Chem 1721 Brief Notes: Chapter 19

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1 Chem 1721 Brief Notes: Chapter 19 Chapter 19: Electrochemistry Consider the same redox reaction set up 2 different ways: Cu metal in a solution of AgNO 3 Cu Cu salt bridge electrically conducting wire Ag Ag+ NO3 - Cu2+ Ag+ Anode Cathode What is the reaction? In the experiment on the left: over time the strip of copper metal decreases in size (and mass) the solution starts clear and colorless, and over time becomes clear and light blue over time a fine grayish-silver powder deposits on the bottom of the beaker ox: Cu (s) Cu 2+ (aq) + 2 e red: Ag + (aq) + 1 e Ag (s) net: Cu + 2 Ag + Cu Ag energy change? Any energy associated with the process goes into changing the temperature of the solution. In the experiment on the right: the Cu/Cu 2+ redox couple and the Ag + /Ag redox couple are separated Cu metal and Ag metal are connected by an electrically conducting wire e transfer from Cu to Ag + occurs through the wire generating a voltage this is an electrochemical cell there are two types of electochemical cells 1. Galvanic cell: a spontaneous chemical reaction occurs that generates a voltage 2. Eletrolytic cell: a nonspontaneous chemical reaction is driven by an applied current Galvanic cells 2 compartments: anode and cathode anode where oxidation occurs; usually shown on the left cathode where reduction occurs; usually shown on the right can write anode and cathode half reactions (parallel to oxidation and reduction half reactions respectively) anode: cathode: net cell: Cu Cu e Ag e Ag Cu + 2 Ag + Cu Ag each compartment must contain an electrically conducting solid; the electrode; where the wire connects in the example above: copper is the anode, silver is the cathode the direction of electron flow in a Galvanic cell is always anode cathode electrons are generated at the anode (where oxidation occurs) electrons are consumed at the cathode (where reduction occurs)

2 battery designations: anode designated because electrons are produced cathode designated + because electrons are consumed compartments are connected by a salt bridge (or porous membrane see the 2 figures below) electrically conducting wire electrically conducting wire salt bridge porous or semi-permeable membrane Anode Cathode Anode Cathode salt bridge (or porous membrane) allows migration of spectator ions between compartments maintain charge neutrality anions flow toward the anode; e s leave the anode + charge builds up; i.e. Cu Cu 2+ cations flow toward the cathode; e s arrive at the cathode positive charge is decreasing; i.e. Ag + Ag short-hand description of a Galvanic cell line notation basically: anode anode compartment cathode compartment cathode for the Cu/Ag + cell: Cu (s) Cu 2+ (aq) Ag + (aq) Ag (s) Consider a Galvanic cell with the following cell reaction: Fe (s) + 2 Fe 3+ (aq) 2 Fe 2+ (aq). Write the anode and cathode half reactions, sketch the set-up of this cell, and write the corresponding line notation. anode: Fe Fe e cathode: Fe 3+ + e Fe 2+ net cell: Fe + 2 Fe 3+ 3 Fe 2+ set up: note: the cathode compartment requires a chemically inert conductor to facilitate the e transfer common chemically inert conductors include Pt (s), C (s) i.e a graphite rod line notation: Fe (s) Fe 2+ (aq) Fe 2+ (aq), Fe 3+ (aq) Pt (s) or other conductor Cell voltage or cell potential; E and E a Galvanic cell produces a particular voltage depending on the exact combination of half reactions (half cells), temperature, pressures of gases, and molar concentration of solutions cell potential, E, measured in volts (V) 1 volt = difference in potential between 2 points 1 J of work is done when 1 Coulomb of charge moves between 2 points (of higher potential to lower potential) differing by 1V 1 V = 1 J/C

3 standard cell potential, E cell potential (in V) when solids and liquids are in their pure form; gasses are at 1 atm pressure; solutions at 1 M concentration E cell = E ox + E red = E anode + E cathode one standard half-cell is designated as a reference and assigned E = 0.00 V standard hydrogen electrode: 2 H + (aq) + 2 e H 2 (g); E = 0.00 V specifically with: [H + ] = 1.00 M, P H2 = 1.00 atm; temp = 298 K consider the following Galvanic cell: anode: Zn Zn e cathode: 2 H e H 2 net cell: Zn + 2 H + Zn 2+ + H 2 ; E cell = 0.76 V if E cathode = 0.00 V then E anode = 0.76 V now consider: anode: Zn Zn e ; E = 0.76 V cathode: Cu e Cu; E =???? net cell: Zn + Cu 2+ Zn 2+ + Cu; E cell = 1.10 V if E cell = E anode + E cathode then E cathode = E cell E anode = 0.34 V Tabulated Standard Reduction Potentials (Appendix I and Table 19.1) all half-reactions written as reductions i.e. oxidizing agent + e reducing agent tabulated in order of increasing E increasing reactivity in direction written increasing tendency for reduction to occur decreasing tendency for oxidation (reverse reaction) to occur stronger reducing agents at top of table; stronger oxidizing agents at bottom of table + value of E indicates a spontaneous half reaction; value of E indicates a nonspontaneous half reaction an electrochemical cell needs a reduction and an oxidation 2 half cells the equation corresponding to the oxidation (anode) half cell must be written in reverse of table when the half-reaction is reversed the sign of E is also reversed when a half reaction is multiplied by some factor to change the stoichiometry you do NOTHING to E Spontaneous cell reactions and E cell a spontaneous reaction has a + E cell Galvanic cells involve spontaneous chemical change E cell must be + consider Galvanic cells based on: Li + + e Li E = 3.04 V Cl e 2 Cl E = 1.36 V AND Pb e Pb E =.13 V 2 PbO 2 + SO 4 + 4H e PbSO H 2 O E = 1.69 V the position of the half-reaction in the Table of Standard Reduction Potentials allows you to predict the anode and cathode of any pair of half cells

4 the half reaction that is lower in the table (larger E ) will be the cathode (reduction) half reaction the half reaction that is higher in the table (smaller E ) will be the anode (oxidation) half reaction remember: must end up with a + E cell for a Galvanic cell relative strengths of oxidizing and reducing agents the lower in the table the stronger the oxidizing agent can cause any half reaction above it to proceed in reverse F 2 is the strongest oxidizing agent listed Li is the strongest reducing agent listed a related concept... the Activity Series of Metals ranks metals as reducing agents metals that like to be oxidized are good reducing agents good reducing agents have small (and frequently negative) E s Activity Series: Li > K > Ba > Ca > Mg > Be > Al > Zn > Fe > Cu > Ag > Au most active metal least active metal best reducing agent worst reducing agent on the list a metal in the Activity Series can reduce any M n+ to the right of it lots of potential questions several examples Consider these half reactions: NO H e NO + 2 H 2 O E = 0.96 V Fe 3+ + e Fe 2+ E = 0.77 V If combined in a Galvanic cell, what will be the net cell reaction and E cell? Can MnO 4 in acidic solution (i.e H + present) oxidize Ni? Ag? Which is a stronger reducing agent, Cr or Mn? Based on the Acitivity Series, can aluminum metal reduce Ca 2+? Ag +? Complete description of a Galvanic cell at this point you should be able to write/identify the anode and cathode half reaction of a cell write the net cell reaction determine E cell, or E anode or E cathode if E cell is given sketch a diagram for a Galvanic cell including: identification of the anode and cathode compartments and their components (i.e electrode and ions in solution); direction of electron flow; salt bridge or porous barrier and ion migration write the line notation that describes a Galvanic cell

5 Galvanic cells, work and free energy the amount of work that a Galvanic cell can do can be quantified w max = nfe cell efficiency = (w act /w max )*100 n = mol e ; F = Faraday constant, C/mol e nf = quantity of charge transferred w act = nfe efficiency = (E/E )*100 free energy change for a reaction in a Galvanic cell is equal to the maximum work that can be achieved ΔG = nfe ΔG = nfe consider again the Cu/Ag Galvanic cell: anode: Cu Cu e ; E = 0.34 V cathode: 2*(Ag e Ag); E = 0.80 V net cell rxn: Cu + 2 Ag + 2 Ag + Cu 2+ ; E cell = 0.46 V Calculate the maximum work that can be done by this cell, ΔG, and K at 25 C. Recall, 1 V = 1 J/C. w max = ΔG = nfe w max = ΔG = (2mol e )(96485 C/mol e )(0.46 J/C) w max = ΔG = 8.9 x 10 4 J, or 89 kj ΔG = RTlnK ln K = (8.9 x 10 4 J)/(8.314 J K 1 mol 1 *298 K) ln K = 36 K = 4.3 x Now consider that this cell operates at only 76% efficiency. Calculate E? efficiency = (E/E )*100 E = (effiency/100)*e E = (.76)(.46V) = 0.35V and the actual work done by the cell... w act = nfe w act = (2mol e )(96485 C/mol e )(0.35 J/C) w act = 6.8 x 10 4 J or 68 kj The Nernst Equation relates E to E for cells operating at nonstandard conditions; [sol n] 1.0 M and P gas 1.0 atm E = E o RT lnq E = E o lnq; at 25 nf n o C for a system at equilibrium, E = 0; Q = K *note a dead battery is at equilibrium E o = RT lnk E o = lnk; at 25 nf n o C Calculate E for a Galvanic cell at 25 C with a Ni cathode in 3.3 M NiCl 2 (aq) and a Tl anode in 0.25 M TlNO 3 (aq). anode: 2(Tl Tl e ) E = 0.34 V cathode: Ni e Ni E = 0.24V as this cell runs, [Ni 2+ ] will decrease (reactant) net cell: 2 Tl + Ni 2+ Ni + 2 Tl + E cell = 0.10 V and [Tl + ] will increase (product) until Q = K E = 0.10 J/C (8.314 J/Kmol)(298 K) (2 mol e)(96485 C/mol e) ln* E = 0.15 V

6 Electrolytic cells reactions in electrolytic cells are nonspontaneous; ΔG is +; E cell is E cell is the minimum voltage required for reaction to occur you need to know: balanced chemical equation to identify mole electrons transferred (n = mol e ) Faraday constant defining quantity of charge transferred per mole of electrons; F = C/mol e current used or required during the electrolysis in amperes; 1 A = 1 C/s electric power in Watts; 1 W = 1 J/s some examples: Determine the time required to plate 85.5 g Zn if 23.0 A passed through a solution of ZnSO 4 (aq). cathode rxn: Zn e Zn logical pathway: g Zn mol Zn mol e C t answer = 1.10 x 10 4 s or 183 min or 3.05 h Determine the current required to plate 2.86 g chromium metal from CrCl 3 (aq) in 2.5 min. cathode rxn: Cr e Cr logical pathway: g Cr mol Cr mol e C; C s = A answer = 106 A

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