HalfReactionBalancing

Net v k1 page 395 module 16 half reaction balancing

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Unformatted text preview: takes longer at the start than the oxidation number method, but half-reactions nearly always supply more than 4 trial coefficients. Which method is best? In general, for easy reactions, try trial and error first. For more complex redox, try oxidation numbers. For very complex redox, try halfreactions. Remember, both redox methods give trial coefficients, not final answers. At the end of both methods, correct the coefficients if needed, using trial and error, until atoms and net charge are the same on both sides. ANSWERS 1. Cu + HNO3 Cu(NO3)2 + NO + H2O Step 1: Break the compounds that ionize into separated ions. Leave out coefficients. Cu2+ + NO3─ + NO + H2O Cu + H+ + NO3─ Step 2: Cross out spectators ions: ©2009 ChemReview.net v. k1 Page 395 Module 16: Half-Reaction Balancing Cu2+ + NO3─ + NO + H2O Cu + H+ + NO3─ NO3─ is a spectator on the right, but not on the left. Some NO3─ on the left changed to NO, and some remained as NO ─. 3 Step 3: Find the two atoms that change oxidation number. Each 0 +1 +5 -2 + + NO ─ Cu + H 3 +2 +2 -2 +1 -2 2+ + NO + H O Cu 2 Copper and nitrogen are the central atoms that change their oxidation number. Step 4: Write the particles containing those two atoms in two separate half-reactions: Cu NO ─ 3 Cu2+ NO Step 5: Complete half-reactions using CA-WHe! Cu 3 e─ + 4 H+ + NO3─ Cu2+ + 2 e─ NO + 2 H2O Step 6: Get the electrons equal in both, then add the two half-reactions. Use a lowest common denominator (LCD) method: the electron coefficient of one as the multiplier of the other: 3 Cu 6 e─ + 8 H+ + 2 NO3─ 3 Cu + 8 H+ + 2 NO3─ 3 Cu2+ + 6 e─ (3x) 2 NO + 4 H2O (2x) 3 Cu2+ + 2 NO + 4 H2O Check: 3 Cu, 8 H, 2 N, 6 O, +6 charge on both sides. Step 7: Plug the trial coefficients into the original equation; finish balancing by trial and error. 3 Cu(NO3)2 + 2 NO + 4 H2O 3 Cu + 8 HNO3 Step 8: Check: 3 Cu, 8 H, 8 N, 24 O, zero net charge on both sides. Balanced. 2. KMnO4 + HNO3 + H2O2 Mn(NO3)2 + O2 + KNO3 + H2O Steps 1 and 2: Break compounds that ionize into separated ions, then cross out the spectators and particles that are the same on both sides. K+ + MnO4─ + H+ + NO3─ + H2O2 Mn2+ + NO3─ + O2 + K+ + NO3─ + H2O Step 3: Find the 2 atoms that change their oxidation number. In peroxides, the O Ox# is ─1. Each +7 -2 +1 + 1 -1 MnO4─ + H+ + H2O2 +2 0 +1 -2 Mn2+ + O2 + H2O Steps 4 and 5: Write the 4 particles containing the 2 atoms that change their oxidation number in separate half-reactions, then balance the half-reactions using the CA-WHe! method. 5 e─ + 8 H+ + MnO4─ H2O2 Mn2+ + 4 H2O O2 + 2 H+ + 2 e─ Step 6: Get electrons equal in both, then add the half-reactions. ©2009 ChemReview.net v. k1 Page 396 Module 16: Half-Reaction Balancing 10 5 e─ + 16 8 H+ + 2 MnO ─ 4 5H O 22 6 16 H+ + 2 MnO ─ + 5 H O 4 22 2 Mn2+ + 8 4 H O 2 5 O + 10 2 H+ + 10 2 e─ 2 5 O2 + 10 H+ + 2 Mn2+ + 8 H2O Check: 2 Mn, 18 O, 16 H atoms on both sides, +4 charge on both sides. Balanced. Step 7: Plug trial coefficients into the original equation; finish balancing by trial and error. Trial:...
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