Problem 36 - General Chemistry: Principles and Modern Applications

Question

Complete and balance these half-equations. (a) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{CO}_{2}\) (acidic solution) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\) (acidic solution) (c) \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{MnO}_{2}\) (basic solution) Indicate whether oxidation or reduction is involved.

Step-by-Step Solution

Answer

The balanced equations are: (a) \(2\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 4\mathrm{CO}_{2} + 2e^-)\); (b) \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14H^{+} + 6e^- \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\); and (c) \(3\mathrm{MnO}_{4}^{-} + 4\mathrm{H}_{2}\mathrm{O} \longrightarrow 3\mathrm{MnO}_{2} + 2\mathrm{OH}^{-} + 4e^-)\). Oxidation is seen in (a) and reduction in (b) and (c)

Step 1: Balancing (a)

First look for the atom that has its oxidation number changed. It's carbon which goes from +3 to +4, therefore it is oxidized. We balance the carbon and oxygen atoms first by placing a 2 in front of \(\mathrm{CO}_{2}\) then we balance the charges using 2 electrons on the left. The balanced equation is \(2\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 4\mathrm{CO}_{2} + 2e^-\)

Step 2: Balancing (b)

The chromium atom is reduced going from +6 to +3 oxidation state. The chromium and oxygen atoms are balanced, then charges are balanced using electrons and hydrogen ions. The resultant balanced equation is \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14H^{+} + 6e^- \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Step 3: Balancing (c)

The manganese atom is reduced from +7 to +4. The manganese and oxygen atoms are balanced first, then charges are balanced using electrons. Finally, hydrogen ions and water molecules are added to neutralize the hydroxide ions in order to produce a balanced basic solution equation. The resultant balanced equation is \(3\mathrm{MnO}_{4}^{-} + 4\mathrm{H}_{2}\mathrm{O} \longrightarrow 3\mathrm{MnO}_{2} + 2\mathrm{OH}^{-} + 4e^-\)

Key Concepts

Understanding Half-Equations

Half-equations are an essential concept in understanding oxidation-reduction reactions, also known as redox reactions. When a chemical reaction involves the transfer of electrons, we separate this process into two parts, each called a half-equation. One half-equation represents the oxidation process – where electrons are lost – and the other represents reduction – where electrons are gained. To create a half-equation, we focus on the species undergoing a change in oxidation state. For example, in the given problems, carbon, chromium, and manganese undergo these changes.

• Oxidation half-equation: Loss of electrons is shown, increasing the oxidation state.
• Reduction half-equation: Gain of electrons is shown, decreasing the oxidation state.

It's crucial to accurately depict the loss or gain of electrons to sustain the balanced nature of the equations in terms of both mass and charge.

Balancing Chemical Equations

Balancing chemical equations is an integral process in ensuring that the number of atoms on the reactant side equals the number on the product side. In redox reactions, this involves a more detailed level of balance considering both atom count and charge. Firstly, we need to balance the atoms of the element undergoing the oxidation or reduction, which often involves adding coefficients to ensure equal numbers on each side of the equation. This is evident in our exercise:

• For carbon in reaction (a), we place a 2 in front of \(\mathrm{CO}_{2}\).
• For the chromium reaction (b), each chromium is balanced by having the formula \(\mathrm{Cr}_{2}\).

After balancing atomic numbers, adjust the charges by adding electrons. Electrons are added to either the left or right side to equalize the charge difference caused by changes in oxidation states. Acidic or basic conditions also influence the balancing process, requiring additional steps such as adding \(\mathrm{H}^{+}\) ions in acidic solutions or \(\mathrm{OH}^{-}\) ions in basic solutions.

Oxidation States in Redox Reactions

Oxidation states, or oxidation numbers, play a pivotal role in understanding and balancing redox reactions. They help track how electrons are transferred in a reaction. Each atom within a molecule is assigned an oxidation state to reflect its degree of oxidation — indicating how 'electron-poor' or 'electron-rich' the atom is. In the provided reactions, the changes in oxidation state are crucial:

• Carbon in \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) increases its oxidation state from +3 to +4, indicating oxidation.
• Chromium in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) decreases from +6 to +3, a reduction process.
• Manganese in \(\mathrm{MnO}_{4}^{-}\) adjusts from +7 to +4, showing reduction.

By understanding changes in oxidation states, we can predict whether a species will undergo oxidation or reduction, guiding the correct balancing of redox equations.