# Chpt 20: Electrochemistry

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2 Cell Potential and Free Energy When both reactants and products are in their standard states, and under constant pressure and temperature conditions where DG o = nfe o DG o is the standard free energy change of the reaction n is the number of mole of electrons transferred in the reaction F is Faraday s constant E o is the standard cell potential Faraday s constant is the quantity of electrical charge on 1 mol of electrons 1 F = 96,500 C/mol e - 1 F = 96,500 J/Vmol e -

3 DG o from Cell emf I Example: For the reaction I 2 (s) + 5Cu 2+ (aq) 2IO 3- (aq) + 5Cu(s) + 12H + (aq) calculate the standard free energy change for the reaction. I 2 (s) 2IO 3- (aq) + 10e - The oxidation reaction is E o red = V The reduction reaction is 5Cu 2+ (aq) +10e - 5Cu(s) E o red = V E o = E o red(reduction) - E o red(oxidation) E o = V - ( V) E o = V from the half-reactions we see that 10 moles of electrons are transferred in the reaction n = 10

4 DG o from Cell emf II Example: For the reaction I 2 (s) + 5Cu 2+ (aq) 2IO 3- (aq) + 5Cu(s) + 12H + (aq) calculate the standard free energy change for the reaction. DG o = nfe o DG o =-(10 mole - )(96,500 J/Vmole - )( V) DG o = +828 kj Since DG o > 0, reaction is non-spontaneous (we could also note this since E o < 0) If reactants and products are not in their standard states, the more general relation can be applied DG = nfe

5 Example A 4 B 1, 2 C 2 D 2, 3 E 1, 4 Over time, you notice that the tin electrode seems to be disappearing while there are deposits forming on the silver electrode. Which of the following is a correct statement? 1. The silver electrode is the cathode and the tin electrode is the anode. 2. Electrons are flowing from the tin electrode to the silver electrode. 3. Nitrate ions are flowing through the salt bridge to the silver solution. 4. The half-reaction occurring at the tin electrode is: Sn e - Sn 2+.

6 Example Chlorine dioxide is used to treat municipal water supplies. 2 NaClO 2 (aq) + Cl 2 (g) 2 ClO 2 (g) + 2 NaCl (aq) Cl 2 + 2e - 2Cl - ClO 2 + e - ClO 2 - Calculate DG o for the reaction. A 78 kj B -39 kj C -78 kj D -156 kj E 39 kj E o red = 1.36 V E o red = V

7 The Nernst Equation I The cell emf will depend reactant/product concentration The dependence of cell emf on concentration can be derived from the dependence of the free energy change on concentration DG = DG o + RTlnQ R is the gas constant (8.314 J/Kmol) T is absolute temperature Q is the reaction quotient, which depends on reactant and product concentrations We can derive an expression for E by replacing DG o with nfe o and DG with nfe

8 The Nernst Equation II Following the substitution -nfe = -nfe o + RTlnQ E = E o - (RT/nF) lnq This relation is the Nernst Equation. The equation is typically expressed in base 10 logarithms log x = ln x E = E o - (2.303RT/nF) logq At standard temperature (T = 298 K) the quantity in parenthesis takes the form E = E o - (0.0592/n) 298 K The Nernst equation can be used in two ways to determine the cell emf under non-standard conditions to find [product] or [reactant] from E

9 Cell emf Under Non-Standard Conditions Example: Calculate the standard emf for a cell that employs the following reaction 2Al(s) + 3I 2 (s) 2Al 3+ (aq) + 6I - (aq) when [Al 3+ ] = 4.0x10-3 M and [I - ] = M from the the standard reduction potentials the cell potential is E o cell = V and n = 6 E = E o - (0.0592/n) logq Q = [Al 3+ ] 2 [I - ] 6 Q = (4.0x10-3 ) 2 (0.010) 6 Q = 1.6 x E = (+2.20V) - (0.0592/6) log (1.6x10-17 ) E = 2.36 V

10 Product Concentration from Cell emf I Example: A voltaic cell is constructed that uses the following reaction and operates at 298 K 2Al(s) + 3Mn 2+ (aq) 2Al 3+ (aq) + 3Mn(s) What is [Al 3+ ] to produce a cell emf of V? Al(s) Al 3+ (aq) + 3e - E o red = V Mn 2+ (aq) + 2e - Mn(s) E o red = V E o = E o red(reduction) - E o red(oxidation) E o = -1.18V - (-1.66V) E o = V

11 Example A galvanic cell is constructed from the following two half reactions Ag + (aq) + e - Ag (s) E o red = 0.80 V H 2 O 2 (aq) + 2H + (aq) + 2e - 2H 2 O(l) E o red = 1.78 V In which scenario is E cell greater than E o cell? 1. [Ag + ] = 1.0 M, [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M 2. [Ag + ] = 2.0 M, [H 2 O 2 ] = 1.0 M, [H + ] = 1.0x10-7 M 3. [Ag + ]= 2.0 M, [H 2 O 2 ] = 0.25 M, [H + ] = 4.0 M A 1 B 2 C 3 D 2, 3 E 1, 3

12 Example A galvanic cell is constructed from the following two half reactions Cu 2+ (aq) + 2e - Cu (s) E o red = 0.34 Ag + + e - Ag (s) E o red = 0.80 The electrodes in this cell are Ag(s) and Cu(s). For which of the following scenarios does the cell potential decrease? 1. CuSO 4 (s) is added to the copper half-cell compartment (assume no volume change) 2. NH 3 (aq) is added to the copper half-cell compartment [Cu 2+ reacts with NH 3 to form Cu(NH 3 ) 4 2+ (aq)] 3. NaCl (s) is added to the silver half-cell compartment [Ag + reacts with Cl - to form AgCl(s)] 4. Water is added to both half-cells until the volumes are doubled A 1, 3 B 2, 3, 4 C 1,4 D 2, 4 E 1, 3, 4

13 Equilibrium Constants for Redox Reactions At equilibrium, DG = 0, Q = K and 0 = E o - (0.0592/n) logk logk = ne o / The above expression allows us to calculate the equilibrium constant for a redox reaction from the standard emf for the reaction Example: What is the equilibrium constant for reaction 2Al(s) + 3Mn 2+ (aq) 2Al 3+ (aq) + 3Mn(s) We have already that determined E o = 0.48V and n = 6 logk = (6)(0.48V)/ logk = K = 4.45 x

14 Corrosion Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound Corrosion of iron (two-step process) I. 2Fe(s) + O 2 (g) + 4H + 2Fe 2+ (aq) + 2H 2 O( ) II. 4Fe 2+ (aq) + O 2 (g)+ 4H 2 O( ) 2Fe 2 O 3 (s) + 8H + (aq) The half reactions for the first step O 2 (g) + 4H + + 4e - 2H 2 O( ) E o red = 1.23 V Fe(s) Fe 2+ (aq) + 2e - E o red = V Salts enhance the corrosion as the ions serve as electrolytes to complete the electrical circuit

15 Corrosion Multiple ways to prevent corrosion. Paint or metal coating on metal alloying Cathodic protection Some metals form a protective oxide layer over its surface and prevent further oxidation Al Zn (coating a metal in zinc is referred to as galvanizing) Stainless steel, a steel alloy with Cr, also forms a protective oxide layer over its surface.

16 Cathodic Protection To prevent corrosion, cathodic protection is sometimes employed. Protect a metal by ensuring it will act as the cathode in an electrochemical cell. A sacrificial anode, typically Zn, is attached to the cathode that is being protected. Compare the following half reactions Fe 2+ (aq) + 2e - Fe(s) E o red = V Zn 2+ (aq) + 2e - Zn(s) E o red = V Mg 2+ (aq) + 2e - Mg(s) E o red = V

17 Electrolysis Electrical energy can be used to induce non-spontaneous redox reactions to occur The electrolysis of water 2H 2 O( ) O 2 (g) + 2H 2 (g) Ox: 2H 2 O( ) O 2 (g) + 4H + + 4e - Red: 4H 2 O( ) + 4e - 2H 2 (g) + 4OH - The cell emf for this reaction is E o = E o red(reduction) - E o red(oxidation) E o = (+1.23 V) E o = 2.06 V Since E o is negative, the reaction in the forward direction is nonspontaneous If work is done on the system (in the form of electrical energy) the reaction can proceed

18 Electrochemical Cell A power supply can provide electrical energy to drive the electrochemical cell Oxidation occurs at the anode The anode is positive since e- are withdrawn from this electrode Reduction occurs at the cathode the cathode is negative since e- are supplied to this electrode

19 Electrolysis and Stoichiometry Example: What volume of H 2 (g) at STP is produced when 5.0 A is passed through a water electrolysis cell for 1 hour? 5 A = 5 C/s # Coulombs = (5 C/s)(1 hr)(3600 s/hr) = 18,000 C mol e - = (18,000 C)/(96,500 C/mol e - ) = mol e - 2H 2 O( ) O 2 (g) + 2H 2 (g) 2 mol H 2 (g) liberated for every 4 mol e - mol H 2 (g) = (0.186mol e - )(2mol H 2 )/(4mol e - ) = mol H 2 (g) V = nrt/p V = (0.093 mol)(0.082 Latm/molK)(298 K)/ (1 atm) = 2.28 L

20 Isolation of Al Aluminum is abundant on Earth (behind O 2 and Si) but until recently was incredibly expensive Aluminum is readily oxidized and is found in oxide compounds Al 3+ (aq) + 3e - Al(s) E o red = V Aluminum cannot be plated out of a solution of aqueous Al 3+ ions since the electrolysis of water would occur first 2 H 2 O + 2e - H 2 + 2OH - E o red = V

21 Isolation of Al Aluminum isolated electrochemically from Al 2 O 3. Cathode: AlF 6 3- (aq) + 3e - Al(s) + 6F - Reduction: 2Al 2 OF F - + C 4AlF CO 2 + 4e - Process consumes a few percent of the power produced in the US annually. CEM 152 SS2011

22 Electrolysis and Electrical Work The change in free energy for a chemical system provides a measure of the maximum useful work extracted from the reaction w max = DG = -nfe We should remember that work done on a system will result in w > 0 since electrolysis involves a non-spontaneous process, E < 0 and hence w > 0 Electrical work is typically expressed as a product of power times time energy (J) = power (J/s) time (s) The unit of electrical power is the watt 1 watt (W) = 1 J/s

23 Electrical Work Example Calculate the number of kwh of electricity required to produce 1 kg of Mg from electrolysis of MgCl 2 if applied emf = 5.0V mol Mg = (1x10 3 g Mg)/(24.3 g/mol) = 41.1 mol mol e - = (41.1 mol Mg)(2 mol e - )/(1 mol Mg) = 82.3 mol e - w = -nfe w = -(82.3 mol e - )(96,500 J/Vmol e - )(-5.0 V) w = 3.97 x 10 7 J 1 kwh = 3.6 x 10 6 J w = (3.97 x 10 7 J)/(3.6 x 10 6 J/kWh) w = 11.0 kwh This is assuming the electrolytic cell is 100 percent efficient

24 Electroplating Electrolytic processes which involve a metal electrode which participates in the cell reaction can be used to deposit the active metal onto another Oxidation at M (active metal) anode M(s) M e - ; E o red < 1.23 V Reduction of M at metallic cathode M e - M(s) ; E o red > The ions from the active metal which go into solution are plated on the cathode electrode (where the reduction occurs) For reference O 2 (g) + 4H + (aq) + 4e - H 2 O (l) E o red = V 2 H 2 O(l) 2e - H 2 (g) + 2OH - E o red = V

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