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Electrochem Homework. 3, 15, 17, 23, 27, 29, 35, 37, 41, 44, 49, 51, 53, 59, 63, 89, 91, 97a, 99, 117. Oxidation Numbers. # of e- gained or lost when forming a compound If ion, actual charge Use the Periodic Table!. General Rules.
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Electrochem Homework 3, 15, 17, 23, 27, 29, 35, 37, 41, 44, 49, 51, 53, 59, 63, 89, 91, 97a, 99, 117
Oxidation Numbers • # of e- gained or lost when forming a compound • If ion, actual charge • Use the Periodic Table!
General Rules • Free uncombined state of atoms -> 0 Ex. Na, K, S8 , Cl2 • Sum of oxidation #’s in neutral compound is zero • Polyatomic ions, oxidation #’s add up to the charge on the ion • F has a -1 ox. state
General Rules Continued • H usually has an ox. state of +1 Exception: NaH where H is -1 • O usually has an ox. State of -2 Exceptions: O-1 and O-1/2 • Use the Periodic Table to find the location of the element
Examples • HNO3 • NO3-1
New Method # of valence electrons - # of unshared electrons Midway number - # of bonding electrons assigned to it* oxidation number * Higher electronegative atom gets the e-
Examples • NH3 • CH3OH
Why do we need a new method? • SO4-2 vs. S2O3-2
Electrochemistry • Study of the relationship between chemical change and electrical work • Application of thermodynamics in everyday applications Ex. Radios, iPods, calculators • Battery houses a spontaneous chemical reaction that releases free energy to produce electricity
Electrochemistry cont. • 2nd type of reaction: Nonspontaneous and absorbs free energy from an external source of electricity Ex. Electroplating, recovery of metals from ores
Electrochemistry cont. • Amount of energy (electrical) consumed or produced in chemical reactions can be accurately measured • All electrochemical reactions involve the transfer of electrons Redox reactions! • The electrochemical cell contains the reacting system
Electrochemistry cont. • Sites of oxidation and reduction separated physically so that reduction occurs at one electrode and oxidation occurs at the other electrode Red Cat and An Ox
Examples of Redox reactions 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
Balancing redox reactions • Write the unbalanced equation • Assign oxidation numbers • Choose coefficients that make the transfer of electrons the same • Balance the charge • Balance remaining elements if necessary
Balancing In Acidic Media • Add H+ and H2O to balance charge Ex. Cr2O7-2 + I- Cr+3 + IO3-
Balancing In Basic Media • Balance charge by adding OH- and H2O Ex. Cr2O7-2 + I- Cr+3 + IO3-
Balance in Acid • MnO4- + Br- MnO2 + BrO2-1
Balance in Base (Drano) • Al + H2O [Al(OH)4]-+ H2
Balance in Acid Cu + NO3- Cu+2 + NO
Half Rxn Method for balancing redox equations • Overall reaction split into two half reactions • Separates oxidation and reduction half reactions • Easy to balance redox reactions occurring in acidic or basic solutions
How? 1. Divide equation into two half reactions 2. Balance the atoms and charges in each half reaction - e- added to left is reduction - e- added to right is oxidation 3. Multiply by integer so e- lost = e- gained 4. Add balanced equations
Ex. Balance in acidic media Ex. Cr2O7-2 + I- Cr+3 + I2
Ex. Balance in basic media MnO4-1 + C2O4-2 MnO2 + CO3-2
Breathalyzer reaction Cr2O7-2+ C2H5OH + H+ Cr+3 + C2H4O + H2O
Equivalents • Acids and Bases Just deal with the number of H+ or OH- being donated Ex. HCl -> 1 equivalent H3PO4 -> 3 equivalents NaOH -> 1 equivalent Ca(OH)2 -> 2 equivalents
Acid Reaction H2SO4 + H2O 2 H+ + SO4-2 1 mol 2 mol 1 mol 98.0 g/mol 2.0 g/mol 96.0 g/mol 2 equivalents 1 equivalent HNO3 – 63.01 g 1 equiv. CH3COOH – 60.03 g 1 equiv. H2SO4 - 49.04 g 1 equivalent of Ca(OH)2 - 37.05 g
Normality • Molarity = moles of solute Liters of solution • Normality = equivalents of solute Liters of solution • N = M x #equivalents
Equivalents and Redox • Not only do equivalents pertain to # H+ or # OH- but also to electrons • Equivalent – mass of oxidizing or reducing substance that gains or loses 1 mole of electrons • 1 mole of electrons equals 1 equivalent of electrons
Equivalents • One equiv. oxidized reacts with one equiv. reduced 1 eq. ox. = 1 eq. red. • If redox equation is balanced – mole ratios can be used in stoichiometry • If NOT balanced – equivalents can be used