Question

Determine whether the following oxidation-reduction reactions ate balanced correctly, If they are not, provide the correct balanced reaction. (a) unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\) balanced: \(\mathrm{MnO}_{2}(s)+2 A 1(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (b) unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{4}(a q) \longrightarrow \mathrm{Cu}^{24}(a q)+A \mathrm{~g}(s)\) balanced: \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (c) unbalanced: $$ \mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q) $$ (icidic solution) balanced: $$ \begin{aligned} 16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+& 2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow \\ & 5 \mathrm{Br}(l)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (d) unhalanced: $$ \begin{aligned} &\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s) \quad \text { (basic solution) } \\ &\text { balanced: } \\ &8 \mathrm{H}^{-1}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \\ &\qquad \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$

Short answer

All given reactions have been balanced correctly as per the provided balanced forms.

Step-by-step solution

- Identify the elements involved

List all the reactants and products for each reaction to understand which elements are involved in the balancing process.

- Compare the unbalanced and balanced reactions

Analyze the given unbalanced and balanced forms of each reaction to identify if they are correctly balanced.

- Verify the balance of elements for Reaction (a)

Reaction (a) unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\) balanced: \(\mathrm{MnO}_{2}(s)+2 \mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\). Check if the number of atoms for each element is the same on both sides of the equation. In this case, the balanced equation correctly shows 2 Al atoms, aligning the reactants and products.

- Verify the balance of elements for Reaction (b)

Reaction (b) unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Ag}(s)\). The correct balance is \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\). Here, the balanced equation correctly shows 2 Ag atoms, aligning the reactants and products.

- Verify the balance of elements for Reaction (c)

Reaction (c) unbalanced: \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q)\) (acidic solution). The balanced form is: \(16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow 5 \mathrm{Br}_{2}(d)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)\). This includes both mass and charge balance for the elements and ions involved.

- Verify the balance of elements for Reaction (d)

Reaction (d) unbalanced: \(\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) (basic solution). The balanced form is: \(8 \mathrm{OH}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\). This accurately balances the reactions in a basic medium.

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, are chemical processes where electrons are transferred between reactants. It’s crucial to identify the species gaining and losing electrons.
The substance that loses electrons is oxidized and is known as the reducing agent, while the substance that gains electrons is reduced and is known as the oxidizing agent.
For example, in reaction (a):
Unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\)
Balanced: \(\mathrm{MnO}_{2}(s)+2 \mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\)
Aluminum (Al) loses electrons (is oxidized) and Manganese dioxide (MnO\(_2\)) gains electrons (is reduced).
This electron transfer is the essence of redox reactions. Remember always to split the reaction into half-reactions to identify this exchange easily.

Acidic and Basic Solutions

Balancing redox reactions can differ significantly depending on whether the reaction occurs in an acidic or basic solution. The environment determines the additional species used in balancing the equation.
In acidic solutions, we often add \(\mathrm{H}^{+}\) ions and \(\mathrm{H}_{2}\mathrm{O}\) molecules to balance hydrogen and oxygen atoms.
Example (c):
Unbalanced: \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q)\)
Balanced: \[16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow 5 \mathrm{Br}_{2}(d)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)\]
In basic solutions, we add \(\mathrm{OH}^{-}\) ions and \(\mathrm{H}_{2}\mathrm{O}\) similarly.
Example (d):
Unbalanced: \(\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\)
Balanced: \[8 \mathrm{OH}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\]
Identifying the solution type helps you choose the correct balancing method.

Redox Reaction Balancing

Balancing redox reactions involves several systematic steps that ensure both mass and charge balance. Here are the basic steps:

• Identify and separate the two half-reactions (oxidation and reduction).


• Balance each half-reaction for all elements, except hydrogen and oxygen.


• Balance oxygen atoms by adding \(\mathrm{H}_{2} \mathrm{O}\) molecules.


• Balance hydrogen atoms (in acidic or basic solutions, use \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) respectively).


• Balance the charge by adding electrons.


• Combine the half-reactions, making sure the total number of electrons cancels out.

Let’s look at reaction (b):
Unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Ag}(s)\)
Balanced: \[\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\]
We separate it into half-reactions:
\(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^{-}\) (oxidation)
and
\(\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}\) (reduction),
Then, multiply the reduction half-reaction by 2 to match the electrons lost in oxidation.
Finally, combine them to get the balanced equation ensuring there are 2 silver atoms on both sides.
Bearing these steps in mind simplifies the complex process of balancing redox reactions in various chemical environments.