Problem 36 - Chemistry Principles and Practice

Question

Balance each of the following redox reactions in basic solution. (a) \(\mathrm{PH}_{3}(\mathrm{~g})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \rightarrow \mathrm{CrO}_{2}^{-}(\mathrm{aq})+\mathrm{P}_{4}(\mathrm{~s})\) (b) \(\mathrm{F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{F}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{~s}) \rightarrow \mathrm{CrO}_{4}^{2-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Combine: \(2PH_3 + 4H_2O \rightarrow P_4 + 6OH^- + CrO_2^- (b) Final: \(F_2 + 2OH^- \rightarrow 2F^- + O_2H_2O (c) (See Step 7, similar breakdown absolutely)

Step 1: Write Down the Half-Reactions

(a) Oxidation: \(\mathrm{PH}_{3} \rightarrow \mathrm{P}_{4}\)Reduction: \(\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{CrO}_{2}^{-}\)(b) Oxidation: \(\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2\)Reduction: \(\mathrm{F}_2 \rightarrow \mathrm{F}^-\)(c) Oxidation: \(\mathrm{H}_2\mathrm{O}_{2} \rightarrow \mathrm{O}_2\)Reduction: \(\mathrm{Cr(OH)}_3 \rightarrow \mathrm{CrO}_{4}^{2-}\)

Step 2: Balance Atoms Other Than O and H in Each Half-Reaction

(a) For oxidation: Balance P with multiples, you get 4 PH₃: \(4\mathrm{PH}_{3} \rightarrow \mathrm{P}_{4}\)(b) For oxidation, H₂O is already balanced:For reduction, F is balanced in \(\mathrm{F}_2 + 2\mathrm{e}^- \rightarrow 2\mathrm{F}^-\)(c) Chromium balances naturally: \(\mathrm{Cr(OH)}_3 \rightarrow \mathrm{CrO}_{4}^{2-}\)

Step 3: Balance Oxygen Atoms

(a) No oxygen in \(\mathrm{PH}_{3}\rightarrow \mathrm{P}_{4}\), but 2 oxygens needed: (For reduction, reduction requires 2 waters to balance oxygens: \(\mathrm{CrO}_{4}^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CrO}_{2}^{-}\)(b) Water gives enough oxygen for the reduction: \(\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 2\mathrm{H}^+\)(c) Add 4 water for Oxidation: \(\mathrm{H}_2\mathrm{O}_{2} \rightarrow \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O}\)For Reduction, add 4 waters: \(\mathrm{Cr(OH)}_3 + 4\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CrO}_{4}^{2-}\)

Step 4: Balance Hydrogen Atoms by Adding H⁺

(a) Add 6 OH⁻ for charge:Oxidation: \(\mathrm{PH}_{3} + 6\mathrm{OH}^- \rightarrow \mathrm{P}_{4} + 6\mathrm{H}_2\mathrm{O}\)(b) For water balance:Oxidation: \(4\mathrm{OH}^- + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{e}^- + 2\mathrm{H}_2\mathrm{O}\)Reduction: \(8\mathrm{OH}^- + \mathrm{F}_2 \rightarrow 2\mathrm{F}^- + 2\mathrm{O}_2\)(c) Add 7 OH⁻:Oxidation: \(\mathrm{H}_2\mathrm{O}_{2} + 7\mathrm{OH}^- \rightarrow \mathrm{O}_2 + 4\mathrm{H}_2\mathrm{O}\)Reduction: \(\mathrm{Cr(OH)}_3 + 7\mathrm{OH}^- \rightarrow \mathrm{CrO}_{4}^{2-} + 4\mathrm{H}_2\mathrm{O}\)

Step 5: Balance Charge with Electrons

(a) Reduction: Add 8 electrons: \(\mathrm{CrO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} + 8\mathrm{e}^- \rightarrow \mathrm{CrO}_2^- + 8\mathrm{OH}^-\)Ox: \(4e^-\): \(\mathrm{PH}_3 \rightarrow \mathrm{P}_4 + 4e^-\)(b) Balance charge: Reduction: \(\mathrm{F}_2 + 2e^- \rightarrow 2\mathrm{F}^-\)Oxidation: \(\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4e^-\)(c) 8 electrons for both:Oxidation: \(2\mathrm{OH}^- \rightarrow \mathrm{O}_2 + 2e^-\)Reduction: \(\mathrm{Cr(OH)}_3 + 8e^- \rightarrow \mathrm{CrO}_{4}^{2-} + 4\mathrm{OH}^-\)

Step 6: Multiply to Equalize Electrons

(a) Multiply oxidation by 2 to equalize electrons:\(2(\mathrm{PH}_3 \rightarrow \mathrm{P}_4 + 4e^- )\)Then solve in step 7.(b) Multiply ox by 2 to match 4 e- transferring:Now:\(2(\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 2e^- + 2\mathrm{H}^+)\)(c) Left equalized next as stated then.

Step 7: Combine Half-Reactions and Cancel Identicals

(a) Combine:\(2\mathrm{PH}_3 + 6\mathrm{OH}^- + 8\mathrm{e}^- \rightarrow \mathrm{P}_4 + 6\mathrm{H}_2\mathrm{O}\) \(\mathrm{CrO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} + 8\mathrm{e}^- \rightarrow \mathrm{CrO}_2^- + 8\mathrm{OH}^-\)Result: \(2\mathrm{PH}_3 + 6\mathrm{OH}^- \rightarrow \mathrm{P}_4 + 6\mathrm{H}_2\mathrm{O} + \mathrm{CrO}_2^- + 8\mathrm{OH}^-\)(b) Final add:\(2\mathrm{F}_2 + 2\mathrm{H}_2\mathrm{O} +4\mathrm{e}^- \rightarrow 4\mathrm{F}^- + 4\mathrm{OH}^-\)(c)\(\mathrm{Cr(OH)}_3 + 4\mathrm{H}_2\mathrm{O} +8\mathrm{OH}^- +2e^- \rightarrow \mathrm{CrO}_4^{2-} + 7\mathrm{H}_2\mathrm{O} + 4\mathrm{OH}^-\) so Cancels backs too potentially then percles.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the Law of Conservation of Mass. When approaching a redox reaction, balancing often involves:

• Identifying the oxidation and reduction processes.
• Writing separate half-reactions for oxidation and reduction.
• Balancing each half-reaction for all elements except oxygen and hydrogen.
• Balancing oxygen atoms by adding water molecules.
• Balancing hydrogen atoms by adding hydrogen ions (H⁺).
• Balancing charge by adding electrons (e⁻).

For a full equation, it's critical to ensure that the total number of electrons lost in oxidation equals those gained in reduction. This may involve multiplying the half-reactions by appropriate factors to equalize the number of electrons. Only after doing so can the half-reactions be combined into one full balanced equation.

Basic Solutions

Balancing redox reactions in basic solutions differs slightly from acidic solutions because we're working with hydroxide ions (OH⁻) instead of hydrogen ions (H⁺). Here's how it's typically done:

• After balancing the redox reaction as if it were in an acidic solution, add OH⁻ to both sides for every H⁺ present.
• The added OH⁻ ions will combine with H⁺ ions on the same side to form water (H₂O).
• Cancel out any identical molecules of water on both sides of the equation.

This conversion ensures that the reaction is balanced as it would appear under basic conditions, where OH⁻ ions are more prevalent than H⁺ ions.

Half-Reactions

Redox reactions can be broken down into two constituent parts known as half-reactions. Each half-reaction represents either the oxidation process, where electrons are lost, or the reduction process, where electrons are gained:

• Oxidation: This part involves the loss of electrons. The species being oxidized increases its oxidation state.
• Reduction: This part involves the gain of electrons. The species being reduced decreases its oxidation state.

In writing and balancing half-reactions: - First, balance the elements involved in the redox change, other than oxygen and hydrogen. - Add water molecules to balance any oxygen atoms. - Add H⁺ ions to balance hydrogen atoms (when initially treated as an acid, then convert using OH⁻ for basic conditions). - Balance out the charge of the reaction by adding electrons accordingly. Combining these half-reactions and ensuring the electrons balance is key to properly representing the redox process as a single, coherent chemical reaction.