AP Chem 4.9: Oxidation-Reduction (Redox) Reactions

Oxidation, reduction, and electrons

Lesson 2.13 introduced oxidation numbers and oxidizing and reducing agents. AP U4.9 uses them to count electrons:

Oxidation氧化: loss of electrons; the oxidation number increases.
Reduction还原: gain of electrons; the oxidation number decreases.
Oxidizing agent氧化剂: the substance that gains electrons; it is itself reduced.
Reducing agent还原剂: the substance that loses electrons; it is itself oxidized.

Oxidation and reduction always happen together: in a reaction, the total number of electrons lost equals the total number gained.

Assigning oxidation numbers: the order of the rules

Apply the rules in this order; when two rules conflict, the earlier one wins:

An element in its free form is 0; a monatomic ion equals its charge.
F in a compound is always −1.
Group 1 metals are +1, and Group 2 metals are +2.
H is usually +1.
O is usually −2.
Cl, Br, and I are usually −1.
Find the one remaining element from “the oxidation numbers add up to the particle’s charge.”

In this order, the special cases come out on their own: O is −1 in H₂O₂ and Na₂O₂ (peroxides); in O₂F₂, F is −1, so O is +1; in LiH, Li is +1, so H is −1 (hydride). In SO₂Cl₂, with O at −2 and Cl at −1, S is +6, the same as in H₂SO₄; in Cu₂O, Cu is +1.

What is oxidized, what is reduced, and how many electrons

Find each element’s oxidation number before and after the reaction: one that goes up is oxidized, and one that goes down is reduced. Multiply by the number of atoms and the coefficient to count electrons:

CO₂(g) + 4H₂(g) → CH₄(g) + 2H₂O(g): C goes from +4 to −4, gaining 8 electrons; 8 H atoms go from 0 to +1, losing 8 electrons in all.
The same element can be both oxidized and reduced: O in $\ce{2H2O2 -> 2H2O + O2}$ , and H in $\ce{Li3N + 2H2 -> LiNH2 + 2LiH}$ (some becomes +1, some becomes −1).

In the reaction below, each of the 2 Mn atoms gains 5 electrons, 10 in all, and each of the 10 Br⁻ ions loses 1, also 10 in all. As written, 10 electrons are transferred:

\ce{2MnO4-(aq) + 10Br-(aq) + 16H+(aq) -> 2Mn^2+(aq) + 5Br2(aq) + 8H2O(l)}

You can decide without writing electrons, too: NO₃⁻ becoming NO means N goes from +5 to +2, a reduction; Fe²⁺ becoming Fe³⁺ is an oxidation.

Half-reactions

Split the oxidation and the reduction into two half-reactions半反应, writing the electrons in the equations:

\ce{Cu(s) -> Cu^2+(aq) + 2e-}

\ce{Ag+(aq) + e- -> Ag(s)}

The first is the oxidation half-reaction (electrons lost), and the second is the reduction half-reaction (electrons gained). To combine them into the overall reaction, multiply each so that electrons lost equal electrons gained, then add them and cancel the electrons: doubling the reduction gives $\ce{Cu + 2Ag+ -> Cu^2+ + 2Ag}$ .

In the same way, when the two half-reactions below are combined, the Fe²⁺ half-reaction is multiplied by 5, so MnO₄⁻ and Fe²⁺ react in a 1 : 5 ratio (the ratio used in the U4.6 titration):

\ce{MnO4-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}

\ce{Fe^2+(aq) -> Fe^3+(aq) + e-}

Assign oxidation numbers first, then analyze reactions:

Try it: oxidation numbers and electron transfer

$\ce{H2SO4}$

H is usually +1
O is usually −2
The remaining element, S, comes from “sum of oxidation numbers = charge”: +6

Result: H +1; S +6; O −2

Check: 2 × (+1) + 1 × (+6) + 4 × (−2) = 0

\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

Element	Before	After	Change
Zn	0	+2	increases: oxidized
Cu	+2	0	decreases: reduced

Oxidizing agent: $\ce{Cu^2+}$ ; Reducing agent: $\ce{Zn}$

Electrons transferred in the equation as written: 2 e⁻ (total lost = total gained)

Oxidation half-reaction (× 1):

\ce{Zn(s) -> Zn^2+(aq) + 2e-}

Reduction half-reaction (× 1):

\ce{Cu^2+(aq) + 2e- -> Cu(s)}

Multiply each half-reaction so the electrons lost equal the electrons gained, then add them and cancel the electrons to get the overall reaction.

Zn is more active than Cu, so Zn gives electrons to Cu²⁺.

\ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)}

Element	Before	After	Change
Cu	0	+2	increases: oxidized
Ag	+1	0	decreases: reduced

Oxidizing agent: $\ce{Ag+}$ ; Reducing agent: $\ce{Cu}$

Electrons transferred in the equation as written: 2 e⁻ (total lost = total gained)

Oxidation half-reaction (× 1):

\ce{Cu(s) -> Cu^2+(aq) + 2e-}

Reduction half-reaction (× 2):

\ce{Ag+(aq) + e- -> Ag(s)}

Multiply each half-reaction so the electrons lost equal the electrons gained, then add them and cancel the electrons to get the overall reaction.

Each Cu loses 2 electrons but each Ag⁺ gains only 1, so Ag⁺ has a coefficient of 2.

Element	Before	After	Change
Mn	+7	+2	decreases: reduced
O	−2	−2	unchanged
Fe	+2	+3	increases: oxidized
H	+1	+1	unchanged

Element	Before	After	Change
Mn	+7	+2	decreases: reduced
O	−2	−2	unchanged
Br	−1	0	increases: oxidized
H	+1	+1	unchanged

\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}

Element	Before	After	Change
H	+1	+1	unchanged
O	−1	−2 and 0	both oxidized and reduced

Oxidizing agent: $\ce{H2O2}$ ; Reducing agent: $\ce{H2O2}$

Electrons transferred in the equation as written: 2 e⁻ (total lost = total gained)

Oxidation half-reaction (× 1):

\ce{H2O2(aq) -> O2(g) + 2H+(aq) + 2e-}

Reduction half-reaction (× 1):

\ce{H2O2(aq) + 2H+(aq) + 2e- -> 2H2O(l)}

Multiply each half-reaction so the electrons lost equal the electrons gained, then add them and cancel the electrons to get the overall reaction.

The same element, O, is both oxidized (−1 → 0) and reduced (−1 → −2): a disproportionation. H₂O₂ is both the oxidizing and the reducing agent.

Element	Before	After	Change
C	+4	−4	decreases: reduced
O	−2	−2	unchanged
H	0	+1	increases: oxidized

Element	Before	After	Change
Li	+1	+1	unchanged
N	−3	−3	unchanged
H	0	+1 and −1	both oxidized and reduced

Which reducing agent is stronger?

The more easily a reducing agent loses electrons, the stronger it is. Active metals are strong reducing agents, and the higher a metal is in the activity series (2.7.2), the stronger it is as a reducing agent: Na is stronger than Zn. Cations that have already lost electrons (such as K⁺ and Na⁺) can hardly lose more, so they are hardly reducing agents at all; H⁺ can only gain electrons, so it is an oxidizing agent, not a reducing agent; and Cl⁻ loses electrons to form Cl₂ only with difficulty, so it is a very weak reducing agent.

Example 1: Two unusual oxidation numbers

Find the oxidation number of S in SO₂Cl₂ and of O in O₂F₂.

Step-by-step solution

$\ce{SO2Cl2}$ : apply the higher-priority rule to O first. There’s no F, metal, or H, so “O is −2” comes first:
$2 \times (-2) = -4$
Then Cl. S and Cl remain; Cl is more electronegative than S, so apply “Cl is −1”:
$2 \times (-1) = -2$
Solve for S from a sum of 0. The oxidation numbers in a neutral compound add up to 0:
$\begin{aligned} x + (-4) + (-2) &= 0 \\ x &= \boxed{+6} \end{aligned}$
$\ce{O2F2}$ : F first. F is the most electronegative element and is always −1 in compounds; that rule ranks above “O is −2”:
$2 \times (-1) = -2$
Solve for O. Only O remains, so the two O atoms must total +2:
$\begin{aligned} 2y + (-2) &= 0 \\ y &= \boxed{+1} \end{aligned}$
Check. O isn’t −2 here because the F rule comes first; +6 for S is its maximum (6 valence electrons), which is reasonable.

Example 2: Mole ratio and electrons from half-reactions

In acidic solution, KMnO₄ solution reacts with FeSO₄ solution. Using the MnO₄⁻ reduction and Fe²⁺ oxidation half-reactions from the previous section, write the overall reaction, and find the mole ratio of MnO₄⁻ to Fe²⁺ and the number of electrons transferred as written.

Step-by-step solution

Decide what is oxidized and reduced. $\ce{Fe^2+}$ → $\ce{Fe^3+}$ : the oxidation number rises as it loses an electron, so it’s oxidized. $\ce{MnO4-}$ → $\ce{Mn^2+}$ : Mn drops from +7 to +2 as it gains electrons, so it’s reduced.
Write the two half-reactions.
$\begin{aligned} &\ce{MnO4- + 8H+ + 5e-} \\ &\ce{-> Mn^2+ + 4H2O} \\[4pt] &\ce{Fe^2+ -> Fe^3+ + e-} \end{aligned}$
Balance the electrons. $\ce{MnO4-}$ gains 5 electrons and $\ce{Fe^2+}$ loses 1, so multiply the Fe half-reaction by 5:
$\ce{5Fe^2+ -> 5Fe^3+ + 5e-}$
Add and cancel the electrons.
$\begin{aligned} &\ce{MnO4-(aq) + 5Fe^2+(aq)} \\ &\ce{+ 8H+(aq)} \\ &\ce{-> Mn^2+(aq) + 5Fe^3+(aq)} \\ &\ce{+ 4H2O(l)} \end{aligned}$
Read off the answers. $\ce{MnO4-}$ : $\ce{Fe^2+}$ = $\boxed{1 : 5}$ , with $\boxed{5}$ electrons transferred as written.
Check the charge.
$\begin{aligned} \text{left} &: -1 + 5(+2) + 8(+1) = +17 \\ \text{right} &: +2 + 5(+3) = +17 \end{aligned}$

Practice: is the named element oxidized, reduced, or neither?

Fe in Fe²⁺ → Fe³⁺
N in NO₃⁻ → NO
Cl in Cl₂ → 2Cl⁻
S in SO₃²⁻ → SO₄²⁻
Cr in Cr₂O₇²⁻ → Cr³⁺
Cr in CrO₄²⁻ → Cr₂O₇²⁻
Se in H₂Se → SeO₄²⁻
Mn in MnO₄⁻ → MnO₂
N in NH₄⁺ → NH₃
O in H₂O₂ → O₂
Cu in Cu²⁺ → Cu⁺

Common mistakes

Mistake 1: “The oxidizing agent is oxidized.” The oxidizing agent causes something else to be oxidized; it gains electrons itself, so it is reduced. The reducing agent loses electrons and is oxidized.

Mistake 2: “O is always −2.” In peroxides (H₂O₂, Na₂O₂), O is −1; in O₂F₂, O is +1, because the F rule comes first. Follow the order of the rules.

Mistake 3: “The electrons transferred equal one atom’s change in oxidation number.” Multiply by the number of atoms and the coefficient too. In the reaction of 2MnO₄⁻ with 10Br⁻, each Mn gains 5 electrons, and 10 electrons are transferred as written.

Quick check

What are the oxidation numbers of S in H₂SO₄ and in SO₂Cl₂?

What is the oxidation number of O in O₂F₂?

For 3Na₂O₂ + S → Na₂SO₄ + 2Na₂O, which statement is correct?

In 2MnO₄⁻ + 10Br⁻ + 16H⁺ → 2Mn²⁺ + 5Br₂ + 8H₂O, how many electrons are transferred as written?

Given MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻, what is the mole ratio of MnO₄⁻ to Fe²⁺ in the overall reaction?

Which of these is the strongest reducing agent?

In CO₂ + 4H₂ → CH₄ + 2H₂O, which elements are oxidized and reduced?

For Li₃N + 2H₂ → LiNH₂ + 2LiH, which statement is correct?

When H₂Se is converted to SeO₄²⁻, what happens to each Se atom?

In Cu + 2Ag⁺ → Cu²⁺ + 2Ag, what is the oxidation half-reaction?

AP-style free response

In acidic solution, KMnO₄ reacts with H₂O₂:

\ce{2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) -> 2Mn^2+(aq) + 5O2(g) + 8H2O(l)}

(a) Give the oxidation numbers of Mn and of O (in H₂O₂) before and after the reaction, and identify the oxidizing and reducing agents.

(b) Write the oxidation and reduction half-reactions, and state what each must be multiplied by to give the overall reaction above.

(c) How many electrons are transferred as written?

(d) A 20.00 mL sample of H₂O₂ solution is titrated with 0.0200 M KMnO₄, requiring 17.60 mL. Calculate the concentration of H₂O₂.

Model answer and scoring

(a) Mn: +7 → +2; O in H₂O₂: −1 → 0. The oxidizing agent is MnO₄⁻ (reduced), and the reducing agent is H₂O₂ (oxidized).

(b) Reduction and oxidation half-reactions:

\ce{MnO4-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}

\ce{H2O2(aq) -> O2(g) + 2H+(aq) + 2e-}

Multiply the reduction by 2 and the oxidation by 5 so each involves 10 electrons; add them, cancel the electrons, and cancel the extra H⁺ on both sides to get the overall reaction.

(c) 10 electrons.

(d) mol MnO₄⁻ = 0.0200 M × 0.01760 L = 3.52 × 10⁻⁴ mol; mol H₂O₂ = 3.52 × 10⁻⁴ × 5/2 = 8.80 × 10⁻⁴ mol; [H₂O₂] = 8.80 × 10⁻⁴ mol ÷ 0.02000 L = 0.0440 M.

Scoring: (a) oxidation numbers of both elements and both agents; (b) both half-reactions and the multipliers; (c) 10; (d) uses the 2 : 5 ratio to get 0.0440 M.

Summary

Oxidation is loss of electrons (oxidation number up); reduction is gain of electrons (oxidation number down). The oxidizing agent is reduced, and the reducing agent is oxidized.
Assign oxidation numbers in the order of the rules: F, Group 1 and 2 metals, H, O, halogens, then solve for the last element from the charge. O is −1 in peroxides and +1 in O₂F₂, and H is −1 in metal hydrides.
Electrons transferred = change per atom × number of atoms × coefficient; the total lost equals the total gained. One element can be both oxidized and reduced.
Multiply two half-reactions so the electrons match, then add them to get the overall reaction; the multipliers set the mole ratio, such as MnO₄⁻ : Fe²⁺ = 1 : 5.

AP tip: When asked how many electrons are transferred, find the elements that change, multiply by the number of atoms and the coefficient, and check with the other element that lost = gained. When comparing reducing agents, say which particle loses electrons more easily.

Oxidation-Reduction (Redox) Reactions氧化还原反应

Learning goals

Before you start

Oxidation, reduction, and electrons

Assigning oxidation numbers: the order of the rules

What is oxidized, what is reduced, and how many electrons

Half-reactions

Which reducing agent is stronger?

Example 1: Two unusual oxidation numbers

Example 2: Mole ratio and electrons from half-reactions

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem