Question

Balance the following redox equations. (a) \(\mathrm{Fe}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{NO}_{2}(g)\) (acidic) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic)

Short answer

Question: Balance the following two redox equations. (a) \(Fe(s)+NO_{3}^{-}(aq) \longrightarrow Fe^{3+}(aq)+NO_{2}(g)\) (acidic) (b) \(Cr(OH)_{3}(s)+O_{2}(g) \longrightarrow CrO_{4}^{2-}(aq)\) (basic) Answer: (a) The balanced redox equation in acidic conditions is: \(Fe(s) + 3NO_{3}^{-}(aq) + 6H^+(aq) \longrightarrow Fe^{3+}(aq) + 3NO_{2}(g) + 3H_2O(l)\) (b) The balanced redox equation in basic conditions is: \(2Cr(OH)_{3}(s) + \frac{3}{2}O_{2}(g) \longrightarrow 2CrO_{4}^{2-}(aq) + 12H_2O(l)\)

Step-by-step solution

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q)\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\)

Balance the atoms other than hydrogen and oxygen

The atoms of elements other than hydrogen and oxygen are already balanced.

Balance the oxygen atoms by adding water molecules

Oxidation: No need to add water since there is no oxygen atom in this half-reaction. Reduction: Add 1 water molecule to balance the oxygen atoms. \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)

Balance the hydrogen atoms by adding hydrogen ions

Oxidation: No need to add hydrogen ions since there is no hydrogen atom in this half-reaction. Reduction: Add 2 H+ ions to balance the hydrogen atoms. \(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)

Balance the charge of the half-reactions by adding electrons

Oxidation: Add 3 electrons to balance the charge. \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3 e^-\) Reduction: Add 1 electron to balance the charge. \(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) + e^- \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)

Make the number of electrons in the two half-reactions equal

Multiply the reduction half-reaction by 3. Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3 e^-\) Reduction: \(3(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) + e^-) \longrightarrow 3(\mathrm{NO}_{2}(g) + \mathrm{H_2O}(l))\) So the new reduction half-reaction is: \(\mathrm{3NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) + 3 e^- \longrightarrow 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\)

Add the two half-reactions together

\(\mathrm{Fe}(s) + 3\mathrm{NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\)

Cancel out any common terms

There are no common terms that can be canceled out. So the balanced redox equation in acidic conditions is: \(\mathrm{Fe}(s) + 3\mathrm{NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic) Follow the same steps 1 to 7 as in (a):

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Cr(OH)}_3(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) Reduction: \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H_2O}(l)\)

Balance the atoms other than hydrogen and oxygen

Oxidation: Add 2 Cr atoms to the product side. \(2\mathrm{Cr(OH)}_3(s) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q)\) Reduction: No need to balance atoms since there are no other elements in this half-reaction.

Balance the oxygen atoms by adding water molecules

Oxidation: Add 6 water molecules to balance the oxygen atoms. \(2\mathrm{Cr(OH)}_3(s) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\) Reduction: Add 4 water molecules to balance the oxygen atoms. \(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{H_2O}(l)\)

Balance the hydrogen atoms by adding hydrogen ions

Oxidation: Add 12 H+ ions to balance the hydrogen atoms. \(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\) Reduction: Add 8 H+ ions to balance the hydrogen atoms. \(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l) + 8\mathrm{H}^+(a q) \longrightarrow \mathrm{H_2O}(l)\)

Balance the charge of the half-reactions by adding electrons

Oxidation: Add 6 electrons to balance the charge. \(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) + 6 e^- \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\) Reduction: Add 4 electrons to balance the charge. \(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l)+ 8\mathrm{H}^+(a q) + 4 e^- \longrightarrow \mathrm{H_2O}(l)\)

Make the number of electrons in the two half-reactions equal

Multiply the reduction half-reaction by 1.5 (or multiply by 3 and then divide by 2). Oxidation: \(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) + 6 e^- \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\) Reduction: \((3/2)(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l)+ 8\mathrm{H}^+(a q) + 4 e^-) \longrightarrow (3/2)\mathrm{H_2O}(l)\) So the new reduction half-reaction is: \(\frac{3}{2}\mathrm{O}_{2}(g) + 6\mathrm{H}_2\mathrm{O}(l)+ 12\mathrm{H}^+(a q) + 6 e^- \longrightarrow 3\mathrm{H_2O}(l)\)

Add the two half-reactions together

\(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) + 6\mathrm{H}_2\mathrm{O}(l)+ 12\mathrm{H}^+(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 9\mathrm{H_2O}(l)\) Now we need one additional step to account for the basic conditions:

Convert hydrogen ions into hydroxide ions

Add 12 OH- ions to both sides of the equation to remove the H+ ions. \(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) + 12\mathrm{OH}^-(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 9\mathrm{H_2O}(l) + 12\mathrm{OH}^-(a q)\) Combine the water and hydroxide ions to form 3 more water molecules. \(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 12\mathrm{H_2O}(l)\) So the balanced redox equation in basic conditions is: \(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 12\mathrm{H_2O}(l)\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, allowing you to understand how substances interact in a reaction. In a balanced chemical equation, the number of each type of atom is conserved on both sides of the equation. This conservation reflects the Law of Conservation of Mass, which states that matter cannot be created or destroyed in an isolated system. In a redox reaction, it is vital to ensure each half-reaction is balanced individually before combining them into one overall balanced equation. Here's how you can approach this:

• Start by listing all the elements involved and tallying their counts on each side of the equation.
• Adjust coefficients (numbers in front of chemical formulas) to balance elements other than oxygen and hydrogen first.
• Balance oxygen atoms by adding water molecules to the side that needs more oxygen atoms.
• Balance hydrogen atoms by adding hydrogen ions (in acidic conditions) or hydroxide ions (in basic conditions).
• Finally, check that the charge is balanced by ensuring the number of electrons lost equals the number gained in the half-reactions.

Oxidation-Reduction

Oxidation-reduction reactions, commonly known as redox reactions, involve the transfer of electrons between substances. In these reactions, reduction refers to the gain of electrons, while oxidation refers to the loss of electrons. Electrons transferred during a redox reaction must be balanced in the chemical equation:

• Identify which species is oxidized (losing electrons) and which is reduced (gaining electrons).
• Write separate oxidation and reduction half-reactions, indicating the transfer of electrons explicitly.
• Balance each half-reaction for atoms and charge, ensuring that electrons are transferred equally in both half-reactions when combined.
• Pay close attention to changes in oxidation states, which help determine how electrons are exchanged among reactants and products.

Understanding these steps is crucial for mastering redox chemistry and solving complex chemical equations efficiently.

Acidic and Basic Conditions

In redox reactions, the conditions of the solution—whether acidic or basic—can alter the balancing process. The presence of either an acidic or basic environment affects how you balance the hydrogen and oxygen in the half-reactions. Here’s a simple breakdown:

• **Acidic Solutions:** In an acidic solution, hydrogen ions (H⁺) are present, making it easier to add H⁺ where needed to balance hydrogen atoms after balancing for oxygen using water molecules.
• **Basic Solutions:** In a basic solution, hydroxide ions (OH⁻) are prevalent. After you balance the redox reaction as if it is in acidic conditions, convert H⁺ ions into H₂O by adding OH⁻ to both sides of the equation, effectively neutralizing the H⁺ ions to adjust for basic conditions.
• This conversion is key to ensuring the reaction accurately reflects the true nature of the solution environment.

Being able to navigate whether a reaction occurs in acidic or basic conditions will ensure more accuracy in balancing equations.

Half-Reaction Method

The half-reaction method is a systematic approach to tackling redox equations. It involves separating the redox reaction into two simpler components, making it easier to balance each part. Here's what you need to do:

• Separate the equation into the oxidation and reduction half-reactions.
• Balance each half-reaction independently to ensure both mass and charge balance.
• First, balance all elements except for oxygen and hydrogen.
• Add water molecules to balance oxygen and hydrogen ions to balance hydrogen in acidic solutions, or hydroxide ions in basic solutions after completing the acid balance.
• Balance the charge discrepancy by calculating the electrons needed and adding them accordingly.
• Once both half-reactions are balanced, adjust the coefficients so that the number of electrons in each half-reaction match.
• Finally, combine the two half-reactions back together, ensuring all electrons cancel out, resulting in a net balanced redox equation.

This method is particularly advantageous for complex redox equations, providing a clear and systematic outline for balancing intricate reactions.