Finding Oxidation Numbers


Finding Oxidation Numbers Example One:  Calcium Phosphide

Find the oxidation numbers for Ca and P in Ca₃P₂. First, the sum of the oxidation numbers must add to zero for a compound. Second, create an algebraic equation based on the compound. Third, the key elements must be substituted into the algebraic equation. The only key element is Ca. Ca has a +2 oxidation number. Therefore, the oxidation for phosphorus(P) must be calculated. The algebraic equation will be 3Ca + 2P =  0. The +2 of Ca replaces the Ca in the algebraic equation. The equation now appears as 3(+2) + 2P = 0. Simplifying gives (+6) + 2P = 0. To solve for the oxidation number for P, -6 is added  to both sides. This gives 2P = -6. Dividing both sides by 2 shows P having a oxidaton number of -3. The -3 is one of the common oxidation numbers for phosphorus. 

Finding Oxidation Numbers Example Two:  Magnesium Phosphate

Find the oxidation numbers for Mg, P and O in Mg₃(PO₄)₂? The key elements are Mg and O. Magnesium has a +2 oxidation number. Oxygen has a -2. Therefore, the oxidation number for P must be calculated. The algebraic equation is 3Mg + 2P +8(O) = 0. The +2 replaces the Mg and the -2 replaces the O in the algebraic equation. The equation now appears as 3(+2) + 2P + 8(-2) = 0. This gives 6 + 2P + ⁻16 = 0. Simplifiying the equation gives 2P + ⁻10 = 0. To solve for the oxidation number for P, 10 is added to both sides. This gives 2P = 10. Dividing both sides by 2 shows phosphorus having an oxidaton number of +5. The +5 is another common oxidation number for phosphorus.     

Finding Oxidation Numbers Example Three:  Calcium Carbonate

Find the oxidation number for each element in CaCO₃. The key elements are Ca and O. Iron has a +2 oxidation number. Oxygen has a -2. Therefore, the oxidation number for C must be calculated. The algebraic equation is Ca + C +3(O) = 0. The +2 replaces the Ca and the -2 replaces the O in the algebraic equation. The equation now appears as ⁺2 + C + 3(-2) = 0. This gives ⁺2 + C + ⁻6 = 0. Simplifiying the equation gives C + ⁻4 = 0. To solve for the oxidation number for C, ⁺4  is added to both sides. This gives C = +4.  The +4 is a common oxidation number for carbon.   




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