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Balancing redox reactions

Balancing redox reactions. Balance oxidation-reduction reactions using redox methods Include: oxidation number method , and half-reaction method. Additional KEY Terms. Oxidation number method. Step 1: Assign oxidation numbers and determine oxidized and reduced. -2. +3. +1. +1. -2.

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Balancing redox reactions

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  1. Balancing redox reactions

  2. Balance oxidation-reduction reactions using redox methods • Include: oxidation number method, and half-reaction method Additional KEY Terms

  3. Oxidation number method Step 1: Assign oxidation numbers and determine oxidized and reduced -2 +3 +1 +1 -2 0 -2 +4 -2 -2 +1 +1 +6 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 Step 2: Calculate number of electrons lost/gained lose 4e- -2 +3 +1 +1 -2 -2 0 +4 -2 -2 +1 +1 +6 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 gains 3e- x 2 atoms = 6e-

  4. Step 3: Add coefficients to balance the number of electrons lost/gained Use common multiple to make them equal lose 4e- x 3 atoms = 12e- 3 2 2 3 gains 3e- x 2 atoms = 6e- x 2 atoms = 12e- So with the coefficients and subscripts you actually have a total of 4 atoms of Cr each gaining 3e-for a total of 12 +3 0 +4 +6 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 Step 4: Balance other atoms by conventional method 4 2 2 3 2 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 3

  5. Balance the following using the oxidation number method: lose 5e- x 3 atoms = 15e- -2 -2 +1 +5 +5 -2 +1 -2 +1 0 +2 2 3 3 5 5 P+ HNO3 + H2O → NO + H3PO4 gains 3e- x 5 atoms = 15e-

  6. Always look at the reactant and product to see if the numbers match– you might need to massage the coefficient to make the atoms equal lose 2e- x 5 atoms = 10e- -2 -2 +1 0 +6 +1 +5 -2 -2 +1 +1 +4 2 2 5 1 H2SeO3 + HClO3 → H2SeO4 + Cl + H2O 5 x 2 atoms = 10e- gains 5e- Since there is already 2 Clon the product side (subscript) I don’t need to add the coefficient to it.

  7. CAN YOU / HAVE YOU? • Balanceoxidation-reduction reactions using redox methods • Include: oxidation number method, and half-reaction method Additional KEY Terms

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