Problem 6 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s}) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(\mathrm{NiO}_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})\) $$\begin{aligned}&\text { (c) } \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \rightarrow\\\&&\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}(\mathrm{aq})\end{aligned}$$ $$\text { (d) } \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{Ag}(\mathrm{s})$$

Step-by-Step Solution

Answer

(a) Cr + 3Fe(OH)₃ + 3OH⁻ → Cr(OH)₃ + 3Fe(OH)₂; (b) NiO₂ + Zn + 2H₂O → Ni(OH)₂ + Zn(OH)₂; (c) 3Fe(OH)₂ + CrO₄²⁻ + 4H₂O → 3Fe(OH)₃ + Cr(OH)₄⁻ + 4OH⁻; (d) N₂H₄ + Ag₂O → N₂ + 2Ag + 2H₂O.

Step 1: Identify Oxidation and Reduction Reactions (a)

For the reaction \( \mathrm{Fe} (\mathrm{OH})_{3} (\mathrm{s}) + \mathrm{Cr} (\mathrm{s}) \rightarrow \mathrm{Cr} (\mathrm{OH})_{3} (\mathrm{s}) + \mathrm{Fe} (\mathrm{OH})_{2} (\mathrm{s}) \), Cr is oxidized from 0 to +3 while Fe is reduced from +3 to +2.

Step 2: Balance Atoms and Charges (a)

Separate the redox processes: \( \mathrm{Cr} (\mathrm{s}) \rightarrow \mathrm{Cr} (\mathrm{OH})_{3} \) and \( \mathrm{Fe} (\mathrm{OH})_{3} \rightarrow \mathrm{Fe} (\mathrm{OH})_{2} \). Balance the hydroxide ions and water to get: \[ \mathrm{Cr} (\mathrm{s}) + 3\mathrm{OH}^{-} \rightarrow \mathrm{Cr} (\mathrm{OH})_{3} + 3e^{-} \] \[ \mathrm{Fe} (\mathrm{OH})_{3} + e^{-} \rightarrow \mathrm{Fe} (\mathrm{OH})_{2} + \mathrm{OH}^{-} \].

Step 3: Combine and Balance Electrons (a)

Multiply the Fe equation by 3 to balance the electrons and combine: \[ \mathrm{Fe} (\mathrm{OH})_{3} + \mathrm{Cr} + 3\mathrm{OH}^{-} \rightarrow \mathrm{Cr} (\mathrm{OH})_{3} + 3\mathrm{Fe} (\mathrm{OH})_{2} \].

Step 4: Identify Oxidation and Reduction Reactions (b)

For the reaction \( \mathrm{NiO}_2 + \mathrm{Zn} \rightarrow \mathrm{Ni} (\mathrm{OH})_2 + \mathrm{Zn} (\mathrm{OH})_{2} \), Zn is oxidized from 0 to +2 and Ni is reduced from +4 to +2.

Step 5: Balance Atoms and Charges (b)

Separate the redox processes: \( \mathrm{Zn} \rightarrow \mathrm{Zn} (\mathrm{OH})_{2} \) and \( \mathrm{NiO}_2 \rightarrow \mathrm{Ni} (\mathrm{OH})_2 \). Balance the reactions: \[ \mathrm{Zn} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Zn} (\mathrm{OH})_2 + 2e^{-} \] \[ \mathrm{NiO}_2 + 2e^{-} + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ni} (\mathrm{OH})_2 + 2\mathrm{OH}^{-} \].

Step 6: Combine and Balance Electrons (b)

Both are balanced in electrons, so combined reaction: \[ \mathrm{NiO}_2 + \mathrm{Zn} + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ni} (\mathrm{OH})_2 + \mathrm{Zn} (\mathrm{OH})_{2} \].

Step 7: Identify Oxidation and Reduction Reactions (c)

For \( \mathrm{Fe} (\mathrm{OH})_2 + \mathrm{CrO}_4^{2-} \rightarrow \mathrm{Fe} (\mathrm{OH})_3 + [\mathrm{Cr} (\mathrm{OH})_4]^{-} \), Fe is oxidized from +2 to +3, and Cr is reduced from +6 to +3.

Step 8: Balance Atoms and Charges (c)

Separate the redox processes: \( \mathrm{Fe} (\mathrm{OH})_2 \rightarrow \mathrm{Fe} (\mathrm{OH})_3 \) and \( \mathrm{CrO}_4^{2-} \rightarrow [\mathrm{Cr} (\mathrm{OH})_4]^{-} \). The reactions balance as: \[ \mathrm{Fe} (\mathrm{OH})_2 + \mathrm{OH}^- \rightarrow \mathrm{Fe} (\mathrm{OH})_3 + e^{-} \] \[ \mathrm{CrO}_4^{2-} + 3e^{-} + 4\mathrm{H}_2\mathrm{O} \rightarrow [\mathrm{Cr} (\mathrm{OH})_4]^{-} + 5\mathrm{OH}^{-} \].

Step 9: Combine and Balance Electrons (c)

Multiply the Fe reaction by 3 to balance electrons: \[ 3\mathrm{Fe} (\mathrm{OH})_2 + \mathrm{CrO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{Fe} (\mathrm{OH})_3 + [\mathrm{Cr} (\mathrm{OH})_4]^{-} + 4\mathrm{OH}^{-} \].

Step 10: Identify Oxidation and Reduction Reactions (d)

For \( \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{Ag}_{2}\mathrm{O} \rightarrow \mathrm{N}_{2} + 2\mathrm{Ag} \), Ag is reduced from +1 to 0, and N is oxidized from -2 to 0.

Step 11: Balance Atoms and Charges (d)

Separate the redox processes: \( \mathrm{N}_{2} \mathrm{H}_{4} \rightarrow \mathrm{N}_{2} \) and \( \mathrm{Ag}_{2}\mathrm{O} \rightarrow 2\mathrm{Ag} \). Balance the reactions: \[ \mathrm{N}_{2} \mathrm{H}_{4} \rightarrow \mathrm{N}_{2} + 4e^{-} + 4\mathrm{H}^{+} \] \[ \mathrm{Ag}_{2} \mathrm{O} + 2e^{-} \rightarrow 2\mathrm{Ag} + \mathrm{O}^{2-} \]. Balance using \( \mathrm{H}_2\mathrm{O} \) and \( \mathrm{OH}^{-} \): \[ 4\mathrm{OH}^{-} + \mathrm{N}_{2} \mathrm{H}_{4} \rightarrow \mathrm{N}_{2} + 4\mathrm{H}_2\mathrm{O} + 4e^{-} \] \[ \mathrm{Ag}_{2} \mathrm{O} + 2e^{-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{Ag} + \mathrm{H}_2\mathrm{O} \].

Step 12: Combine and Balance Electrons (d)

Multiply the Ag reaction by 2 to balance electrons: \[ 4\mathrm{OH}^{-} + \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{Ag}_{2}\mathrm{O} \rightarrow \mathrm{N}_{2} + 2\mathrm{Ag} + 3\mathrm{H}_2\mathrm{O} \].

Key Concepts

Basic Solution

Redox reactions often take place in different types of solutions, and "basic solution" specifically means that the environment has more hydroxide ions (OH^-) than hydrogen ions (H⁺). This basicity affects the balancing of redox reactions. To keep everything electrically neutral while balancing redox reactions in basic solutions,
extra hydroxide ions may need to be added. These steps ensure that the charges and the number of atoms are balanced on both sides of the equation. Inevitably, this turns to water (H₂O) when hydroxide interacts with hydrogen ions that come from species in its various forms throughout the reaction.

• Always start by balancing atoms other than hydrogen and oxygen.
• Then, balance oxygen atoms by adding water molecules to the side lacking oxygen.
• Complete the balancing by adjusting hydrogen atoms using hydroxide ions.
• The ions added will interact with water to further assist in balancing.

This approach makes sure that the reaction fits within the boundaries of a basic solution.

Oxidation States

Oxidation states serve as a tool to track electron transfer in redox reactions. An atom's oxidation state is a hypothetical charge that it would have if the bonds were completely ionic. Changes in oxidation states help identify which elements undergo reduction or oxidation.

• An increase in oxidation state indicates oxidation (loss of electrons).
• A decrease in oxidation state indicates reduction (gain of electrons).

For example, in the reaction \( \mathrm{Fe(OH)}_3 + \mathrm{Cr}(s) \rightarrow \mathrm{Cr(OH)}_3 + \mathrm{Fe(OH)}_2 \), chromium's oxidation state increases from 0 to +3 (oxidation), and iron's oxidation state decreases from +3 to +2 (reduction). Thus, tracking these changes helps us to efficiently separate and balance half-reactions.

Electron Transfer

Understanding electron transfer is crucial as it is the essence of a redox reaction. In these reactions, electrons move from one species (the reducing agent) to another (the oxidizing agent). This transfer is what makes balancing redox equations unique and sometimes complex.
By focusing on which element donates electrons and which one gains them, we can simplify reactions into two separate half-reactions that represent the separate oxidation and reduction sequences. Electron balancing ensures that both sides of the redox equation reflect the conservation of charge and mass.

• Identify the species that lose electrons (undergo oxidation).
• Recognize the species that gain electrons (undergo reduction).
• Ensure the number of electrons lost is equal to the number of electrons gained, ensuring electrical neutrality.

This aspect simplifies understanding how substances change during reactions by focusing on electron movement.

Half-Reactions

Breaking down a redox reaction into half-reactions is an effective method to manage the complexities of balancing equations. Each half-reaction represents either an oxidation or a reduction process and must be balanced separately before combining them into the final equation.

• Oxidation half-reaction: This reaction shows the species losing electrons. For instance, \( \mathrm{Cr}(s) \rightarrow \mathrm{Cr(OH)}_3 \) includes the loss of electrons by chromium.
• Reduction half-reaction: This is where something gains electrons. An example is \( \mathrm{Fe(OH)}_3 + e^{-} \rightarrow \mathrm{Fe(OH)}_2 \), which shows iron gaining electrons.

Ensure both reactions independently balance in terms of mass and charge, then check that each half-reaction transfers equal numbers of electrons. Only after that point can they be summed to display the full redox reaction.
This breakdown approach simplifies understanding and ensures the reaction is correctly balanced according to oxidation-reduction principles. It also allows for meticulous tracking of each component involved, which is invaluable in complex reactions.