Chemistry Week 11

Chemistry Week 11 - Chemical Processes and Reactions...

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Chemical Processes and Reactions TEXTBOOK READING : BLB-10 , Chapter 20. 1-2. Practice PROBLEMS: (Ch 20) 18, 20 (b) Reduction/Oxidation (Redox) Reactions – electrons are transferred away from one species (undergoing oxidation) to another species (undergoing reduction). To identify reactions as redox reactions, we must evaluate oxidation states of elements in chemical reactions. Rules for Determining Oxidation States (Purely “ionic” treatment; use electronegativities…) (1) = 0 for elements, e.g., C(s), O 2 (g), N 2 (g), etc. (2) = +1 for alkali metals (Li, Na, K, Rb, Cs) in compounds (unusual exception: there are reports of Cs encapsulated by large cages in nonaqueous solvents – J. Dye, Michigan St.) (3) = +2 for alkaline-earth metals (Be, Mg, Ca, Sr, Ba) in compounds (4) = 1 for F in all compounds (no exceptions here) (5) = 2 for O in compounds, except for peroxides (O 2 2 : = 1) and superperoxides (O 2 : = 1/2) (6) = 1 for halogens (Cl, Br, I) in compounds, except when combined with F or O (7) = +1 for hydrogen combined with nonmetals; 1 for hydrogen combined with metals, e.g. +1 in H 2 O, NH 3 , H 2 S, etc.; 1 in LiH, LiAlH 4 , etc. (8) Sum of oxidation states in chemical formula = total charge. Examples of Redox reactions (we will focus on aqueous solutions): Co 3+ (aq) + Sn 2+ (aq) ⎯→ Co 2+ (aq) + Sn 4+ (aq) Oxidation States : Reactants -- Co = +3, Sn = +2 Products -- Co = +2, Sn = +4 Therefore, Sn loses electrons (oxidation); Co gains electrons (reduction). During this reaction, the ions change oxidation states. As the chemical equation is written, mass balance is satisfied, but charge balance is not. To understand how such redox reactions will take place, it is important to make sure such reactions are balanced for mass and charge and that we identify which species loses electrons and which species gains electrons. We use the method of half-reactions : REDUCTION : Co 3+ (aq) + e ⎯→ Co 2+ (aq) (gains electrons) OXIDATION : Sn 2+ (aq) ⎯→ Sn 4+ (aq) + 2e (loses electrons) To balance this reaction, we can add these two half-reactions, but we must eliminate the “free electrons” as these do not enter the final balanced chemical equation. To do this, we must multiply the REDUCTION half-reaction by 2. The final reaction, therefore, is
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2 Co 3+ (aq) + Sn 2+ (aq) ⎯→ 2 Co 2+ (aq) + Sn 4+ (aq) Since we consider Redox reactions in aqueous media, we must also consider the effect of pH, i.e., whether a reaction occurs under acidic ([H + ] > [OH ]) or basic ([H + ] < [OH ]) conditions: Acidic Conditions : use H + (aq) to obtain charge balance; then use H 2 O(l) for mass balance Cu(s) + NO 3 (aq) ⎯→ Cu 2+ (aq) + NO 2 (g) Oxidation States : Reactants -- Cu = 0, O = 2, N = +5 Products -- Cu = +2, O = 2, N = +4 Therefore, Cu loses electrons (oxidation); N gains electrons (reduction) REDUCTION : NO 3 (aq) + e ⎯→ NO 2 (g) 2 H + (aq) + NO 3 (aq) + e ⎯→ NO 2 (g) (charge balanced) 2 H + (aq) + NO 3 (aq) + e ⎯→ NO 2 (g) + H 2 O(l) (mass balanced) 4 H + (aq) + 2 NO 3 (aq) + 2 e ⎯→ 2 NO 2 (g) + 2 H 2 O(l) OXIDATION : Cu(s) ⎯→ Cu 2+ (aq) + 2 e To combine these two half-reactions, we must multiply the REDUCTION half-reaction by 2, before adding: Cu(s) + 2 NO 3 (aq) + 4 H + (aq) ⎯→ Cu 2+
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This note was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

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Chemistry Week 11 - Chemical Processes and Reactions...

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