Question

$$ \begin{array}{l}{2 \mathrm{Mn}^{2+}(\mathrm{aq})+8 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+10 \mathrm{Hg}^{2+}(\mathrm{aq}) \rightarrow} \\ {2 \mathrm{MnO}_{4}-(\mathrm{aq})+16 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{Hg}_{2}^{2+}(\mathrm{aq})}\end{array} $$

Short answer

In the given redox equation, Mn is oxidized from an oxidation state of +2 to +7, and Hg is reduced from an oxidation state of +2 to +1. Two half-reactions can be written to represent these changes: Oxidation half-reaction: \(2 \text{Mn}^{2+} (\text{aq}) + 2 \text{O}_{2} (\text{g}) + 10 \text{H}^{+} (\text{aq}) + 10 e^{-} \rightarrow 2 \text{MnO}_4^{-} (\text{aq}) + 8 \text{H}^{+} (\text{aq})\) Reduction half-reaction (multiplied by 5 to balance electrons): \(10 \text{Hg}^{2+} (\text{aq}) + 10 e^{-} + 10 \text{H}^{+} (\text{aq}) + 5 \text{H}_{2} (\text{g}) \rightarrow 5 \text{Hg}_2^{2+} (\text{aq})\) Combining the balanced half-reactions, we get the original redox equation.

Step-by-step solution

Identify the elements undergoing redox and their initial oxidation states

To determine the elements undergoing redox, we first have to assign the oxidation numbers of each element in the given equation. For this purpose, we'll use basic rules of assigning oxidation numbers: - Oxidation state of free elements (not combined with any other element) is 0. - Oxidation state of H is usually +1. - Oxidation state of O is usually -2. - The sum of oxidation states for all atoms in a molecule or ion must equal the total charge on that molecule or ion. Assigning the oxidation states in the given equation, we get: \(2 \text{Mn}^{2+} (\text{aq}) + 8 \text{H}_2 \text{O} (\text{l}) + 10 \text{Hg}^{2+} (\text{aq}) \rightarrow 2 \text{MnO}_4^- (\text{aq}) + 16 \text{H}^+ (\text{aq}) + 5 \text{Hg}_2^{2+} (\text{aq})\) Mn: +2 to +7 (oxidized) Hg: +2 to +1 (reduced) Thus, Mn and Hg are undergoing redox.

Write the half-reactions

We'll now separate the redox reaction into two half-reactions: one for oxidation and one for reduction. Oxidation half-reaction: Mn: \(2 \text{Mn}^{2+} (\text{aq}) + 2 \text{O}_{2} (\text{g}) + 10 \text{H}^{+} (\text{aq}) \rightarrow 2 \text{MnO}_4^{-} (\text{aq}) + 8 \text{H}^{+} (\text{aq})\) Reduction half-reaction: Hg: \(2 \text{Hg}^{2+} (\text{aq}) + 2 \text{H}^{+} (\text{aq}) + \text{H}_{2} (\text{g}) \rightarrow \text{Hg}_2^{2+} (\text{aq})\)

Balance the half-reactions for charge

For each half-reaction, we need to balance the charges by adding electrons to the appropriate side. Oxidation half-reaction: \(2 \text{Mn}^{2+} (\text{aq}) + 2 \text{O}_{2} (\text{g}) + 10 \text{H}^{+} (\text{aq}) + 10 e^{-} \rightarrow 2 \text{MnO}_4^{-} (\text{aq}) + 8 \text{H}^{+} (\text{aq})\) Reduction half-reaction: \(2 \text{Hg}^{2+} (\text{aq}) + 2 e^{-} + 2 \text{H}^{+} (\text{aq}) + \text{H}_{2} (\text{g}) \rightarrow \text{Hg}_2^{2+} (\text{aq})\)

Combine the half-reactions

Now, we'll combine the two half-reactions to go back to our original balanced redox equation. To do this, first, make sure the same number of electrons are present in both half-reactions. The lowest common multiple of 10 (electrons in the oxidation half-reaction) and 2 (electrons in the reduction half-reaction) is 10. So, we will need to multiply the entire reduction half-reaction by 5 to make the electrons equal before combining. \(2 \text{Mn}^{2+} (\text{aq}) + 8 \text{H}_2 \text{O} (\text{l}) + 10 \text{Hg}^{2+} (\text{aq}) \rightarrow 2 \text{MnO}_4^{-} (\text{aq}) + 16 \text{H}^{+} (\text{aq}) + 5 \text{Hg}_2^{2+} (\text{aq})\) And that is the step-by-step breakdown of the given exercise.

Key concepts

Oxidation Numbers

Understanding oxidation numbers is the cornerstone of grasping redox reactions. An oxidation number is a value assigned to an element in a chemical compound that represents its actual or hypothetical charge if the compound was composed of ions.

Determining oxidation numbers helps us identify which elements are oxidized (increase in oxidation number) or reduced (decrease in oxidation number) during a reaction. Here's a straightforward guideline to follow:

• Elements in their elemental form have an oxidation number of 0.
• The oxidation number of hydrogen is generally +1 when bonded to nonmetals and -1 when bonded to metals.
• Oxygen typically has an oxidation number of -2, except in peroxides where it's -1 or in compounds with fluorine where it can vary.
• For a neutral compound, the sum of oxidation numbers must equal 0; for a polyatomic ion, it must equal the ion's charge.

Using these rules, you can calculate the change in oxidation numbers to find which elements are involved in the redox process, just as we observed in the exercise with Mn and Hg.

Half-Reactions

The breakdown of a redox reaction into its two complementary processes – oxidation and reduction – is done using half-reactions. Each half-reaction shows how electrons are lost or gained by reactants. Writing half-reactions allows us to focus on the actual redox changes that are happening and balance them separately.

In the exercise, we separate the overall reaction into an oxidation half-reaction and a reduction half-reaction. This not only makes it easier to balance but also helps us understand the flow of electrons from one species to another – the heart of any redox reaction. Once each half-reaction is balanced for mass and charge, we can then recombine them to give the balanced overall equation.

Balancing Redox Equations

The step-by-step solution mirrors the intricate process of balancing redox equations. This crucial skill ensures that the number of elements and charges are equal on both sides of the equation – a fundamental principle known as the conservation of mass and charge. For students, it's essential to:

• Assign oxidation numbers to understand which elements are oxidized and reduced.
• Separate the reaction into half-reactions representing the oxidation and reduction.
• Balance each half-reaction for atoms and charges, adding water, hydrogen ions, or hydroxide ions where appropriate.
• Add electrons to balance charges in each half-reaction.

We also ensure the number of electrons lost in the oxidation half-reaction is equal to the number gained in the reduction half-reaction. This can involve multiplying half-reactions by appropriate factors before combining them to cancel out the electrons and produce a balanced equation.

Oxidation States

When discussing redox reactions, it's vital to differentiate between oxidation numbers and oxidation states. The terms are often used interchangeably but they represent slightly different concepts.

The oxidation state can be thought of as the theoretical charge an atom would have if its bonds were completely ionic. The exercise improvement advice emphasizes the importance of recognizing changes in the oxidation state for identifying redox reactions. When an atom's oxidation state increases during a reaction, it undergoes oxidation; when it decreases, it undergoes reduction.

Mastering the concept of oxidation states will not only assist you in balancing redox equations but will also deepen your understanding of the electron transfer and changes occurring during a reaction.