Lecture 7

Lecture 7 - Summary • Electrochemistry describes...

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Unformatted text preview: Summary • Electrochemistry describes the rela3onship between chemical change and electrical work. •  Electrochemical processes absorb or release energy • It involves movement of electrons from one chemical species to another through an oxida3on ­reduc3on reac3on.. • Redox (oxida3on –reduc3on) reac3on involves transfer of electrons from a reducing agent to an oxidizing agent. • The half reac3on method of balancing divides the overall reac3on into the half reac3ons that are balanced separately before recombining them again. The spontaneous reaction between zinc and copper(II) ion Change of oxida3on number 0 +2 0 +1 +2 0 Zn + Cu2+ Zn2+ + Cu +2 0 Zn + 2H+ Zn2+ + H2(g) +7 +3 +2 +4 MnO4− + C2O42 ­ Mn2+ + CO2 +6  ­1 Cr2O72- + I- +3 0 Cr3+ + I2 Electrochemical Cells •  Two types of electrochemical cells are based on redox reac3ons. •  In a voltaic (Galvanic) cell, a spontaneous reac3on (dG < 0) generates electricity and work is done by the system on the surrounding. •  In an electroly3c cell, the surrounding supply electricity which drives a non ­spontaneous reac3on (dG > 0). In this case work is done on the system by the surrounding. General characteristics of voltaic and electrolytic cells. Voltaic Cell •  •  1)  2)  3)  4)  The reac3on always runs spontaneously in the direc3on that produces a posi3ve cell poten3al. Four things for a complete descrip3on. Nature of all the components ­ electrodes and ions Designa3on of anode and cathode Direc3on of flow Cell Poten3al A voltaic cell based on the zinc-copper reaction Construc3on of Voltaic Cell •  •  •  •  •  •  •  •  •  •  In spontaneous redox reac/ons, electrons are transferred and energy is released. Work can be done form the released energy when the electrons flow through an external device. The oxida/on occurs at the anode. The reduc/on occurs at the cathode. The charges in each cell would not be balanced once electrons flow from anode to cathode . A salt bridge, usually a U ­shaped tube that contains a salt solu/on, is used to keep the charges balanced. –  Ca/ons move toward the cathode. –  Anions move toward the anode. In the cell, electrons leave the anode and flow through the wire to the cathode. As the electrons leave the anode, the ca/ons formed dissolve into the solu/on in the anode compartment. As the electrons reach the cathode, ca/ons in the cathode are aDracted to the now nega/ve cathode. The electrons are taken by the ca/on, and the neutral metal is deposited on the cathode. A voltaic cell using inactive electrodes •  Ac3ve electrodes themselves par3cipate are the components of redox reac3on (example, Zn/Cu electrodes) •  In many redox reac3ons reac3ng species are not capable of serving as electordes. Inac3ve electrodes are used in these cases. •  In I ­/MnO4 ­ reac3on neither species can serve as an electrode. •  Graphite and pla3num are sued as inac3ve electrodes. Notation for a Voltaic Cell Components of anode compartment Components of cathode compartment (oxidation half-cell) (reduction half-cell) Phase of lower oxidation state Phase of higher oxidation state Phase of higher oxidation state Phase of lower oxidation state Phase boundary between half-cells Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite inert electrodes Describing a Voltaic Cell with Diagram and Notation PROBLEM: PLAN: Draw a 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. 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). SOLUTION: Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) Measurement of a standard cell potential Cell Poten3al •  Oxidizing agent pulls the electron. •  Reducing agent pushes the electron. •  The push or pull (“driving force”) is called the cell poten3al Ecell •  Ecell >0 for a spontaneous process •  Also called the electromo3ve force (emf) •  Unit is the volt(V) •  = 1 joule of work/coulomb of charge •  Measured with a voltmeter Determining an unknown Eohalf-cell with the standard reference (hydrogen) electrode. Calculating an Unknown Eohalf-cell from Eocell PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br -(aq) Eocell = 1.83 V Calculate Eobromine, given Eozinc = -0.76 V. PLAN: The reaction is spontaneous as written since the Eocell is (+). Zinc is being oxidized and is the anode. Therefore, the Eobromine can be found using Eocell = Eocathode - Eoanode. SOLUTION: Anode: Zn(s) Zn2+(aq) + 2e- EoZn as Zn2+(aq) + 2e- E = +0.76 V Zn(s) is -0.76 V Eocell = Eocathode - Eoanode = 1.83 = Eobromine - (-0.76) Eobromine = 1.83 + (-0.76) = 1.07 V 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 Eocell. • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) + stronger reducing agent Cu2+(aq) Zn2+(aq) stronger weaker oxidizing agent oxidizing agent + Cu(s) weaker reducing agent Sample Problem 21.4 Writing Spontaneous Redox Reactions PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate Eocell for each. (1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e(3) MnO2(s) + 4H+(aq) + 2ePLAN: NO(g) + 2H2O(l) N2H5+(aq) Mn2+(aq) + 2H2O(l) Eo = 0.96 V Eo = -0.23 V Eo = 1.23 V Put the equations together in varying combinations so as to produce (+) Eocell for each combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of Eo. Balance the number of electrons gained and lost without changing the Eo. In ranking the strengths, compare the combinations in terms of Eocell. Sample Problem 21.4 Writing Spontaneous Redox Reactions continued (2 of 3) SOLUTION: (a) (1) NO3-(aq) + 4H+(aq) + 3e- Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (1) NO3-(aq) + 4H+(aq) + 3e(2) N2H5+(aq) (A) x4 4NO(g) + 3N2(g) + 8H2O(l) (B) 2NO(g) + 3MnO2(s) + 4H+(aq) Eo = -0.96 V Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) + 4H+(aq) + 2e- Eocell = 1.19 V x3 NO3-(aq) + 4H+(aq) + 3e- (3) MnO2(s) + 4H+(aq) + 2e(1) NO(g) + 2H2O(l) Eo = +0.23 V NO(g) + 2H2O(l) N2(g) + 5H+(aq) + 4e- 4NO3-(aq) + 3N2H5+(aq) + H+(aq) Rev (1) NO(g) + 2H2O(l) NO(g) + 2H2O(l) Eo = 0.96 V x2 Eo = 1.23 V Eocell = 0.27 V Mn2+(aq) + 2H2O(l) x 3 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) Writing Spontaneous Redox Reactions Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) + 4H+(aq) + 2e(2) N2H5+(aq) Mn2+(aq) + 2H2O(l) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) + 4H+(aq) + 2e(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) E0 = +0.23 V E0 = 1.23 V E0cell = 1.46 V Mn2+(aq) + 2H2O(l) x 2 N2(g) + 2Mn2+(aq) + 4H2O(l) 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+ ...
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