Experiment 6 Synthesis and Analysis of a Magical Green Crystal Part Deux: Oxalate Content Analysis by Redox Titration Using a Vile Purple Fluid CH 204 Fall 2006 Dr. Brian AndersonBut first… Last week: Synthesis Metal complex coordination compounds Oxidation/Reduction (Redox) reactions Calculating limiting reagent, theoretical yield, and percent yield This week: Oxidation-Reduction (Redox) chemistryWhat is redox chemistry? Moving electrons between different atoms: Cu 2+ (aq) + Z n(s) Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + 2 e Cu(s) Cu 2+ gains electrons. Cu 2+ is REDUCED. Zn(s) 2 e + Zn 2+ (aq) Zn(s) loses electrons. Zn(s) is OXIDIZED.Our redox reaction We will use MnO 4 to oxidize the oxalate ligands surrounding the Fe 3+ . The carbon in the oxalate ions will be oxidized, and the oxalate will change from C 2 O 4 2 to CO 2 (g). MnO 4 – (aq) + C 2 O 4 2– (aq) CO 2 (g) + Mn 2+ (aq) Reduction OxidationBalancing redox reactions Separate the overall equation into two half-reactions. For each half-reaction: 1. Balance the main atom. 2. Add H 2 O to balance O. 3. Add H + to balance H. 4. Balance the charge using electrons. When you’re done, add the two half-reactions and cancel out the electrons. Let’s go to the doc cam and try one...Oxidation half-reaction Oxidation of C 2 O 4 2 to CO 2 is simple enough: (Remember, half-reactions do not include the other reactant) C 2 O 4 2 → 2CO 2 + 2e Reduction half-reaction The oxidizing agent, MnO 4 - , gets reduced to Mn 2+ MnO 4 – → Mn 2+ + ??? Balance Mn Balance O using H 2 O Balance H using H + Balance charge using e –Add the two half reactions First multiply the equations in order to balance out the electrons: C 2 O 4 2 → 2CO 2 + 2e × 5 8H + + MnO 4 + 5e → Mn 2+ + 4H 2 O × 2 5C 2 O 4 2 → 10CO 2 + 10e 16H + + 2MnO 4 + 10e → 2Mn 2+ + 8H 2 O The equation for the overall reaction is: 16H + + 2MnO 4 + 5C 2 O 4 2 → 10CO 2 + 2Mn 2+ + 8H 2 OAlways balance in acidic solution As easy as 1-2-4. 1) Balance the oxidized/reduced atoms 2) Balance oxygens using H 2 O 3) Balance hydrogens using H + 4) Balance charge using e-What if the solution is basic? Here’s what Whitten, Davis, Peck, and Stanley say: To balance in a basic solution, for each O needed (1) Add two OH - to the side needing O and (2) Add one H 2 O to the other side Then, for each H needed, (1) Add one H 2 O to the side needing H and (2) Add one OH - to the other side.Here’s what the current book suggests “In basic solution, balance O by using H 2 O; then balance H by adding H 2 O to the side of each half reaction that needs H and adding OH - to the other side. “When we add . . . OH - . . . → . . . H 2 O . . . to a half-reaction, we are effectively adding one H atom to the right. When we add . . . H 2 O . . . → . . . OH - . . . We are effectively adding one H atom to the left. Note that one H 2 O molecule is added for each H atom needed.”Do it the E-Z way instead Balance the equation in acidic solution, and if it’s supposed to be in basic solution, just add enough OH - to both sides to get rid of all the H + . Just like this...Permanagante is reduced to manganese (IV) oxide in basic solution: MnO 4 → MnO 2 Balance O using H 2 O: MnO 4 → MnO 2 + 2H 2 O Balance H using H + : MnO 4 + 4H + → MnO 2 + 2H 2 O Balance charge using e-: MnO 4 + 3e + 4H + → MnO 2 + 2H 2 O Add one OH - for every H+. Add OH - to both sides! MnO 4 + 3e + 4H + + 4OH → MnO 2 + 2H 2 O + 4OH Combine waters and delete redundant waters: MnO 4 + 3e + 2H 2 O → MnO 2 + 4OH Balancing redox reactions review • Separate the reactants into half reactions. • Balance the main atom. • Balance the half-reactions using H 2 O to balance O, then use H + to balance H. Balance the charge with electrons. • Add the two half-reactions – electrons must cancel. • If necessary, convert acidic solution to basic by adding OH - to both sides and crossing out spectator water molecules.Today: Sample prep and three titrations Land mine! 1:1 mixture of ethanol/water means mix them together in a beaker BEFORE you pour them in! The KMnO 4 solution is already standardized and ready to go. Make sure you record the concentration: 0.0377 M.Quiz next week Balance redox reaction Balance redox reaction in acidic solution Question about Experiment 6 lab
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