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Chapter 21 Electrochemistry: Fundamentals

Chapter 21 Electrochemistry: Fundamentals. Key Points About Redox Reactions. Oxidation (electron loss ) always accompanies reduction (electron gain). The oxidizing agent is reduced , and the reducing agent is oxidized.

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Chapter 21 Electrochemistry: Fundamentals

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  1. Chapter 21 Electrochemistry: Fundamentals Key Points About Redox Reactions Oxidation (electron loss) always accompanies reduction (electron gain). The oxidizing agent is reduced, and the reducing agent is oxidized. The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

  2. Zn(s) + 2H+(aq)Zn2+(aq) + H2(g) A summary of redox terminology. OXIDATION Zn loses electrons. One reactant loses electrons. Zn is the reducing agent and becomes oxidized. Reducing agent is oxidized. The oxidation number of Zn increases from 0 to + 2. Oxidation number increases. REDUCTION Other reactant gains electrons. Hydrogen ion gains electrons. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidizing agent is reduced. The oxidation number of H decreases from +1 to 0. Oxidation number decreases.

  3. Electrochemical cell Oxidation half-reaction X X+ + e- Oxidation half-reaction A- A + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ e- B Overall (cell) reaction X + Y+ X+ + Y; DG < 0 Overall (cell) reaction A- + B+ A + B; DG > 0 General characteristics of voltaic and electrolytic cells. VOLTAIC / GALVANIC CELL ELECTROLYTIC CELL System does work on its surroundings Energy is released from spontaneous redox reaction Energy is absorbed to drive a nonspontaneous redox reaction Surroundings(power supply) do work on system(cell)

  4. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction Cu2+(aq) + 2e- Cu(s) Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) A voltaic cell based on the zinc-copper reaction.

  5. Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l) A voltaic cell using inactive electrodes.

  6. Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) inert electrode Notation for a Voltaic Cell components of anode compartment (oxidation half-cell) components of cathode compartment (reduction half-cell) phase of lower oxidation state phase of lower oxidation state phase of higher oxidation state phase of higher oxidation state phase boundary between half-cells Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) , Mn2+(aq) | graphite

  7. PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. e- Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Cr Ag K+ NO3- Reduction half-reaction Ag+(aq) + e- Ag(s) Cr3+ Ag+ Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Diagramming Voltaic Cells PLAN: Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). SOLUTION: Voltmeter salt bridge Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

  8. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction 2H3O+(aq) + 2e- H2(g) + 2H2O(l) Overall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) Determining an unknown E0half-cell with the standard reference (hydrogen) electrode.

  9. PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V anode: Zn(s) Zn2+(aq) + 2e- E = +0.76 E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V Calculating an Unknown E0half-cell from E0cell Calculate E0bromine given E0zinc = -0.76V PLAN: The reaction is spontaneous as written since the E0cell is (+). Zinc is being oxidized and is the anode. Therefore the E0bromine can be found using E0cell = E0cathode - E0anode. SOLUTION: E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76) E0bromine = 1.86 - 0.76 = 1.07 V

  10. F2(g) + 2e- 2F-(aq) Cl2(g) + 2e- 2Cl-(aq) MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) Ag+(aq) + e- Ag(s) Fe3+(g) + e- Fe2+(aq) O2(g) + 2H2O(l) + 4e- 4OH-(aq) strength of reducing agent Cu2+(aq) + 2e- Cu(s) strength of oxidizing agent N2(g) + 5H+(aq) + 4e- N2H5+(aq) Fe2+(aq) + 2e- Fe(s) 2H2O(l) + 2e- H2(g) + 2OH-(aq) Na+(aq) + e- Na(s) Li+(aq) + e- Li(s) Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 2H+(aq) + 2e- H2(g) -0.23 -0.44 -0.83 -2.71 -3.05

  11. Example: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Writing Spontaneous Redox Reactions • By convention, electrode potentials are written as reductions. • When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. • The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell. • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. stronger reducing agent stronger oxidizing agent weaker oxidizing agent weaker reducing agent

  12. PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0cell for each. (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = 0.96V (2) N2(g) + 5H+(aq) + 4e- N2H5+(aq) E0 = -0.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) E0 = 1.23V Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength (b) Rank the relative strengths of the oxidizing and reducing agents: PLAN: Put the equations together in varying combinations so as to produce (+) E0cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0. In ranking the strengths, compare the combinations in terms of E0cell.

  13. E0 = 0.96V Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23V (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (A) 4NO3-(aq) + 3N2H5+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l) (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- Rev (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- E0 = -0.96V E0 = 1.23V X2 (B) 2NO(g) + 3MnO2(s) + 4H+(aq) 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (a) E0cell = 1.19V X4 X3 E0cell = 0.27V X3

  14. Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23V E0 = 1.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l) Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) E0cell = 1.46V X2 (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO3- > N2 reducing agents: N2H5+ > NO (B): oxidizing agents: MnO2 >NO3- reducing agents: NO > Mn2+ (C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+

  15. Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO2 > NO3- > N2 Reducing agents: N2H5+ > NO > Mn2+

  16. Summary • A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge. • The salt bridge provides ions to maintain the charge balance when the cell operates. • Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell. • The output of a cell is called cell potential (Ecell) and is measured in volts. • When all substances are in standard states, the cell potential is the standard cell potential (Eocell). • Ecell equals Ecathode minus Eanode,Ecell = Ecathode - Eanode. • Conventionally, the half cell potential refers to its reduction half-reaction. • Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking the oxidizing agent or reducing agent. • Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones. • Spontaneous reaction is indicated negative ∆G and positive ∆E, ∆G = - nF∆E. • We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.

  17. strength as reducing agents Relative Reactivities (Activities) of Metals Li K Ba Ca Na can displace H from water 1. Metals that can displace H from acid Mg Al Mn Zn Cr Fe Cd 2. Metals that cannot displace H from acid can displace H from steam 3. Metals that can displace H from water Co Ni Sn Pb can displace H from acid 4. Metals that can displace other metals from solution H2 Cu Hg Ag Au cannot displace H from any source

  18. By substituting standard state values into E0cell, we get E0cell = (0.0592V/n) log K (at 298 K) The interrelationship of DG0, E0, and K. Reaction at standard-state conditions DG0 K E0cell DG0 < 0 > 1 > 0 spontaneous 0 1 0 at equilibrium > 0 < 1 < 0 nonspontaneous DG0 = -nFEocell DG0 = -RT lnK E0cell K E0cell = -RT lnK nF

  19. RT Ecell = E0cell - ln Q nF 0.0592 log Q Ecell = E0cell - n The Effect of Concentration on Cell Potential DG = DG0 + RT ln Q -nF Ecell = -nF Ecell + RT ln Q Nernst equation • When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell • When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell • When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell

  20. Calculating K and DG0 from E0cell PROBLEM: Lead can displace silver from solution: Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s) As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DG0 at 298 K for this reaction. Pb2+(aq) + 2e- Pb(s) Ag+(aq) + e- Ag(s) Ag+(aq) + e- Ag(s) E0cell = 0.592V log K n n x E0cell (2)(0.93V) = 0.592V 0.592V PLAN: Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide SOLUTION: E0 = -0.13V Anode E0 = 0.80V Cathode E0cell = E0cathode – E0anode = 0.93V 2X E0cell = - (RT/n F) ln K DG0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V) log K = DG0 = -1.8x102kJ K = 2.6x1031

  21. PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm Calculate Ecell at 298 K. H2 E0 = 0.00V P x [Zn2+] 2H+(aq) + 2e- H2(g) Q = H2 E0 = -0.76V Zn2+(aq) + 2e- Zn(s) [H+]2 E0 = +0.76V Zn(s) Zn2+(aq) + 2e- (0.30)(0.010) Q = 0.0592V Ecell = E0cell - (2.5)2 log Q n Using the Nernst Equation to Calculate Ecell PLAN: Find E0cell and Q in order to use the Nernst equation. SOLUTION: Determining E0cell : Q = 4.8x10-4 Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V

  22. Diagramming Voltaic Cells Ag K+ NO3- Reduction half-reaction Ag+(aq) + e- Ag(s) Ag+ Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Zn bar in a Zn(NO3)2 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Zn electrode is negative relative to the Ag electrode. PROBLEM: Identify the redox reactions Write each half-reaction. Associate the (-)(Zn) pole with the anode (oxidation) and the (+) (Ag) pole with the cathode (reduction). PLAN: SOLUTION: e- Voltmeter Oxidation half-reaction Zn2+(aq) + 2e- Zn(s) salt bridge Zn Zn2+ Anode Cathode Overall (cell) reaction Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s) Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)

  23. -wmax -Ecell = charge charge = n F n = # mols e- F = Faraday constant Free Energy and Electrical Work If there is no current flows, the potential represents the maximum work the cell can do. If there is no current flows, no energy is lost to heat the cell component. DG a -Ecell DG = wmax = charge x (-Ecell) DG = - n F Ecell In the standard state All components are at standard state. DG0 = - n F E0cell F = 96,485 C/mol DG0 = - RT ln K 1V = 1J/C E0cell = - (RT/n F) ln K F = 9.65x104J/V*mol E0cell = - (0.05916/n) log K at RT

  24. RT Ecell = E0cell - ln Q nF 0.0592 log Q Ecell = E0cell - n The Effect of Concentration on Cell Potential Cell operates with all components at standard states. Most cells are starting at Non-standard state. DG = DG0 + RT ln Q -nF Ecell = -nF Ecell + RT ln Q Nernst equation • When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell forward reaction • When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell Equilibrium • When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reversereaction

  25. PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm H2 RT nF P x [Zn2+] 2H+(aq) + 2e- H2(g) Q = H2 Zn2+(aq) + 2e- Zn(s) [H+]2 (0.30)(0.010) Q = 0.0592 Ecell = E0cell - 0.0592 (2.5)2 log Q log Q Ecell = E0cell - n n Sample Problem Using the Nernst Equation to Calculate Ecell 2H+(aq) + Zn (s) H2(g) + Zn2+ (aq) Calculate Ecell at 298 K. PLAN: Find E0cell and Q in order to use the Nernst equation. Ecell = E0cell - ln Q SOLUTION: Determining E0cell : E0 = 0.00V cathode E0 = -0.76V anode Ecell0 = E0c-E0a = 0.00-(-0.76)V = 0.76 V Q = 4.8x10-4 Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V

  26. Summary • A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge. • The salt bridge provides ions to maintain the charge balance when the cell operates. • Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell. • The output of a cell is called cell potential (Ecell) and is measured in volts. • When all substances are in standard states, the cell potential is the standard cell potential (Eocell). • Ecell equals Ecathode minus Eanode,Ecell = Ecathode - Eanode. • Conventionally, the half cell potential refers to its reduction half-reaction. • Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking the oxidizing agent or reducing agent. • Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones. • Spontaneous reaction is indicated negative ∆G and positive ∆E, ∆G = -nF∆E. • We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.

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