Electrochemistry (Ch 17) - AP Chemistry

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4/1/2011 GALVANIC CELLS • Consider a galvanic cell based on the reaction Al 3+ (aq) + Mg(s) Al(s) + Mg 2+ (aq) – The half-reactions are • Al 3+ + 3 e - Al E° = -1.66 V • Mg 2+ + 2 e - Mg E° = -2.37 V – Give the balanced cell reaction and calculate E° for the cell. • 2 Al 3+ (aq) + 3 Mg(s) 2 Al(s) + 3 Mg 2+ (aq) • E° cell = 0.71V LINE NOTATION • Line notation is handy for representing electrochemical cells. – The anode components are listed on the left and the cathode components are listed on the right, separated by double vertical lines (indicating the salt bridge or porous disk). • For example, the cell on the previous slide would be: Mg(s) | Mg 2+ (aq) || Al 3+ (aq) | Al(s) – Phase difference (boundary) is indicated by a single vertical line • Notice that the substance constituting the anode is far left and cathode is far right – Commas are used to separate ionic species in the same solution Pt(s)|ClO 3- (aq), ClO 4- (aq), H + (aq) ||H + (aq), MnO 4- (aq), Mn 2+ (aq)|Pt(s) Ch 17 - Electrochemistry 19 Ch 17 - Electrochemistry 20 COMPLETE DESCRIPTION OF A GALVANIC CELL • NOTE: A cell will always run spontaneously in the direction that produces a positive cell potential (E° cell ) • A complete description of a galvanic cell includes: – The cell potential (always positive for a galvanic cell where E° cell = E° cat – E° an ) 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 . – Designation of the anode and cathode – The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half-reaction is a conducting solid. DESCRIPTION OF A GALVANIC CELL • Describe completely the galvanic cell based on the following halfreactions under standard conditions: Ag + + e - Ag Fe 3+ + e - Fe 2+ E° = 0.80 V E° = 0.77 V – Ag + (aq) + Fe 2+ (aq) Fe 3+ (aq) + Ag(s) E° cell = 0.03 V – Electrons flow from Fe 2+ to Ag + – Anode: Fe 2+ compartment Cathode: Ag + compartment – Pt(s)|Fe 2+ (aq), Fe 3+ (aq)||Ag + (aq)|Ag(s) Ch 17 - Electrochemistry 21 Ch 17 - Electrochemistry 22 CELL POTENTIAL, ELECTRICAL WORK, AND FREE ENERGY Section 17.3 CELL POTENTIAL AND WORK • The work that can be done by the electrons transferred through a wire depends on the “push” (thermodynamics) behind the electrons. – This electromotive force (emf) is defined in terms of a potential difference (in volts) between two points in the circuit (1 V = 1 J/C) – Cell potential (E ) and work (w) have opposite signs: w max = –qE max • In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum Ch 17 - Electrochemistry 23 Ch 17 - Electrochemistry 24 4
4/1/2011 CELL POTENTIAL AND WORK • Suppose a certain galvanic cell has a maximum potential (at zero current) of 2.50 V – In a particular experiment, 1.33 moles of electrons were passed through this cell at an average actual potential of 2.10 V. The actual work done is: w = -qE • 1 mole of electrons has a charge of 96,485 C (called a “faraday”, F = 96,485 C/mol) – Actual work: w = -(1.33 mol)(96485 C/mol)(2.10 J/C) = -2.69 x 10 5 J – Maximum work: w max = - (1.33)(96485)(2.50) = -3.21 x 10 5 J – The efficiency of this cell is: WORK AND FREE ENERGY • Since w max = ∆G – For a galvanic cell, w max = -qE max = ∆G – Since q = nF – Then, ∆G = -qE max = -nFE max – Or ∆G = -nFE – For standard conditions: ∆G° = -nFE° Ch 17 - Electrochemistry 25 Ch 17 - Electrochemistry 26 ∆G = -nFE° • This equation states that the maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. – Provides an experimental means to obtain ∆G (as you will do in LAB) – Confirms that a galvanic cell will run in the direction that gives a positive value for E cell (positive E cell corresponds to a negative ∆G ) CALCULATING ∆G° • Using the Standard Reduction Potentials Table, calculate ∆G° for the reaction: • Is this reaction spontaneous Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq) – E cell = 0.78 V – ∆G = -1.5 x 10 5 J (SPONTANEOUS) Ch 17 - Electrochemistry 27 Ch 17 - Electrochemistry 28 PREDICTING SPONTANEITY • Using the Standard Reduction Potentials Table, predict whether 1 M HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution. DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION Section 17.4 – E° cell = -0.54 V (not spontaneous under standard conditions) – General rule: Oxidizing agents (reactants) higher on the list will oxidize reducing agents (products) lower on the list • Ex: H + will react with Fe but not Ag • The best REDUCING AGENT is Li, the best OXIDIZING AGENT is F 2 Ch 17 - Electrochemistry 29 Ch 17 - Electrochemistry 30 5
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Electrochemistry (Ch 17) - AP Chemistry

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