Electrochemistry Essentials: Assigning Oxidation Numbers with Confidence

What an Oxidation Number Is (and What It Is Not)

An oxidation number (oxidation state) is a bookkeeping value assigned to an atom to keep track of electron ownership in compounds and ions. It is not a measured charge in most covalent molecules; instead, it is a consistent accounting tool that helps you (1) identify which atom is oxidized or reduced and (2) balance redox equations later.

When assigning oxidation numbers, treat bonds as if the more electronegative atom “owns” the bonding electrons. The oxidation number is the charge the atom would have under that assumption.

1) Prioritized Rule List (Use in This Order)

Rule 1: Free elements have oxidation number 0

Any element in its standard elemental form has oxidation number 0: Na(s), O2(g), Cl2(g), P4(s), S8(s).

Rule 2: Monatomic ions have oxidation number equal to their charge

Examples: Na+ is +1, Mg2+ is +2, Cl− is −1, Al3+ is +3.

Rule 3: Group 1 and Group 2 metals in compounds

• Group 1 (Li, Na, K, …) is almost always +1 in compounds.
• Group 2 (Mg, Ca, …) is almost always +2 in compounds.

Rule 4: Fluorine is always −1 in compounds

Fluorine is the most electronegative element, so it takes −1 in its compounds (no common exceptions in typical general chemistry contexts).

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Rule 5: Oxygen is usually −2 (but watch the common exceptions)

• Most oxides and oxyanions: O = −2 (e.g., H2O, SO4^2−).
• Peroxides (O–O single bond): O = −1 (e.g., H2O2, Na2O2).
• Superoxides (rare in intro problems): average O = −1/2 (e.g., KO2).
• With fluorine: oxygen can be positive (e.g., OF2 has O = +2 because F is −1).

Rule 6: Hydrogen is usually +1 (but can be −1 in metal hydrides)

• In acids and most covalent compounds: H = +1 (e.g., HCl, H2O, NH3).
• In metal hydrides: H = −1 (e.g., NaH, CaH2).

Rule 7: Halogens (Cl, Br, I) are usually −1, except when bonded to O or F

In many salts: Cl = −1 (e.g., NaCl). In oxyanions (like ClO3−) or compounds with fluorine, halogens can be positive.

Rule 8: The sum of oxidation numbers equals the overall charge

• For a neutral compound, the sum is 0.
• For a polyatomic ion, the sum equals the ion charge.

This rule is the “equation” you solve after applying the fixed-value rules above.

2) Worked Examples (Progressing in Difficulty)

Example A: Simple salt, NaCl

Step 1: Identify fixed oxidation numbers. Group 1 metal Na is +1.

Step 2: Compound is neutral, so sum is 0. Therefore Cl must be −1.

Answer: Na = +1, Cl = −1.

Example B: Magnesium chloride, MgCl2

Mg (Group 2) is +2. Two Cl atoms must sum to −2, so each Cl is −1.

Answer: Mg = +2, Cl = −1.

Example C: Polyatomic ion, sulfate SO4²⁻

Step 1: Oxygen is usually −2. With 4 oxygens: total for O is 4 × (−2) = −8.

Step 2: Let sulfur be x. Sum equals ion charge: x + (−8) = −2.

Step 3: Solve: x = +6.

Answer: S = +6, O = −2.

Example D: Peroxide, hydrogen peroxide H2O2

Step 1: Recognize peroxide pattern (O–O). In peroxides, O = −1.

Step 2: Hydrogen is usually +1 in covalent compounds. Two H gives +2 total.

Step 3: Two O at −1 each gives −2 total. Sum is 0 (neutral molecule), consistent.

Answer: H = +1, O = −1.

Example E: Permanganate, KMnO4

Step 1: K is Group 1, so K = +1.

Step 2: Oxygen is usually −2. Four O gives −8 total.

Step 3: Let Mn be x. Neutral compound: (+1) + x + (−8) = 0.

Step 4: Solve: x = +7.

Answer: K = +1, Mn = +7, O = −2.

Example F: Ammonium ion, NH4⁺

Step 1: Hydrogen is usually +1. Four H gives +4 total.

Step 2: Let N be x. Sum equals +1: x + (+4) = +1.

Step 3: Solve: x = −3.

Answer: N = −3, H = +1.

Example G: Nitrate ion, NO3⁻

O is −2 each, total −6. Let N be x. x + (−6) = −1, so x = +5.

Answer: N = +5, O = −2.

Example H: Iron(III) oxide, Fe2O3

O is −2 each, total −6. Neutral compound: 2(Fe) + (−6) = 0, so 2(Fe) = +6 and Fe = +3.

Answer: Fe = +3, O = −2.

3) Troubleshooting Patterns (Common Places People Slip)

Peroxides vs. oxides: check for O–O

• Oxide (no O–O bond implied): O = −2 (e.g., Na2O).
• Peroxide (contains O–O): O = −1 (e.g., Na2O2, H2O2).

Practical tip: if the formula is M2O2 for an alkali metal or MO2 for an alkaline earth metal, suspect a peroxide and test it with the sum rule.

Hydrides vs. acids: decide whether H is +1 or −1

• If H is bonded to a metal (especially Group 1/2): treat as hydride, H = −1 (e.g., NaH, CaH2).
• If H is in an acid or bonded to nonmetals: H = +1 (e.g., HCl, H2S, NH3).

Quick check: in NaH, Na is almost certainly +1, so H must be −1 to make the compound neutral.

Unusual oxidation states in oxyanions: the central atom can get very positive

In oxyanions, oxygen contributes a large negative total (usually −2 each), so the central atom often becomes strongly positive to match the ion charge. Examples you should expect to see:

• NO3−: N = +5
• SO4^2−: S = +6
• MnO4−: Mn = +7
• Cr2O7^2−: Cr = +6

Pattern: if there are many oxygens, do not be surprised by +5, +6, or +7 on the central atom.

Halogens in oxyanions: halogens can be positive

Chlorine is −1 in Cl−, but in ClO3− oxygen is −2 each (−6 total), so chlorine must be +5 to reach −1 overall. When halogens are bonded to oxygen, oxygen “wins” electron ownership, pushing the halogen oxidation number upward.

Sanity checks you can do in seconds

• Does the sum match the overall charge?
• Are fixed-rule elements (Group 1/2, F, usually O and H) consistent with their standard values?
• If you got a fractional oxidation number, is it a known case (superoxide) or did you miss a peroxide/hydride exception?

4) Practice Set (Oxidation Numbers + Identify the Changing Atom in a Redox Reaction)

Instructions: For each item, (a) assign oxidation numbers for the specified atoms (or all atoms if you prefer), and (b) write one brief sentence identifying which element changes oxidation state in the reaction (and whether it is oxidized or reduced). You do not need to balance the equations here; focus on oxidation numbers.

Part A: Oxidation numbers only (warm-up)

• 1) CaCl2
• 2) Al2O3
• 3) NO2−
• 4) CO3^2−
• 5) H2SO3
• 6) Na2O2 (state whether it is oxide or peroxide)
• 7) NH3
• 8) Cr2O7^2−
• 9) OF2
• 10) NaH (state whether it is hydride or not)

Part B: Oxidation numbers + identify what changes in the redox reaction

• 1) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• 2) 2Na(s) + Cl2(g) → 2NaCl(s)
• 3) Fe2+(aq) + Ce4+(aq) → Fe3+(aq) + Ce3+(aq)
• 4) 2H2O2(aq) → 2H2O(l) + O2(g)
• 5) 2I−(aq) + Br2(aq) → I2(s) + 2Br−(aq)
• 6) 2Mg(s) + O2(g) → 2MgO(s)
• 7) ClO−(aq) + Cl−(aq) → Cl2(g) + O2−(aq) (treat oxygen as −2 in oxide form; focus on chlorine changes)
• 8) NH4+(aq) + NO2−(aq) → N2(g) + 2H2O(l) (assign oxidation numbers to N in each nitrogen-containing species)

Answer format suggestion (keep it brief)

Example response style: KMnO4: K = +1, Mn = +7, O = −2. In the reaction, Mn changes from +7 to +2, so Mn is reduced.