Redox Reactions and Oxidation Numbers

Overview of Reactions in Aqueous Solution

Chapter 4 focuses on three key types of chemical reactions that occur in aqueous solutions:
- Precipitation reactions.
- Acid-base reactions.
- Oxidation-reduction reactions (frequently abbreviated as "redox" reactions).
The specific focus of this discussion is on understanding the mechanics, identification, and information contained within oxidation-reduction reactions.

Fundamentals of Oxidation-Reduction (Redox) Reactions

Definition: Oxidation-reduction reactions are essentially electron transfer reactions. This means that in every redox process, one reactant loses electrons and another reactant gains those same electrons.
The Magnesium-Oxygen Example:
- Consider the balanced chemical equation representing the reaction between magnesium and oxygen to form magnesium oxide: $2Mg + O_2 \rightarrow 2MgO$
- In the product magnesium oxide ( $MgO$ ), which is an ionic compound, the constituent elements carry specific charges:
  - Magnesium has a charge of $+2$ .
  - Oxygen has a charge of $-2$ .
- Interpretation of Charges:
  - A positive charge (cation) indicates the loss of electrons.
  - A negative charge (anion) indicates the gain of electrons.
- In this reaction, magnesium is losing electrons while oxygen is gaining them.

The Half-Reaction Method

Redox reactions can be analyzed by examining the "half-reaction" for each reactant individually, moving from the reactant side (left) to the product side (right).
Oxidation Half-Reaction:
- Represents the substance that loses electrons.
- For magnesium: $2Mg \rightarrow 2Mg^{2+} + 4e^-$
- There are two moles of magnesium. Since each magnesium becomes $+2$ , a total of four electrons ( $4e^-$ ) are lost.
- The substance that loses electrons is said to be oxidized.
Reduction Half-Reaction:
- Represents the substance that gains electrons.
- For oxygen: $O_2 + 4e^- \rightarrow 2O^{2-}$
- The four electrons lost by magnesium are gained by the oxygen molecules.
- The substance that gains electrons is said to be reduced.
Overall Reaction:
- The sum of the two half-reactions provides the overall redox process.
- Reactant side: $2Mg + O_2 + 4e^-$
- Product side: $2Mg^{2+} + 2O^{2-} + 4e^-$
- In the final balanced redox equation, the electrons are never shown. Because there are four moles of electrons on both the reactant and product sides, they cancel out, resulting in the net equation: $2Mg + O_2 \rightarrow 2MgO$ .

Identifying Oxidized and Reduced Species

Oxidized: The species that loses electrons.
Reduced: The species that gains electrons.
Reducing Agent: The substance that is oxidized. It acts as a reducing agent because its presence and subsequent loss of electrons are necessary for the other substance to be reduced.
Oxidizing Agent: The substance that is reduced. It acts as an oxidizing agent because it facilitates the oxidation of the other substance by accepting its electrons.
Example: Zinc and Copper Sulfate:
- Initial setup: A solid bar of zinc ( $Zn$ ) placed in a blue solution of copper sulfate ( $CuSO_4$ ).
- Reaction: $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$
- Analysis using charges:
  - Reactant Zinc ( $Zn$ ) has a charge of $0$ .
  - In $CuSO_4$ , Copper ( $Cu$ ) is $+2$ and Sulfate ( $SO_4$ ) is $-2$ .
  - In $ZnSO_4$ , Zinc ( $Zn$ ) is $+2$ and Sulfate ( $SO_4$ ) is $-2$ .
  - Product Copper ( $Cu$ ) has a charge of $0$ .
- Conclusion: Zinc goes from $0$ to $+2$ (loses $2e^-$ , oxidized, reducing agent). Copper goes from $+2$ to $0$ (gains $2e^-$ , reduced, oxidizing agent).

Oxidation Numbers: Definitions and Rules

Definition: An oxidation number is the charge an atom would have in a molecule or an ionic compound if electrons were completely transferred. It can be positive, negative, or zero (neutral).
The Seven Guidelines for Assigning Oxidation Numbers:
1. Free Elements: Atoms in their elemental state (uncombined with other elements) always have an oxidation number of $0$ . Examples: $Na, Be, K, Pb, H_2, O_2, P_4$ .
2. Monatomic Ions: The oxidation number is equal to the charge of the ion. Examples: $Li^+$ is $+1$ ; $Fe^{3+}$ is $+3$ ; $O^{2-}$ is $-2$ .
3. Oxygen: Usually $-2$ . (Note: In the specific case of superoxide, the oxidation number is $-1/2$ . In $H_2O_2$ , it is $-1$ , though these exceptions are secondary to the general rule).
4. Hydrogen: Usually $+1$ , except when bonded to metals in binary compounds (where it is $-1$ ).
5. Group 1A and 2A Metals: Elements in Group 1A (alkali metals) are always $+1$ . Elements in Group 2A (alkaline earth metals) are always $+2$ . Fluorine and Chlorine are usually $-1$ .
6. Sum of Oxidation Numbers:
  - In a neutral molecule, the sum of all oxidation numbers must equal $0$ .
  - In a polyatomic ion, the sum must equal the net charge of the ion.
7. Non-Integer States: Oxidation numbers do not have to be integers. Fractions (like $-1/2$ for oxygen in surperoxides) are acceptable and possible.

Practical Examples of Calculating Oxidation Numbers

Example A: $Li_2O$
- Lithium ( $Li$ ) is in Group 1A, so its oxidation number is $+1$ .
- Oxygen ( $O$ ) follows the general rule and is $-2$ .
Example B: $HNO_3$ (Nitric Acid)
- Hydrogen is $+1$ . Oxygen is $-2$ .
- Equation for neutral molecule: $1(H) + 1(N) + 3(O) = 0$
- Substitution: $1(+1) + N + 3(-2) = 0 \rightarrow 1 + N - 6 = 0$
- Result: $N = +5$ .
Example C: $Cr_2O_7^{2-}$ (Dichromate Ion)
- Oxygen is $-2$ . The total charge is $-2$ .
- Equation: $2(Cr) + 7(O) = -2$
- Substitution: $2(Cr) + 7(-2) = -2 \rightarrow 2Cr - 14 = -2$
- Calculation: $2Cr = +12 \rightarrow Cr = +6$ .

The Five Types of Redox Reactions

Combination Reaction:
- Two or more reactants combine to form a single product.
- Example: $2Al + 3Br_2 \rightarrow 2AlBr_3$
- Note: Coefficients from balancing are ignored when calculating individual oxidation numbers.
Decomposition Reaction:
- One reactant breaks down into two or more products.
- Example: $2KClO_3 \rightarrow 2KCl + 3O_2$
Combustion Reaction:
- A substance reacts with oxygen ( $O_2$ ), usually producing a flame.
- Example: $S + O_2 \rightarrow SO_2$
- Example: $2Mg + O_2 \rightarrow 2MgO$
Displacement Reaction:
- An atom or ion in a compound is replaced by an atom of another element ( $A + BC \rightarrow AC + B$ ).
- Hydrogen Displacement: A metal displaces hydrogen from water or acid. Example: $Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2$ .
- Metal Displacement: A metal displaces another metal in a compound. Example: $Ni + Pb(NO_3)_2 \rightarrow Ni(NO_3)_2 + Pb$ .
- Halogen Displacement: A more reactive halogen displaces a less reactive one (Activity: F_2 > Cl_2 > Br_2 > I_2). Example: $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$ .
Disproportionation Reaction:
- The same element is simultaneously oxidized and reduced within the same reaction.
- Example: $Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O$
  - Chlorine starts at $0$ .
  - In $ClO^-$ , Chlorine is $+1$ (oxidized).
  - In $Cl^-$ , Chlorine is $-1$ (reduced).
- Example: $2NO_2 + H_2O \rightarrow HNO_2 + HNO_3$
  - In $NO_2$ , Nitrogen is $+4$ .
  - In $HNO_2$ , Nitrogen is $+3$ (reduced).
  - In $HNO_3$ , Nitrogen is $+5$ (oxidized).

Activity Series of Metals

The activity series is a list of metals arranged by their ability to displace hydrogen or other metals.
Metals at the very top are the most reactive and can displace hydrogen from cold water.
Metals in the middle can displace hydrogen from steam or acids.
Metals at the bottom cannot displace hydrogen at all.
Halogen Reactivity: Fluorine is the most reactive, while Iodine is the least reactive.

Chemistry in Action: The Breath Analyzer

Law enforcement uses redox chemistry to detect blood alcohol levels via a breath analyzer.
The device contains potassium dichromate ( $K_2Cr_2O_7$ ), where Chromium ( $Cr$ ) is in the $+6$ oxidation state (typically orange/red).
When an individual breathes into the device, any alcohol (ethanol) present reacts with the dichromate.
The reaction reduces the Chromium from $Cr^{6+}$ to $Cr^{3+}$ .
This chemical change results in a visible color shift to green, indicating the presence of alcohol.