Lecture 9

Lecture 9 - Summary • Red ­Ox reac/ons involve transfer of electrons between reac/ng species;(reducing agent transfer electron

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Unformatted text preview: Summary • Red ­Ox reac/ons involve transfer of electrons between reac/ng species; (reducing agent transfer electron to oxidizing agent). • Flow of electron is associated with electrical work. • The redox reac/on can be separated into half reac/ons; the oxida/on half ­ reac/on and the reduc/on half reac/on. • Electrical energy can be extracted through an electrochemical cell where the two half reac/ons are separated. • In voltaic cells, spontaneous reac/ons generate electrical energy. • Output of electrical energy in a voltaic cell is propor/onal to the difference in the electrical poten/al of two half reac/ons (electrodes). This difference is known as electrical (cell) poten/al or electromo/ve force • Ecell > 0 for a spontaneous process • When electrochemical reac/ons occur at the standard states, the poten/al difference is referred to as standard cell poten/al (E0cell). Poten/al at each half cell at standard state is called the standard electrode poten/al (E0half ­cell) •  E0half ­cell values for different half ­cell are determined with respect to a standard reference half cell. This reference half cell poten/al at standard state is fixed as 0. • Rela/ve strengths of oxidizing and reducing agents can be determined from the list of E0 half cell values. More posi/ve the E0 value stronger it is as an oxidizing agent. More nega/ve is the E0 value more stronger it is as an reducing agent. •  Change in nega/ve free energy is propor/onal to posi/ve Ecell poten/al ΔG = -nFEcell ΔGo = -nFEocell Eocell = -RT/nF (ln K) Nernst Equa/on E = E° − RT nF ln Q Figure 21.11 The relation between Ecell and log Q for the zinc-copper cell. Concentra/on Cells •  The Nernst equa/on implies that a cell could be created that has the same substance at both electrodes. ° •  For such a cell, Ecell would be 0, but Q would not. •  Therefore, as long as the concentra/ons are different, E will not be 0. Figure 21.12 A concentration cell based on the Cu/Cu2+ half-reaction. Sample Problem 21.7 PROBLEM: Calculating the Potential of a Concentration Cell A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0.010 M AgNO3; in half-cell B, electrode B dips into 4.0 x 10-4 M AgNO3. What is the cell potential at 298.15 K? Which electrode has a positive charge? Figure 21.13 The laboratory measurement of pH: Example of a concentration cell Electrochemical Processes in Batteries Alkaline battery Silver button battery Lead-acid battery Nickel-metal hydride battery Lithium-ion battery Hydrogen fuel cell. The corrosion of iron The effect of metal-metal contact on the corrosion of iron Faster corrosion Corrosion •  Rus/ng  ­ spontaneous oxida/on. •  Most structural metals have reduc/on poten/als that are less posi/ve than O2 . •  Fe → Fe+2 +2e ­ Eº= 0.44 V •  O2 + 2H2O + 4e ­ → 4OH ­ Eº= 0.40 V •  Fe+2 + O2 + H2O → Fe2O3 + H+ •  Reac/ons happens in two places. Preven/ng Corrosion •  Coa/ng to keep out air and water. •  Galvanizing  ­ Pu]ng on a zinc coat •  Has a lower reduc/on poten/al, so it is more easily oxidized. •  Alloying with metals that form oxide coats. •  Cathodic Protec/on  ­ A_aching large pieces of an ac/ve metal like magnesium that get oxidized instead. ...
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This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.

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