Gián án Chapter 18 Electrochemistry

Chapter 18 Electrochemistry 2011, NKMB Co., Ltd. Chemistry, Julia Burdge, 2 st Ed. McGraw Hill. Mr. Truong Minh Chien ; losedtales@yahoo.com http://tailieu.vn/losedtales http://mba-programming.blogspot.com 2 Redox Reaction • one or more elements change oxidation number  all single displacement, and combustion,  some synthesis and decomposition • always have both oxidation and reduction  split reaction into oxidation half-reaction and a reduction half-reaction • aka electron transfer reactions  half-reactions include electrons • oxidizing agent is reactant molecule that causes oxidation  contains element reduced • reducing agent is reactant molecule that causes reduction  contains the element oxidized Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 3 Oxidation & Reduction • oxidation is the process that occurs when  oxidation number of an element increases  element loses electrons  compound adds oxygen  compound loses hydrogen  half-reaction has electrons as products • reduction is the process that occurs when  oxidation number of an element decreases  element gains electrons  compound loses oxygen  compound gains hydrogen  half-reactions have electrons as reactants Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 4 Rules for Assigning Oxidation States • rules are in order of priority 1. free elements have an oxidation state = 0  Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 1. monatomic ions have an oxidation state equal to their charge  Na = +1 and Cl = -1 in NaCl 1. (a) the sum of the oxidation states of all the atoms in a compound is 0  Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0 Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 5 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion  N = +5 and O = -2 in NO 3 – , (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds  Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds  Mg = +2 in MgCl 2 Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 6 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below  nonmetals higher on the table take priority Nonmetal Oxidation State Example F -1 CF 4 H +1 CH 4 O -2 CO 2 Group 7A -1 CCl 4 Group 6A -2 CS 2 Group 5A -3 NH 3 Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 7 Oxidation and Reduction • oxidation occurs when an atom’s oxidation state increases during a reaction • reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 8 Oxidation–Reduction • oxidation and reduction must occur simultaneously  if an atom loses electrons another atom must take them • the reactant that reduces an element in another reactant is called the reducing agent  the reducing agent contains the element that is oxidized • the reactant that oxidizes an element in another reactant is called the oxidizing agent  the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 9 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H + → 3 S + 2 NO + 4 H 2 O MnO 2 + 4 HBr → MnBr 2 + Br 2 + 2 H 2 O Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 10 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H + → 3 S + 2 NO + 4 H 2 O MnO 2 + 4 HBr → MnBr 2 + Br 2 + 2 H 2 O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 ox agred ag +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction oxidation reduction red agox ag Chemistry, Julia Burdge, 2 nd e., McGraw Hill. [...]... half-reactions 7) check Tro, Chemistry: A Molecular Approach 13 Ex 18. 3 – Balance the equation: I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s) in basic solution I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s) Assign Oxidation States Separate into halfreactions ox: I−(aq) → I2(aq) red: MnO − → MnO red: 4 (aq) 2(s) Tro, Chemistry: A Molecular Approach 14 Ex 18. 3 – Balance the equation: I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s)... OH− (aq) in base, then H2O neutralize adding H+ 4 H2O(aq) + MnO4−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−(aq) the H+ with OHMnO4−(aq) + 2 H2O(l) → MnO2(s) + 4 OH−(aq) Tro, Chemistry: A Molecular Approach 15 Ex 18. 3 – Balance the equation: I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s) in basic solution Balance ox: 2 I−(aq) → I2(aq) + 2 e− Halfred: MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−(aq) reactions by charge... e− } x3 red: MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−(aq) }x2 ox: 6 I−(aq) → 3 I2(aq) + 6 e− red: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−(aq) Tro, Chemistry: A Molecular Approach 16 Ex 18. 3 – Balance the equation: I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s) in basic solution Add the ox: 6 I−(aq) → 3 I2(aq) + 6 e− Halfred: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−(aq) reactions − −... Electric Current Flowing Directly Between Atoms Tro, Chemistry: A Molecular Approach 24 Electric Current Flowing Indirectly Between Atoms Tro, Chemistry: A Molecular Approach 25 Electrochemical Cells • electrochemistry is the study of redox reactions that produce or require an electric current • the conversion between chemical energy and electrical energy is carried out in an electrochemical cell • spontaneous... 29 Current and Voltage • the number of electrons that flow through the system per second is the current  unit = Ampere  1 A of current = 1 Coulomb of charge flowing by each second  1 A = 6.242 x 1 018 electrons/second  Electrode surface area dictates the number of electrons that can flow • the difference in potential energy between the reactants and products is the potential difference  unit = . Chapter 18 Electrochemistry 2011, NKMB Co., Ltd. Chemistry, Julia Burdge, 2 st Ed 6) add half-reactions 7) check Tro, Chemistry: A Molecular Approach 14 Ex 18. 3 – Balance the equation: I − (aq) + MnO 4 − (aq) → I 2(aq) + MnO 2(s) in