Electrochemistry Brown, LeMay Ch 20 AP Chemistry

2 20.1: Oxidation-reduction reactions “Redox”: rxns with a change in oxidation number At least one element’s oxidation number will increase and one will decrease Consists of two half-reactions: oxidation: loss of e-; oxidation number increases reduction: gain of e-; oxidation number decreases (“reduces”) Oxidizing agent or oxidant: causes another substance to be oxidized; gains e- Reducing agent or reductant: causes another substance to be reduced; loses e-

3 Ex: Determine the half-reactions. Cr 2 O 7 2- (aq) + Cl 1- (aq) → Cr 3+ (aq) + Cl 2 (g) Reduction half-reaction: Cr 2 O 7 2- (aq) → Cr 3+ (aq) Oxidizing agent Oxidation half-reaction: Cl 1- (aq) → Cl 2 (g) Reducing agent

4 20.2: Balancing Redox Reactions “Method of Half-Reactions” steps: 1.Assign oxidation numbers to all species; based on this, split the half-reactions of reduction and oxidation. 2.For each, balance all elements except H and O. 3.Balance oxygen by adding H 2 O to the opposite side. 4.Balance hydrogen by adding H + to the opposite side. 5.Balance charges by adding e- to side with overall positive charge. 6.Multiply each half-reaction by an integer so there are an equal number of e- in each. 7.Add the half reactions; cancel any species; check final balance.

5 Ex: Balance the following reaction, which takes place in acidic solution : Cr 2 O 7 2- (aq) + Cl 1- (aq) → Cr 3+ (aq) + Cl 2 (g) Reduction: Cr 2 O 7 2- (aq) → Cr 3+ (aq) H 2 O (l) 14 H + (aq) + 6 e- + Oxidation: Cl - (aq) → Cl 2 (g) 6 Cl - (aq) → 3 Cl 2 (g) + 6 e e- 3 [ ]

6 6 e H + (aq) + Cr 2 O 7 2- (aq) → 2 Cr 3+ (aq) + 7 H 2 O (l) 6 Cl - (aq) → 3 Cl 2 (g) + 6 e- 14 H + (aq) + Cr 2 O 7 2- (aq) + 6 Cl - (aq) → 2 Cr 3+ (aq) + 7 H 2 O (l) + 3 Cl 2 (g) Since occurs in acidic solution, H + appears as a reactant If reaction occurs in basic solution, balance using same method, but neutralize H + with OH - as a last step.

7 Ex: In basic solution, MnO 4 - (aq) + Br - (aq) → MnO 2 (s) + BrO 3 - (aq) Reduction:MnO 4 - (aq) → MnO 2 (s) MnO 4 - (aq) → MnO 2 (s) + 2 H 2 O (l) 4 H + (aq) + MnO 4 - (aq) → MnO 2 (s) + 2 H 2 O (l) 3 e- + 4 H + (aq) + MnO 4 - (aq) → MnO 2 (s) + 2 H 2 O (l)

8 Oxidation:Br - (aq) → BrO 3 - (aq) 3 H 2 O (l) + Br - (aq) → BrO 3 - (aq) 3 H 2 O (l) + Br - (aq) → BrO 3 - (aq) + 6 H + (aq) 3 H 2 O (l) + Br - (aq) → BrO 3 - (aq) + 6 H + (aq) + 6 e- 2[3 e- + 4 H + (aq) + MnO 4 - (aq) → MnO 2 (s) + 2 H 2 O (l)] 6 e- + 8 H + (aq) + 2 MnO 4 - (aq) → 2 MnO 2 (s) + 4 H 2 O(l) 3 H 2 O (l) + Br - (aq) → BrO 3 - (aq) + 6 H + (aq) + 6 e- 2 H + (aq) + 2 MnO 4 - (aq) + Br - (aq) → 2 MnO 2 (s) + BrO 3 - (aq) + H 2 O (l)

9 Neutralize with OH - : 2 H + (aq) + 2 MnO 4 - (aq) + Br - (aq) + 2 OH - (aq) → 2 MnO 2 (s) + BrO 3 - (aq) + H 2 O (l) + 2 OH - (aq) 2 H 2 O (l) + 2 MnO 4 - (aq) + Br - (aq) → 2 MnO 2 (s) + BrO 3 - (aq) + H 2 O (l) + 2 OH - (aq) Simplify: H 2 O (l) + 2 MnO 4 - (aq) + Br - (aq) → 2 MnO 2 (s) + BrO 3 - (aq) + 2 OH - (aq)

: Voltaic (or Galvanic) Cells Device that utilizes energy released in a spontaneous electrochemical reaction by directing e- transfer along an external pathway, rather than directly between reactants Electrodes: metals (or graphite) connected by an external circuit Anode: half-cell where oxidation occurs Cathode: half-cell where reduction occurs Luigi Galvani (1737–1798) Alessandro Volta (1745–1827)

11 Ex: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Oxidation half-cell: Zn (s) → Zn 2+ (aq) + 2 e- Reduction half-cell: Cu 2+ (aq) + 2 e- → Cu (s) anode Zn Cu (-) (+) salt bridge, saturated with NaNO 3 (aq) cathode e- flow Na 1+ NO 3 1- Zn 2+ (aq) Cu 2+ (aq) * Notation: Anode| Anode product|Salt |Cathode|Cathode product Zn(s)|Zn 2+ (aq, 1 M)|NaNO 3 (saturated)|Cu 2+ (aq, 1 M)|Cu(s)

Ex: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

: Electromotive force (emf or E ) Potential difference between cathode and anode that causes a flow of e-. –The result of the difference in electric fields produced at each electrode Also called cell potential or cell voltage, E cell Units = Volt(1 V = 1 J / 1 C) Standard conditions (25°C, 1 atm, 1 M): E º cell E º cell is unique to each set of half-cells involved Charles-Augustin de Coulomb (1736–1806)

14 Standard Reduction Potentials Assigned to each reduction reaction E º cell = E º reduction (cathode) - E º reduction (anode) All E º red are referenced to standard hydrogen electrode (S.H.E.) : 2 H+ (aq, 1 M) + 2 e- → H 2 (g, 1 atm) E º reduction = 0 V

15 Ex: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Anode: Zn (s) → Zn 2+ (aq) + 2 e- –This is oxidation, but E º reduction is recorded as the reduction reaction: Zn 2+ (aq) + 2 e- → Zn (s), where E º reduction = V Cathode: Cu 2+ (aq) + 2 e- → Cu (s) E º reduction = V E º cell = E º reduction (cathode) - E º reduction (anode) = 0.34 V – ( V) E º cell = V

16 Positive voltages = reaction is spontaneous (cell will produce voltage,  G is -) Changing the stoichiometric coefficients of a half-cell does not affect the value of E º, since potential is a measure of the energy per electrical charge. However, having more moles means the reaction will continue for a greater time.

17 Using the Standard Reduction Potential table When E º red is very negative, reduction is difficult (but oxidation is easy): Li 1+ (aq) + e- → Li (s) E º red = V –Li (s) oxidizes easily, so it is the strongest reducing agent ; lower on the Standard Reduction Potential chart signifies the anode. When E º red is very positive, reduction is easy: F 2 (g) + 2 e- → 2 F 1- E º red = V –F 2 reduces easily, so it is the strongest oxidizing agent ; higher on the chart signifies the cathode. Predicting metal “activity series”: most reactive metals are near bottom

: Spontaneity of Redox Reactions Gibbs free energy and emf:  G o = - n F E o n = # of moles of e- transferred Faraday’s constant, F = 96,500 C/mol of e- 1 F = 1 mole of e-, which has a charge of 96,500 C Units = (mol e-)(C/mol e-)(V) = (mol e-)(C/mol e-)(J/C) = J As E º cell increases,  Gº decreases (becomes more likely to be spontaneous) Michael Faraday ( )

: Effect of Concentration on E cell The Nernst Equation:  G =  G o + RT ln Q and  G o = - n F E o (R = J/(molK)) - n F E = - n F E º + RT ln Q Walther Nernst ( )

20 At equilibrium,  G = 0 GoGo K E o cell Reaction Negative> 1PositiveSpontaneous 010Equilibrium Positive< 1NegativeNon- spontaneous since T º = 298 K

Equilibrium composition measurements Calorimetry and entropy: S 0 and  S 0 Electrical measurements: E 0 and E 0 cell 3 Thermodynamic Quantities:

22 * E 0 cell of cells Model of a portion of a cell membrane shows a K + channel (blue) and a Na + channel (red). The area outside the cell (top) is rich in Na + and low in K +. Inside the cell, the fluids are relatively rich in K + and low in Na +.

: Electrolytic cells Electrical energy is used to cause a non- spontaneous reaction to occur. Current (I): a measure of the charge (Q) per unit time (t) Units = amps or amperes (A) 1 A = 1 C / 1 s André-Marie Ampère (1775 – 1836)

24 Ex: Na + (l) + Cl - (l) → Na (l) + Cl 2 (g) Oxidation half-cell: 2 Cl - (l) → Cl 2 (g) + 2 e- Reduction half-cell: Na + (l) + e- → Na (l) anode Pt (-)(+) cathode e- flow NaCl (l) Cl 2 (g)Na (l) In a voltaic cell : anode is (-), cathode is (+). In an electrolytic cell : anode is (+), cathode is (-).

: Quantitative Electrolysis Ex: A current of A is passed through an electrolytic cell containing molten NaCl for 1.5 hours. Write the electrode reactions. Calculate the mass of products formed at each electrodes Oxidation: 2 Cl - (l) → Cl 2 (g) + 2 e - Reduction: Na + (l) + e - → Na (l) Strategy: calculate the total charge of e- transferred, and use F to convert to moles.

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27 Electrolysis of Aqueous Mixtures If the voltage on an aqueous mixture of CuCl 2 and ZnCl 2 in an electrolytic cell is slowly increased, what products will form at each electrode? oWhat exists in solution that could be reduced? Cu 2+, Zn 2+, and H 2 O (not Cl - ) Cu e - → CuEº red = 0.34 V Zn e - → ZnEº red = V 2 H 2 O + 2 e - → H OH - Eº red = V oSince Eº red (Cu) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the cathode): Cu (s) > Zn (s) > H 2 (g)

28 If the voltage on an aqueous mixture of CuCl 2 and ZnCl 2 in an electrolytic cell is slowly increased, what products will form at each electrode? oWhat exists in solution that could be oxidized? Cl - and H 2 O (not Cu 2+ and Zn 2+ ) 2 H 2 O → O H e - Eº ox = V 2 Cl - → Cl e - Eº ox = V oSince Eº ox (H 2 O) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the anode): O 2 (g) > Cl 2 (g)

29 Electroplating The anode is a silver bar, and the cathode is an iron spoon. Anode: Ag (s) → Ag + (aq) Cathode: Ag + (aq) → Ag (s)