Chemical reactions Revision Notes

The Role of Balanced Chemical Equations (Higher)

A balanced chemical equation provides the essential molar ratios between reactants and products. For example, in the reaction 2H₂ + O₂ → 2H₂O, it indicates that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. These coefficients (the numbers in front of the chemical formulas) are critical for all quantitative calculations. They represent the smallest whole-number ratio of particles involved. Without a correctly balanced equation, any mass calculations will be inaccurate. Ensure the equation is balanced for both atoms and charges (if applicable for ionic reactions).

Calculating Moles from Mass (Higher)

The number of moles of a substance can be calculated using its mass and its relative formula mass (Mr). The formula is: moles = mass / Mr. The relative formula mass (Mr) is determined by adding the relative atomic masses (Ar) of all the atoms present in the chemical formula. For example, for H₂O, Mr = (2 × Ar of H) + (1 × Ar of O). For instance, if you have 18 g of water, and Mr of water is 18, then moles of water = 18 g / 18 g/mol = 1 mol. This step is usually the first calculation when reacting a known mass.

Using Mole Ratios (Higher)

Once you have the moles of a known substance, you use the mole ratio from the balanced chemical equation to find the moles of an unknown substance. This ratio is expressed by the coefficients in the balanced equation. If the equation is A + 2B → C, and you know the moles of A, you can find the moles of B by multiplying the moles of A by 2, or the moles of C by multiplying the moles of A by 1 (assuming a 1:1 ratio for A and C). This step links the known quantity to the unknown quantity through the stoichiometry of the reaction. It is a direct proportional relationship.

Calculating Mass from Moles (Higher)

The final step is to convert the calculated moles of the unknown substance back into a mass. This is done by rearranging the moles formula: mass = moles × Mr. You will need to calculate the relative formula mass (Mr) for the unknown substance. For example, if you calculated that you need 2 moles of oxygen (O₂) and its Mr is 32, then the mass of oxygen needed is 2 mol × 32 g/mol = 64 g. This provides the final answer in the desired unit of mass, typically grams.

Understanding Ion Charges

Ions are atoms or molecules that have gained or lost electrons, resulting in a net electrical charge. Common metal ions from Group 1 (e.g., Na+, K+) have a +1 charge, Group 2 (e.g., Mg2+, Ca2+) have a +2 charge, and Group 3 (e.g., Al3+) have a +3 charge. Common non-metal ions from Group 7 (e.g., Cl-, Br-) have a -1 charge, and Group 6 (e.g., O2-, S2-) have a -2 charge. Some common polyatomic ions include nitrate (NO3-), sulfate (SO4^2-), and carbonate (CO3^2-).

The Principle of Charge Balance

In any stable ionic compound, the total positive charge from the metal cations must exactly balance the total negative charge from the non-metal anions. This ensures that the compound has no overall electrical charge. This principle is fundamental to correctly deducing chemical formulae. If the charges are not balanced, the formula represents an unstable or incorrect arrangement of ions.

Steps to Deduce a Formula

To deduce the formula of an ionic compound:

1. Write down the symbol and charge for each ion.
2. Determine the lowest common multiple (LCM) of the magnitudes of the charges.
3. Divide the LCM by the magnitude of each ion's charge to find the number of each ion needed.
4. Write the formula using subscripts to indicate the number of each ion. If only one ion is needed, do not write '1'. If polyatomic ions are needed in groups of more than one, enclose the polyatomic ion in parentheses before adding the subscript.

Examples of Deducing Formulae

For Calcium Oxide: Calcium (Ca2+), Oxide (O2-). Charges are +2 and -2. They balance 1:1. Formula: CaO. For Lithium Fluoride: Lithium (Li+), Fluoride (F-). Charges are +1 and -1. They balance 1:1. Formula: LiF. For Aluminium Oxide: Aluminium (Al3+), Oxide (O2-). LCM of 3 and 2 is 6. Need two Al3+ (2 x +3 = +6) and three O2- (3 x -2 = -6). Formula: Al2O3. For Potassium Sulfate: Potassium (K+), Sulfate (SO4^2-). LCM of 1 and 2 is 2. Need two K+ (2 x +1 = +2) and one SO4^2- (1 x -2 = -2). Formula: K2SO4.

The pH Scale Basics

The pH scale measures the relative acidity or alkalinity of an aqueous solution. It ranges from 0 to 14. A pH of 7 indicates a neutral solution, such as pure water. Solutions with a pH below 7 are acidic, while solutions with a pH above 7 are alkaline (or basic). The closer the pH is to 0, the stronger the acid. The closer the pH is to 14, the stronger the alkali. This scale helps us classify various substances based on their chemical properties related to hydrogen ion concentration.

Acids and Alkalis

Acids are substances that have a pH less than 7. Examples include lemon juice (pH ~2), vinegar (pH ~3), and hydrochloric acid (pH 0-1). These solutions contain a higher concentration of hydrogen ions (H⁺). Alkalis (or bases) are substances that have a pH greater than 7. Examples include baking soda solution (pH ~9), soap (pH ~10), and sodium hydroxide (pH 13-14). These solutions contain a higher concentration of hydroxide ions (OH⁻). Neutral solutions, like pure water, have a balanced concentration of H⁺ and OH⁻ ions.

Strength of Acids and Alkalis

The strength of an acid or alkali is indicated by how far its pH value is from 7. A strong acid has a very low pH (e.g., pH 1 or 2), meaning it fully dissociates in water to produce a high concentration of H⁺ ions. A weak acid (e.g., pH 5 or 6) only partially dissociates, producing fewer H⁺ ions. Similarly, a strong alkali has a very high pH (e.g., pH 13 or 14), fully dissociating to produce many OH⁻ ions. A weak alkali (e.g., pH 8 or 9) partially dissociates, producing fewer OH⁻ ions.

Ions Present in Aqueous Solutions

When an ionic compound is dissolved in water, the solution contains ions from the dissolved compound and ions from water itself. Water dissociates to produce a small concentration of hydrogen ions (H+) and hydroxide ions (OH-). Therefore, an aqueous solution of an ionic compound contains: metal cations (from the compound), H+ ions (from water), non-metal anions (from the compound), and OH- ions (from water). These four types of ions compete to be discharged at the electrodes during electrolysis.

Reactions at the Cathode (Negative Electrode)

At the cathode, positive ions (cations) move towards the electrode and gain electrons (reduction). There is a competition between the metal cations and hydrogen ions (H+). The less reactive ion will be discharged. If the metal is more reactive than hydrogen (e.g., potassium, sodium, calcium, magnesium, aluminium, zinc, iron, lead), then hydrogen gas will be produced. If the metal is less reactive than hydrogen (e.g., copper, silver, gold, platinum), then the metal itself will be deposited. For example, in copper(II) chloride solution, copper is less reactive than hydrogen, so copper metal is produced.

Reactions at the Anode (Positive Electrode)

At the anode, negative ions (anions) move towards the electrode and lose electrons (oxidation). There is a competition between the non-metal anions and hydroxide ions (OH-). If halide ions (Cl-, Br-, I-) are present in a reasonably concentrated solution, then the corresponding halogen gas will be produced. For example, chloride ions will produce chlorine gas. If other anions like sulfate (SO42-) or nitrate (NO3-) are present, or if halide ions are very dilute, then oxygen gas will be produced from the hydroxide ions.

Summary of Products for Aqueous Electrolysis

To summarise: At the cathode, if the metal is more reactive than hydrogen, then hydrogen gas is produced. If the metal is less reactive than hydrogen, the metal is produced. At the anode, if concentrated halide ions (Cl-, Br-, I-) are present, the halogen is produced. If no halides, or if they are dilute, or if other anions (like sulfate, nitrate) are present, then oxygen gas is produced from the hydroxide ions. This selective discharge is crucial for predicting the products of electrolysis.