Extraction of Valences before REDOX Reactions

1. Compute for the total valences of each extreme element of the compound. If subscript is available, multiply it with the existing valence of the element.

Eg.

K₂Cr₂O₇ thus, K₂¹⁺Cr₂O₇^-2 

 

Subscript is available so multiply it with the original valence.

K₂⁺¹⁽²⁾Cr₂O₇^{-2 (7)} thus, the total valences will be K₂⁺²Cr₂O₇^-14

2. Subtract the valences (disregard the signs). The difference will be the oxidation state of the middle element. When subscript is present divide the difference using it. The quotient will be the final valence of the middle element. Divide the remaining elements with their subscript to return to their original oxidation.

Eg.

K₂⁺²Cr₂O₇^-14 from the +2 valence of Potassium and -14 valence of Oxygen, subtract them producing a difference, irregardless of the signs, of +12 which is the valence of Chromium.

K₂^+2/2Cr₂^+12/2O₇^-14/7 the quotient will be +6

thus, K₂⁺¹Cr₂⁺⁶O₇^-2 

 

3. To check if the valences are correctly given, the positive signs and the negative signs should be equaled to zero when added in their final oxidation state.

Eg.

K₂⁺²Cr₂⁺¹²O₇^-14           

  (+2)+(+12)+(-14) = 0

4. The final valence of each element extracted from step 2 will be the final oxidation state to be used in the REDOX Reactions. The element that reacts to gain or lose electrons shall be prioritized.

Eg.

Cr₂⁶⁺ + 6e 2Cr⁺³