ELECTROCHEMISTRY Chapter 9 Electric automobile ELECTROCHEMISTRY Chapter 9

Why Study Electrochemistry? Batteries Corrosion Industrial production of chemicals such as Cl2, NaOH, F2 and l Biological redox reactions The heme group

TRANSFER REACTIONS Atom transfer Electron transfer HCl (g) + H2O (l)  Cl- (aq) + H3O+ (aq) Electron transfer Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) - 2 e- 2 x +1 e-

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.

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.

Demo: Balanced Chemical Equation Total Net Ionic Equation Zn(s) + CuSO4(aq)  _______________ ACTIVITY SERIES OF METALS Balanced Chemical Equation Total Net Ionic Equation Net Ionic Equation

Let’s Examine Reaction… Zn (s) + Cu+2 + SO4-2  Zn+2 + SO4-2 + Cu(s) Zn (s) + Cu+2  Zn+2 + Cu(s) LEO GER Lost 2 e Gained 2 e Zn – lost 2 e  OXIDAZED Cu – gained 2 e  REDUCED

DEFINITION Reactions in which one reactant is oxidized (lose e-) and the other reactant is reduced (gains e-) are called oxidation-reduction reaction, or REDOX NOTE: e- gained by one reactant are the e- lost by the other one

To understand electrochemistry, we need to be able to keep track of electrons, we do so by assigning oxidation numbers.

SUMMARY: Redox Reactions The charge on each atom in any uncombined element is always ZERO [Cu(s); Br2(g)…] In a redox reaction, electrons are transferred from one reactant to another reactant Oxidation – lose e-  Reducing agent Reduction – gain e-  Oxidizing agent

Rules for assigning Oxidation Numbers  pg.658 Table 1 Examples: Cl2 Na2O Fe2O3 PbSO4 KNO2 Fe(NO3)3 Supply the oxidation number of the underlined element in the following formulas; a) Zn3(PO4)2, b) NaNO2, c) SnBr2, d) HSbO2, e) Mg(MnO4)2, f) NH4NO3

Identify the reactant oxidized and the reactant reduced in each of the following equations. Which of the following equations represent redox reactions? Pg. 662 (#19, 20)

Direct Redox Reactions Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) 2Al (s) + 3Cu2+  2 Al3+ + 3 Cu (s)

Electroplating

Indirect Redox Reactions A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. Electron transfer Oxidation Reduction 11_battry.mov 21mo2an2.mov Ions

How to balance for both charge and mass ? Balancing Equations Cu (s) + Ag+ (aq)  Cu2+ (aq) + Ag (s) How to balance for both charge and mass ? Step 1: Identify the oxidation and reduction HALF-REACTIONS: OX: Cu  Cu2+ + 2e- RED: Ag+ + e-  Ag Step 2: Balance each HALF-REACTION for charge and mass (done) 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  Cu2+ + 2e- Reduction (Oxidizing agent) 2 Ag+ + 2 e-  2 Ag Step 4: Add half-reactions to give the overall equation. Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2Ag (s)

Balancing Equations (2) Balance the following in acid solution— VO2+ + Zn  VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn  Zn2+ Red VO2+  VO2+ Step 2: Balance each half-reaction for mass. Red 2 H+ + VO2+  VO2+ + H2O Add H2O on O-deficient side and add H+ on other side for H-balance.

Balancing Equations (3) Step 3: Balance half-reactions for charge. Ox Zn  Zn2+ + 2e- Red e- + 2 H+ + VO2+  VO2+ + H2O Step 4: Multiply by an appropriate factor to balance the electron transfer in OX. and RED. Ox Zn  Zn2+ + 2e- Red 2e- + 4 H+ + 2 VO2+  2 VO2+ + 2 H2O Step 5: Add half-reactions Zn + 4 H+ + 2 VO2+  Zn2+ + 2 VO2+ + 2 H2O

Tips on Balancing Equations Never add O2, O atoms, or O2- to balance oxygen. Never add H2 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.

Electrochemical Cells An apparatus in which a redox reaction occurs by transferring electrons through an external connector. VOLTAIC CELL Product favored reaction chemical reaction  electric current Batteries are voltaic cells ELECTROLYTIC CELL Reactant favored reaction electric current  chemical reaction

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)  Zn2+(aq) + 2e- Cu2+ is reduced and is the oxidizing agent Cu2+(aq) + 2e-  Cu(s)

CHEMICAL CHANGE  ELECTRIC CURRENT (2) Oxidation: Zn(s)  Zn2+(aq) + 2e- Reduction: Cu2+(aq) + 2e-  Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s) Electrons are transferred from Zn to Cu2+, but there is no useful electric current.

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

ANODE OXIDATION CATHODE REDUCTION • Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments. This maintains electrical neutrality.

Electrochemical Cell Electrons move from anode to cathode in the wire. Anions & cations move through the salt bridge. Electrochemical Cell

Standard Notation for Electrochemical Cells Phase boundary Salt bridge Phase boundary ANODE Zn / Zn2+ // Cu2+ / Cu CATHODE Cathode electrode Anode electrode Active electrolyte in reduction half-reaction Active electrolyte in oxidation half-reaction OXIDATION REDUCTION

CELL POTENTIAL, E Zn  Zn2+ + 2e- 2e- + Cu2+ Cu ANODE CATHODE 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 [Zn2+] and [Cu2+] = 1.0 M.

CELL POTENTIAL, Eo STANDARD CELL POTENTIAL, Eo For Zn/Cu, voltage is 1.10 V at 25°C and when [Zn2+] and [Cu2+] = 1.0 M. This is the STANDARD CELL POTENTIAL, Eo Eo is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 °C.

Eo and DGo Michael Faraday 1791-1867 Eo is related to DGo, the free energy change for the reaction. DGo = - n F Eo F = Faraday constant = 9.6485 x 104 J/V•mol n = the number of moles of electrons transferred. Discoverer of electrolysis magnetic props. of matter electromagnetic induction benzene and other organic chemicals Zn / Zn2+ // Cu2+ / Cu n = 2 n for Zn/Cu cell ?

For a product-favored reaction Reactants  Products Eo and DGo (2) DGo = - n F Eo For a product-favored reaction battery or voltaic cell: Chemistry  electric current Reactants  Products DGo < 0 and so Eo > 0 (Eo is positive) For a reactant-favored reaction - electrolysis cell: Electric current  chemistry Reactants  Products DGo > 0 and so Eo < 0 (Eo is negative)

Calculating Cell Voltage Balanced half-reactions can be added together to get the overall, balanced equation. Anode: 2 I-  I2 + 2e- Cathode: 2 H2O + 2e-  2 OH- + H2 Net rxn: 2 I- + 2 H2O  I2 + 2 OH- + H2 If we know Eo for each half-reaction, we can calculate Eo for the net reaction.

STANDARD CELL POTENTIALS, Eo Can’t measure half- reaction Eo directly. Therefore, measure it relative to a standard HALF CELL: the Standard Hydrogen Electrode (SHE). 2 H+(aq, 1 M) + 2e- H2(g, 1 atm) Eo = 0.0 V

Zn/Zn2+ versus H+/H2 Zn/Zn2+ half-cell combined with a SHE. Eo for the cell is +0.76 V

Eo for Zn/Zn2+ half-cell Overall reaction is reduction of H+ by Zn metal. Zn(s) + 2 H+ (aq)  Zn2+ + H2(g) Eo = +0.76 V Therefore, Eo for Zn  Zn2+ (aq) + 2e- is ?? +0.76 V.

Standard REDUCTION potentials Zn  Zn2+ (aq) + 2e- Eo = +0.76 V Q. Relative to H2 is Zn a (better/worse) reducing agent ? A. Zn is a better reducing agent than H2. What is Eo for the reverse reaction ? Zn2+ + 2e-  Zn The value for the REDUCTION 1/2-cell is the negative of that for the OXIDATION 1/2-cell: Zn  Zn2+ (aq) + 2e- Eo = +0.76 V THUS Zn2+ + 2e-  Zn Eo = -0.76 V

Cu/Cu2+ and H2/H+ Cell Eo = +0.34 V

Overall reaction is reduction of Cu2+ by H2 gas. Cu/Cu2+ half cell Eo Overall reaction is reduction of Cu2+ by H2 gas. Cu2+ (aq) + H2(g)  Cu(s) + 2 H+(aq) Measured Eo = +0.34 V Therefore, Eo for Cu2+ + 2e-  Cu is ?? +0.34 V

Zn/Cu Electrochemical Cell What is Eo for the Zn/Cu cell (Daniel’s cell) ?? Anode, negative, source of electrons Cathode, positive, sink for electrons Anode: Zn(s)  Zn2+(aq) + 2e- Eo = +0.76 V Cathode: Cu2+(aq) + 2e-  Cu(s) Eo = +0.34 V Net: Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s) Eo = Eo(anode) + Eo(cathode) = 0.76 + 0.34 = +1.10 V

Uses of Eo 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.

STANDARD REDUCTION POTENTIALS Oxidizing ability of ion Half-Reaction Eo (Volts) Reducing ability of element Cu2+ + 2e-  Cu + 0.34 2 H+ + 2e-  H2 0.00 Zn2+ + 2e-  Zn -0.76 BEST Oxidizing agent ? ? Cu2+ BEST Reducing agent ? ? Zn

Standard Redox Potentials, Eo Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H+ and Cu2+. H2 can reduce Cu2+ but not Zn2+ Cu cannot reduce H+ or Zn2+. Use tabulated reduction potentials to analyse spontaneity of ANY REDOX REACTION

Determining Eo for a Voltaic Cell Cd  Cd2+ + 2e- or Cd2+ + 2e-  Cd Fe  Fe2+ + 2e- or Fe2+ + 2e-  Fe

Eo for Fe/Cd Cell • Fe is a better reducing agent than Cd LHS species is better oxidizing agent RHS species is better reducing agent Cd 2+ + 2e-  Cd -0.40 Fe 2+ + 2e-  Fe -0.44 • Fe is a better reducing agent than Cd Cd 2+ is a better oxidizing agent than Fe 2+ Overall reaction as written is spontaneous: Fe + Cd 2+  Cd + Fe 2+ Eo = +0.04 V The reverse reaction is not spontaneous: Cd + Fe 2+  Fe + Cd 2+ Eo = -0.04 V