Electrochemistry

Chapter 19

2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

19.1

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

0 0 2+ 2-

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:

1. Write the balanced half-reactions involved. 2. Calculate the number of moles of electrons that

were transferred. 3. Calculate the number of moles of substance

that was produced/consumed at the electrode. 4. 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 cellGalvanic 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.

• 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.

Galvanic Cell

•To obtain a useful current, To obtain a useful current, we separate the oxidizing we separate the oxidizing and reducing agents so that and reducing agents so that electron transfer occurs thru electron transfer occurs thru an external wire. an external wire.

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

CHEMICAL CHANGE --->CHEMICAL CHANGE --->ELECTRIC CURRENTELECTRIC CURRENT

CHEMICAL CHANGE --->CHEMICAL CHANGE --->ELECTRIC CURRENTELECTRIC CURRENT

This is accomplished in a This is accomplished in a GALVANICGALVANIC or or VOLTAICVOLTAIC cell. cell.

A group of such cells is called a A group of such cells is called a batterybattery..

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

19.2

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

Standard Electrode Potentials

19.3

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

2e- + 2H+ (1 M) H2 (1 atm)

Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

Shorthand Notation for Cells

Shorthand Notation for: Zn° + Cu+2 Zn+2 + Cu°

Anode Cathodeelectrode

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 CathodeFe°  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°