17-Nov-97Electrochemistry (Ch. 21)1 ELECTROCHEMISTRY Chapter 21 Electric automobile redox reactions electrochemical cells electrode processes construction notation cell potential and  G o standard reduction potentials (E o ) non-equilibrium conditions (Q) batteries corrosion

17-Nov-97Electrochemistry (Ch. 21)2 TRANSFER REACTIONS Atom transfer HCl (g) + H 2 O (l)  Cl - (aq) + H 3 O + (aq) Electron transfer Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) L oss of E lectrons = O XIDATION ( LEO ) G ain of E lectrons = R EDUCTION ( GER ) - 2 e- 2 x +1 e-

17-Nov-97Electrochemistry (Ch. 21)3 Electron Transfer Reactions Electron transfer reactions are oxidation- reduction or redox reactions. Redox reactions can result in : – generation of an electric current, or – be caused by imposing an electric current. When external electric current is involved, this field of chemistry is called ELECTROCHEMISTRY.

17-Nov-97Electrochemistry (Ch. 21)4 Terminology for Redox Reactions OXIDATION—loss of electron(s) by a species; increase in oxidation number. REDUCTION—gain of electron(s); decrease in oxidation number. OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized.

17-Nov-97Electrochemistry (Ch. 21)5 Direct Redox Reactions Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) 11_CuAg.mov 21mo2an1.mov 2Al (s) + 3Cu 2+  2 Al Cu (s)

17-Nov-97Electrochemistry (Ch. 21)6 Indirect Redox Reactions A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. 11_battry.mov 21mo2an2.mov Oxidation Reduction Electron transfer Ions

17-Nov-97Electrochemistry (Ch. 21)7 Balancing Equations Cu (s) + Ag + (aq)  Cu 2+ (aq) + Ag (s) Step 3: Multiply each half-reaction by a factor that makes the reducing agent supply as many electrons as the oxidizing agent requires - ELECTRON TRANSFER NUMBER (2 here) Oxidation (Reducing agent) Cu  Cu e- Reduction (Oxidizing agent) 2 Ag e-  2 Ag Step 4: Add half-reactions to give the overall equation. Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2Ag (s) How to balance for both charge and mass ? Step 1: Identify the oxidation and reduction HALF-REACTIONS: OX: Cu  Cu e- RED: Ag + + e-  Ag Step 2: Balance each HALF-REACTION for charge and mass (done)

17-Nov-97Electrochemistry (Ch. 21)8 Balance the following in acid solution— VO Zn  VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn  Zn 2+ RedVO 2 +  VO 2+ Step 2:Balance each half-reaction for mass. OxZn  Zn 2+ Red2 H + + VO 2 +  VO 2+ + H 2 O Add H 2 O on O-deficient side and add H + on other side for H-balance. Balancing Equations (2)

17-Nov-97Electrochemistry (Ch. 21)9 Step 3:Balance half-reactions for charge. OxZn  Zn e- Rede- + 2 H + + VO 2 +  VO 2+ + H 2 O Step 4:Multiply by an appropriate factor to balance the electron transfer in OX. and RED. OxZn  Zn e- Red 2e- + 4 H VO 2 +  2 VO H 2 O Step 5:Add half-reactions Zn + 4 H VO 2 +  Zn VO H 2 O Balancing Equations (3)

17-Nov-97Electrochemistry (Ch. 21)10 Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen. Never add H 2 or H atoms to balance hydrogen. Be sure to write the correct charges on all the ions. Check your work at the end to make sure mass and charge are balanced.

17-Nov-97Electrochemistry (Ch. 21)11 Why Study Electrochemistry? Industrial production of chemicals such as Cl 2, NaOH, F 2 and l Biological redox reactions The heme group Batteries Corrosion

17-Nov-97Electrochemistry (Ch. 21)12 Electrochemical Cells An apparatus in which a redox reaction occurs by transferring electrons through an external connector. Batteries are voltaic cells ELECTROLYTIC CELL Reactant favored reaction electric current  chemical reaction VOLTAIC CELL Product favored reaction chemical reaction  electric current

17-Nov-97Electrochemistry (Ch. 21)13 CHEMICAL CHANGE  ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ is reduced and is the oxidizing agent Cu 2+ (aq) + 2e-  Cu(s)

17-Nov-97Electrochemistry (Ch. 21)14 Oxidation: Zn(s)  Zn 2+ (aq) + 2e- Reduction: Cu 2+ (aq) + 2e-  Cu(s) Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Electrons are transferred from Zn to Cu 2+, but there is no useful electric current. CHEMICAL CHANGE  ELECTRIC CURRENT (2)

17-Nov-97Electrochemistry (Ch. 21)15 To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. CHEMICAL CHANGE  ELECTRIC CURRENT (2) This is accomplished in a VOLTAIC cell. (also called GALVANIC cell) A group of such cells is called a battery.

17-Nov-97Electrochemistry (Ch. 21)16 Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments. This maintains electrical neutrality. ANODE OXIDATION CATHODE REDUCTION

17-Nov-97Electrochemistry (Ch. 21)17 Electrochemical Cell Electrons move from anode to cathode in the wire. Anions & cations move through the salt bridge. 1!_cell.mov 21mo4an1.mov

17-Nov-97Electrochemistry (Ch. 21)18 Standard Notation for Electrochemical Cells ANODE Zn / Zn 2+ // Cu 2+ / Cu CATHODE OXIDATION Anode electrode Active electrolyte in oxidation half-reaction Cathode electrode Active electrolyte in reduction half-reaction Salt bridge Phase boundary REDUCTION

17-Nov-97Electrochemistry (Ch. 21)19 CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. Zn  Zn e- ANODE 2e- + Cu 2+  Cu CATHODE

17-Nov-97Electrochemistry (Ch. 21)20 This is the STANDARD CELL POTENTIAL, E o E o is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25  C. CELL POTENTIAL, E o For Zn/Cu, voltage is 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M.

17-Nov-97Electrochemistry (Ch. 21)21 E o and  G o E o is related to  G o, the free energy change for the reaction.  G o = - n F E o F = Faraday constant = x 10 4 J/Vmol n = the number of moles of electrons transferred. Michael Faraday Discoverer of electrolysis magnetic props. of matter electromagnetic induction benzene and other organic chemicals n for Zn/Cu cell ? n = 2 Zn / Zn 2+ // Cu 2+ / Cu

17-Nov-97Electrochemistry (Ch. 21)22 For a reactant-favored reaction - electrolysis cell: Electric current  chemistry Reactants  Products  G o > 0 and so E o < 0 (E o is negative) For a product-favored reaction – battery or voltaic cell: Chemistry  electric current Reactants  Products  G o 0 (E o is positive) E o and  G o (2)  G o = - n F E o

17-Nov-97Electrochemistry (Ch. 21)23 Calculating Cell Voltage If we know E o for each half-reaction, we can calculate E o for the net reaction. Balanced half-reactions can be added together to get the overall, balanced equation. Anode:2 I -  I 2 + 2e- Cathode:2 H 2 O + 2e-  2 OH - + H 2 Net rxn: 2 I H 2 O  I OH - + H 2

17-Nov-97Electrochemistry (Ch. 21)24 STANDARD CELL POTENTIALS, E o Can’t measure half- reaction E o directly. Therefore, measure it relative to a standard HALF CELL: the S tandard H ydrogen E lectrode ( SHE ). 2 H + (aq, 1 M) + 2e- H 2 (g, 1 atm) E o = 0.0 V

17-Nov-97Electrochemistry (Ch. 21)25 Zn/Zn 2+ versus H + /H 2 Zn/Zn 2+ half-cell combined with a SHE. E o for the cell is V

17-Nov-97Electrochemistry (Ch. 21)26 E o for Zn/Zn 2+ half-cell Zn(s) + 2 H + (aq)  Zn 2+ + H 2 (g) E o = V Therefore, E o for Zn  Zn 2+ (aq) + 2e- is ?? V. Overall reaction is reduction of H + by Zn metal.

17-Nov-97Electrochemistry (Ch. 21)27 Standard REDUCTION potentials What is E o for the reverse reaction ? Zn e-  Zn The value for the REDUCTION 1/2-cell is the negative of that for the OXIDATION 1/2-cell: Zn  Zn 2+ (aq) + 2e- E o = V THUS Zn e-  Zn E o = V Zn  Zn 2+ (aq) + 2e- E o = V Q. Relative to H 2 is Zn a (better/worse) reducing agent ? A. Zn is a better reducing agent than H 2.

17-Nov-97Electrochemistry (Ch. 21)28 E o = V Cu/Cu 2+ and H 2 /H + Cell

17-Nov-97Electrochemistry (Ch. 21)29 Cu 2+ (aq) + H 2 (g)  Cu(s) + 2 H + (aq) Measured E o = V Therefore, E o for Cu e-  Cu is ?? Cu/Cu 2+ half cell E o Overall reaction is reduction of Cu 2+ by H 2 gas V

17-Nov-97Electrochemistry (Ch. 21)30 Zn/Cu Electrochemical Cell Anode: Zn(s)  Zn 2+ (aq) + 2e- E o = V Cathode, positive, sink for electrons Anode, negative, source of electrons What is E o for the Zn/Cu cell (Daniel’s cell) ?? E o = E o ( anode ) + E o ( cathode ) = = V Net: Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Cathode: Cu 2+ (aq) + 2e-  Cu(s) E o = V

17-Nov-97Electrochemistry (Ch. 21)31 Uses of E o Values This shows we can a) decide on relative ability of elements to act as reducing agents (or oxidizing agents) b) assign a voltage to a half-reaction that reflects this ability.

17-Nov-97Electrochemistry (Ch. 21)32 BEST Reducing agent ? ? STANDARD REDUCTION POTENTIALS Half-Reaction E o (Volts) Cu e-  Cu Oxidizing ability of ion Reducing ability of element 2 H + + 2e-  H Zn e-  Zn BEST Oxidizing agent ? ? Cu 2+ Zn

17-Nov-97Electrochemistry (Ch. 21)33 Standard Redox Potentials, E o Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H + and Cu 2+. H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+. Use tabulated reduction potentials to analyse spontaneity of ANY REDOX REACTION

17-Nov-97Electrochemistry (Ch. 21)34 Determining E o for a Voltaic Cell Cd  Cd e- or Cd e-  Cd Fe  Fe e- or Fe e-  Fe

17-Nov-97Electrochemistry (Ch. 21)35 Fe is a better reducing agent than Cd E o for Fe/Cd Cell Cd e-  Cd Fe e-  Fe RHS species is better reducing agent LHS species is better oxidizing agent Cd 2+ is a better oxidizing agent than Fe 2+ Overall reaction as written is spontaneous: Fe + Cd 2+  Cd + Fe 2+ E o = V The reverse reaction is not spontaneous: Cd + Fe 2+  Fe + Cd 2+ E o = V