Electrochemistry Batteries. Batteries Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal.

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Presentation transcript:

Electrochemistry Batteries

Batteries Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal grid in sulfuric acid: PbO 2 (s) + SO 4 2- (aq) + 4H + (aq) + 2e -  PbSO 4 (s) + 2H 2 O(l) Anode: Pb: Pb(s) + SO 4 2- (aq)  PbSO 4 (s) + 2e -

Batteries Lead-Acid Battery The overall electrochemical reaction is PbO 2 (s) + Pb(s) + 2SO 4 2- (aq) + 4H + (aq)  2PbSO 4 (s) + 2H 2 O(l) for which E  cell = E  red (cathode) - E  red (anode) = ( V) - ( V) = V. Wood or glass-fiber spacers are used to prevent the electrodes form touching.

Batteries Lead-Acid Battery

Batteries Alkaline Battery Anode: Zn cap: Zn(s)  Zn 2+ (aq) + 2e - Cathode: MnO 2, NH 4 Cl and C paste: 2NH 4 + (aq) + 2MnO 2 (s) + 2e -  Mn 2 O 3 (s) + 2NH 3 (aq) + 2H 2 O(l) The graphite rod in the center is an inert cathode. For an alkaline battery, NH 4 Cl is replaced with KOH. Anode: Zn powder mixed in a gel: Zn(s)  Zn 2+ (aq) + 2e - Cathode: reduction of MnO 2.

Batteries Alkaline Battery

Batteries Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. On Apollo moon flights, the H 2 -O 2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen: 2H 2 O(l) + O 2 (g) + 4e -  4OH - (aq) Anode: 2H 2 (g) + 4OH - (aq)  4H 2 O(l) + 4e -

Batteries Fuel Cells

Corrosion Corrosion of Iron Since E  red (Fe 2+ ) < E  red (O 2 ) iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l). Anode: Fe(s)  Fe 2+ (aq) + 2e -. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3.xH 2 O(s). Oxidation occurs at the site with the greatest concentration of O 2.

Corrosion Corrosion of Iron

Corrosion Preventing the Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc. Zinc protects the iron since Zn is the anode and Fe the cathode: Zn 2+ (aq) +2e -  Zn(s), E  red = V Fe 2+ (aq) + 2e -  Fe(s), E  red = V With the above standard reduction potentials, Zn is easier to oxidize than Fe.

Corrosion Preventing the Corrosion of Iron

Corrosion To protect underground pipelines, a sacrificial anode is added. The water pipe is turned into the cathode and an active metal is used as the anode. Often, Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg(s), E  red = V Fe 2+ (aq) + 2e -  Fe(s), E  red = V

Corrosion Preventing the Corrosion of Iron

Electrolysis Electrolysis of Aqueous Solutions Nonspontaneous reactions require an external current in order to force the reaction to proceed. Electrolysis reactions are nonspontaneous. In voltaic and electrolytic cells: –reduction occurs at the cathode, and –oxidation occurs at the anode. –However, in electrolytic cells, electrons are forced to flow from the anode to cathode. –In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.)

Electrolysis Electrolysis of Aqueous Solutions

Electrolysis Example, decomposition of molten NaCl. Cathode: 2Na + (l) + 2e -  2Na(l) Anode: 2Cl - (l)  Cl 2 (g) + 2e -. Industrially, electrolysis is used to produce metals like Al. Electrolysis with Active Electrodes Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating.

Electrolysis Electrolysis with Active Electrodes

Electrolysis Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO 4 : Anode: Ni(s)  Ni 2+ (aq) + 2e - Cathode: Ni 2+ (aq) + 2e -  Ni(s). Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

Electrolysis Quantitative Aspects of Electrolysis We want to know how much material we obtain with electrolysis. Consider the reduction of Cu 2+ to Cu. –Cu 2+ (aq) + 2e -  Cu(s). –2 mol of electrons will plate 1 mol of Cu. –The charge of 1 mol of electrons is 96,500 C (1 F). –Since Q = It, the amount of Cu can be calculated from the current (I) and time (t) taken to plate.

Electrolysis Electrical Work Free-energy is a measure of the maximum amount of useful work that can be obtained from a system. We know If work is negative, then work is performed by the system and E is positive. The emf can be thought about as a measure of the driving force for a redox process.

Electrolysis Electrical Work In an electrolytic cell and external source of energy is required to force the reaction to proceed. In order to drive the nonspontaneous reaction the external emf must be greater than E cell. From physics: work has units watts: 1 W = 1 J/s. Electric utilities use units of kilowatt-hours: