Problem 44 - Chemistry The Central Science

Question

An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q),(\mathbf{b})\) acidic aqueous \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to aqueous \(\mathrm{Cr}^{3+}\) (\mathbf{c} ) \text { aqueous } \mathrm { Hg } _ { 2 } ( \mathrm { NO } _ { 3 } ) _ { 2 } \text { to mercury metal. Write balanced } equations for these reactions.

Step-by-Step Solution

Answer

The balanced chemical equations for the three given reactions are: a) \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 6 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4} + 8 \mathrm{H}_{2}\mathrm{O}\) b) \(1 \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{K}^{+} + 3 \mathrm{SO}_{4}^{2-} + 7 \, \mathrm{H_2O}\) c) \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Hg} + \mathrm{SO}_4^{2-} + 4 \mathrm{NO}_3^-\)

Step 1: Write the unbalanced equation for each reaction

First, we have to write the unbalanced chemical equations representing the reactions. a) \(\mathrm{KMnO}_4 + \mathrm{SO}_2 \rightarrow \mathrm{MnSO}_4\) b) \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + \mathrm{SO}_2 \rightarrow \mathrm{Cr}^{3+} + \mathrm{K}^{+}\) c) \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 \rightarrow \mathrm{Hg} + \mathrm{NO}_3^-\)

Step 2: Balance the atoms in each reaction

Next, we will balance the atoms in the chemical equations using the oxidation-reduction method. a) Oxidation half-reaction: \(\mathrm{SO}_{2} \rightarrow \mathrm{SO}_{4}^{2-}\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) Multiply the oxidation half-reaction by 2: \(2 \mathrm{SO}_{2} \rightarrow 2 \mathrm{SO}_{4}^{2-}\) Multiply the reduction half-reaction by 5: \(5 \mathrm{MnO}_{4}^{-} \rightarrow 5 \mathrm{Mn}^{2+}\) Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions, if needed: \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 2 \cdot 2 \mathrm{H}^{+} \rightarrow 5 \mathrm{Mn}^{2+} + 2 \mathrm{SO}_{4}^{2-} + 5 \mathrm{K}^{+} + 8-5 \, \mathrm{H}^{+}\) \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 6 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4} + 8 \mathrm{H}_{2}\mathrm{O}\) b) Oxidation half-reaction: \(\mathrm{SO}_2 \rightarrow \mathrm{SO}_4^{2-}\) Reduction half-reaction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}\) Multiply the oxidation half-reaction by 3: \(3 \mathrm{SO}_{2} \rightarrow 3 \mathrm{SO}_{4}^{2-}\) No multiplication needed for the reduction half-reaction. Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\): \(1 \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{K}^{+} + 3 \mathrm{SO}_{4}^{2-} + 7 \, \mathrm{H_2O}\) c) Oxidation half-reaction: \(\mathrm{SO}_2 \rightarrow \mathrm{SO}_4^{2-}\) Reduction half-reaction: \(\mathrm{Hg}^{2+} \rightarrow \mathrm{Hg}\) No multiplication needed for the oxidation half-reaction. Multiply the reduction half-reaction by 2: \(2 \mathrm{Hg}^{2+} \rightarrow 2 \mathrm{Hg}\) Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions, if needed: \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Hg} + \mathrm{SO}_4^{2-} + 4 \mathrm{NO}_3^-\) The three balanced equations are: a) \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 6 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4} + 8 \mathrm{H}_{2}\mathrm{O}\) b) \(1 \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{K}^{+} + 3 \mathrm{SO}_{4}^{2-} + 7 \, \mathrm{H_2O}\) c) \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Hg} + \mathrm{SO}_4^{2-} + 4 \mathrm{NO}_3^-\)

Key Concepts

chemical equations

Chemical equations are used to represent chemical reactions by displaying the reactants on one side and the products on the other side, separated by an arrow. This shows the transformation of substances and helps to understand the process of the reaction.
In the context of our exercise, the chemical equations for the reduction reactions describe how sulfur dioxide (\(\mathrm{SO}_2\) in aqueous form reacts with other substances like \(\mathrm{KMnO}_4\), \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\), and mercury nitrate). Writing these equations is the fundamental first step to solving any chemical reaction problem.
A proper chemical equation is crucial because:

• It gives a clear visual of the substances involved.
• It specifies the physical states of each substance (solid, liquid, gas, aqueous).
• It indicates the direction of reaction, showing how substances transform.

Understanding chemical equations sets the stage for further analysis, like balancing the reaction and determining oxidation states.

balancing reactions

Balancing chemical equations is essential because it ensures the law of conservation of mass is obeyed. According to this principle, matter cannot be created or destroyed.
In a balanced chemical reaction, the number of atoms for each element in the reactants must equal the number of atoms for that element in the products.
This process involves adjusting coefficients, the numbers placed before molecules to achieve balance.
To balance the reactions from the exercise, we use the method of balancing charges and adjusting molecule coefficients. Consider these tips:

• Start by balancing atoms that appear in only one reactant and one product.
• Balance complex molecules, such as permanganate, after simpler substances.
• Ensure that both mass and charge are balanced in the complete reaction.

Balancing the reactions assures that the chemical equations accurately represent the reaction happening in reality.

half-reaction method

The half-reaction method is a critical tool for balancing oxidation-reduction reactions, also known as redox reactions. These reactions involve the change in oxidation states of involved species.
This method breaks the overall equation into two separate half-reactions: oxidation and reduction.
In the context of the exercise, separating each reaction helps in tracking electron transfer easily. This is the step where the electrons lost in oxidation balance with those gained in reduction.
Here is a simple overview of the half-reaction method:

• Identify the substances that are oxidized and reduced.
• Write separate half-reactions for oxidation and reduction.
• Balance atoms and charges in each half-reaction.
• Multiply appropriately to equalize the number of electrons transferred.
• Combine the half-reactions to form the final balanced equation.

Using the half-reaction method makes it significantly easier to balance complex redox reactions compared to the inspection method.

oxidation states

Oxidation states, or oxidation numbers, indicate the degree of oxidation of an atom in a chemical compound. The concept is fundamental in understanding redox reactions as it shows the quantity of electrons lost or gained.
In redox reactions, one species gets oxidized (increase in oxidation state) while another gets reduced (decrease in oxidation state).
To determine oxidation states in a compound, consider these simplified rules:

• For elemental state, the oxidation state is zero (e.g., \(\mathrm{N}_2\), \(\mathrm{O}_2\)).
• For ions, the oxidation state equals the charge of the ion (e.g., \(\mathrm{Na}^+\) is +1).
• Group 1 metals always have +1, and Group 2 metals have +2 oxidation states in compounds.
• The sum of oxidation states in a neutral molecule must be zero, and for a charged ion, it must equal the ion's charge.

In the exercise, understanding of these states helps track which elements are oxidized or reduced, setting the foundation for using the half-reaction balancing method effectively.