Stoichiometry and Redox Reactions

Stoichiometry in the context of redox reactions involves the calculation of the relative quantities of reactants and products involved in these reactions, ensuring that the number of electrons lost in oxidation matches the number of electrons gained in reduction. Let’s learn more in detail.

1.0Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes in which the oxidation states of atoms are changed. These reactions are characterized by the transfer of electrons between chemical species. Here's how they work:

1. Oxidation: This involves the loss of electrons by a molecule, atom, or ion. In this process, the oxidation state of the substance increases.
2. Reduction: Conversely, this involves the gain of electrons by a molecule, atom, or ion, leading to a decrease in the oxidation state of the substance.

A simple way to remember this is the mnemonic "OIL RIG": Oxidation Is Loss, Reduction Is Gain.

2.0Types of Redox Reaction

Combination Reactions

In combination redox reactions, two or more reactants combine to form a single product. These reactions often involve elements gaining or losing electrons to form a compound. An example is the reaction between hydrogen and oxygen to form water:

                 2H₂   +  O₂  →   2H₂O

Decomposition Reactions

Decomposition reactions involve a single compound breaking down into two or more simpler substances. These reactions can involve the release of electrons by one of the breakdown products. For instance, the decomposition of hydrogen peroxide to form water and oxygen:

                   2H₂O₂      →   2H₂O  +  O₂

Displacement Reactions

Displacement reactions occur when an element in a compound is replaced by another element. There are two types:

Single Displacement: An element displaces another element in a compound, typically involving a metal displacing another from its compound. For example:

 Zn  +  CuSO_4​    →   ZnSO₄ + Cu

Double Displacement: Two elements in different compounds swap places. This type is not always a redox reaction unless it involves a change in oxidation states.

Disproportionation (Dismutation) Reactions

In disproportionation reactions, a single element undergoes both oxidation and reduction in the same chemical reaction. This is a special type of redox reaction where an element in one oxidation state is transformed into two different products, each in a different oxidation state. An example is the reaction of hydrogen peroxide, catalyzed by iodine:

                 2H₂O₂  →   2H₂O  +  O₂

3.0Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. It is based on the conservation of mass and the concept of moles. Here we will discuss Important Basic Concepts of Stoichiometry-

1. Moles: The mole is a unit used to express amounts of a chemical substance, defined as containing exactly 6.022×10²³ particles (atoms, molecules, ions, or electrons).
2. Molar Mass: The mass of one mole of a substance, usually expressed in grams per mole (g/mol), which is numerically equivalent to the atomic or molecular mass in atomic mass units (amu).
3. Avogadro's Number: 6.022×10²³, the number of particles in one mole of any substance.
4. Balanced Chemical Equation: A chemical equation in which the number of atoms for each element is the same on both the reactant and product sides, ensuring the law of conservation of mass is obeyed.

Further we will understand steps of redox stoichiometry Calculations, Stoichiometric Coefficient and Limiting reactant.

What are the steps of redox stoichiometry Calculations?

1. Write the Balanced Equation: Ensure the chemical equation is balanced, with equal numbers of each type of atom on both sides of the equation.
2. Convert Mass to Moles: If dealing with mass, convert it to moles using the molar mass of the substances involved.
3. Use Mole Ratios: Apply the mole ratios derived from the balanced equation to find out how many moles of other substances are produced or consumed in the reaction.
4. Convert Moles Back to Desired Units: Convert the mole amounts back into grams or another unit if necessary, using the molar mass.

Stoichiometric Coefficient

The stoichiometric coefficient of a substance in a chemical equation represents the ratio in which chemicals react and are produced. It is the number placed in front of the chemical formulas in a balanced chemical equation, indicating the number of moles of each substance involved in the reaction. For instance, in the reaction:

2H₂ + O₂ → 2H₂O

The coefficients 2, 1, and 2 for hydrogen, oxygen, and water, respectively, indicate that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Limiting Reactant

The limiting reactant in a chemical reaction is the reactant that is entirely consumed and thus determines the maximum amount of product that can be formed. This concept is crucial when the reactants are not present in the exact ratios dictated by the balanced chemical equation.

Identifying the Limiting Reactant

1. Calculate the mole ratio of the reactants provided in the reaction to the stoichiometric ratio in the balanced equation.
2. Determine which reactant has the least amount in terms of the stoichiometric ratio. This is your limiting reactant, as it will be completely used up first, stopping the reaction.

4.0Solved Examples

Let’s learn with the help of examples

Example 1: Balanced Chemical Equation:

CH₄  +  2O₂  →  CO₂  +  2H₂O

Steps:

1. Calculate moles of methane:

• Molar mass of CH₄ is approximately 16 g/mol.
• Moles of CH₄ = Mass / Molar Mass =16 g/mol​  = 1 mole.

2. Use mole ratios from the balanced equation:

• The equation shows that 1 mole of methane produces 2 moles of water.
• Moles of H_2​O produced = 1 mole CH₄​ × 2 moles H₂O / mole CH₄= 2 moles H₂​O.

3. Convert moles of water to grams:

• Molar mass of H₂O is approximately 18 g/mol.
• Mass of H₂O = Moles × Molar Mass = 2 moles × 18 g/mol = 36 grams.

Example 2:

Consider the reaction:

4NH₃  +  5O₂  →  4NO +  6H₂O

Suppose we have 10 moles of NH₃ (ammonia) and 10 moles of O₂ (oxygen):                                       

• The mole ratio of NH₃ to O₂ in the balanced equation is 4:5.
• For 10 moles of NH₃, you would need 5/4×10=12.5 moles of O₂.
• Since only 10 moles of O₂​ are available, O₂ is the limiting reactant.