Question

Chromium(II) sulfate, \(\mathrm{CrSO}_{4^{\prime}}\) is a reagent that has been used in certain applications to help reduce carbon-carbon double bonds \((\mathrm{C}=\mathrm{C})\) in molecules to single bonds ( \(\mathrm{C}-\mathrm{C}\) ). The reagent can be prepared via the following reaction. $$\begin{array}{c} 4 \mathrm{Zn}(\mathrm{s})+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{ZnSO}_{4}(\mathrm{aq})+2 \mathrm{CrSO}_{4}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$

Short answer

The chemical reaction involves the reactants zinc (Zn), \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) that react to produce \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \). The equation is balanced, having equal numbers of each atom on both sides.

Step-by-step solution

Identify the reactants and products

The reagents for this reaction are zinc (Zn), \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \). The products of this reaction are \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \)

Balance the equation

The key to balancing the equation is to ensure that the number of atoms for each element is the same on both sides of the equation. In the given equation, we can see that this is already balanced correctly. For example, looking at the zinc atom, we have 4 zinc atoms on the left side and 4 zinc atoms that are part of \( \mathrm{ZnSO}_{4} \) on the right side.

Interpret the equation

This equation represents the chemical reaction where zinc, \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) react to produce \( \mathrm{ZnSO}_{4} \), \( \mathrm{CrSO}_{4} \), \( \mathrm{K}_{2} \mathrm{SO}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \). \( \mathrm{CrSO}_{4} \) is the reagent in focus and it is used to reduce double carbon bonds to single bonds in molecules.

Key concepts

Oxidation-Reduction Reactions

In chemistry, oxidation-reduction reactions, also known as redox reactions, are processes where electrons are transferred between chemical species. One substance gets oxidized (loses electrons), while the other substance is reduced (gains electrons). This creates a balance, as electrons do not appear from nowhere or disappear into nothing.
During redox reactions, the oxidation states of involved elements change, which is a core part of understanding these reactions.

• Oxidation: The loss of electrons, leading to an increase in oxidation state. In the exercise, zinc ( \( \mathrm{Zn} \)) gets oxidized to zinc sulfate (\( \mathrm{ZnSO}_4 \)).
• Reduction: The gain of electrons, leading to a decrease in oxidation state. Here, the chromium in \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) is reduced, forming \( \mathrm{CrSO}_4 \).

Redox reactions are crucial in many applications, including biological systems and industrial processes.
The step-by-step balancing described in the solution is essential to ensure mass and charge are conserved in these reactions.
Understanding these fundamentals can aid in grasping the practical applications of redox reactions.

Chromium Compounds

Chromium is a transition metal known for its diverse oxidation states ranging from -2 to +6, which gives rise to numerous compounds with varied properties and uses.
The most common of these compounds include chromates (\( \mathrm{CrO}_4^{2-} \)) and dichromates (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)), with chromium typically being in the +6 oxidation state in these ions.
Chromium's flexibility in oxidation states, especially the movement between +3 and +6, is useful in redox reactions, as seen in the exercise.

• Chromium +3: Often stable and found in compounds like \( \mathrm{CrSO}_4 \). This form is less toxic compared to +6 compounds.
• Chromium +6: Present in chromates and dichromates, often used in industrial processes, but more hazardous.

In the given reaction, chromium is used in the form of potassium dichromate, \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \), a powerful oxidizing agent reduced to \( \mathrm{CrSO}_4 \).
This flexibility in changing oxidation states makes chromium compounds valuable in applications requiring stable oxidizing or reducing conditions.

Carbon-Carbon Bond Reduction

Reducing carbon-carbon double bonds (\( \mathrm{C}=\mathrm{C} \)) to single bonds (\( \mathrm{C}-\mathrm{C} \)) involves removing the double bond's pi-electrons, resulting in a single bond.
This type of reduction is significant in organic chemistry, especially in the synthesis and modification of organic molecules.
The procedure often requires a reducing agent to donate electrons to the molecule, forming more stable single bonds.
In this exercise, \( \mathrm{CrSO}_4 \) acts as such a reducing agent.

• It supplies electrons to the pi bond, thereby breaking it.
• The electrons are incorporated into the molecule, converting it from a more reactive double-bonded state to a less reactive single-bonded form.

Using chromium(II) compounds, such as \( \mathrm{CrSO}_4 \), is advantageous because they can stabilize the intermediate states that form during the reduction process.
This allows for controlled and selective reactions, making these agents useful in complex synthetic pathways.