Question

Balance each of the following half-reactions. a. \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)\) c. \(\mathrm{Br}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)\) d. \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}(s)\)

Short answer

The short version of the balanced half-reactions are: a. \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q) + 2 \mathrm{e^-}\) b. \(\mathrm{Fe}^{3+}(a q) + \mathrm{e^-} \rightarrow \mathrm{Fe}^{2+}(a q)\) c. \(2 \mathrm{Br}^-(a q) \rightarrow \mathrm{Br}_{2}(l) + 2 \mathrm{e^-}\) d. \(\mathrm{Fe}^{2+}(a q) + 2 \mathrm{e^-} \rightarrow \mathrm{Fe}(s)\)

Step-by-step solution

Determine the change in oxidation numbers

In this reaction, the oxidation number of copper (Cu) in the solid state is 0, and in the Cu²⁺ ion, the oxidation number of Cu is +2.

Balance the atoms

There's only one Cu atom on each side, so the atoms are already balanced.

Balance the charge

The left side of the reaction has a charge of 0 and the right side has a charge of +2. To balance the charge, we need to add 2 electrons (e⁻) to the right side of the reaction. The balanced half-reaction is: \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q) + 2 \mathrm{e^-}\) #b. Balancing Fe³⁺(aq) → Fe²⁺(aq)#

Determine the change in oxidation numbers

In this reaction, the oxidation number of iron (Fe) changes from +3 in Fe³⁺ to +2 in Fe²⁺.

Balance the atoms

There's only one Fe atom on each side, so the atoms are already balanced.

Balance the charge

The left side of the reaction has a charge of +3 and the right side has a charge of +2. To balance the charge, we need to add one electron (e⁻) to the right side of the reaction. The balanced half-reaction is: \(\mathrm{Fe}^{3+}(a q) + \mathrm{e^-} \rightarrow \mathrm{Fe}^{2+}(a q)\) #c. Balancing Br⁻(aq) → Br2(l)#

Determine the change in oxidation numbers

In this reaction, the oxidation number of bromine (Br) changes from -1 in the Br⁻ ion to 0 in the liquid Br₂ molecule.

Balance the atoms

To balance the Br atoms, we need to have two bromine atoms on both sides of the reaction. So, we multiply the Br⁻ ion by 2: 2Br⁻(aq) → Br₂(l)

Balance the charge

The left side of the reaction has a total charge of -2 and the right side has a charge of 0. To balance the charge, we need to add two electrons (2 e⁻) to the right side of the reaction. The balanced half-reaction is: \(2 \mathrm{Br}^-(a q) \rightarrow \mathrm{Br}_{2}(l) + 2 \mathrm{e^-}\) #d. Balancing Fe²⁺(aq) → Fe(s)#

Determine the change in oxidation numbers

In this reaction, the oxidation number of iron (Fe) changes from +2 in the Fe²⁺ ion to 0 in the solid Fe atom.

Balance the atoms

There's only one Fe atom on each side, so the atoms are already balanced.

Balance the charge

The left side of the reaction has a charge of +2 and the right side has a charge of 0. To balance the charge, we need to add 2 electrons (e⁻) to the left side of the reaction. The balanced half-reaction is: \(\mathrm{Fe}^{2+}(a q) + 2 \mathrm{e^-} \rightarrow \mathrm{Fe}(s)\)

Key concepts

Oxidation Number

The oxidation number, also known as the oxidation state, is a useful tool in chemistry that helps us keep track of electrons during chemical reactions. It is essentially a charge assigned to an atom, which helps us understand the transfer of electrons. This is crucial in redox reactions where one substance is oxidized and another is reduced.

• Oxidation involves an increase in oxidation number. This means that the element loses electrons. For example, when copper (Cu) goes from a state of 0 in solid form to +2 in the ion \(Cu^{2+}(aq)\), it has lost electrons.
• Reduction involves a decrease in oxidation number, meaning the element gains electrons. For instance, in the conversion of \(Fe^{3+}(aq)\) to \(Fe^{2+}(aq)\), the iron atom gains an electron, reducing its oxidation state by 1.

Understanding oxidation numbers can thus provide insights into how electrons are transferred in reactions, allowing for the balancing of overall charges in chemical equations.

Half-Reactions

In a redox reaction, half-reactions are a way of breaking down the process into oxidation and reduction parts. Each half-reaction shows either the loss or gain of electrons. This is particularly useful for understanding how the electrons are transferred between reactants.

• The oxidation half-reaction shows how electrons are lost, often including the substance being oxidized and the products formed.
• The reduction half-reaction displays how electrons are gained, illuminating the substance being reduced and the resultant products.

Analyzing half-reactions helps in balancing chemical equations, as they allow us to observe the changes in oxidation states separately. For example, in the reaction \(2Br^-(aq) \rightarrow Br_2(l) + 2e^-\), this half-reaction clearly indicates the oxidation of \(Br^-\) to \(Br_2\) along with the electrons transferred.

Electron Balance

Electron balance is a key step in the process of balancing redox reactions. It ensures that the number of electrons lost in oxidation matches the number gained in reduction, thereby upholding the principle of conservation of charge.

• Start by identifying how many electrons are involved in each half-reaction. Electrons lost must be equal to electrons gained for the reaction to be balanced.
• For example, when balancing \(Fe^{2+}(aq) \rightarrow Fe(s)\), electrons are added to one side to make sure total charges on both sides are equivalent.

Successfully balancing electrons ensures that the chemical equation reflects conservation laws and depicts the actual process occurring during the reaction.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They display the reactants on the left side and the products on the right, with an arrow in between indicating the direction of the reaction. Writing correct chemical equations is crucial for understanding chemical processes.

• In a balanced chemical equation, the number of atoms of each element is the same on both sides. Balancing is essential as it respects the law of conservation of mass.
• For example, in balancing \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(aq) + 2 \mathrm{e^-}\), both atoms and charges must be taken into account.

Understanding and balancing chemical equations is a fundamental skill in chemistry that aids in predicting the products and understanding the nature of the reaction.