What is the oxidation state assigned to Mn in potassium permanganate, KMnO4? (Recall that answers must be numbers, that is, F is -1 not 1- in HF.)
Answer:
CHM-1045 Lecture Notes - Lecture 7: Redox, Oxidation State, Nonmetal
17 Nov 2017
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We used the oxidation rule to find oxidation number. No2- = x + 2(-2)= -1, x-4=-1, x=+3, n= +3, o=-2. Nh3= x + 3(1)= 0, x+3=0, x=-3, n=-3, h=+1. Alo2-= x + 2(-2)=-1, x-4=-1, x= +3, al=+3, o=-2. Mno4- + fe+2 ----- fe+3 + mn+2: write the reduction and the oxidation reaction separate. Mno4- ------- mn+2 + 4h2o: add h+ to balance hydrogen. Mno4- + 8h+ --------- mn+2 + 4h2o: add electrons to balance charge. 5e+ mno4- + 8h+ --------- mn+2 + 4h2o: multiply by whole number to make number of electrons equal (fe+2------------fe+3 + e) multiply by 5. 5fe+2---------5fe+3 + 5e: add both reaction together. 5e+ mno4- + 8h+ --------- mn+2 + 4h2o. = 5fe+2 + 5e+ mno4- +8h+---------5fe+3 +5e + mn+2 + 4h2o: cancel the electrons. 5fe+2 + 5e+ mno4- +8h+---------5fe+3 +5e + mn+2 + 4h2o. 5fe+2 + mno4- + 8h+---------5fe+3 + mn+2 + 4h2o: substrate atoms that appear on both side.