### 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 +12What 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 formula.

**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.

Ten Examples

Problems 11-25

Return to Redox menu