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Redox Reactions & Electrochemical Cells
I. Balancing Redox Reactions
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I. Balancing Redox Reactions
STEP 1. Split Reaction into 2 Half-Reactions STEP 2. Balance Elements Other than H & O STEP 3. Balance O by Inserting H2O into eqns. as necessary STEP 4. Balance H with H+ or H2O (see 4a, 4b) STEP 5. Balance Charge by Inserting Electrons as needed STEP 6. Multiply Each 1/2 Reaction by Factor needed to make no. of Electrons in each 1/2 Reaction Equal STEP 7. Add Eqns. & Cancel Out Duplicate terms, where possible
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I. Balancing Redox Reactions (continued)
STEP 4a. In ACID: Balance H by Inserting H+, as needed STEP 4b. In BASE: Balance H by (i) inserting 1 H2O for each missing H & (ii) inserting same no. of OH- on OTHER SIDE OF REACTION as H2Os added in (i)
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I. Balancing Redox Reactions (continued)
Example Complete and Balance Following Reaction: CuS (s) + NO3 - (aq) Cu2+(aq) + SO42- (aq) + NO (g) STEP1. Split into 2 Half-Reactions a CuS Cu SO42- b NO NO
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I. Balancing Redox Reactions (continued)
STEP 2. Balance Elements Other than H & O Already O.K. !
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I. Balancing Redox Reactions (continued)
STEP 3. Balance O by inserting H2O into equations as necessary a.3 CuS + 4H2O Cu2+ + SO42- b.3 NO NO + 2H2O
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I. Balancing Redox Reactions (continued)
STEP 4. ACIDIC, so Balance H by inserting H+ as needed a4. CuS + 4H2O Cu2+ + SO H+ b4. NO H NO + 2H2O
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I. Balancing Redox Reactions (continued)
STEP 5. Balance Charge by inserting Electrons, where necessary a5. CuS + 4H2O Cu2+ + SO H+ + 8e- b5. NO H+ + 3e NO + 2H2O
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I. Balancing Redox Reactions (continued)
STEP 6. Multiply each Eqn. by factor to make No. of Electrons in Each 1/2 Reaction the Same a6. Multiply by 3x 3CuS + 12H2O 3Cu SO H e- b6.Multiply by 8x 8NO H+ + 24e NO + 16H e-
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I. Balancing Redox Reactions (continued)
STEP 7. Add Eqns. and Cancel Out Duplicated Terms (a7 + b7) H+ 3CuS + 12H2O + 8NO H e- 3Cu2+ + 3SO H+ + 8NO +16 H2O e- 4H2O
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I. Balancing Redox Reactions (continued)
So, the final, balanced reaction is: 3CuS(s) + 8 NO3-(aq) + 8H+ (aq) Cu2+(aq) + 3 SO42-(aq) + 8NO(g) H2 O(l)
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Checking mass balance and charge balance in Equation
L.H.S 3 x Cu 3 x S 8 x N 24 x O 8 x H (8 x 1-) + (8 x H+) = 0 R.H.S. 3 x Cu 3 x S 8 x N 24 x O 8 x H (3 x 2+ )+(3 x 2- ) = 0
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Redox Reactions in Electrochemistry
Two Types of Electrochemical Cells: 1. Galvanic 2. Electrolytic Galvanic Cell - Converts a Chemical Potential Energy into an Electrical Potential to Perform Work Electrolytic Cell- Uses Electrical Energy to Force a Chemical Reaction to happen that would not otherwise occur
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Anode and Cathode in Electrochemistry
ANODE - Where OXIDATION takes place (-e-) CATHODE - Where REDUCTION takes place (+e-)
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Electrochemistry and the Metals Industry
Many Electrochemical Processes are used Commercially for Production of Pure Metals: e.g. Al Manufacture (by electrolysis of Al2O3) Mg Manufacture (by electrolysis of MgCl2) Na Manufacture (by electrolysis of NaCl)
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Electrolylitic Production of Al using the HALL CELL
(major plant in ALCOA, TN Al2O3 dissolved in molten cryolite (Na3AlF6) at 950 0C (vs C for pure Al2O3) Graphite Anodes (+) Steel case C lining (Cathode) (-) Al Al2O3 in molten Na3AlF6 Al Molten Al
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Hall Cell for Al Manufacture
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2 Al2O3 (sln) + 3C (s) 4 Al (l) + 3CO2 (g)
Hall Cell Process Reaction: 2 Al2O3 (sln) + 3C (s) 4 Al (l) + 3CO2 (g) Location of Hall cell plant in E. Tennessee through availability of inexpensive Hydroelectric power. Process uses 50,000 – 100,000 A.
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