1. Galvanic Cells Electrochemistry = the interchange of chemical and electrical energy = used constantly in batteries, chemical instruments, etc… I.Galvanic Cells A.Definitions 1)Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced) 2)Oxidation = loss of electron(s) to become more positively charged 3)Reduction = gain of electron(s) to become more negatively charged B.Using Redox Reactions to generate electric current (moving electrons) 1)8H + (aq) + MnO 4 - (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) a)Fe 2+ is oxidized and MnO 4 - is reduced b)Half Reaction = oxidation or reduction process only Reduction: 8H + + MnO 4 - + 5e - Mn 2+ + 4H 2 O Oxidation: 5(Fe 2+ Fe 3+ + e-) Sum = Redox Rxn

2. C.Balancing Redox Equations: Half-Reaction Method in Acidic Solution –MnO 4 - (aq) + Fe 2+ (aq) -----> Fe 3+ (aq) + Mn 2+ (aq) (acidic solution) 1)Identify and write equations for the two half-reactions a.MnO 4 - -----> Mn 2+ (this is the reduction half-reaction) (+7)(-2) (+2) b.Fe 2+ -----> Fe 3+ (this is the oxidation half-reaction) (+2) (+3) 2.Balance each half-reaction a.Add water if you need oxygen b.Add H + if you need hydrogen (since we are in acidic solution) c.Balance the charge by adding electrons d.MnO 4 - -----> Mn 2+ + 4H 2 O e.8H + + MnO 4 - -----> Mn 2+ + 4H 2 O (+7) (+2) f.5e- + 8H + + MnO 4 - -----> Mn 2+ + 4H 2 OBalanced! g.Fe 2+ -----> Fe 3+ + 1e- Balanced!

3. 3.Equalize the number of electrons in each half-reaction and add reactions a.5(Fe 2+ -----> Fe 3+ + 1e-) = 5Fe 2+ -----> 5Fe 3+ + 5e- b.5e- + 8H + + MnO 4 - -----> Mn 2+ + 4H 2 O c.5Fe 2+ + 8H + + MnO 4 - ------> 5Fe 3+ + Mn 2+ + 4H 2 O d.Species (including e-) on each side cancel out (algebra) e.Check that the charges and elements all balance: DONE! 4.Example: H + + Cr 2 O 7 2- + C 2 H 5 OH -----> Cr 3+ + CO 2 + H 2 O D.Balancing Redox Equations: Half-Reaction Method in Basic Solution 1)Follow the Acidic Solution Method until you have the final balance eqn 2)H + can’t exist in basic solution, so add enough OH - to both sides to turn all of the H + in H 2 O 3)Example: Ag + CN - + O 2 -----> Ag(CN) 2 - (basic solution) a.Ag + CN - -----> Ag(CN) 2 - (oxidation half-reaction) b.Becomes: Ag + 2CN - -----> Ag(CN) 2 - + 1e- Balanced c.O 2 -----> (reduction half-reaction) d.Becomes: 4e- + 4H + + O 2 -----> 2H 2 OBalanced

4. e.Equalize the electrons in each half-reaction and add reactions f.4(Ag + 2CN - -----> Ag(CN) 2 - + 1e-) g.Becomes: 4Ag + 8CN - -----> 4Ag(CN) 2 - + 4e- h.Add to: 4e- + 4H + + O 2 -----> 2H 2 O i.Gives: 4Ag + 8CN - + 4H + + O 2 -----> 4Ag(CN) 2 - + 2H 2 O DONE! j.Add OH- ions to both sides to remove H + ions k.4Ag + 8CN - + 4H + + O 2 + 4OH - -----> 4Ag(CN) 2 - + 2H 2 O + 4OH - l.4Ag + 8CN - + 4H 2 O + O 2 -----> 4Ag(CN) 2 - + 2H 2 O + 4OH - m.Cancel water molecules appearing on both sides of the equation 4Ag + 8CN - + 2H 2 O + O 2 -----> 4Ag(CN) 2 - + 4OH - n.Check that everything balances REALLY DONE!

5. 4)In solution: a)Fe and MnO 4 - collide and electrons are transferred b)No work can be obtained; only heat is generated 5)In separate compartments, electrons must go through a wire = Galvanic Cell a)Generates a current = moving electrons from Fe 2+ side to MnO 4 - side b)Current can produce work in a motor c)Salt Bridge = allows ion flow without mixing solutions (Jello-like matrix)

6. d)Chemical reactions occur at the Electrodes = conducting solid dipped into the solution i)Anode = electrode where oxidation occurs (production of e-) ii)Cathode = electrode where reduction occurs (using up e-) E.Cell Potential 1)Think of the Galvanic Cell as an oxidizing agent “pulling” electrons off of the reducing agent. The “pull” = Cell Potential a)  cell = Cell Potential = Electromotive Force = emf b)Units for  cell = Volt = V 1 V = 1 Joule/1 Coulomb 2)Voltmeter = instrument drawing current through a known resistance to find V Potentiometer = voltmeter that doesn’t effect V by measuring it

7. II.Standard Reduction Potentials A.Standard Hydrogen Electrode 1)When measuring a value, you must have a standard to compare it to 2)Cathode = Pt electrode in 1 M H + and 1 atm of H 2 (g) Half Reaction: 2H + + 2e - H 2 (g)  1/2 = 0 3)We will use this cathode to find  cell of other Half Reactions

8. B.What is the  1/2 of Zn o Zn 2+ + 2e - 1)Cathode = SHE Anode = Zn(s) in 1 M Zn 2+ (aq) (standard states) 2)Voltmeter reads  cell = 0.76 V 3)Define   cell =  cell at standard states so everyone can compare data 4)   cell =   (H + H 2 ) +   (Zn o Zn 2+ ) +0.76 V = 0.00 V + x 6)x =   (Zn o Zn 2+ ) = +0.76 V 7)Combine known values in other Galvanic Cells to determine other   1/2 values a) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) b) Anode: Zn Zn 2+ + 2e - c) Cathode: Cu 2+ + 2e - Cu d)   cell = 1.10 V =    Zn) +   (Cu 2+ ) e) 1.10 V = +0.76 V +   (Cu 2+ ) f)   (Cu 2+ + 2e - Cu) = +0.34 V

9. 7)Standard Reduction Potentials can be found in your text appendices a)Always given as a reduction process b)All solutes are 1M, gases = 1 atm 8)Combining Half Reactions to find Cell Potentials a)Reverse one of the half reactions to an oxidation; this reverses the sign of  1/2 b)Don’t need to multiply for coefficients = Intensive Property (color, flavor) c)Example: 2Fe 3+ (aq) + Cu o 2Fe 2+ (aq) + Cu 2+ (aq) i.Fe 3+ + e - Fe 2+  1/2 = +0.77 V ii.Cu 2+ + 2e - Cu o  1/2 = +0.34 V iii.Reverse of (ii) added to (i) = -0.34 V + +0.77 V = +0.43 V =  1/2

10. C.Line Notation = shorthand way to draw Galvanic Cells 1)2Al 3+ (aq) + 3Mg(s) 2Al(s) + 3Mg 2+ (aq) 2)Line Notation: Mg(s) | Mg 2+ (aq) || Al 3+ (aq) | Al(s) a)Left side = Anode, Right side = Cathode b)| = phase change, || = salt bridge c)Far left = anodic electrode material, Far right = cathodic electrode material 3)2MnO 4 - (aq) + 6H + (aq) + 5ClO 3 - (aq) 2Mn 2+ (aq) + 3H 2 O(l) + 5ClO 4 - (aq) a)Anode: ClO 3 - + H 2 O ClO 4 - + 2H + + 2e - b)Cathode: MnO 4 - +5e - + 8H + Mn 2+ + 4H 2 O c)Pt(s) | ClO 3 - (aq), ClO 4 - (aq), H + (aq) || H + (aq), MnO 4 - (aq), Mn 2+ (aq) | Pt(s) ClO 3 -, ClO 4 -, H + Mn 2+, H +, MnO 4 - Pt(s)

11. D.Direction of electron flow in a cell 1)Cell always runs in a direction to produce a positive  cell 2)Fe 2+ + 2e - Fe o  1/2 = -0.44 V MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O  1/2 = +1.51 V 3)We put the cell together to get a positive potential a)5(Fe o Fe 2+ + 2e - )  1/2 = +0.44 V b)2(MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O)  1/2 = +1.51 V 16H + (aq) + 2MnO 4 - (aq) + 5Fe o (s) 2Mn 2+ (aq) + 5Fe 2+ (aq) + 8H 2 O(l)  cell = 1.95V 4)Line Notation: Fe(s) | Fe 2+ (aq) || MnO 4 - (aq), Mn 2+ (aq), H + (aq) | Pt(s)

12. E.The complete description of a Galvanic Cell 1)Items to include in the description a)Cell potential (always +) and the balanced overall reaction b)Direction of electron flow c)Designate the Anode and the Cathode d)Identity of the electrode materials and the ions present with concentration 2)Example: Completely describe the Galvanic Cell based on these reactions Ag + + e- Ag  o = +0.80 V Fe 3+ + e- Fe 2+  o = +0.77 V Ag + (aq) + Fe 2+ (aq) Ag o (s) + Fe 3+ (aq)  o cell = +0.03 V Pt(s) | Fe 2+ (aq), Fe 3+ (aq) || Ag + (aq) | Ag(s)