Lecture15, 2-13-08 Nernst Equation

Lecture15, 2-13-08 Nernst Equation - Cell EMF Cell EMF...

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Unformatted text preview: Cell EMF Cell EMF Oxidizing and Reducing Agents Oxidizing and Reducing Agents Cell Potential Determine the cell potential and the spontaneous reaction for a voltaic cell with electrodes Cu 2+ /Cu and Fe 3+ /Fe 2+ Fe +3 (aq) + e- Fe +2 (aq) E = 0.77 V Cu +2 (aq)+2e- Cu(s) E = 0.34 V Cu(s) Cu +2 (aq)+2e- E = -0.34 V 2Fe +3 (aq) + 2e- 2Fe +2 (aq) E = 0.77 V Cu(s) + 2Fe 3+ Cu 2+ + 2Fe 2+ E o cell = 0.43V Line Notation Cu(s) + Fe +3 (aq) Cu +2 (aq) + Fe +2 (aq) s o l i d Aqueous Aqueous solid Anode on the left Cathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are aqueous, a platinum electrode is indicated. Cu(s) Cu +2 (aq) Fe +2 (aq),Fe +3 (aq) Pt(s) Line Notation Galvanic Cell Zn Zn 2+ (anode) porous plate Cu 2+ Zn 2+ Cu 2+ Cu (cathode) e- 1.10 V Start Here End Here Tell the story. What you see during your trip from the anode to the cathode? Zn(s) Zn 2+ (aq, 1M) Cu 2+ (aq, 1M) Cu(s) 1.00 M 1.00 M Practice Completely describe the voltaic cell based on the following half-reactions under standard conditions. MnO 4- + 8 H + +5e- Mn +2 + 4H 2 O E = 1.51 Fe +3 +3e- Fe(s) E = 0.036V Spontaneity of Redox Reactions Spontaneity of Redox Reactions Consider a silver-nickel voltaic cell E 1/2 (Ag + /Ag) = 0.80 V; E 1/2 (Ni 2+ /Ni) = -0.28 V E (cell) = E 1/2 (Ag + /Ag) -...
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Lecture15, 2-13-08 Nernst Equation - Cell EMF Cell EMF...

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