of a battery and is known as a galvanic or voltaic cell

30. The metals are known as electrodes ANODE - Where OXIDATION takes place Think of them alphabetically: Anode, Cathode. Oxidation, Reduction CATHODE - Where REDUCTION takes place Generally: When drawing cell, anode is on the left, cathode is on the right but this is not always true!

31. Remember the activity series??  This is what the activity series compares: the ability for an element to be reduced or oxidized  What metals have the greatest reducing power? (are able to give electrons most readily?). These are at the top of the list  All are compared to hydrogen as the standard (although H 2 is not a metal, it is readily “displaced” from compounds by many metals).

33.  If we separate two 1/2 cells physically, then provide a conduit through which the electrons travel from one cell to the other, the electrons flow and a current is produced

35.  Over time the reaction will stop. Why?  Charge build-up in the cells will stop the electron flow.  Salt bridge: allows charge neutrality in each cell to be maintained. Salt bridge/porous disk: allows for ion migration so that the solutions will remain neutral.

36. Example: Zinc as the anode: A piece of Zn metal is immersed in ZnSO 4 (aq) Atoms of the zinc metal lose electrons and become zinc ions in solution. The zinc electrode loses mass Half-reaction : Zn → Zn +2 + 2e- (oxidation half-reaction) Oxidation always occurs at the anode.

37. Copper is the cathode: A piece of Cu metal is immersed in CuSO 4 (aq) Copper ions in solution gain electrons to add copper atoms to the copper metal. The half-reaction: Cu +2 + 2e- → Cu The copper electrode gains mass This is a reduction half-reaction. Reduction always occurs at the cathode.

38. The electrons move from the anode to the cathode through the external circuit. Zn +2 go into solution. SO 4 -2 ions in solution move from the cathode to anode through the salt bridge. The cell continues to operate as long as there is a potential energy difference between the half-cells.

39. Cell Notation A shorthand method of representing a galvanic cell: Zn(s) | Zn 2+ (aq),1M; SO 4 2- (1M) || Cu 2+ (aq), 1M; SO 4 2- (1M)| Cu(s)  Begin at the solid zinc anode. A single vertical line indicates phase change from solid zinc anode to aqueous zinc sulfate solution. The double vertical line indicates the salt bridge between the anode and the cathode. A single vertical line indicates a phase change from aqueous copper (II) solution in the cathode to the solid copper cathode.

40. Oxidation half-reaction Zn(s) Salt bridge Zn 2+ Cu 2+ Na + Zn Cu SO 4 2– Zn 2+ (aq) + 2e – Voltmeter e–e– Anode (–) (+) REVIEW!!!

41. Zn 2+ Zn Oxidation half-reaction Zn(s) Salt bridge Zn 2+ Cu 2+ Na + Zn Cu SO 4 2– Zn 2+ (aq) + 2e – Voltmeter e–e– 2e – lost per Zn atom oxidized Anode (–) (+) e–e–

42. Zn 2+ Zn Oxidation half-reaction Reduction half-reaction Cu 2+ (aq) + 2e – Zn(s) Salt bridge Zn 2+ Cu 2+ Na + Zn Cu SO 4 2– Zn 2+ (aq) + 2e – Cu(s) Voltmeter e–e– e–e– 2e – lost per Zn atom oxidized Anode (–) Cathode (+) e–e–

43. Cu 2+ e–e– Cu 2e – gained per Cu 2+ ion reduced Zn 2+ Zn Oxidation half-reaction Reduction half-reaction Cu 2+ (aq) + 2e – Zn(s) Salt bridge Anode (–) Cathode (+) Zn 2+ Cu 2+ Na + Zn Cu SO 4 2– Zn 2+ (aq) + 2e – Cu(s) Voltmeter e–e– e–e– 2e – lost per Zn atom oxidized e–e–

44. Cu 2+ e–e– Cu 2e – gained per Cu 2+ ion reduced Zn 2+ Zn Oxidation half-reaction Reduction half-reaction Overall (cell) reaction Zn(s) + Cu 2+ (aq) Cu 2+ (aq) + 2e – Zn(s) Salt bridge Zn 2+ Cu 2+ Na + Zn Cu SO 4 2– Zn 2+ (aq) + 2e – Cu(s) Zn 2+ (aq) + Cu(s) Voltmeter e–e– e–e– Anode (–) Cathode (+) 2e – lost per Zn atom oxidized e–e–

45. Galvanic cell animation  Galvanic cell Galvanic cell  http://www.mhhe.com/physsci/chemistry/essentialche mistry/flash/galvan5.swf

46. Standard Hydrogen Electrode (SHE CELL)  Calculates the potential for an electrochemical cell : A standard electrode is needed to provide a reference point. Standards: Concentration 1 M Partial pressure of gases 1 atm Temperature 25˚C (298K).

47. SHE continued The reaction show reduction potential compared to hydrogen H 2 (g) → 2H+(aq) + 2e- A platinum electrode provides a surface on which the hydrogen gas can be in contact with the hydrogen ions (aq) The hydrogen electrode is always placed as the negative electrode of a cell and is written as: The potential of this cell is assigned the value of 0 V. The potentials for other cells are then relative to this value. If E is positive, the metal is less reactive than hydrogen (More easily reduced) If E is negative, the metal is more reactive than hydrogen (More easily oxidized )

48. The potential difference of the zinc half-cell may be found by comparison. Zinc has a greater tendency to ionize than does hydrogen gas so when the circuit is closed, the electrons will flow down the potential gradient from the zinc strip to the platinum electrode. This difference in potential is what causes the electrons to flow. This is known as the electromotive force (emf).

50. Electromotive force The potential difference that drives the flow of electrons in a galvanic cell May be measured with a voltmeter and found by adding the potentials of the oxidation and reduction half-cells Going back to the zinc and copper half cells: Use the standard reduction potential table to find the potentials for the half-cells and calculate the voltage for the zinc-copper cell above. All values given are for reductions: Calculations Zn → Zn +2 + 2e- +0.76 Volts (oxidized, so cell potential reversed in sign) Cu +2 + 2e- → Cu +0.34 Volts Total = +1.10 Volts

51. Calculating cell potentials  Example: An electrochemical cell is created using gold and magnesium half-cells. Determine what is the anode, cathode and overall voltage of the cell  Look at chart. Which metal is more easily reduced?  The more positive, the more readily reduced: Au +3 + 3e- → Au E = +1.50  The more negative, the more readily oxidized: Mg → Mg +2 + 2e- E = +2.37 (reverse to show oxidation, the sign is reversed)  Add the two for total voltage: +1.50 + +2.37 = +3.87  A positive voltage means the reaction is spontaneous!

52. Electrolysis  Non spontaneous reactions forced to occur through an outside energy source  Anode positive (connected to positive end of battery) and Cathode negative (connected to negative end of battery)  Molten NaCl: Na+ accepts e-‘s at cathode and Cl- release e-‘s at anode  Movement of ions maintains current in cell.  Half reactions:  Na + + e- → Na(s)-2. 71V NOT Spontaneous!  Cl- → ½ Cl 2 + e- -1.36V  Overall: Na+ + Cl- → Na + ½ Cl 2

53.  Forces a chemical system into non-equilibrium states  For NaCl in solution: H 2 O more easily reduced than Na+ so cathode reaction: 2H 2 O + 2e- → H 2 (g) +2OH-

54.  electrolysis animation electrolysis animation  http://www.sepuplhs.org/high/hydrogen/electrolysis_si m.html

55. Faraday’s Laws:  Found chemical changes related to the quantity of the electricity and the nature of the substances in the reaction  Mass formed or consumed proportional to charge or amount of energy consumed  mass formed or consumed proportional to atomic or molecular weight  mass inversely proportional to the number of electrons /mole needed to change oxidation state  96,487 Coulombs of charge is generated by 1 mole of electrons (1 Faraday)  Electricity is measured in amperes  Charge per time or 1 Coulomb/second  Used to determine the amount of substance formed at each electrode due to an applied electric current.

56. Examples  How much product will be produced when a current of 0.452 amperes is passed through a cell of molten calcium chloride for 1.50 hours?  What are the half reactions?  Ca +2 + 2e- → Ca (l)  2Cl- → Cl 2 + 2e-  Quantities formed depends on the number of electrons that pass (charge)  C = 1.50h x 60min/h x 60sec/min x 1A/sec x.452A = 2.44 x 10 3 Coulombs  Mass calcium: 2.44 x 10 3 Coulombs x 1mole/96,500C x 1mole Ca/2 mol e- x 40.1gCa/1mol Ca =.507 g Ca  Mass chlorine: 2.44x 10 3 C x 1mole e-/96,500C x 1 mol Cl 2 /2mol e- x 70.9g Cl 2 /mol =.896gCl 2  Practice:  How many grams of barium are produced from molten barium chloride by applying 0.500A for 30 minutes?