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 2+ + 2 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 2+ + 2 e Ag + + 1 e Ag Cu + 2 Ag + Cu 2+ + 2 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)
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 2+ + 2 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
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 2+ + 2 e cathode: 2 H + + 2 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 2+ + 2 e ; E = 0.76 V cathode: Cu 2+ + 2 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 2 + 2 e 2 Cl E = 1.36 V AND Pb 2+ + 2 e Pb E =.13 V 2 PbO 2 + SO 4 + 4H + + 2 e PbSO 4 + 2 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
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 3 + 4 H + + 3 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
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, 96485 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 2+ + 2 e ; E = 0.34 V cathode: 2*(Ag + + 1 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 10 15 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.02569 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 =.02569 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 + + 1 e ) E = 0.34 V cathode: Ni 2+ + 2 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* 0.252 3.3 E = 0.15 V
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 = 96485 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 2+ + 2 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 3+ + 3 e Cr logical pathway: g Cr mol Cr mol e C; C s = A answer = 106 A