Question

Balance the following oxidation-reduction equations. $$\text { (a) } \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{S}(\mathrm{s})$$ $$\begin{aligned} &\text { (b) } \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{Ag}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$

Short answer

The balanced equations are: \n(a) \(3O_{2}(g) + 4Fe_{2}S_{3}(s) + 12H_{2}O \rightarrow 8Fe(OH)_{3}(s) + 3S(s) + 12H^{+} + 12e^{-}\)

Step-by-step solution

Separate half-reactions

Identify the species that is being oxidized (losing electrons) and the species being reduced (gaining electrons) and write down these half-reactions for each equation. For example, for the first equation \(Fe_{2}S_{3}\) is oxidized to \(Fe(OH)_{3}\) and \(O_{2}\) is reduced to \(S\). Hence we obtain:\(Fe_{2}S_{3} \rightarrow 2Fe(OH)_{3}\) and\(O_{2} + H_{2}O \rightarrow S\)

Balance mass

First, balance all atoms except for hydrogen and oxygen in each half-reaction. Then, balance oxygen by adding H2O molecules to the side lacking oxygen. After that, balance hydrogen by adding H+ ions to the side lacking hydrogen. For example, for the first equation we obtain:\(2Fe_{2}S_{3} + 12H_{2}O \rightarrow 4Fe(OH)_{3}\) and\(O_{2} + 2H_{2}O \rightarrow S + 4H^{+}\)

Balance charges

Add electrons to the side of the half-reaction that has a greater positive charge. For the first equation we get:\(2Fe_{2}S_{3} + 12H_{2}O \rightarrow 4Fe(OH)_{3} + 24e^{-}\) and\(O_{2} + 2H_{2}O + 8e^{-} \rightarrow S + 4H^{+}\)

Combine half-reactions

Once the charges and atoms are balanced in each half-reaction, multiply the half-reactions by the smallest possible integers such that the number of electrons in the two half-reactions are equal when added together. Then add the reactions together, simplifying where necessary.As a result, we get the balanced equation for the first reaction:\(3O_{2}(g) + 4Fe_{2}S_{3}(s) + 12H_{2}O \rightarrow 8Fe(OH)_{3}(s) + 3S(s) + 12H^{+} + 12e^{-}\)

Repeat for the rest equations

Repeat the same steps for the remaining reactions.\nOne can apply the same methods to balance the equations (b) and (c).

Key concepts

Electron Transfer

In redox reactions, electron transfer is at the very heart of the process. Simply put, electrons move from one chemical species to another.

• When a substance loses electrons, it is oxidized.
• When a substance gains electrons, it is reduced.

During the balancing process, recognizing who is "losing" and "gaining" these electrons helps in identifying the half-reactions. For instance, Fe in \(Fe_2S_3\) loses electrons to become \(Fe(OH)_3\) in one half-reaction. Meanwhile, \(O_2\) gains electrons to be reduced to S in another half-reaction. Tracking these electron transfers ensures the equation captures everything happening in the reaction, aligning with the conservation of charge.

Oxidation States

Understanding oxidation states is key to identifying how electrons shift during a reaction. Oxidation states, or oxidation numbers, help us keep track of electron distribution in atoms.

• Oxidation involves an increase in oxidation state.
• Reduction involves a decrease in oxidation state.

For example, in the transition from \(Fe_2S_3\) to \(Fe(OH)_3\), the oxidation state of iron changes, indicating electron loss. Similarly, for \(O_2\) transforming into sulfur, the decrease in oxidation state suggests gaining electrons. Recognizing and assigning correct oxidation states lets us precisely determine what's oxidized and what's reduced. This step is vital because it clarifies the electron exchange that is happening within the reaction.

Half-Reactions

Half-reactions break a full redox reaction into two parts, making analysis and balancing simpler.

• Oxidation half-reaction: Highlights the species that lose electrons.
• Reduction half-reaction: Focuses on the species that gain electrons.

Each half is balanced with respect to both mass and charge. For example, the oxidation half-reaction \(Fe_2S_3 \rightarrow 2Fe(OH)_3\) shows iron's oxidation while the reduction half-reaction like \(O_2 + H_2O \rightarrow S\) outlines sulfur's formation. By analyzing in halves, you ensure both aspects are properly accounted for before combining them again. Each half indicates the electrons involved in the reaction, helping perfect balance in both mass and charge.

Balancing Equations

Balancing redox equations might seem tricky at first, but it boils down to a few key steps. First, you balance the atoms involved apart from hydrogen and oxygen.

• Balance other elements first.
• Address oxygen by adding \(H_2O\) as needed.
• For hydrogen consistency, add \(H^+\) ions appropriately in solution.

Next, charges are balanced by proper assignment of electrons. Electrons are added based on excess positive charge to neutralize each half-reaction. Once the half-reactions are balanced for atoms and charge, you then combine them. Matching electron numbers allows the final reaction to be charge-balanced. This detailed method for balancing ensures both conservation of mass and charge in redox reactions, leading to correctly resolved chemical equations.