Question

Balance these redox reactions by inspection. a) \(\mathrm{Na}+\mathrm{F}_{2} \rightarrow \mathrm{NaF}\) b) \(\mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \rightarrow \mathrm{Al}+\mathrm{H}_{2} \mathrm{O}\)

Short answer

a) \(2\mathrm{Na} + \mathrm{F}_2 \rightarrow 2\mathrm{NaF}\); b) \(\mathrm{Al}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{Al} + 3\mathrm{H}_2\mathrm{O}\).

Step-by-step solution

Identify Compounds and Their Forms

For reaction (a), identify sodium (\(\mathrm{Na}\)) and fluorine (\(\mathrm{F}_2\)) as reactants. On the product side, we have sodium fluoride (\(\mathrm{NaF}\)). For reaction (b), aluminum oxide (\(\mathrm{Al}_2\mathrm{O}_3\)) and hydrogen gas (\(\mathrm{H}_2\)) are the reactants, with aluminum (\(\mathrm{Al}\)) and water (\(\mathrm{H}_2\mathrm{O}\)) as products.

Check Elements with Single Occurrence

In reaction (a), check the elements present: sodium and fluorine. Each appears only once on both sides. In reaction (b), identify that aluminum and oxygen atoms appear once uniquely, but hydrogen is on both sides in more than one compound.

Balance the Elements in Reaction (a)

Reaction (a) is \(\mathrm{Na} + \mathrm{F}_2 \rightarrow \mathrm{NaF}\). To balance this, observe that there are two fluorine atoms in \(\mathrm{F}_2\) but only one fluorine in \(\mathrm{NaF}\). Place a coefficient of 2 in front of \(\mathrm{NaF}\), then also place a 2 in front of \(\mathrm{Na}\) to balance the sodium:\[2\mathrm{Na} + \mathrm{F}_2 \rightarrow 2\mathrm{NaF}\].

Balance the Elements in Reaction (b)

For \(\mathrm{Al}_2 \mathrm{O}_3 + \mathrm{H}_2 \rightarrow \mathrm{Al} + \mathrm{H}_2 \mathrm{O}\), start with aluminum. \(\mathrm{Al}_2\mathrm{O}_3\) has 2 aluminum atoms, so place a 2 in front of \(\mathrm{Al}\). Write the partially balanced reaction:\[\mathrm{Al}_2\mathrm{O}_3 + \mathrm{H}_2 \rightarrow 2\mathrm{Al} + \mathrm{H}_2\mathrm{O}\].

Balance Oxygen and Recheck Reaction (b)

In \(\mathrm{Al}_2\mathrm{O}_3\), there are 3 oxygen atoms. Put a 3 in front of \(\mathrm{H}_2\mathrm{O}\):\[\mathrm{Al}_2\mathrm{O}_3 + \mathrm{H}_2 \rightarrow 2\mathrm{Al} + 3\mathrm{H}_2\mathrm{O}\].

Balance Hydrogen in Reaction (b)

From the 3 molecules of \(\mathrm{H}_2\mathrm{O}\), we have 6 hydrogen atoms. Therefore, put a 3 in front of \(\mathrm{H}_2\) to balance the hydrogen:\[\mathrm{Al}_2\mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{Al} + 3\mathrm{H}_2\mathrm{O}\].

Verify the Balance

Check that both reactions now have balanced atoms on both sides. For reaction (a), each element (Na and F) appears equally on both sides. For reaction (b), aluminum has 2 atoms, oxygen has 3 atoms, and hydrogen has 6 atoms on both sides.

Key concepts

Chemical Equations

Chemical equations are like recipes for chemical reactions. They tell us which substances start off or react (the reactants) and which substances are formed or produced (the products). Each side of a chemical equation must have the same number of each type of atom. This is known as balancing the equation.

For example, in the unbalanced equation for sodium reacting with fluorine, \(\mathrm{Na} + \mathrm{F}_2 \rightarrow \mathrm{NaF}\), we start by identifying the atoms involved: sodium (Na) and fluorine (F). There are two fluorine atoms on the left but only one in the sodium fluoride (NaF) on the right. This is why we adjust by adding coefficients, so we put a 2 in front of \(\mathrm{NaF}\), which also requires us to put a 2 in front of \(\mathrm{Na}\). Now both sides are balanced with two sodium and two fluorine atoms.

• Reactants are the starting substances.
• Products are the substances formed.
• Balance equations by making sure each atom type has equal numbers on both sides.

Stoichiometry

Stoichiometry allows chemists to relate the quantities of reactants and products in a chemical reaction. It involves using balanced equations to calculate the amount of reactants needed or products formed. Balancing chemical equations is the first step in stoichiometry because it sets the correct proportions needed to carry out calculations.

When balancing the equation for the reaction \(\mathrm{Al}_2 \mathrm{O}_3 + \mathrm{H}_2 \rightarrow \mathrm{Al} + \mathrm{H}_2 \mathrm{O}\), stoichiometry helps determine how much of each substance is involved. First, balance aluminum by noting \(\mathrm{Al}_2\) requires 2 aluminum atoms, hence a 2 is placed in front of \(\mathrm{Al}\). Match the amount of oxygen next and, finally, the hydrogen atoms. Stoichiometry also allows us to predict how much water we'd get (3 molecules) if we start with 2 aluminum oxides.

• It's about the ratio of compounds in reactions.
• Helps in calculating reactant and product amounts.
• Relies on balanced chemical equations for accurate predictions.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between substances. Oxidation is the process of losing electrons, while reduction is gaining electrons. Recognizing which substances are oxidized and which are reduced is key to understanding redox reactions.

In the reaction \(\mathrm{Al}_2 \mathrm{O}_3 + 3\mathrm{H}_2 \rightarrow 2\mathrm{Al} + 3\mathrm{H}_2 \mathrm{O}\), aluminum is reduced as it gains electrons to form aluminum metal, and hydrogen is oxidized as it reacts with oxygen to form water.

• The substance undergoing oxidation loses electrons.
• The substance undergoing reduction gains electrons.
• Balancing redox reactions takes into account electron transfer.

Understanding redox reactions involves checking how oxidation states change across the reaction and ensuring everything, including charge and mass, is balanced.