Question

Balance each of the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method. a. \(\mathrm{Mg}(s)+\mathrm{Hg}^{2+}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Hg}_{2}^{2+}(a q)\) b. \(\mathrm{NO}_{3}^{-}(a q)+\mathrm{Br}^{-}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{Br}_{2}(l)\) c. \(\mathrm{Ni}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{NO}_{2}(g)\) d. \(\mathrm{ClO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{Cl}_{2}(g)\)

Short answer

#tag_title#Step 1: Write Half-Reactions# #tag_content#Oxidation: \(\mathrm{Ni}(s) \rightarrow \mathrm{Ni}^{2+}(aq)\) Reduction: \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}_{2}(g)\) #tag_title#Step 2: Balance atoms other than O and H# #tag_content#Both half-reactions are already balanced for the elements other than O and H. #tag_title#Step 3: Balance O atoms using H2O# #tag_content#Reduction: \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}_{2}(g) + H_2O(l)\) #tag_title#Step 4: Balance H atoms using H+# #tag_content#Reduction: \(\mathrm{NO}_{3}^{-}(aq) + 2H^+(aq) \rightarrow \mathrm{NO}_{2}(g) + H_2O(l)\) #tag_title#Step 5: Balance charge by adding electrons# #tag_content#Oxidation: \(\mathrm{Ni}(s) \rightarrow \mathrm{Ni}^{2+}(aq) + 2e^-\) Reduction: \(\mathrm{NO}_{3}^{-}(aq) + 2H^+(aq) + 2e^- \rightarrow \mathrm{NO}_{2}(g) + H_2O(l)\) #tag_title#Step 6: Combine the half-reactions and simplify# #tag_content#\(\mathrm{Ni}(s) + \mathrm{NO}_{3}^{-}(aq) + 2H^+(aq) \rightarrow \mathrm{Ni}^{2+}(aq) + \mathrm{NO}_{2}(g) + H_2O(l)\) The balanced equation for reaction c is: \(\mathrm{Ni}(s) + \mathrm{NO}_{3}^{-}(aq) + 2H^+(aq) \rightarrow \mathrm{Ni}^{2+}(aq) + \mathrm{NO}_{2}(g) + H_2O(l)\) d. \(\mathrm{ClO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{Cl}_{2}(g)\) #tag_title#Step 1: Write Half-Reactions# #tag_content#Oxidation: \(\mathrm{2Cl}^{-}(aq) \rightarrow \mathrm{Cl}_{2}(g)\) Reduction: \(\mathrm{ClO}_{4}^{-}(aq) \rightarrow \mathrm{ClO}_{3}^{-}(aq)\) #tag_title#Step 2: Balance atoms other than O and H# #tag_content#Both half-reactions are already balanced for the elements other than O and H. #tag_title#Step 3: Balance O atoms using H2O# #tag_content#Reduction: \(\mathrm{ClO}_{4}^{-}(aq) \rightarrow \mathrm{ClO}_{3}^{-}(aq) + H_2O(l)\) #tag_title#Step 4: Balance H atoms using H+# #tag_content#Reduction: \(\mathrm{ClO}_{4}^{-}(aq) + 2H^+(aq) \rightarrow \mathrm{ClO}_{3}^{-}(aq) + H_2O(l)\) #tag_title#Step 5: Balance charge by adding electrons# #tag_content#Oxidation: \(\mathrm{2Cl}^{-}(aq) \rightarrow \mathrm{Cl}_{2}(g) + 2e^-\) Reduction: \(\mathrm{ClO}_{4}^{-}(aq) + 2H^+(aq) + 2e^- \rightarrow \mathrm{ClO}_{3}^{-}(aq) + H_2O(l)\) #tag_title#Step 6: Combine the half-reactions and simplify# #tag_content#\(\mathrm{ClO}_{4}^{-}(aq) + 2H^+(aq) + 2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{ClO}_{3}^{-}(aq) + H_2O(l) + \mathrm{Cl}_{2}(g)\) The balanced equation for reaction d is: \(\mathrm{ClO}_{4}^{-}(aq) + 2H^+(aq) + 2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{ClO}_{3}^{-}(aq) + H_2O(l) + \mathrm{Cl}_{2}(g)\)

Step-by-step solution

Write Half-Reactions#

Oxidation: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq)\) Reduction: \(2\mathrm{Hg}^{2+}(aq) \rightarrow \mathrm{Hg}_{2}^{2+}(aq)\)

Balance atoms other than O and H#

Both half-reactions are already balanced for the elements other than O and H.

Balance O atoms using H2O#

There are no O atoms in either half-reaction.

Balance H atoms using H+#

There are no H atoms in either half-reaction.

Balance charge by adding electrons#

Oxidation: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^-\) Reduction: \(2\mathrm{Hg}^{2+}(aq) + 2e^- \rightarrow \mathrm{Hg}_{2}^{2+}(aq)\)

Combine the half-reactions and simplify#

\(2\mathrm{Hg}^{2+}(aq) + \mathrm{Mg}(s) \rightarrow \mathrm{Hg}_{2}^{2+}(aq) + \mathrm{Mg}^{2+}(aq)\) The balanced equation for reaction a is: \(2\mathrm{Hg}^{2+}(aq) + \mathrm{Mg}(s) \rightarrow \mathrm{Hg}_{2}^{2+}(aq) + \mathrm{Mg}^{2+}(aq)\) b. \(\mathrm{NO}_{3}^{-}(a q)+\mathrm{Br}^{-}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{Br}_{2}(l)\)

Write Half-Reactions#

Oxidation: \(\mathrm{2Br}^{-}(aq) \rightarrow \mathrm{Br}_{2}(l)\) Reduction: \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g)\)

Balance atoms other than O and H#

Both half-reactions are already balanced for the elements other than O and H.

Balance O atoms using H2O#

Reduction: \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g) + 2H_2O(l)\)

Balance H atoms using H+#

Reduction: \(\mathrm{NO}_{3}^{-}(aq) + 4H^+(aq) \rightarrow \mathrm{NO}(g) + 2H_2O(l)\)

Balance charge by adding electrons#

Oxidation: \(\mathrm{2Br}^{-}(aq) \rightarrow \mathrm{Br}_{2}(l) + 2e^-\) Reduction: \(\mathrm{NO}_{3}^{-}(aq) + 4H^+(aq) + 2e^- \rightarrow \mathrm{NO}(g) + 2H_2O(l)\)

Combine the half-reactions and simplify#

\(\mathrm{NO}_{3}^{-}(aq) + 4H^+(aq) + 2\mathrm{Br}^{-}(aq) \rightarrow \mathrm{NO}(g) + 2H_2O(l) + \mathrm{Br}_{2}(l)\) The balanced equation for reaction b is: \(\mathrm{NO}_{3}^{-}(aq) + 4H^+(aq) + 2\mathrm{Br}^{-}(aq) \rightarrow \mathrm{NO}(g) + 2H_2O(l) + \mathrm{Br}_{2}(l)\) c. \(\mathrm{Ni}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{NO}_{2}(g)\)

Key concepts

Half-Reaction Method

The half-reaction method is a systematic approach used to balance redox reactions, particularly in aqueous solutions. This method involves breaking down a redox reaction into two separate equations—one for oxidation and one for reduction.

In each half-reaction, you separately balance the atoms and the charges. Here's a step-by-step look at how it works:

• Oxidation half-reaction: This equation represents the process where a substance loses electrons. For instance, in the first given exercise, the half-reaction for magnesium undergoing oxidation is: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^-\)
• Reduction half-reaction: This equation depicts the gaining of electrons by a substance. Consider the half-reaction for mercury ions undergoing reduction: \(2\mathrm{Hg}^{2+}(aq) + 2e^- \rightarrow \mathrm{Hg}_{2}^{2+}(aq)\)

By balancing these two half-reactions individually, we ensure every atom and charge are accounted for before combining them back into a single balanced equation.

Acidic Solution

Redox reactions in acidic solutions have an additional layer of complexity. In such environments, we must account for the presence of hydrogen ions (\(H^+\)) and water (\(H_2O\)) in the balancing process. This makes it possible to balance the oxygen and hydrogen atoms involved in the reactions.

Here’s the step-by-step approach when working with acidic solutions:

• Oxygen atoms: Balance these atoms by adding \(H_2O\) molecules to the side deficient in oxygen. For example, converting \(NO_3^-\) to \(NO\) requires adding water: \(\mathrm{NO_3}^-(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H_2O}(l)\)
• Hydrogen atoms: Add \(H^+\) ions to balance the hydrogen atoms. Using the same reaction, we add \(H^+\) to balance: \(\mathrm{NO_3}^-(aq) + 4H^+(aq) \rightarrow \mathrm{NO}(g) + 2H_2O(l)\)

Balancing redox reactions in acidic solutions ensures that the changes in oxidation state are correctly reflected and managed.

Oxidation and Reduction

Oxidation and reduction, also known as redox processes, are complementary reactions where the transfer of electrons occurs. Understanding these processes is crucial for balancing redox reactions.

Here's what you need to know about each:

• Oxidation: This involves the loss of electrons by a molecule, atom, or ion. As a result, the oxidation state of the substance increases. For instance, \(\mathrm{Br}^-\) ions lose electrons and form \(\mathrm{Br}_2\) in the process: \(\mathrm{2Br}^-(aq) \rightarrow \mathrm{Br}_2(l) + 2e^-\)
• Reduction: The gain of electrons characterizes this process, reducing the oxidation state. Using the example of nitrate ions, \(\mathrm{NO_3}^-(aq)\) is reduced to \(\mathrm{NO}(g)\) \(\rightarrow \mathrm{NO_3}^-(aq) + 2e^- \rightarrow \mathrm{NO}(g)\). Reduction can also involve subsequent formation of covalent bonds involving oxygen or other electronegative atoms.

Redox reactions encompass both oxidation and reduction, ensuring the total transfer of electrons remains balanced.

Electron Balance

Electron balance ensures that the number of electrons lost during oxidation equals the number gained during reduction in a redox reaction. Achieving electron balance is crucial for the complete reaction to be properly balanced.

Here’s how to ensure electron balance in any redox reaction:

• Identify electron transfers: Look at both the oxidation and reduction half-reactions to determine the number of electrons transferred. For example, in the oxidation reaction of \(\mathrm{Mg}(s)\), two electrons are lost: \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^-\).
• Add or adjust coefficients: Manipulate the stoichiometric coefficients in half-reactions to equalize electron transfers. This means adjusting numbers until the electrons in each half-reaction match. In the reduction of \(\mathrm{Hg}^{2+}(aq)\), two electrons combine with \(\mathrm{Hg}^{2+}(aq)\) to form \(\mathrm{Hg}_{2}^{2+}(aq)\): \(2\mathrm{Hg}^{2+}(aq) + 2e^- \rightarrow \mathrm{Hg}_{2}^{2+}(aq)\).

By carefully adjusting and combining the half-reactions, you ensure the redox equation balances perfectly without any disparity in electron transfer.