Chapter%2012%20Electrochemistry0

Chapter%2012%20Electrochemistry0 - Chapter 12...

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Chapter 12: Electrochemistry I. Electrochemistry A. the use of spontaneous chemical reactions to produce electricity B. the use of electricity to drive nonspontaneous reactions forward C. Half Reactions Oxidation Pneumonic Device a. Loss of electrons (e - ) O xidation b. Gain of Oxygen I s c. Loss of 2H + L osing Reduction a. Gain of electrons R eduction b. Loss of Oxygen I s c. Gain of 2H + G aining II. Balancing Half Reactions A. Oxidation States An element is always zero [ex: F (s), Ag (s)] Diatomic molecules are always zero [ex: I 2 , F 2 , Cl 2 , O 2 , H 2 , N 2 , Br 2 ] Flouride is always -1 Oxygen is always -2 (unless w/ peroxide H 2 O 2 ) Hydrogen is always +1 (unless connected to a metal or metalloid) [ex: AlH 3 H=-1] B. Acidic solution Step 1: Assign oxidation numbers Step 2: Break into half reactions Step 3: Balance nonhydrogen/oxygen atoms Step 4: Balance oxygen with H 2 O Step 5: Balance hydrogen with H + Step 6: Balance electrons (e - ) Step 7: Combine both half reactions o Flip one half reactions if necessary to cancel e - C. Basic solution Step 1: Assign oxidation numbers Step 2: Break into half reactions Step 3: Balance nonhydrogen/oxygen atoms Step 4: Balance oxygen with H 2 O Step 5: Balance hydrogen by adding 1 H 2 O per hydrogen o Double H 2 O on the other side o Be careful not to add the same amount of H 2 O Step 6: Add the same amount of OH - as H 2 O to the other side Step 7: Balance electrons (e - ) Step 8: Combine both half reactions D. Acidic solution 1
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Ex1: MnO 4 - + Fe 2+ Mn 2+ + Fe 3+ Step 1: Assign oxidation numbers +7 -8 +2 +2 +3 MnO 4 - + Fe 2+ Mn 2+ + Fe 3+ Step 2: Break into half reactions Step 3: Balance nonhydrogen/oxygen atoms MnO 4 - Mn 2+ Fe 2+ Fe 3+ Step 4: Balance oxygen with H 2 O MnO 4 - Mn 2+ + 4H 2 O Fe 2+ Fe 3+ Step 5: Balance hydrogen with H + 8H + + MnO 4 - Mn 2+ + 4H 2 O Fe 2+ Fe 3+ Step 6: Balance electrons (e - ) 5e - + 8H + + MnO 4 - Mn 2+ + 4H 2 O Fe 2+ Fe 3+ e - Step 7: Combine both half reactions Flip one half reaction to cancel e - 5e - + 8H + + MnO 4 - Mn 2+ + 4H 2 O 5 x [Fe 2+ Fe 3+ + e - ] 8H + + MnO 4 - + 5Fe 2+ 5 Fe 3+ + Mn 2+ + 4H 2 O E. Basic Solution Ex2: MnO 4 - + Br - MnO 2 + BrO 3 - Step 1: Assign oxidation numbers
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This note was uploaded on 03/12/2010 for the course CHEM6B Chem 6B taught by Professor Johnson during the Spring '10 term at UCSD.

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Chapter%2012%20Electrochemistry0 - Chapter 12...

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