ELECTROCHEMISTRY Chapter 19, 4.9

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parameter, q extensive) = nfecell (theory) or rearranging to the Nernst equation André-Marie Ampère Michael Faraday Ecell = E o cell (RT/nF) ln Q where E o cell = ΔG o / nf = E o cell (0.

therefore only Q (reaction quotient) can change the voltage Oxidation - Reduction Reactions (4.9) oxidation reduction Balancing Redox Equations (half-reaction method,19.

1 ELECTROCHEMISTRY Chapter 19, 4.9 Overview of an Electrochemical Process at Constant T and P ΔG = ΔG o + RT ln Q = welec (maximum) Note: I below stands for current measured in amperes = qecell = ItEcell (exp lab) (E intensive parameter, q extensive) = nfecell (theory) or rearranging to the Nernst equation André-Marie Ampère Michael Faraday Ecell = E o cell (RT/nF) ln Q where E o cell = ΔG o / nf = E o cell ( /n) log Q (at 25 o C) 1. n determined by balancing redox equation 2. E o cell determined by E o of species being oxidized and E o of species being reduced (standard reduction potentials) 3. therefore only Q (reaction quotient) can change the voltage Oxidation - Reduction Reactions (4.9) oxidation reduction Balancing Redox Equations (half-reaction method,19.2) oxidation numbers, ON (in order of priority, 4.9) 1. Sum of the oxidation numbers of atoms in a neutral species is zero; in an ion the sum equals the ion charge. 2. Alkali metals (Group I) ON = 1, alkaline earth metals (Group II) ON = 2, Group III usually have ON = Fluorine ON = -1 always. Other halogens usually have ON = -1 except in compounds with oxygen or other halogens when the oxidation number can be positive (follows the trend in electronegativity). 4. Hydrogen has ON = 1 except in metal hydrides when the oxidation number is negative. 5. Oxygen usually has ON = -2 except in compounds with fluorine when the oxidation number can be positive and in compounds containing the O-O bond. For peroxides ON = -1.

neutralize H + with OH - *for basic reactions neutralize H + with OH - Basically two ways to balance: 1.

2 - 2 - Steps in Balancing Oxidation-Reduction Equations by Half-Reaction Method (19.2) (use oxidation numbers and include electrons in half-reactions) (omit step d if oxidation numbers used with electrons) *for basic reactions neutralize H + with OH - *for basic reactions neutralize H + with OH - Basically two ways to balance: 1. start with electrons in half-reactions determined from oxidation numbers OR 2. use electrons to balance charge after all elements are balanced *include these steps for a basic reaction EX 1. Balance in acidic solution MnO 4- (aq) + Fe(s) Mn 2+ (aq) + Fe 2+ (aq)

Balance HNO 2 NO 3 - + NO Electrochemical Cells

3 - 3 - EX 2. Balance in basic solution Ag(s) + CN - (aq) + O 2(g) Ag(CN) 2(aq) disproportionation (p. 1081) EX 3. Balance HNO 2 NO NO Electrochemical Cells (19.3, Figure 19.2) half-cells electrodes anode cathode electrolyte salt bridge FIG I - Terms Used in a Galvanic (Voltaic) Cell

- 4 - types of electrochemical cells ΔG = welec(max) = nfecell 1. spontaneous: ΔG < 0, Ecell> 0 2.

22) Luigi Galvani Alessandro Volta a) spontaneous: galvanic (voltaic) b) nonspontaneous: electrolytic

2+ (aq) + Sn(s) Cu(s) + Sn 2+ (aq) OXIDATION half-reaction: Cu(s) Cu 2+ (aq) + 2e - REDUCTION half-reaction:

vertical lines species in same phase separated by commas types of electrodes active electrode involved in

(oxidation) notation: Zn Zn 2+ inert electrode not involved in electrode half-reaction (inert solid

solution Fe 3+ (aq) + e - Fe 2+ (aq) (reduction) notation: Fe 3+, Fe 2+ Pt electrodes involving metals and

AgCl Ag electrodes involving gases a gas is bubbled over an inert electrode ex: H 2 gas over Pt electrode H

cathode half-cell (ordered way spontaneous reaction goes) electrodes appear on the far left (anode) and far

4 - 4 - types of electrochemical cells ΔG = welec(max) = nfecell 1. spontaneous: ΔG < 0, Ecell> 0 2. nonspontaneous: ΔG > 0, Ecell < 0 FIG II- Electrochemical Cells (Figure 19.22) Luigi Galvani Alessandro Volta a) spontaneous: galvanic (voltaic) b) nonspontaneous: electrolytic OXIDATION half-reaction: Sn(s) Sn 2+ (aq) + 2e - REDUCTION half-reaction: Cu 2+ (aq) + 2e - Cu(s) OVERALL: Cu 2+ (aq) + Sn(s) Cu(s) + Sn 2+ (aq) OXIDATION half-reaction: Cu(s) Cu 2+ (aq) + 2e - REDUCTION half-reaction: Sn 2+ (aq) + 2e - Sn(s) OVERALL: Sn 2+ (aq) + Cu(s) Sn(s) + Cu 2+ (aq) electrochemical cell notation (cell diagram or line notation, p. 897) Cell Diagram or Line Notation for Galvanic Cells half-cell notation different phases separated by vertical lines species in same phase separated by commas types of electrodes active electrode involved in electrode half-reaction (most metal electrodes) ex: Zn 2+ /Zn metal electrode Zn(s) Zn 2+ (aq) + 2e - (oxidation) notation: Zn Zn 2+ inert electrode not involved in electrode half-reaction (inert solid conductors which serve to contact solution with external electrical circuit) ex: Pt electrode in Fe 3+ /Fe 2+ solution Fe 3+ (aq) + e - Fe 2+ (aq) (reduction) notation: Fe 3+, Fe 2+ Pt electrodes involving metals and their slightly soluble salts ex: Ag/AgCl electrode AgCl(s) + e - Ag(s) + Cl - (aq) (reduction) notation: Cl - AgCl Ag electrodes involving gases a gas is bubbled over an inert electrode ex: H 2 gas over Pt electrode H 2(g) 2H + (aq) + 2e - (oxidation) notation: Pt H 2 H + cell notation anode half-cell written to left of cathode half-cell (ordered way spontaneous reaction goes) electrodes appear on the far left (anode) and far right (cathode) in notation salt bridges represented by double vertical lines ex: Zn Zn 2+ and Fe 3+, Fe 2+ Pt half-cells Zn(s) Zn 2+ (aq) + 2e - (anode, oxidation) Fe 3+ (aq) + e - Fe 2+ (aq) [ 2] (cathode, reduction) Zn + 2Fe 3+ Zn Fe 2+ notation: Zn Zn 2+ Fe 3+, Fe 2+ Pt

5 - 5 - Cell Potential, Electrical Work, and Gibbs Free Energy (19.3, 19.5) electrical work electric current EX 4. A 6 volt battery delivers 1.25 amperes for 1.5 hours. How much work does the battery do? Electrolysis (19.8, stoichiometry of spontaneous/nonspontaneous reactions) EX 5. A galvanic (voltaic) electrochemical cell whose overall reaction is Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) oxidizes metallic copper at the anode dissolving the metal Cu(s) Cu 2+ (aq) + 2 e - and reduces silver ions in solution by plating them out as metallic silver at the cathode. Ag + (aq) + e - Ag(s) a) The cell delivers 0.1 A. How long does it take to dissolve 5.00 g of copper at the anode? (M Cu = ) b) An external power source is connected to the cell and the cell is run electrolytically. If the cell produces A for 101 minutes, how much copper is deposited and how much silver dissolves? (M Ag= g/mol)

- 6 - EX 6. An aqueous solution of a palladium salt is electrolyzed by 1.50 A for one hour which produces 2.9779 g of Pd at the cathode.

4) standard states stable form (allotrope) at P = 1 atm and specified T (usually 25 C) pure solid pure liquid gas ideal gas behavior 1 molar (1 M)

halfreaction so conveniently define a zero for the scale standard reduction potential, E o by international convention, E o for the reduction of H + (aq)

6 - 6 - EX 6. An aqueous solution of a palladium salt is electrolyzed by 1.50 A for one hour which produces g of Pd at the cathode. What is the oxidation state of Pd? (M Pd = g/mol) Standard Electrode Potentials (19.4) standard states stable form (allotrope) at P = 1 atm and specified T (usually 25 C) pure solid pure liquid gas ideal gas behavior 1 molar (1 M) solution ideal solution behavior standard cell potential, E o there is no way to separately measure the potential of an oxidation or reduction halfreaction so conveniently define a zero for the scale standard reduction potential, E o by international convention, E o for the reduction of H + (aq) to give H2(g) is taken to be zero at all temperatures: 2 H + (aq, 1 M) + 2 e - H2(g, P = 1 atm) RED: 2H + (aq) + 2e - H 2(g) => H + (aq, 1 M) H 2(g, 1 atm) Pt OX: H 2(g) 2H + (aq) + 2e - => Pt H 2(g, 1 atm) H + (aq, 1 M) FIG IV SHE/Zinc FIG V SHE/Silver

Find E o (Ag 2CrO 4 Ag) for the electrode and determine E o for the following reaction if E o (Fe 3+ Fe 2+

7 - 7 - EX 7. Will Ag + oxidize Zn metal or will Zn 2+ oxidize metallic silver? strength of oxidizing agent strength of reducing agent EX 8. Find E o (Ag 2CrO 4 Ag) for the electrode and determine E o for the following reaction if E o (Fe 3+ Fe 2+ ) = V and ΔG o = kj for the reaction: K 2CrO 4(aq) + 2 Ag(s) + 2 FeC1 3(aq) Ag 2CrO 4(s) + 2 FeC1 2(aq) + 2 KCl(aq) Disproportionation and Oxidizing/Reducing Agents E o E o

- 8 - Cell Potential and Concentration (19.

05916/n) log Q (25 o C); E o cell = E o cathode E o anode Half-Cells EX 9.

500 atm b) H + + e - 1 / 2 H 2; [H + ] = 0.00100 M, P H2 = 0.500 atm c) [H + ] = 1.00 10-14 M, P H2 = 1.

8 - 8 - Cell Potential and Concentration (19.6) Walther Nernst The Nernst Equation Ecell = E o cell (RT/nF) ln Q = E o cell ( /n) log Q (25 o C); E o cell = E o cathode E o anode Half-Cells EX 9. Determine E for the hydrogen electrode: 2 H e - H 2 a) [H + ] = M, P H2 = atm b) H + + e - 1 / 2 H 2; [H + ] = M, P H2 = atm c) [H + ] = M, P H2 = 1.00 atm d) rewrite hydrogen electrode equation in base, [OH - ] = 1.00 M, P H2 = 1.00 atm EX 10. For the half reaction, Cu e - Cu, where, E o = V what is the reduction potential of the half-cell when [Cu 2+ ] = 0.10 M?

- 9 - Cells: EX 11. If [Cu 2+ ] = 0.10 M and [Ag + ] = 0.20 M, what is the voltage of an electrochemical cell based upon the couple: E o (Cu 2+ Cu) = 0.340 V and E o (Ag + Ag) = 0.800 V? EX 12.

9 - 9 - Cells: EX 11. If [Cu 2+ ] = 0.10 M and [Ag + ] = 0.20 M, what is the voltage of an electrochemical cell based upon the couple: E o (Cu 2+ Cu) = V and E o (Ag + Ag) = V? EX 12. An electrochemical cell is based upon the couple: E o (Cu 2+ Cu) = V and E o (Ag + Ag) = V. If the voltage of the cell were V when [Cu 2+ ] = 0.10 M, what is the silver ion concentration? EX 13. If the cell voltage is V what is the ph of the cathode compartment for the following cell where E o (Zn 2+ Zn) = V? Zn(s) Zn 2+ (aq, 1.00 M) H + (aq,? M) H 2(g, 1 atm) Pt(s)

10 Cell Potential, Free Energy, and the Equilibrium Constant (19.5) G Reaction Parameters at Standard State K E cell Reaction at standardstate conditions < 0 = 0 > 0 > 1 = 1 < 1 > 0 = 0 < 0 spontaneous at equilibrium nonspontaneous E cell K RT E cell = ln K nf EX 14. What is the equilibrium constant for the cell of example 11? EX 15. If E o (Cr 3+ Cr) = V and E o (Zn 2+ Zn) = V what is the equilibrium constant for 2 Cr 3+ (aq) + 3 Zn(s) 2 Cr(s) + 3 Zn 2+ (aq) solubility product, Ksp

- 11 - EX 16. From the following reaction cell where Ecell = 0.397 V Pt H 2(1 atm) H + (1.00 M) Cl - (l.00 10-3 M), Ag + (? M) AgCl(s) Ag(s) determine the value of K sp(agcl).

11 EX 16. From the following reaction cell where Ecell = V Pt H 2(1 atm) H + (1.00 M) Cl - (l M), Ag + (? M) AgCl(s) Ag(s) determine the value of K sp(agcl). Relevant standard reduction potentials are 2 H + (aq) + 2 e - H 2(g) E o = V Ag + (aq) + e - Ag(s) E o = V EX 17. Determine the value of K sp(agscn) given that AgSCN(s) + e - Ag(s) + SCN - (aq) Ag + (aq) + e - Ag(s) E o = V E o = V

- 12 - Concentration Cells - Q wins! (p.

The cell is no longer able to do work once the concentrations are equal. Corrosion (19.

Fe(s) Fe 2+ (aq) + 2e - -E o = 0.447 V Cu(s) Cu 2+ (aq) + 2e - -E o = 0.

12 Concentration Cells - Q wins! (p. 912) Ecell > 0 as long as the half-cell concentrations are different. The cell is no longer able to do work once the concentrations are equal. Corrosion (19.9) 4 Fe 2+ (aq) + O 2(g) + (4 + 2x) H 2O(l) 2 Fe 2O 3 xh 2O(s) + 8 H + (aq) rust OXIDATION (at anode) Fe(s) Fe 2+ (aq) + 2e - -E o = V Cu(s) Cu 2+ (aq) + 2e - -E o = V Zn(s) Zn 2+ (aq) + 2e - -E o = V cathodic protection: use a sacrificial anode REDUCTION (at cathode) O 2(g) + 4 H + (aq) + 4e - 2H 2O(l) E o = V acidic conditions O 2(g) + 2 H 2O (l) + 4e - 4 OH - (aq) E o = V basic conditions

The signal is due to opening and closing of sodium channels in the axon s membrane, allowing Na + to enter or not enter the cell. Entering Na + causes an increase in charge inside the cell.

13 Nerve Transmission (p. 914) A nerve cell (neuron) can be likened to a receiver, transmitter, and transmission line between the two. Dendrites receive im-pulse signals from a neighboring neuron. Transmission of electrical signals within the receiving neuron proceeds from dendrites to synapse along the length of the axon. The signal is due to opening and closing of sodium channels in the axon s membrane, allowing Na + to enter or not enter the cell. Entering Na + causes an increase in charge inside the cell. At rest, the inside of a neuron is slightly negative (~ -70 mv) due to a higher concentration of K + ions inside (concentration gradient balanced by a potential difference across the membrane, Einside - Eoutside which can be calculated by modifying the Nernst equation) E = E o + ( /n) log Q K + Na + inside 20 x y outside x 10 y where n is the charge of the ion. An impulse opens Na + channels. When stimulated past a threshold Na + rushes into the axon, causing a region of positive charge (depolarization) within the axon and a potential of ~30 mv across the membrane. The channels open in one direction along the axon as an opened channel needs a rest period before it can reopen. The region of positive charge causes nearby Na + channels to close. Just after the Na + channels close, K + channels open and K + exits the axon (repolarization). The membrane is brought back to its negative resting potential by Na + /K + pumps in the cell membrane. This process continues as a chain-reaction along the axon. The influx of sodium ions depolarises the axon, and the outflow of potassium ions repolarises it.

ELECTROCHEMISTRY Chapter 14

ELECTROCHEMISTRY Chapter 14 Basic Concepts: Overview of Electrochemical Process at Constant T, P (14-1) ΔG = ΔG o + RT ln Q = w elec (maximum) = qe = ItE (exp) (E intensive parameter, q extensive) = nfe