Question

Balance each of the following oxidation-reduction chemical reactions. a. \(\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\) b. \(\operatorname{MgO}(s)+\mathrm{C}(s) \rightarrow \mathrm{Mg}(s)+\mathrm{CO}(g)\) c. \(\operatorname{Sr}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Sr}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{Co}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{CoCl}_{2}(a q)+\mathrm{H}_{2}(g)\)

Short answer

The short answers of the balanced redox reactions are: a. \(P_{4}(s) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s)\) b. \(MgO(s) + C(s) \rightarrow Mg(s) + CO(g)\) c. \(Sr(s) + 2H_{2}O(l) \rightarrow Sr(OH)_{2}(aq) + H_{2}(g)\) d. \(Co(s) + 2HCl(aq) \rightarrow CoCl_{2}(aq) + H_{2}(g)\)

Step-by-step solution

Identify the oxidation and reduction half-reactions

The first step is to identify the oxidation and reduction half-reactions. In this case, we can see that \(\mathrm{P}\) is gaining oxygen, while \(\mathrm{O}\) is losing it: Oxidation half-reaction: \(\mathrm{P}_{4} \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}\) Reduction half-reaction: \(\mathrm{O}_{2} \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}\)

Balance atoms other than O and H

In this case, since there are no hydrogen atoms, we only need to worry about balancing the phosphorus atoms, which are already balanced in both half-reactions.

Balance oxygen atoms

In the oxidation half-reaction, there are 10 oxygen atoms in the product: \(\mathrm{P}_{4} \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}\) In the reduction half-reaction, we need to add 10 oxygen atoms to the product side to balance them: \(\mathrm{O}_{2} \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}\)

Balance charges

Both half-reactions are neutral, so there is no need to balance charges.

Combine half-reactions

Finally, we can add the two half-reactions together to get the complete balanced equation: \(\mathrm{P}_{4} + 5\mathrm{O}_{2} \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}\) The balanced equation for reaction a is: \(\mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \rightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)\) ##Reaction b: \(\operatorname{MgO}(s)+\mathrm{C}(s) \rightarrow \mathrm{Mg}(s)+\mathrm{CO}(g)\)## For the next three reactions, we will follow the same steps as the first reaction.

Identify the oxidation and reduction half-reactions

Oxidation half-reaction: \(\mathrm{C} \rightarrow \mathrm{CO}\) Reduction half-reaction: \(\mathrm{MgO} \rightarrow \mathrm{Mg}\)

Balance atoms other than O and H

Atoms of carbon and magnesium are already balanced in both half-reactions.

Balance oxygen atoms

The oxidation half-reaction is already balanced for oxygen, and for the reduction half-reaction, we need to add one oxygen atom to the product side: \(\mathrm{MgO} \rightarrow \mathrm{Mg}+\mathrm{O}\)

Balance charges

There are no charges to balance in this reaction.

Combine half-reactions

Combining the half-reactions gives the complete balanced equation: \(\mathrm{C}+\mathrm{MgO} \rightarrow \mathrm{Mg}+\mathrm{CO}\) The balanced equation for reaction b is: \(\mathrm{MgO}(s) + \mathrm{C}(s) \rightarrow \mathrm{Mg}(s) + \mathrm{CO}(g)\) ##Reaction c: \(\operatorname{Sr}(s) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{Sr}(\mathrm{OH})_{2}(a q) + \mathrm{H}_{2}(g)\)##

Identify the oxidation and reduction half-reactions

Oxidation half-reaction: \(\mathrm{Sr} \rightarrow \mathrm{Sr}(\mathrm{OH})_{2}\) Reduction half-reaction: \(\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{2}\)

Balance atoms other than O and H

The strontium and hydrogen atoms are already balanced in both half-reactions.

Balance oxygen atoms

There are two oxygen atoms in the oxidation half-reaction product, so we need to balance the oxygen atoms in the reduction half-reaction: \(\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{2}+\mathrm{O}_{2}\)

Balance charges

There are no charges to balance in this reaction.

Combine half-reactions

Combining the half-reactions gives the complete balanced equation: \(\mathrm{Sr}+\mathrm{H}_{2}\mathrm{O}+\mathrm{O}_{2} \rightarrow \mathrm{Sr}(\mathrm{OH})_{2}+\mathrm{H}_{2}\) However, we can simplify the reaction by removing the oxygen molecules on both sides. Thus, the balanced equation for reaction c is: \(\mathrm{Sr}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{Sr}(\mathrm{OH})_{2}(a q) + \mathrm{H}_{2}(g)\) ##Reaction d: \(\mathrm{Co}(s) + \mathrm{HCl}(a q) \rightarrow \mathrm{CoCl}_{2}(a q) + \mathrm{H}_{2}(g)\)##

Identify the oxidation and reduction half-reactions

Oxidation half-reaction: \(\mathrm{Co} \rightarrow \mathrm{CoCl}_{2}\) Reduction half-reaction: \(\mathrm{HCl} \rightarrow \mathrm{H}_{2}\)

Balance atoms other than O and H

The atoms of cobalt and hydrogen are already balanced in both half-reactions, as well as the chlorine atoms in the oxidation half-reaction.

Balance chlorine atoms

In the reduction half-reaction, we need to balance the chlorine atoms: \(\mathrm{2HCl} \rightarrow \mathrm{H}_{2}+\mathrm{2Cl}\)

Balance charges

There are no charges to balance in this reaction.

Combine half-reactions

Combining the half-reactions gives the complete balanced equation: \(\mathrm{Co}+2\mathrm{HCl} \rightarrow \mathrm{CoCl}_{2}+\mathrm{H}_{2}\) The balanced equation for reaction d is: \(\mathrm{Co}(s) + 2\mathrm{HCl}(a q) \rightarrow \mathrm{CoCl}_{2}(a q) + \mathrm{H}_{2}(g)\)

Key concepts

Chemical Equation Balancing

Balancing a chemical equation is like aligning both sides of a balance. You want to ensure that each type of atom appears in equal numbers on both sides of the equation. This process stems from the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. For instance, consider the reaction between phosphorus and oxygen forming phosphorus pentoxide. Initially, count the atoms of each element. Adjust coefficients to make sure that each element has the same number on both sides. Remember that coefficients multiply the entire molecule they precede.

• Identify each element on both sides of the equation.
• Adjust coefficients to balance the number of atoms.
• Cross-check to ensure all elements are balanced.

Balancing becomes easier with practice, and using the half-reaction method can simplify complex reactions.

Half-Reaction Method

The half-reaction method is a systematic process often used to balance oxidation-reduction (redox) reactions. This technique breaks down the overall reaction into two parts - oxidation and reduction half-reactions. Each part focuses on electron transfer, which is central to redox reactions. To balance a redox reaction using this method, follow these steps:

• Divide the reaction into two half-reactions, identifying what is oxidized and what is reduced.
• Balance all atoms except oxygen and hydrogen.
• Balance oxygen atoms by adding water molecules where needed.
• Balance hydrogen using hydrogen ions (H⁺).
• Equalize the electron transfer in both half-reactions by multiplying them by appropriate coefficients.
• Finally, add the half-reactions together, ensuring that electrons are cancelled out.

This method is particularly useful for reactions in aqueous solutions involving electron transfer.

Redox Chemistry

Redox chemistry is a vital part of chemistry that deals with reactions where electrons are transferred between substances. "Redox" is short for reduction-oxidation, where 'reduction' refers to the gain of electrons, while 'oxidation' means the loss of electrons. To identify a redox reaction, look for changes in the oxidation states of atoms throughout the reaction. Recognizing how these changes relate to electron flow is crucial:

• Oxidation involves the increase in oxidation state due to loss of electrons.
• Reduction involves the decrease in oxidation state through the gain of electrons.

Understanding these concepts helps in identifying the reduction and oxidation agents: the substance being reduced is the oxidizing agent, while the one being oxidized is the reducing agent.

Electron Transfer in Reactions

Electron transfer is the core of redox reactions and is crucial for processes like energy conversion in batteries and biological systems. In a redox reaction, electrons move from one reactant to another. This movement is pivotal as it drives the reaction forward:

• In oxidation, electrons are donated to another substance.
• During reduction, a substance accepts electrons.

Understanding electron transfer helps you track energy changes during a reaction. It’s essential to remember that electrons cannot be isolated in such reactions; they are always transferred from one atom or molecule to another. When you comprehend this electron dance, balancing redox reactions becomes intuitive, as you’re really just ensuring fair exchanges in an electron economy.