### Redox Rules

What is the oxidation number of . . .

Problems 1 - 10

Ten Examples

Problems 11-25

Return to Redox menu

**Problem #1:** N in NO_{3}¯

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

**Problem #2:** C in CO_{3}^{2}¯

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

**Problem #3:** Cr in CrO_{4}^{2}¯

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 Cr_{2}O_{7}^{2}¯

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 Cr_{2} being +12. Each Cr atom is considered individually.

**Problem #5:** Fe in Fe_{2}O_{3}

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 VO_{2}^{+}

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

**Problem #8:** V in VO^{2+}

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 MnO_{4}¯

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

**Problem #10:** Mn in MnO_{4}^{2}¯

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 K_{2}ReH_{9}

We know that potassium always takes on a +1 oxidation number. Eliminate the potassium to get this:
ReH_{9}^{2}¯

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 NaReO_{4}.

Ten Examples

Problems 11-25

Return to Redox menu