Oxidation-Reduction Reactions and Solution Stoichiometry

Oxidation-Reduction Reactions

Introduction to Redox Reactions

Oxidation-reduction (redox) reactions are fundamental chemical processes involving the transfer of electrons between species. These reactions are essential in both biological and industrial contexts, such as cellular respiration and metal extraction.

• Oxidation: The loss of electrons by a substance.

• Reduction: The gain of electrons by a substance.

• Oxidation and reduction always occur together in a redox reaction.

• Redox reactions are often summarized by the mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) or LEO the lion says GER (Lose Electrons = Oxidation, Gain Electrons = Reduction).

Example: In the reaction between calcium and oxygen to form calcium oxide:

• Calcium is oxidized (loses electrons).

• Oxygen is reduced (gains electrons).

Oxidation Numbers

Definition and Purpose

To identify redox processes, chemists assign an oxidation number (or oxidation state) to each element in a compound or ion. This is a formalism for electron bookkeeping and does not necessarily represent actual charges on atoms.

• Oxidation numbers help determine which atoms are oxidized and which are reduced in a reaction.

Rules for Assigning Oxidation Numbers

• Atoms in their elemental form have an oxidation number of zero. Example: $\text{O}_2$, $\text{H}_2$, $\text{Na}$ all have oxidation number 0.

• The oxidation number of a monatomic ion equals its charge. Example: $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1.

• Nonmetals usually have negative oxidation numbers, but exceptions exist:

  • Oxygen: Usually -2, except in peroxides (e.g., $\text{H}_2\text{O}_2$) where it is -1.

  • Hydrogen: +1 when bonded to nonmetals, -1 when bonded to metals.

  • Fluorine: Always -1. Other halogens: usually -1, unless combined with oxygen (oxyanions), where they can be positive.

• The sum of oxidation numbers in a neutral compound is zero.

• The sum of oxidation numbers in a polyatomic ion equals the ion's charge.

• When assigning oxidation numbers, count every atom, including those indicated by subscripts.

Examples: Assigning Oxidation Numbers

• In which compound is the oxidation state of oxygen -1?

  • (a) $\text{P}_2\text{O}_5$ : O = -2

  • (b) $\text{H}_2\text{O}$ : O = -2

  • (c) $\text{H}_2\text{SO}_4$ : O = -2

  • (d) $\text{H}_2\text{O}_2$ : O = -1 (peroxide)

  • (e) $\text{KCH}_3\text{COO}$ : O = -2

• Find the oxidation state of the boldfaced element:

  • (a) $\text{P}_2\text{O}_5$: $2x + 5(-2) = 0 \implies x = +5$

  • (b) $\text{NaH}$: H = -1

  • (c) $\text{Cr}_2\text{O}_7^{2-}$: $2x + 7(-2) = -2 \implies x = +6$

  • (d) $\text{SnBr}_4$: $x + 4(-1) = 0 \implies x = +4$

  • (e) $\text{BaO}_2$: O = -1 (peroxide)

Displacement Reactions and Activity Series

Displacement (Single Replacement) Reactions

In displacement reactions, an element reacts with a compound and displaces another element from it. This is common in redox chemistry, especially with metals and acids.

• An ion in solution is replaced by another element.

• Active metals above hydrogen in the activity series will react with acids to produce hydrogen gas.

• Less active (noble) metals do not react with acids under standard conditions.

Activity Series of Metals

The activity series ranks metals by their tendency to be oxidized (lose electrons). A metal higher in the series will displace a metal ion lower in the series from solution.

Metal	Relative Activity
Potassium (K)	Most easily oxidized
Sodium (Na)	Very active
Magnesium (Mg)	Active
Zinc (Zn)	Moderately active
Iron (Fe)	Less active
Hydrogen (H)	Reference
Copper (Cu)	Less easily oxidized
Silver (Ag)	Very unreactive
Gold (Au)	Least easily oxidized

Additional info: The full activity series includes more metals, but this table summarizes the main trend.

Examples of Displacement Reactions

• Reaction of magnesium with iron(II) chloride:

  • Molecular: $\text{Mg}(s) + \text{FeCl}_2(aq) \rightarrow \text{MgCl}_2(aq) + \text{Fe}(s)$

  • Net ionic: $\text{Mg}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Fe}(s)$

• Reaction of aluminum with hydrobromic acid:

  • Molecular: $2\text{Al}(s) + 6\text{HBr}(aq) \rightarrow 2\text{AlBr}_3(aq) + 3\text{H}_2(g)$

  • Net ionic: $2\text{Al}(s) + 6\text{H}^+(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{H}_2(g)$

• Reaction of zinc with copper(II) sulfate:

  • Molecular: $\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$

  • Net ionic: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

  • In this reaction: Zinc is oxidized (loses electrons), copper(II) ion is reduced (gains electrons).

Concentration of Solutions

Molarity and Solution Calculations

The concentration of a solution is a measure of the amount of solute dissolved in a given quantity of solvent. Molarity (M) is the most common unit of concentration in chemistry.

• Molarity (M): $\text{M} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$

• Molarity can be used as a conversion factor between moles and liters in stoichiometric calculations.

Example: Calculating Molarity

• Problem: What is the molarity of a solution made by dissolving 23.4 g of sodium sulfate ($\text{Na}_2\text{SO}_4$) in enough water to make 125 mL of solution?

• Step 1: Calculate moles of $\text{Na}_2\text{SO}_4$:

  • Molar mass of $\text{Na}_2\text{SO}_4$ = 142 g/mol

  • Moles = $\frac{23.4\ g}{142\ g/mol} = 0.165\ mol$

• Step 2: Convert volume to liters:

  • 125 mL = 0.125 L

• Step 3: Calculate molarity:

  • $\text{M} = \frac{0.165\ mol}{0.125\ L} = 1.32\ M$

Summary Table: Key Concepts

Concept	Definition/Rule	Example
Oxidation	Loss of electrons	$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
Reduction	Gain of electrons	$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$
Oxidation Number	Bookkeeping charge assigned to atoms	O in $\text{H}_2\text{O}_2$ is -1
Molarity	Moles of solute per liter of solution	$\text{M} = \frac{0.165}{0.125} = 1.32\ M$