Problem 49 - Chemistry and Chemical Reactivity

Question

Write balanced equations for the following half-reactions. (a) \(\mathrm{UO}_{2}^{+}(\mathrm{aq}) \rightarrow \mathrm{U}^{4+}(\mathrm{aq})\) (acid solution) (b) \(\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})\) (acid solution) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq}) \rightarrow \mathrm{N}_{2}(\mathrm{g})\) (basic solution) (d) \(\mathrm{ClO}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})\) (basic solution)

Step-by-Step Solution

Answer

(a) \( \mathrm{UO}_2^+ + 2\mathrm{H}^+ + \mathrm{e}^- \rightarrow \mathrm{U}^{4+} + \mathrm{H}_2\mathrm{O} \); (b) \( \mathrm{ClO}_3^- + 6\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow \mathrm{Cl}^- + 3\mathrm{H}_2\mathrm{O} \); (c) \( \mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^- \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^- \); (d) \( \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^- \rightarrow \mathrm{Cl}^- + 2\mathrm{OH}^- \).

Step 1: Oxidation and Reduction Determination (a)

Determine the oxidation state of uranium in \(\mathrm{UO}_2^+(\mathrm{aq})\) and \(\mathrm{U}^{4+}(\mathrm{aq})\).- In \(\mathrm{UO}_2^+\): - Assume oxygen is \(-2\) and solve: \(U + (-2)\times2 = +1\). Thus, \(U = +5\).- In \(\mathrm{U}^{4+}\): - \(U\) is \(+4\).So \(\mathrm{UO}_2^+\) reduces to \(\mathrm{U}^{4+}\), indicating a reduction reaction.

Step 2: Balancing the Half-Reaction (a)

Write the electrons to balance the change in oxidation states. The uranium goes from +5 to +4, which means it gains one electron:\[ \mathrm{UO}_2^+ + 2\mathrm{H}^+ + \mathrm{e}^- \rightarrow \mathrm{U}^{4+} + \mathrm{H}_2\mathrm{O} \] This equation is balanced in terms of mass (atoms) and charge.

Step 3: Oxidation and Reduction Determination (b)

For \(\mathrm{ClO}_3^-\) to \(\mathrm{Cl}^-\):- In \(\mathrm{ClO}_3^-\): - Assume oxygen as \(-2\), do: \(Cl + (-2)\times3 = -1\). Thus, \(Cl = +5\).- In \(\mathrm{Cl}^-\): - \(Cl\) is \(-1\).This is a reduction as chlorine reduces from +5 to -1.

Step 4: Balancing the Half-Reaction (b)

Write the electrons to balance the change in oxidation states and use water and \(\mathrm{H}^+\) to balance the half-reaction in acidic medium:\[ \mathrm{ClO}_3^- + 6\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow \mathrm{Cl}^- + 3\mathrm{H}_2\mathrm{O} \]The equation is balanced in terms of atoms and charge.

Step 5: Oxidation and Reduction Determination (c)

For \(\mathrm{N}_2\mathrm{H}_4\) to \(\mathrm{N}_2\):- In \(\mathrm{N}_2\mathrm{H}_4\): - Hydrazine oxidation state: Nitrogen at \(-2\).- In \(\mathrm{N}_2\): - \(N\) is \(0\).This is an oxidation as nitrogen increases from -2 to 0.

Step 6: Balancing the Half-Reaction (c)

Balance the equation in terms of mass and charge, paying attention to the electrons gained or lost and balancing additional \(\mathrm{OH}^-\) in a basic solution:\[ \mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^- \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^- \] Make sure the equation is charge- and mass-balanced.

Step 7: Oxidation and Reduction Determination (d)

For \(\mathrm{ClO}^-\) to \(\mathrm{Cl}^-\):- In \(\mathrm{ClO}^-\): - \(Cl\) at \(+1\), \(O\) at \(-2\).- In \(\mathrm{Cl}^-\): - \(Cl\) is \(-1\).This is a reduction.

Step 8: Balancing the Half-Reaction (d)

To balance in a basic solution:- Use electrons, water, and hydroxide ions.\[ \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^- \rightarrow \mathrm{Cl}^- + 2\mathrm{OH}^- \]Ensure balance in terms of charge and atoms.

Key Concepts

Oxidation States

In redox reactions, identifying oxidation states is crucial to determine which species undergoes oxidation and which undergoes reduction. Oxidation states, or numbers, provide us with insights into the electron transfer that occurs during a reaction. Generally, atoms in their elemental form have an oxidation state of zero. Different atoms may have standard oxidation states depending on their usual chemical bonding.Let's break it down further:

• Oxygen usually carries an oxidation state of -2.
• Hydrogen generally has an oxidation state of +1.

These common rules help determine the oxidation states of other atoms within a compound.For instance, consider uranium in \(\mathrm{UO}_2^+\). By assuming oxygen is \(-2\), you can solve for uranium's oxidation state: \[ U + (-2)\times2 = +1 \]This results in uranium having an oxidation state of +5. Similarly, identifying the oxidation state helps us track electron movement, indispensable for writing balanced redox equations.

Half-Reaction Balancing

Balancing redox equations through half-reaction methods is a systematic approach to ensure both mass and charge are preserved. This method breaks the entire reaction into two halves: one depicts oxidation while the other represents reduction. Each half-reaction is balanced separately by ensuring that electrons, atoms, and charges are accounted for, then combined back to give the balanced redox equation.To balance:

• Identify the oxidation and reduction half-reactions.
• Balance atoms other than oxygen and hydrogen.
• Balance oxygen using water molecules.
• Balance hydrogen using hydrogen ions (\(\mathrm{H}^+\)).
• Finally, balance the charge by adding electrons.

Once each half is balanced, make sure the number of electrons lost in the oxidation half equals those gained in the reduction half before combining them. This methodically allows each reaction to maintain its electron parity, keeping the integrity of the equation intact.

Acidic and Basic Solutions

Redox equations behave differently in acidic and basic solutions due to the nature of the medium. Acidic environments utilize hydrogen ions (\(\mathrm{H}^+\)) to balance redox reactions, while in basic environments, hydroxide ions (\(\mathrm{OH}^-\)) are employed.In acidic solutions:

• Balance oxygen atoms with water \(\mathrm{H}_2\mathrm{O}\).
• Balance hydrogen atoms using \(\mathrm{H}^+\).
• Balance charges finally by adding electrons following the change in oxidation numbers.

For basic solutions:

• After balancing the reaction as if it were in acidic solution, add \(\mathrm{OH}^-\) to both sides to neutralize \(\mathrm{H}^+\), converting them into water (\(\mathrm{H}_2\mathrm{O}\)).
• Balance additional charges involved due to extra \(\mathrm{OH}^-\).

This ensures the balance of atoms and charge is appropriate to the solution's environment, allowing the reaction to proceed congruently with the chemical context.