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Balancing redox reactions

Balancing redox reactions. Two methods for balancing redox reactions: Oxidation number method Half-reaction method. Balance redox equations using the oxidation number method. Balance redox equations in acidic and basic solutions using the half reaction method.

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Balancing redox reactions

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  1. Balancing redox reactions

  2. Two methods for balancing redox reactions: Oxidation number method Half-reaction method. • Balance redox equations using the oxidation • number method. • Balance redox equations in acidic and basic • solutions using the half reaction method.

  3. Why we balance electrons lost and gained in redox equations. x 2 atoms = 2e- gains 1e- 0 0 +1 -1 2 Cu + Cl2 → Cu+ + Cl- 2 2 x 2 = 2e- loses 1e- Have to make sure the number of atomsand the number of electronstransferred are equal.

  4. Step 1: Assign oxidation numbers. -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 – watch the number of atoms. 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-

  5. Step 3: Balance atoms that lose / gain electrons. – Use lowest common multiple. – Massage numbers to make atoms equal. lose 4e- x 3 = 12e- 3 2 2 3 gains 3e- x 2 atoms = 6e- x 2 = 12e- +3 0 +4 +6 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 Step 4: Balance others by conventional method. 4 2 2 3 2 K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 3

  6. lose 5e- x 3 = 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 = 15e-

  7. lose 2e- x 5 = 10e- -2 -2 +1 0 +6 +1 +5 -2 -2 +1 +1 +4 2 5 1 H2SeO3 + HClO3 → H2SeO4 + Cl2 + H2O 5 x 2 = 10e- gains 5e- Step 3: Balance atoms that lose / gain electrons. – Use lowest common multiple. –Massage numbers to make atoms equal.

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