Balancing redox reactions

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

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

K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 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- K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 +4 +6 +3 2 3 3 2 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 Step 4: Balance other atoms by conventional method 2 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 2 3 3 4 2

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

H2SeO3 + HClO3 → H2SeO4 + Cl + H2O 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- +1 +4 -2 +1 +5 -2 +1 +6 -2 +1 -2 5 H2SeO3 + HClO3 → H2SeO4 + Cl + H2O 2 5 1 2 gains 5e- x 2 atoms = 10e- Since there is already 2 Cl on the product side (subscript) I don’t need to add the coefficient to it.

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