Question

Balance the following equations in basic solution: a) \(\mathrm{PbO}_{2}+\mathrm{KCl} \longrightarrow \mathrm{KClO}+\mathrm{KPb}(\mathrm{OH})_{3}\) b) \(\mathrm{KMnO}_{4}+\mathrm{KIO}_{3} \longrightarrow \mathrm{MnO}_{2}+\mathrm{KIO}_{4}\) c) \(\mathrm{K}_{2} \mathrm{MnO}_{4} \longrightarrow \mathrm{MnO}_{2}+\mathrm{KMnO}_{4}\)

Short answer

a) PbO2 + KCl -> KClO + KPb(OH)3 b) KMnO4 + KIO3 -> MnO2 + KIO4 c) K2MnO4 -> MnO2 + KMnO4

Step-by-step solution

Title - Write the Unbalanced Equations

The first step is to write the given unbalanced equations: (a) \(\text{PbO}_{2}+\text{KCl} \to \text{KClO}+\text{KPb}(\text{OH})_{3}\) (b) \(\text{KMnO}_{4}+\text{KIO}_{3} \to \text{MnO}_{2}+\text{KIO}_{4}\) (c) \(\text{K}_{2}\text{MnO}_{4} \to \text{MnO}_{2}+\text{KMnO}_{4}\)

Title - Assign Oxidation States

Identify the oxidation states of each element in the reactants and products to determine the changes in oxidation states and identify which elements are oxidized and which are reduced: a) Pb in \(\text{PbO}_{2}\) changes from +4 (in \(\text{PbO}_{2}\)) to +2 (in \(\text{KPb(OH)_{3}}\)), Cl in \(\text{KCl}\) changes from -1 (in \(\text{KCl}\)) to +1 (in \(\text{KClO}\)). b) Mn in \(\text{KMnO}_{4}\) changes from +7 to +4 (in \(\text{MnO}_{2}\)), I in \(\text{KIO}_{3}\) changes from +5 to +7 (in \(\text{KIO}_{4}\)). c) Mn in \(\text{K}_{2}\text{MnO}_{4}\) changes from +6 to +4 (in \(\text{MnO}_{2}\)) and +7 (in \(\text{KMnO}_{4}\)).

- Balance the Atoms in Each Half-Reaction

Each reaction consists of oxidation and reduction half-reactions. We will balance each half-reaction separately for atoms other than H and O: a) Oxidation: \(\text{KCl} \to \text{KClO}\) Reduction: \(\text{PbO}_{2} \to \text{KPb}(\text{OH})_{3}\). b) \(\text{KMnO}_{4} \to \text{MnO}_{2}\) and \(\text{KIO}_{3} \to \text{KIO}_{4}\). c) \(\text{K}_{2}\text{MnO}_{4} \to \text{MnO}_{2}\) and \(\text{K}_{2}\text{MnO}_{4} \to \text{KMnO}_{4}\).

- Balance Oxygen Atoms Using Water

For each half-reaction, balance the number of oxygen atoms by adding water molecules: a) Oxidation: \(\text{KCl} + \text{H}_{2}\text{O} \to \text{KClO} + 2\text{H}^{+}\) Reduction: \(\text{PbO}_{2} + 3\text{H}_{2}\text{O} \to \text{KPb}( \text{OH})_{3} + 3\text{OH}^{-}\). b) Reduction: \(\text{KMnO}_{4} \to \text{MnO}_{2} + 4\text{H}_{2}\text{O}\) Oxidation: \(\text{KIO}_{3} \to \text{KIO}_{4} + \text{H}_{2}\text{O}\). c) Oxidation: \(\text{K}_{2}\text{MnO}_{4} \to \text{MnO}_{2} + 2\text{H}_{2}\text{O}\) Reduction: \(\text{K}_{2}\text{MnO}_{4} \to \text{KMnO}_{4} + \text{H}_{2}\text{O}\).

- Balance Hydrogen Atoms Using Hydroxide Ions

To balance hydrogen atoms, add \(\text{OH}^{-}\) ions: a) Oxidation: \(\text{KCl} + \text{H}_{2}\text{O} + 2\text{OH}^{-} \to \text{KClO} + 2\text{H}_{2}\text{O}\) Reduction: \(\text{PbO}_{2} + 3\text{H}_{2}\text{O} + 3\text{OH}^{-} \to \text{KPb}(\text{OH})_{3} + 3\text{OH}^{-}\). b) Reduction: \(\text{KMnO}_{4} \to \text{MnO}_{2} + 4\text{H}_{2}\text{O} + 4\text{OH}^{-}\) Oxidation: \(\text{KIO}_{3} + 2\text{OH}^{-} \to \text{KIO}_{4} + \text{H}_{2}\text{O}+ 2\text{H}^{+} + 2\text{OH}^{-}\). c) Oxidation: \(\text{K}_{2}\text{MnO}_{4} + 2\text{OH}^{-} \to \text{MnO}_{2} + 2\text{H}_{2}\text{O} \) Reduction: \(\text{K}_{2}\text{MnO}_{4} + \text{OH}^{-} \to \text{KMnO}_{4} + \text{H}_{2}\text{O}\).

- Combine Half-Reactions and Balance Charge

Combine the balanced half-reactions from each equation, ensuring the number of electrons lost in oxidation equals the number gained in reduction: a) Overall: \(\text{PbO}_{2} + \text{KCl} \to \text{KClO} + \text{KPb}(\text{OH})_{3}\). b) Overall: \(\text{KMnO}_{4} + \text{KIO}_{3} \to \text{MnO}_{2} + \text{KIO}_{4}\). c) Overall: \(\text{K}_{2}\text{MnO}_{4} \to \text{MnO}_{2} + \text{KMnO}_{4}\).

- Verify Atoms and Charge Balance

Double-check the final balanced equations to ensure that all atoms and charge are balanced: a) \(\text{PbO}_{2} + \text{KCl} \to \text{KClO} + \text{KPb}(\text{OH})_{3}\) b) \(\text{KMnO}_{4} + \text{KIO}_{3} \to \text{MnO}_{2} + \text{KIO}_{4}\) c) \(\text{K}_{2}\text{MnO}_{4} \to \text{MnO}_{2} + \text{KMnO}_{4}\)

Key concepts

oxidation states

Oxidation states are a way to keep track of electrons in chemical reactions. They tell us how many electrons an atom has gained, lost, or shared with another atom. In a chemical equation, you can assign oxidation states to each element to see which elements are getting oxidized (losing electrons) and which are getting reduced (gaining electrons). For example, in the reaction \(\mathrm{PbO}_{2} + \mathrm{KCl} \longrightarrow \mathrm{KClO} + \mathrm{KPb}(\mathrm{OH})_{3}\), lead (\(Pb\)) changes its oxidation state from \(+4\) in \(\mathrm{PbO}_{2}\) to \(+2\) in \(\mathrm{KPb}(\mathrm{OH})_{3}\). Chlorine (\(Cl\)) changes its oxidation state from \(-1\) in \(\mathrm{KCl}\) to \(+1\) in \(\mathrm{KClO}\). This tells us that \(Pb\) is being reduced and \(Cl\) is being oxidized. Similarly, analyzing other reactions can help identify which atoms are undergoing oxidation and which are undergoing reduction.

reduction half-reactions

Half-reactions are a way of splitting a redox reaction into two parts. Reduction half-reactions show the part of the overall equation where a substance gains electrons. For instance, in the equation \(\mathrm{PbO}_{2} \longrightarrow \mathrm{KPb}(\mathrm{OH})_{3}\), the lead (\(Pb\)) gains electrons, so this is our reduction half-reaction. Creating half-reactions for each element helps focus on one change at a time, making it easier to balance the full equation later.

oxidation half-reactions

Just like reduction half-reactions, oxidation half-reactions show the part of the reaction where a substance loses electrons. Take for example \(\mathrm{KCl} \longrightarrow \mathrm{KClO}\). Chlorine (\(Cl\)) is oxidized as it goes from an oxidation state of \(-1\) to \(+1\). Writing down oxidation and reduction half-reactions separately helps in understanding and tackling the changes in each element before combining them to balance the full equation.

balancing atoms

Balancing atoms means ensuring that there are the same number of each type of atom on both sides of the equation. Once the oxidation and reduction half-reactions are written, you balance all atoms except oxygen (O) and hydrogen (H) first. In the example \(\mathrm{KMnO}_{4} + \mathrm{KIO}_{3} \longrightarrow \mathrm{MnO}_{2} + \mathrm{KIO}_{4}\), you would see that Mn and I atoms need balancing first. Mn changes from \(+7\) in \(\mathrm{KMnO}_{4}\) to \(+4\) in \(\mathrm{MnO}_{2}\) and \(I\) changes from \(+5\) in \(\mathrm{KIO}_{3}\) to \(+7\) in \(\mathrm{KIO}_{4}\). These need to be sorted before you move on to balancing oxygen and hydrogen.

oxygen and hydrogen balance

Balancing oxygen (O) and hydrogen (H) are usually done in a couple of steps. First, you balance the oxygen atoms by adding water molecules (\(\mathrm{H}_{2}\mathrm{O}\)). For example, when balancing \(\mathrm{KIO}_{3} \longrightarrow \mathrm{KIO}_{4}+ \mathrm{H}_{2}\mathrm{O}\), you add water to the appropriate side to balance oxygen atoms. Next, you balance hydrogen atoms by adding hydroxide ions (\(\mathrm{OH}^{-}\)) when working in basic solutions. Let's say you balanced oxygen and now you end up with excess hydrogen on one side; you balance it by adding \(\mathrm{OH}^{-}\). This is like solving puzzles with each small step bringing you nearer to a completely balanced sport in the equation.