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°
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