Question

Write plausible half-equations to represent each of the following in basic solution. (a) oxidation of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) to \(\mathrm{FeO}_{4}^{2-}\) (b) reduction of \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) to silver metal

Short answer

(a) \( \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s}) \rightarrow \mathrm{FeO}_{4}^{2-} + 1e^- + 4\mathrm{OH}^- + 4\mathrm{H}_2\mathrm{O} \)\n (b) \( \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} + 1e^- \rightarrow \mathrm{Ag} \)

Step-by-step solution

Writing Half-Reactions

Write down the unbalanced half-reactions: \n (a) \( \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s}) \rightarrow \mathrm{FeO}_{4}^{2-} \), \n (b) \( \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} \rightarrow \mathrm{Ag} \)

Balancing Atoms and Charges

Check if we have equal number of each type of atoms on each side of the equation except hydrogen and oxygen. Since there are already 4 oxygen atoms and 1 iron atom on both sides in reaction (a), no changes are needed here.\nFor reaction (b), there is already 1 silver atom on both sides, hence no changes.\nNext, balance the charges by adding electrons (e-) to the side with more positive overall charge. Hence\n (a) \( \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s}) \rightarrow \mathrm{FeO}_{4}^{2-} + 1e^- \), \n (b) \( \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} + 1e^- \rightarrow \mathrm{Ag} \)

Balancing Hydrogen and Oxygen Atoms

For statement (a), balance the oxygen atoms by adding water molecules (H2O) on the side with less oxygen. Then balance hydrogen by adding a suitable number of hydroxide ions (OH-) on the side with less hydrogen. For statement (b), there's no need to balance hydrogen or oxygen here since there are none present, so this reaction is already complete.\nSo, \n(a) \( \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s}) \rightarrow \mathrm{FeO}_{4}^{2-} + 1e^- + 4\mathrm{OH}^- + 4\mathrm{H}_2\mathrm{O}\)

Key concepts

Half-Reactions in Basic Solution

When we're looking at redox reactions, it's often easier to split them into two parts – one for oxidation and one for reduction. These are called half-reactions. In basic solutions, like with our example of the oxidation of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) to \(\mathrm{FeO}_{4}^{2-}\), we need to take into account the presence of hydroxide ions \(\mathrm{OH}^-\).

After writing down the unbalanced half-reaction, it's important to then balance any atoms that are not oxygen or hydrogen. Following this, we proceed to balance the oxygen atoms by adding \(\mathrm{H}_2\mathrm{O}\) molecules where needed, and then we balance hydrogen by adding \(\mathrm{OH}^-\) ions.

Finally, charges must be balanced as well – this is commonly done by adding electrons \(e^-\) to the more positive side thus ensuring that the total charge is the same on both sides of the equation. In the example given, for the oxidation half-reaction, after balancing the oxygen with water molecules, \(\mathrm{OH}^-\) ions were added to balance hydrogen and lastly an electron to balance the charge, leading to the balanced half-reaction in a basic solution.

Oxidation and Reduction Equations

Redox reactions consist of two key processes: oxidation, where a substance loses electrons, and reduction, where it gains electrons. Remember this handy phrase: 'LEO says GER' – Lose Electrons Oxidation, Gain Electrons Reduction.

In our examples, the oxidation process is seen with iron moving from \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) to \(\mathrm{FeO}_{4}^{2-}\), where it loses electrons. Conversely, the reduction process involves \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) gaining electrons to form silver metal. The balancing of these equations requires ensuring that the overall number of electrons lost in oxidation matches the number gained in reduction.

For the reduction half-reaction given, no balancing of hydrogen or oxygen was necessary because they weren't present in the reaction – it was already balanced with the addition of only an electron. The balancing of oxidation and reduction equations is crucial because only then can they be combined to represent the full, balanced redox reaction.

Balancing Charges and Atoms in Chemistry

In chemistry, it's a fundamental principle that in a balanced equation, both mass and charge are conserved. Meaning, the number of each type of atom and the overall charge must be equal on both sides of the equation.

When balancing charges, if you have a surplus of positive charge, you add electrons to that side because electrons have a negative charge. This helps to neutralize the imbalance. As for atoms, the 'Atoms First' approach suggests balancing all atoms other than oxygen and hydrogen first, then moving on to balance these two using water \(\mathrm{H}_2\mathrm{O}\) and hydroxide ions \(\mathrm{OH}^-\) in a basic solution.

In our initial oxidation reaction, after balancing iron and oxygen atoms, we added one electron to balance the charge. This step-by-step approach makes sure the resulting equations reflect the conservation of mass and charge, which is crucial for the equations to make sense both mathematically and chemically.