Electrochemistry Chapter 19

Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-) 19.1

Amps, Time, Coulombs, Faradays, and Moles of Electrons Three equations relate these quantities: amperes x time = Coulombs 96,485 coulombs = 1 Faraday 1 Faraday = 1 mole of electrons The thought process for interconverting between amperes and moles of electrons is: amps & time   Coulombs   Faradays   moles of electrons

Calculating the Quantity of Substance Produced or Consumed To determine the quantity of substance either produced or consumed during electrolysis given the time and a known current flow these steps: Write the balanced half-reactions involved. Calculate the number of moles of electrons that were transferred. Calculate the number of moles of substance that was produced/consumed at the electrode. Convert the moles of substance to desired units of measure.

Example:    A 40.0 amp current flowed through molten iron(III) chloride for 10.0 hours (36,000 s).  Determine the mass of iron and the volume of chlorine gas (measured at 25oC and 1 atm) that is produced during this time. Write the half-reactions that take place at the anode and at the cathode. anode (oxidation):  2 Cl-   Cl2(g) + 2 e- cathode (reduction)  Fe3+ + 3 e-    Fe(s) Calculate the number of moles of electrons.

Calculate the moles of iron and of chlorine produced using the number of moles of electrons calculated and the stoichiometries from the balanced half-reactions.  According to the equations, three moles of electrons produce one mole of iron and 2 moles of electrons produce 1 mole of chlorine gas.

Calculate the mass of iron using the molar mass and calculate the volume of chlorine gas using the ideal gas law (PV = nRT).

Electrolytic cell Galvanic cell (also called voltaic cell) uses chemical reaction to produce electrical energy (flow of electrons). When zinc metal is placed in CuSO4 solution, the following reaction takes place: Zn(s)  +  CuSO4(aq)              ZnSO4(aq)  +  Cu(s) Oxidation:   Zn(s)              Zn+2 + 2e-1 Reduction:  Cu+2 + 2e-1              Cu Overall:      Zn(s) + Cu+2              Zn+2 + Cu(s)

Electrolytic cell Electrons will not flow in the following apparatus: Why not? The circuit is not complete. There must be a continuous flow of charge for the electrons to flow.

Galvanic Cell But if the reaction is carried out using a salt bridge to complete the circuit and maintain charge neutrality, electrons are transferred from Zn° to Cu+2 through a wire producing electrical energy.

CHEMICAL CHANGE ---> ELECTRIC CURRENT 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 GALVANIC or VOLTAIC cell. A group of such cells is called a battery. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

Galvanic Cell

Anode Cathode Oxidation occurs Reduction occurs Electrons produced Electrons are consumed Has negative sign (-) Has positive sign (+) Anions migrate toward Cations migrate toward

Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode 19.2

Standard Electrode Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) 19.3

Shorthand Notation for Cells Shorthand Notation for: Zn° + Cu+2 Zn+2 + Cu° Anode Cathode electrode electrode

Shorthand Notation for Cells Write shorthand notation for:   Fe(s) + 2Fe+3(aq) 3Fe+2(aq) Fe Fe2+ + 2e- = oxidation (anode) Fe3+ + 1e- Fe2+ = reduction (cathode) Anode Cathode Fe°  Fe+2      Fe+3   Fe+2

Shorthand Notation for Cells Write shorthand notation for:   2Ag+1(aq)  +  Ni(s)         2Ag(s) + Ni+2(aq) Ni°  Ni+2    Ag+1  Ag°