CHAPTER 17: ELECTROCHEMISTRY. Big Idea 3
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1 CHAPTER 17: ELECTROCHEMISTRY Big Idea 3
2 Electrochemistry Conversion of chemical to electrical energy (discharge). And its reverse (electrolysis). Both subject to entropic caution: Convert reversibly to keep systems at equilibrium and convert all available chemical work (ΔG) to and from the equivalent electrical work (QΔV). Electrons from REDOX reactions.
3 Review: RedOx Equations The e are visible in ½ reactions. 3 H 2 O 2 3 O H e + 2 Au e 2 Au 2 Au H 2 O 2 3 O H Au
4 Galvanic Cell Device in which chemical energy is changed to electrical energy. Uses a spontaneous redox reaction to produce a current that can be used to do work.
5
6 Galvanic Cells One ½ cell rxn. occurs in each compartment. Zn Zn e in the anode. Cu e Cu in cathode. But not without a connection.
7 Ion Salt Bridge But even with a connection of the electrodes, no current flows. We need to allow neutrality in the solutions with a salt bridge to shift counterions.
8 Galvanic Cells & Redox Review: Balancing RedOx Reactions Cr 2 O 2-7 (aq) à 2Cr 3+ (aq) SO 2-3 (aq) à SO 2-4 (aq)
9 Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The pull, or driving force, on the electrons is called the cell potential ( E o ), or the electromotive force (emf) of the cell. Unit of electrical potential is the volt (V). Ø 1 joule of work per coulomb of charge transferred.
10 Galvanic Cells The cell potential (always positive for a galvanic cell where E cell = E (cathode) E (anode) and the balanced cell reaction. The direction of electron flow, obtained by inspecting the half reactions and using the direction that gives a positive E cell.
11 Galvanic Cell All half-reactions are given as reduction processes in standard tables. Table M, 1atm, 25 C When a half-reaction is reversed, the sign of E is reversed. When a half-reaction is multiplied by an integer, E remains the same. A galvanic cell runs spontaneously in the direction that gives a positive value for E cell.
12 Galvanic Cell Potential Example E cell = E (cathode) E (anode) Fe 3+ (aq) + Cu(s) Cu 2+ (aq) + Fe 2+ (aq) Fe 3+ + e Fe 2+ E = 0.77 V Cu e Cu E = 0.34 V
13 Calculating E cell Std Conditions E cell = E reduction + E oxidation Zn(s) + Cu 2+ (aq) à Zn 2+ (aq) + Cu(s)
14 Line Notation Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines which indicated salt bridge or porous disk. The concentration of aqueous solutions should be specified in the notation when known. Example: Mg(s) Mg 2+ (aq) Al 3+ (aq) Al(s) Mg Mg e (anode) Al e Al (cathode)
15 Maximum Cell Potential Directly related to the free energy difference between the reactants and the products in the cell. ΔG = nfe E o cell = standard state cell potential F = Faraday's constant (96,485 C/mole e - ) n = number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell
16 Calculate the ΔG o Calculate the ΔG o for the reaction: Cu 2+ (aq) + Fe (s) à Cu (s) + Fe 2+ (aq)
17 Predicting Spontaneity Predict whether 1 M HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution. à Find the E o cell for the reaction
18 Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted the actual work realized is always less than the calculated maximum.
19 Concentration Cell
20 Nernst Equation The relationship between cell potential and concentrations of cell components At 25 C: E = E log Q n ( ) E = log n K ( )
21 Nernst Equation E cell = cell potential at non-standard state conditions E o cell = standard state cell potential Q = reaction quotient for the reaction. aa + bb à cc + dd,
22 Nernst Equation Example E o cell is 0.48 V for the galvanic cell based on the reaction : 2 Al + 3 Mn 2+ à 2 Al Mn Where the concentrations are as follows, [Mn 2+ ] = 0.50 M, and [Al 3+ ] = 1.50 M E = E log Q n ( )
23 Nernst Example #2 Describe the cell based on the following conditions: VO H + + e - à VO 2+ + H 2 O E o = 1.00V Zn e - à Zn E o = -0.76V
24 Dead Battery A dead battery occurs when the cell has reached equilibrium and there is no longer any chemical driving force to push the electrons through the wire. At equilibrium the two components have the same free energy, ΔG = 0
25 Electrolysis Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
26 Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time quantity of charge moles of electrons moles of analyte grams of analyte
27 Stoichiometry of Electrolysis current and *me quan*ty of charge Coulombs of charge = amps (C/s) seconds (s) quan*ty of charge moles of electrons mol e = Coulombs of charge 1 mol e 96,485 C
28 Stoichiometry of Electrolysis An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate g of the metal from a solution containing M(NO3)3. What is the metal?
29 Commercial Electrolysis Processes Production of aluminum Purification of metals Metal plating Electrolysis of sodium chloride Production of chlorine and sodium hydroxide
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