Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A summary of redox terminology. Figure 21.1 Key Points About Redox Reactions Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) OXIDATION •Oxidation (electron loss) always accompanies reduction (electron gain). One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. •The oxidizing agent is reduced, and the reducing agent is oxidized. Oxidation number increases. The oxidation number of Zn increases from x to +2. •The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent. Other reactant gains electrons. REDUCTION Hydrogen ion gains electrons. Oxidizing agent is reduced. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidation number decreases. The oxidation number of H decreases from +1 to 0. 21-3 21-4 3 4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Half-Reaction Method for Balancing Redox Reactions Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. 1. Divide the reaction into half-reactions - •Each reaction is balanced for mass (atoms) and charge. Determine the O.N.s for the species undergoing redox. •One or both are multiplied by some integer to make the number of electrons gained and lost equal. +6 -1 Cr2O72-(aq) + I-(aq) •The half-reactions are then recombined to give the balanced redox equation. Cr2O72I Advantages: •The separation of half-reactions reflects actual physical separations in electrochemical cells. •The half-reactions are easier to balance especially if they involve acid or base. •It is usually not necessary to assign oxidation numbers to those species not undergoing change. - Cr3+ I2 +3 0 Cr3+(aq) + I2(aq) Cr is going from +6 to +3 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction 14H+(aq) + Cr2O72net: +12 6e- + 14H+(aq) + Cr2O72- 21-5 5 Cr3+(aq) + I2(aq) 2 Cr3+ net: +6 2 Cr3+ + 7H2O(l) Add 6e - to left. + 7H2O(l) 21-6 6 1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Balancing Redox Reactions in Acidic Solution 6e- + 14H+(aq) + Cr2O722 I continued Balancing Redox Reactions in Basic Solution + 7H2O(l) 2 Cr3+ - Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. + 2e I2 - 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) Cr(+6) is the oxidizing agent and I(-1) is the reducing agent. + 14OH (aq) 3. Multiply each half-reaction by an integer, if necessary 2 I- I2 + 2e- 2 Cr3+ 2- 7H2O + Cr2O72- + 6 I- + 7H2O(l) 3 I2 + 6e- 14H (aq) + Cr2O7 (aq) + 6 I (aq) + - 2Cr3+ + 3I2 + 7H2O + 14OH- Reconcile the number of water molecules. 4. Add the half-reactions together 6 I- + 14OH-(aq) 14H2O + Cr2O72- + 6 I- X3 6e- + 14H+ + Cr2O72- 2Cr3+(aq) + 3I2(s) + 7H2O(l) - 2Cr3+ + 3I2 + 14OH- Do a final check on atoms and charges. 2Cr (aq) + 3I2(s) + 7H2O(l) 3+ Do a final check on atoms and charges. 21-7 21-8 7 8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.2 Sample Problem 21.1: The redox reaction between dichromate ion and iodide ion. PROBLEM: Cr2O7 2- I- Balancing Redox Reactions by the Half-Reaction Method Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4-(aq) + C2O42-(aq) PLAN: Cr3+ + I2 Proceed in acidic solution and then neutralize with base. SOLUTION: MnO4+7 MnO44H+ + MnO4- 21-9 9 MnO2(s) + CO32-(aq) MnO2 +4 MnO2 C2O42+3 C2O42- MnO2 + 2H2O C2O42- + 2H2O +3e- +2e- CO32+4 2 CO322CO32- + 4H+ 21-10 10 2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.3 Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method VOLTAIC CELL System Energydoes is released work on from its spontaneous surroundings redox reaction continued: 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 2CO32- + 4H+ + 2e- 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 2CO32- + 4H+ + 2e- 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 8H+ + 2MnO4- +6e3C2O42- + 6H2O 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 3C2O42- + 6H2O 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 6CO32- + 12H+ + 6e- 2MnO2+ 4H2O 6CO32- + 12H+ + 6e- Oxidation half-reaction X X+ + e - 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH- + 4OH2MnO2(s) + 6CO32-(aq) + 2H2O(l) 21-11 21-12 11 Reduction half-reaction B++ eB Overall (cell) reaction X + Y+ X+ + Y; DG < 0 Overall (cell) reaction A- + B+ A + B; DG > 0 12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The spontaneous reaction between zinc and copper(II) ion. Figure 21.5 A voltaic cell based on the zinc-copper reaction. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Zn(s) + Cu2+(aq) 13 Oxidation half-reaction AA + e- Reduction half-reaction Y++ e- Y Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 21-13 ELECTROLYTIC CELL Surroundings(power Energy is absorbed tosupply) drive a nonspontaneous redox reaction do work on system(cell) X3 X2 Figure 21.4 General characteristics of voltaic and electrolytic cells. Reduction half-reaction Cu2+(aq) + 2eCu(s) Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Zn2+(aq) + Cu(s) 21-14 14 3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.6 Notation for a Voltaic Cell components of anode compartment components of cathode compartment (oxidation half-cell) (reduction half-cell) phase of lower phase of higher oxidation state oxidation state phase of higher oxidation state A voltaic cell using inactive electrodes. Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l) phase of lower oxidation state phase boundary between half-cells Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Examples: Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l) inert electrode 21-15 21-16 15 16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.2: PROBLEM: PLAN: Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Diagramming Voltaic Cells Why Does a Voltaic Cell Work? 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. The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. Identify the oxidation and reduction reactions and write each halfreaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). e- SOLUTION: Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Ecell > 0 for a spontaneous reaction Voltmeter salt bridge Cr 1 Volt (V) = 1 Joule (J)/ Coulomb (C) Ag K+ NO3- Reduction half-reaction Ag+(aq) + eAg(s) Cr3+ Ag + Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) 21-17 17 Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) 21-18 18 4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.7 Table 21.1 Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) Common alkaline battery 1.5 Lead-acid car battery (6 cells = 12V) 2.0 Calculator battery (mercury) 1.3 Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070 Determining an unknown E0half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Overall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) 21-19 21-20 19 20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PROBLEM: Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Calculating an Unknown E0half-cell from E0cell Sample Problem 21.3: Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V Calculate E0bromine given E0zinc = -0.76V SOLUTION: E0Zn E0cell = E0cathode E0bromine 21-21 Zn2+(aq) + 2e anode: Zn(s) as - Zn2+(aq) E0anode + 2e- = 1.83 = - strength of oxidizing agent 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. E = +0.76 Zn(s) is -0.76V E0bromine - (-0.76) = 1.86 - 0.76 = 1.07 V F2(g) + 2e2F-(aq) Cl2(g) + 2e2Cl-(aq) MnO2(g) + 4H+(aq) + 2eMn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3eNO(g) + 2H2O(l) Ag+(aq) + eAg(s) Fe3+(g) + eFe2+(aq) O2(g) + 2H2O(l) + 4e4OH-(aq) Cu2+(aq) + 2eCu(s) 2H+(aq) + 2eH2(g) N2(g) + 5H+(aq) + 4eN2H5+(aq) Fe2+(aq) + 2eFe(s) 2H2O(l) + 2eH2(g) + 2OH-(aq) Na+(aq) + eNa(s) Li+(aq) + eLi(s) E0(V) +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 -0.23 -0.44 -0.83 -2.71 -3.05 strength of reducing agent PLAN: 21 Reduction half-reaction 2H3O+(aq) + 2eH2(g) + 2H2O(l) 21-22 22 5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength Writing Spontaneous Redox Reactions •By convention, electrode potentials are written as reductions. PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0cell for each. •When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. (b) Rank the relative strengths of the oxidizing and reducing agents: (1) NO3-(aq) + 4H+(aq) + 3e- •The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell. (2) N2(g) + 5H+(aq) + 4e(3) MnO2(s) •When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Zn(s) Example: Cu2+(aq) + stronger reducing agent Zn2+(aq) stronger oxidizing agent + weaker oxidizing agent PLAN: Cu(s) weaker reducing agent 21-23 + E0 = 0.96V 2e- Mn2+(aq) E0 = -0.23V + 2H2O(l) E0 = 1.23V 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. 21-24 23 24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (a) (1) (1) NO3-(aq) + 4H+(aq) + 3e- Rev (2) N2H5+(aq) NO3-(aq) 4H+(aq) + (2) N2H5+(aq) (A) N2(g) + + 3e- N2(g) + 5H+(aq) + 4e- (1) NO(g) + 2H2O(l) E0 (B) 2NO(g) + 3MnO2(s) + 4H+(aq) Rev (2) N2H5+(aq) X4 = +0.23V E0cell = 1.19V (2) N2H5+(aq) (C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) + + 3e- Mn2+(aq) + 2H2O(l) X2 E0cell E0 = 1.23V E0cell = 1.46V Mn2+(aq) + 2H2O(l) X2 N2(g) + 2Mn2+(aq) + 4H2O(l) (b) Ranking oxidizing and reducing agents within each equation: E0 = -0.96V E0 E0 = +0.23V Mn2+(aq) + 2H2O(l) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e- X3 Mn2+(aq) + 2H2O(l) 4H+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e- 4NO(g) + 3N2(g) + 8H2O(l) NO3-(aq) (3) MnO2(s) +4H+(aq) + 2e- 21-25 + 4e- NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) +4H+(aq) + 2e- Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) NO(g) + 2H2O(l) E0 = 0.96V 5H+(aq) NO(g) + 2H2O(l) 4NO3-(aq) + 3N2H5+(aq) + H+(aq) Rev (1) NO(g) + 2H2O(l) 25 +4H+(aq) NO(g) + 2H2O(l) N2H5+(aq) = 1.23V (A): oxidizing agents: NO3- > N2 reducing agents: N2H5+ > NO = 0.27V (B): oxidizing agents: MnO2 > NO3- reducing agents: NO > Mn2+ (C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+ X3 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) 21-26 26 6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) Relative Reactivities (Activities) of Metals 1. Metals that can displace H from acid A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of 2. Metals that cannot displace H from acid Oxidizing agents: MnO2 > NO3- > N2 Reducing agents: N2H5+ > NO > Mn2+ 3. Metals that can displace H from water 4. Metals that can displace other metals from solution 21-27 21-28 27 28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The reaction of calcium in water. Free Energy and Electrical Work Oxidation half-reaction Ca(s) Ca2+(aq) + 2e- Reduction half-reaction 2H2O(l) + 2eH2(g) + 2OH-(aq) DG a -Ecell -Ecell = -wmax DG = wmax = charge x (-Ecell) DG = -n F Ecell charge In the standard state charge = n F n = #mols eF = Faraday constant F = 96,485 C/mol e- 29 DG0 = - RT ln K E0cell = - (RT/n F) ln K 1V = 1J/C F = 9.65x104J/V*mol e- Overall (cell) reaction Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq) 21-29 DG0 = -n F E0cell 21-30 30 7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.9 The interrelationship of DG0, E0, and K. DG DG 0 DG0 = -nFEocell 0 K E 0 cell PROBLEM: Reaction at standard-state conditions <0 >1 >0 0 1 0 at equilibrium >0 <1 <0 nonspontaneous spontaneous 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: DG0 = -RT lnK 0 cell 2X E0cell = E0cell = (0.0592V/n) log K (at 298 K) K log K = E0cell = -RT lnK nF 21-31 Pb2+(aq) + 2eAg+(aq) + e- Pb(s) Ag(s) E0 = -0.13V E0 = 0.80V E0 = 0.13V E0 = 0.80V Pb(s) Pb2+(aq) + 2eAg+(aq) + eAg(s) 0.592V n n x E0cell 0.592V log K = E0cell = 0.93V DG0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V) (2)(0.93V) 0.592V K = 2.6x1031 DG0 = -1.8x102kJ 21-32 31 32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.6: The Effect of Concentration on Cell Potential [Zn2+] = 0.010M -nF Ecell = -nF Ecell + RT ln Q Ecell = E0cell - RT nF ln Q SOLUTION: Determining E0cell : E0cell 21-33 0.0592 2H+(aq) + 2e- H2(g) E0 = 0.00V Zn2+(aq) + 2e- Zn(s) E0 Zn(s) •When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell - PH 2 = 0.30atm PLAN: Find E0cell and Q in order to use the Nernst equation. •When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = Ecell = [H+] = 2.5M Calculate Ecell at 298 K. •When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell E0cell Using the Nernst Equation to Calculate Ecell 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: DG = DG0 + RT ln Q 33 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. By substituting standard state values into E0cell, we get E Calculating K and DG0 from E0cell Sample Problem 21.5: Ecell = E0cell - log Q n Ecell = 0.76 - Zn2+(aq) 0.0592V n + 2e- log Q P x [Zn2+] H2 [H+]2 = -0.76V E0 = +0.76V (0.0592/2)log(4.8x10-4) Q= Q= (0.30)(0.010) (2.5)2 Q = 4.8x10-4 = 0.86V 21-34 34 8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 21.10 Figure 21.11 A concentration cell based on the Cu/Cu2+ half-reaction. The relation between Ecell and log Q for the zinc-copper cell. Oxidation half-reaction Cu(s) Cu2+(aq, 0.1M) + 2e- Cu Reduction half-reaction 1.0M) + 2eCu(s) 2+(aq, Overall (cell) reaction Cu2+(aq,1.0M) Cu2+(aq, 0.1M) 21-35 21-36 35 36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 21.7: PROBLEM: PLAN: Calculating the Potential of a Concentration Cell A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3. What is the cell potential at 298 K? Which electrode has a positive charge? E0cell will be zero since the half-cell potentials are equal. Ecell is calculated from the Nernst equation with half-cell A (higher [Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+. SOLUTION: Ag+(aq, 0.010M) half-cell A Ecell = E0cell - 0.0592V 1 log [Ag+] Ag+(aq, 4.0x10-4M) half-cell B dilute [Ag+]concentrated Ecell = 0 V -0.0592 log 4.0x10-2 = 0.0828V Half-cell A is the cathode and has the positive electrode. 21-37 37 9