Lecture+k - Ch 12 - Oxidation-Reduction recap oxidation...

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Ch 12 - Oxidation-Reduction recap Redox reactions Exchange of electrons (or atoms) between species e reduced in reaction oxidized in reaction oxidation agent (oxidant) reducing agent (reductant) Standard reduction potentials E o red For a half reaction To balance redox reactions, balance atoms, electrons, protons/hydroxides, calculate reaction potential (E o cell = E o anode + E o cathode ) 0.03 V salt bridge >> > < ANODE CATHODE Pt(s) Ag(s) Fe 2+ /Fe 3+ Ag + Pt | Fe2+, Fe3+ || Ag+ | Ag E cell º= E Fe2+ Fe3+ º+ E Ag+ Ag º = 0.77 V + 0.80 V = 0.03 V
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Reduction Potential Conventions 1. Standard conditions for electrochemical half-reactions are the same as in thermodynamics: 25 ° C, 1 atm for gases and 1 M for solutions 2. Half-reactions are as reductions in standard half-cell potentials E º. E º are measured relative to that for the reduction of H + (aq) to H 2 (g) which is defined a 0 V. Substances easier to reduce than H + (aq) have positive values of E º while substances harder to reduce have negative values. 3. In a galvanic cell the more positive E º occurs as a reduction at the cathode and the more negative E º occurs as an oxidation at the anode. 4. Therefore, the standard potential for a galvanic cell E cell º = the more positive standard E º minus the more negative E º. A positive value of E cell º indicates the reaction is spontaneous. 5. Consistent with 4 and 5, the standard cell potential is E cell º= E canode º- E cathode º 6. The more positive the value of the E º the more likely the substance is an oxidizing agent (and thus reduced upon reaction), while the more negative the value of the E º the more likely the substance is a reducing agent (and thus oxidized upon reaction).
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Cell Notation Question: Enter the balanced chemical equation that describes the electrochemical cell that is represented by the cell notation as shown. Fe (s) | Fe 3+ (aq) || Sn 4+ (aq), Sn 2+ (aq) | Pt (s) Solution: The reaction to the left of the double line represents the anode of the cell. At the anode an oxidation is occurring. Write the oxidation half reaction. Fe (s) Fe 3+ + 3 e - The cathode is represented on the right side of the equation. Reduction occurs at the cathode. Write the half reaction for the reduction. Sn 4+ + 2 e - Sn 2+ Use the elements in the above two reactions to write an unbalanced equation. Fe (s) + Sn 4+ (aq) Fe 3+ (aq) + Sn 2+ (aq) Balance the equation to ensure that the charges are equal on both sides of the equation. 2 Fe (s) + 3 Sn 4+ (aq) 2 Fe 3+ (aq) + 3 Sn 2+ (aq)
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Cell Potential (Rx potential) Question: The standard emf (electromotive force) of a galvanic cell described below is 1.03. Determine the value (V) of the standard half cell potential for the AuCl4-/Au couple. (Standard reduction potential Fe 3+ Fe, -0.04 V) Fe(s) + AuCl 4 - (aq) Fe 3+
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This note was uploaded on 11/04/2009 for the course CHEM Chem 1C taught by Professor Farmer during the Spring '09 term at UCL.

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Lecture+k - Ch 12 - Oxidation-Reduction recap oxidation...

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