BalancingRedoxReactions - Balancing Redox Reactions...

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Balancing Redox Reactions (Wiediger) 1. Identify atoms involved in oxidation and reduction by assigning oxidation numbers. 2. Write the two half reactions (one for oxidation, one for reduction). 3. Balance number of atoms involved in redox . 4. Determine electrons for each half reaction. 5. Balance number of electrons between half reactions by multiplying everything in half reaction by appropriate number. 6. Add half reactions and cancel out electrons. If all atoms are balanced in reaction, then you are done. If not, proceed to step 7. (On next page.) Example: Cr 3+ (aq) + Cl aq) Cr (s) + Cl 2 (g) 1. Chromium is being reduced from 3+ to 0 and chlorine is being oxidized from 1– to 0. 2. Cr 3+ (aq) Cr (s) (Reduction) Cl (aq) Cl 2 (g) (Oxidation) 3. Cr 3+ (aq) Cr (s) 2 Cl (aq) Cl 2 (g) (The formation of Cl 2 will require 2Cl .) 4. The one chromium atom is gaining three electrons and the two chlorine atoms are giving off 2 elections (one per chlorine atom). Cr 3+ (aq) + 3e Cr (s) 2Cl (aq) Cl 2 (g) + 2e 5. To have an equal number of electrons in each half reaction, you must scale the reactions up to 6 electrons. To do this, multiply everything in the upper half reaction by 2 and multiply everything in the lower half reaction by 3. 2 Cr 3+ (aq) + 6 e 2 Cr (s) 6 Cl (aq) 3 Cl 2 (g) + 6 e 6. Adding half reactions means just putting everything on the reactants side together and putting everything on the products side together and canceling out anything that appears on both sides (must be exactly the same on both sides). 2Cr 3+ (aq) + 6e + 6Cl (aq) 2Cr (s) + 3Cl 2 (g) + 6e Cancel out 6 electrons. 2Cr 3+ (aq) + 6Cl (aq) 2Cr (s) + 3Cl 2 (g) There are 2 chromium atoms and 6 chlorine atoms on both sides, thus balancing is complete. NOTE:
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This note was uploaded on 04/17/2008 for the course CHEM 121b taught by Professor Wiediger during the Spring '08 term at Southern Illinois University Edwardsville.

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BalancingRedoxReactions - Balancing Redox Reactions...

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