Ch14-TCF

Ch14-TCF - 1 Ch. 14 Fundamentals of Electrochemistry...

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1 Ch. 14 Fundamentals of Electrochemistry • Electrochemistry (corrosion, galvanic and electrolytic cells) • Electrochemical deposition processes; e.g ., electroplating • Electrochemical energy systems (batteries, fuel cells, photovoltaics). • Electrochemical analytical methods , e.g., potentiometry and voltammetry • Advanced sensors, e.g., chem-FET s • Modern approaches, e.g., ultramicroelectrodes, modified electrodes, and scanning probe methods
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2 Basic Concepts I. Definitions Redox ( red uction- ox idation) reactions involve the transfer of electrons from one species to another oxidation : removal of electrons reduction : addition of electrons oxidizing agent (oxidant) takes electrons ( i.e. is reduced) reducing agent (reductant) gives electrons ( i.e. is oxidized)
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3 Balancing redox reactions Example 1: (See Appendix D) (1) Write the unbalanced redox reaction. H 2 O 2 + Fe 2+ Fe 3+ + H 2 O (in acid) (2) Separate the reaction into its two half reactions (don’t worry about electrons yet). H 2 O 2 H 2 O (reduction) Fe 2+ Fe 3+ (oxidation)
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4 (3) Balance the atoms other than H and O in each half- reaction. H 2 O 2 H 2 O (nothing to balance) Fe 2+ Fe 3+ (already balanced) (4) Add H 2 O to balance O and then add H + to balance H. H 2 O 2 + 2H + 2H 2 O Fe 2+ Fe 3+ (5) Add e to each half-reaction to balance the charge. 2e + H 2 O 2 + 2H + 2H 2 O Fe 2+ Fe 3+ + e
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5 (6) Multiply one or both half-reactions by whole numbers to make the number of electrons equal. 2e + H 2 O 2 + 2H + 2H 2 O (reduction) 2(Fe 2+ Fe 3+ + e ) (oxidation) (7) Add the two half-reactions together and cancel the electrons and any other fragments. H 2 O 2 + 2Fe 2+ + 2H + 2Fe 3+ + 2H 2 O (total reaction) (8) Verify that the atoms balance and the charges balance. (9) If in basic solution, use OH to cancel any H + .
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6 Example 2: (1) Write the unbalanced redox reaction. Bi(OH) 3 + SnO 2 2 Bi 0 + SnO 3 2 (in base) (2) Separate the reaction into its two half reactions (don’t worry about electrons yet). Bi(OH) 3 Bi (reduction) SnO 2 2 SnO 3 2 (oxidation)
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7 (3) Balance the atoms other than H and O in each half- reaction. Bi(OH) 3 Bi (nothing to balance) SnO 2 2 SnO 3 2 (nothing to balance) (4) Add H 2 O to balance O . .. Bi(OH) 3 Bi + 3H 2 O H 2 O + SnO 2 2 SnO 3 2 … and then add H + to balance H 3H + + Bi(OH) 3 Bi + 3H 2 O H 2 O + SnO 2 2 SnO 3 2 + 2H +
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8 (5) Add e to each half-reaction to balance the charge. 3e + 3H + + Bi(OH) 3 Bi + 3H 2 O (reduction) H 2 O + SnO 2 2 SnO 3 2 + 2H + + 2e (oxidation) (6) Multiply one or both half-reactions by whole numbers to make the number of electrons equal.
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Ch14-TCF - 1 Ch. 14 Fundamentals of Electrochemistry...

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