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Rules for Assigning Oxidation Numbers
- The oxidation number of an element in any elementary substance is zero.
For example, the oxidation number of chlorine in Cl 2 , phosphorus in P 4 , and sulfur in S 8 is
The oxidation number of an element in a monatomic ion is equal to the charge on that ion.
Compound Ions Oxidation No. NaCl Na+^ + Cl –^ – Al 2 O 3 Al 3+^ + O2–^ –
- Certain elements have the same oxidation number in almost all their compounds.
Group 1 always forms +1 ions: Li +^ , Na+^ , K+^ , etc. oxidation number + Group 2 always forms +2 ions: Mg2+^ , Ca2+^ , etc. oxidation number + Fluorine always has an oxidation number of –1. Oxygen has oxidation number of –2 except in peroxides, O 2 2–^ , (examples: H 2 O 2 , Na 2 O 2 ) and in superoxides, O 2 –^ , (example: KO 2 ) where it has oxidation numbers of –1 and –½, respectively. Hydrogen has oxidation number of +1 except in hydrides, H–^ , (examples: NaH, CaH 2 ) where it has an oxidation number of –1.
- The sum of the oxidation numbers of all the atoms in a neutral species is zero; in an ion, it is equal to the charge of that ion. Li 3 N: ox. no. of Li +^ is +1 ClO 2 –^ : ox. no. of O is – 3(+1) + ox. no. of N = 0 ox. no. of Cl + 2(–2) = – ox. no. of N = –3 ox. no. of Cl – 4 = – ox. no. of Cl = +
Rules for Balancing Redox Equations
We will use the example:
Cu(s) + NO 3 –^ (aq) Cu2+^ (aq) + NO 2 (g) (acidic solution)
1. Split the equation into two half-equations, one for oxidation and one for reduction.
Remember: oxidation is an increase in oxidation number and reduction is a decrease in oxidation number. Using the rules for determining oxidation numbers (ox. no.): Cu + NO 3 –^ Cu2+^ + NO 2 ox. no. Cu=0 N=+5 Cu=+2 N=+ O=–2 O=– reduction half equation: NO 3 –^ NO 2 (ox. no. of N decreases: +5 +4) oxidation half-equation: Cu Cu2+^ (ox. no. of Cu increases: 0 +2) (Notice that we have left out H+^ and H 2 O for acidic solution or OH–^ and H 2 O for basic solution for now. We will add these in later as we need them.)
2. Balance one of the half-equations with respect to both atoms and charge.
First we balance the oxidation half-equation since it is easier.
(a) Balance the atoms of the element being oxidized. The atoms of Cu are already balanced. Cu Cu 2+
(b) Balance oxidation number by adding electrons. For an oxidation half-equation, we add electrons to the right. Since the oxidation number of copper increases from 0 to +2, we add two electrons to the right.
(d) Balance hydrogen by adding H 2 O molecules. Since we have two hydrogen atoms on the left and none on the right, we add one H 2 O molecule to the right. NO 3 –^ + 2 H+^ + e–^ NO 2 + H 2 O
(e) Check to make sure that oxygen is balanced. If it is, the half-equation is almost certainly balanced correctly with respect to both mass (atoms) and charge. There are three oxygen atoms on the left and three on the right. Oxygen is balanced.
4. Combine the two half-equations in such a way as to eliminate electrons.
We generally do this by multiplying each half-equation by the number of electrons in the other half-equation:
1 × oxidation eq.: Cu Cu2+^ + 2 e– 2 × reduction eq.: 2 NO 3 –^ + 4 H+^ + 2 e–^ 2 NO 2 + 2 H 2 O Cu(s) + 2 NO 3 –^ (aq) + 4 H+^ (aq) Cu2+^ (aq) + 2 NO 2 (g) + 2 H 2 O
This equation is now balanced: one Cu atom is on both sides of the equation, two N atoms are on both sides, six O atoms are on both sides, and four H atoms are on both sides. The total charge is +2 on both sides of the equation.
** **Simplifying Balanced Equations ****
- For an equation in acidic solution , there will sometimes be H 2 O molecules and H+^ ions on both sides of the equation. We want to cancel out the excess amounts of these. The equation we just balanced doesn’t have any excess H 2 O molecules or H+^ ions, but in the equation 2MnO 4 –^ (aq) + 5HSO 3 –^ (aq) + 5H 2 O + 16H+^ (aq) 2Mn2+^ (aq) + 5SO 4 2–^ (aq) + 8H 2 O + 15H+^ (aq)
there are both excess H 2 O molecules and H+^ ions. After canceling five H 2 O molecules and fifteen H+^ ions from both sides of the equation 2MnO 4 –^ (aq) + 5HSO 3 –^ (aq) + 5H 2 O + 16H+^ (aq) 2Mn2+^ (aq) + 5SO 4 2–^ (aq) + 8H 2 O + 15H+^ (aq) we have 2MnO 4 –^ (aq) + 5HSO 3 –^ (aq) + H+^ (aq) 2Mn2+^ (aq) + 5SO 4 2–^ (aq) + 3H 2 O
In basic solution , there will sometimes be H 2 O molecules and OH–^ ions on both sides of the equation. We want to cancel as many of them as possible. For example, in the equation 3ClO–^ (aq) + 2NO(g) + 8OH–^ (aq) + 3H 2 O 3Cl –^ (aq) + 2NO 3 –^ (aq) + 6OH–^ (aq) + 4H 2 O we can cancel six OH–^ ions and three H 2 O molecules 3ClO–^ (aq) + 2NO(g) + 8OH–^ (aq) + 3H 2 O 3Cl –^ (aq) + 2NO 3 –^ (aq) + 6OH–^ (aq) + 4H 2 O resulting in 3ClO–^ (aq) + 2NO(g) + 2OH–^ (aq) 3Cl –^ (aq) + 2NO 3 –^ (aq) + H 2 O
- In a few cases, we find that the final equation doesn’t have the smallest whole number coefficients possible. In the equation 6 Cl 2 (g) + 6 H 2 O 10 Cl –^ (aq) + 2 ClO 3 –^ (aq) + 12 H+^ (aq) we divide each coefficient by two to obtain the smallest whole number coefficients: 3 Cl 2 (g) + 3 H 2 O 5 Cl –^ (aq) + ClO 3 –^ (aq) + 6 H+^ (aq)
