Redox Rules
What is the oxidation number of . . .
Problems 1 - 10

Ten Examples

Problems 11-25

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Problem #1: N in NO3¯

The O is −2 and three of them makes −6. Since −1 is left over, the N must be +5

Problem #2: C in CO32¯

The O is −2 and three of them makes −6. Since −2 is left over, the C must be +4

Problem #3: Cr in CrO42¯

The O is −2 and four of them makes −8. Since −2 must be left over, the Cr must be +6

Problem #4: Cr in Cr2O72¯

The O is −2 and seven of them makes −14. Since −2 is left over, the two Cr must be +12.

What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr2 being +12. Each Cr atom is considered individually.

Problem #5: Fe in Fe2O3

The O is −2 and three of them makes −6. Each Fe must then be +3

Problem #6: Pb in PbOH+

The O is −2 and the H is +1. In order to have a +2 for the formula, the Pb must be a +2.

Problem #7: V in VO2+

Two O gives a total of −4. To make the formula have a +1 charge, the V must be +5

Problem #8: V in VO2+

The V is +4. This comes from the one O being −2 and the fact that a +2 must be present on the ion.

Problem #9: Mn in MnO4¯

Four O totals to −8. The Mn is +7, leaving −1 left over.

Problem #10: Mn in MnO42¯

Four O totals to −8. The Mn is +6, leaving −2 left over.

Bonus Problem: What is the oxidation number of Re in potassium nonahydridorhenate, which has the formula K2ReH9

We know that potassium always takes on a +1 oxidation number. Eliminate the potassium to get this:
ReH92¯

That looks very much like a hydride, when hydrogen takes on a +1 oxidation number.

The fact that there are nine of them means Re must have an oxidation number of +7.

Another compound with Re = +7 is NaReO4.


Ten Examples

Problems 11-25

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