Balancing redox reactions

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Presentation transcript:

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.

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

K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 Step 1: Assign oxidation numbers. +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 – watch the number of atoms. 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: Balance atoms that lose / gain electrons. – Use lowest common multiple. – Massage numbers to make atoms equal. lose 4e- x 3 = 12e- K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 +4 +6 +3 2 3 3 2 gains 3e- x 2 atoms = 6e- x 2 = 12e- Step 4: Balance others 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 lose 5e- x 3 = 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 = 15e-

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