CR - Reactions Revisited: Metathesis and Redox (Lesson)

Reactions Revisited: Metathesis and Redox

For AP Chemistry, we will recategorize the reactions types you reviewed in the earlier lesson of this module into two broad groups: Metathesis reactions and oxidation/reduction reactions. 

Metathesis Reactions

Metathesis Reactions are double displacement reactions along with other types:

• Acid-base neutralization
• Salt formation
• Hydrocarbon reactions other than oxidation (rare in AP Chemistry)

This type of reaction has a 'driving force" - a reason - the reaction will occur that involves removing at least one pair of ions from the solution. This removal can include:

1. Formation of a gas. Gases may form directly in a double displacement reaction or from the decomposition of one of the products. The gases will bubble off or evolve from the solution. 

Some common gases formed include H₂S, CO₂, SO₂, and NH₃.  

Gas Formations

COMMON GASES	PRODUCTS THAT BECOME GASES
H2S	Any sulfide ion plus any acid form H2S and a salt
CO2	Any carbonate ion plus any acid will form CO2, HOH, and a salt
SO2	Any sulfite ion plus any acid form SO2, HOH, and a salt
NH3	Any ammonium ion plus any soluble strong hydroxide react upon heating to form NH3 (g), HOH, and a salt.

2. 2. Formation of a molecule. The formation of molecules in solution removes ions from the solution and the reaction happens. Weak electrolytes will join together to form these molecules. Water formation in an acid/base neutralization is a great illustration of this driving force.

Weak electrolytes are species that stay as molecules in solution, rather than breaking into ions. Weak acids and nitrogen bases like ammonia are the most common weak electrolytes. 

Weak Acids/Weak Bases

WEAK ACIDS	WEAK BASES
Formic Acid (HCOOH) Acetic Acid (CH3COOH) Benzoic Acid (C6H5COOH) Hydrofluoric Acid (HF) Phosphoric Acid (H3PO4) Sulfurous Acid (H2SO3) Carbonic Acid (H2CO3) Nitrous Acid (HNO2)	Ammonia (NH3) Trimethylamine (N(CH3)3) Methylamine (CH3NH2)

Actually, it is easier to remember the strong acids than the extensive list of weak acids. A silly way to remember the strong acid list goes "hickle, hyber, hi, HoNo, Hicklo, H₂SO₄."

3. Formation of a precipitate. A precipitate is an insoluble substance (solid) formed by the reaction. This happens when ions bond so strongly the solvent cannot pull them apart.  This removes a cation and an anion from the solution.  Using a table of solubility will let you predict which reactant ions are likely to join to form a precipitant. 

Redox (Short for Oxidation and Reduction)

When electrons are transferred, the oxidation numbers of at least two elements must change. 

• synthesis
• decomposition
• single displacement
• combustion
• other

There are many redox reactions that fall into the 'other' or atypical category. 

To determine if a reaction is a redox reaction, look at the reactants to see if there is both an oxidizing agent (a substance that tends to gain electrons) and a reducing agent (a substance that tends to lose electrons). If the problem states that the reaction occurs in either an acidic or basic solution, the reaction is most likely redox: hydrogen ions, hydroxide ions, and water should be included as appropriate. 

The following charts show the common oxidizing and reducing agents and their products. Have these tables handy as you write chemical reactions. 

Redox Half Reactions

In order to determine if a particular reaction is a redox reaction and be able to write the half reactions, you must look at the oxidation numbers of each element. If an element undergoes a change in oxidation number, a redox reaction has occurred.  

Let's start by learning how to assign oxidation numbers. The following set of rules is helpful in assigning oxidation numbers. These rules actually need to be applied in the order listed. If there is a contradiction, follow the rule with the lowest number.

When balancing the net ionic forms of redox reactions, focus on oxidation state, charge, and accounting for all particles. The method is sometimes called the half-reaction method.

Follow these steps to balance and write redox half reactions:

1. Assign oxidation numbers to each atom in the equation.
2. Predict the products using a chart or likely behavior of species given as reactants. Likely behavior for example would include potassium ions (and sodium ions) as a reactant. K⁺ will generally not be reduced to elemental potassium if anything else could be reduced.
3. Spectator ions (those not involved in the redox process) need to be eliminated.
4. Identify what is being reduced as well as what is being oxidized.  Write two equations, one showing the reduction and the other showing the oxidation.
5. Use coefficients to balance the half reactions for
   • charges
   • elements, oxygen, hydrogen
   • electrons lost and gained
6. Add half reactions back together. 

Example One: Fe⁺² ions are added to a potassium permanganate solution under acidic conditions.

• A source of H⁺ ions is needed for this reaction to happen
• Checking the list of common oxidations and reductions, you find that permanganate is commonly reduced. 
• In permanganate, MnO₄^2-, the manganese has a +7 oxidation number. In reduction, this becomes the manganese +2 ion. It is reduced from +7 to +2 by gaining 5 electrons. 
• Checking the oxidation common reactions, you can see that multivalent cations oxidize by going from the lower oxidation number to the higher state, releasing electrons. This is noted as metalous $\longrightarrow$ metallic form. So iron represents the oxidation here going from Fe²⁺ to Fe³⁺.

Now for balancing these reactions. Our first obvious problem is that oxygen is present in permanganate, but missing as a product. To resolve this, it's time to make use of the 'acidic conditions' described in the prompt.

Acids contribute hydrogen ions to solution, and those H⁺ ions will take up the oxide ions from the polyatomic ion and form water. Since there are four oxide ions needing to join with hydrogen ions, adding 8 H⁺ as a reactant will allow 4 H₂O to form as a product. 

Now to balance the half reactions to account for electrons. We see that 5 electrons are being consumed in the reduction, and the source must be the oxidation reaction. Here, each iron II ion donates one e- for the reduction to happen.  So to balance we need a total of 5 electrons - the five from the oxidation are the five going to the reduction.

Double check the reactions are now balanced for charge. The reduction reactants charge totals +2 (-1 + -5 + +8) and match the net +2 charge on the product side.  The oxidation reactant charge totals +10 (5 x -2 = -10). So now the half reactions can be added back together to the net ionic equation. 

Example Two: Cadmium sulfide is reacted with iodine and hydrogen chloride to produce cadmium chloride, hydrogen iodide, and sulfur. 

CdS + I₂ + HCl --> CdCl₂ + HI + S

Step 1: Oxidation numbers

Cd = +2  S = -2  I2 = 0  H = +1   Cl = -1  Cd = +2   Cl = -1  H = +1  I = -1  S = 0

(elements in their atomic or molecular form have oxidation number zero)

Step 2: 

Sulfur goes from being a -2 reactant to zero as a product.  This is an oxidation (more positive)

Iodine goes from being a zero reactant to -1 as a product. This is a reduction (more negative)

Step 3:

Oxidation: the sulfur releases two electrons in the reaction that can be used in the reduction.

Reduction: the two iodine atoms each gain one electron, for a total of two needed from the oxidation.

Step 4:

ox:  S^2- --> S + 2e-

red: I₂ + 2e- --> 2 I^-1

Steps 5 and 6:  these two equations need to be combined so that the electrons donated by the oxidation to the reduction are equal in number, and removed since they are found on both sides. In this example the electrons do balance.

S^2- + I₂ + 2e- --> S + 2e- + 2 I₂

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