Question

Balance the following half-reactions: (a) (acidic) \(\mathrm{VO}^{2+}(a q) \rightarrow \mathrm{V}^{3+}(a q)\) (b) (basic) \(\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Ni}_{2} \mathrm{O}_{3}(s)\) (c) (acidic) \(\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{NO}_{2}(a q)\) (d) (basic) \(\mathrm{Br}_{2}(a q) \rightarrow \mathrm{BrO}_{3}^{-}(a q)\)

Short answer

(a) Vo2+ + 2H+ + e^- → V3+ + H2O; (b) 2Ni(OH)2 + 2OH^- → Ni2O3 + 2H2O + 2e^-; (c) NO3^- + 2H+ + e^- → NO2 + H2O; (d) Br2 + 12OH^- → 2BrO3^- + 6H2O + 10e^-.

Step-by-step solution

Identify Changes in Oxidation States (a)

In the half-reaction \(\mathrm{VO}^{2+}(a q) \rightarrow \mathrm{V}^{3+}(a q)\), vanadium's oxidation state changes from +4 to +3. So, 1 electron is gained, indicating a reduction.

Balance Charge by Adding Electrons (a)

Add one electron to the right side to balance the charge: \[\mathrm{VO}^{2+} + 2\mathrm{H}^+ + \mathrm{e}^- \rightarrow \mathrm{V}^{3+} + \mathrm{H}_2\mathrm{O}\].

Identify Changes in Oxidation States (b)

For \(\mathrm{Ni}( ext{OH})_2 \rightarrow \mathrm{Ni}_2\mathrm{O}_3\), nickel's oxidation state goes from +2 in Ni(OH)\(_2\) to +(3/2) in Ni\(_2\)O\(_3\). Each nickel changes by +1/2.

Balance Electrons (b)

Since the oxidation number per Ni changes by +1/2, multiply the entire half-cell by 2 to account for full electron gain: \[2\mathrm{Ni}( ext{OH})_2 + 2\mathrm{OH}^- \rightarrow \mathrm{Ni}_2\mathrm{O}_3 + 2\mathrm{H}_2\mathrm{O} + 2\mathrm{e}^-\].

Analyze Changes in Oxidation States (c)

\(\text{NO}_3^- \rightarrow \text{NO}_2\): In NO\(_3^-\), nitrogen is +5, while in NO\(_2\), it is +4. Nitrogen gains 1 electron, indicating reduction.

Balance Electrons and Protons (c)

Add one electron to the right side, protons, and water to balance. \[\text{NO}_3^- + 2\text{H}^+ + \text{e}^- \rightarrow \text{NO}_2 + \text{H}_2\text{O}\].

Assess Oxidation States (d)

For \(\text{Br}_2 \rightarrow \text{BrO}_3^-\), Br\(_2\) is 0, and in BrO\(_3^-\), Br is +5. Each bromine atom undergoes a 5-electron increase, involving 10 electrons for \(\text{Br}_2\).

Balance Electrons (d)

Add OH\(^-\) to make the basic environment. \[\text{Br}_2 + 12\text{OH}^- \rightarrow 2\text{BrO}_3^- + 6\text{H}_2\text{O} + 10\text{e}^- \].

Key concepts

Oxidation State Change

Understanding how oxidation states change is crucial in balancing half-reactions. Oxidation states help us track the transfer of electrons between atoms. In a chemical reaction, when an atom's oxidation state increases, it loses electrons. Conversely, if its oxidation state decreases, it gains electrons.

Consider the example of vanadium in the half-reaction \( \mathrm{VO}^{2+}(aq) \rightarrow \mathrm{V}^{3+}(aq) \). Here, vanadium's oxidation state decreases from +4 to +3, indicating it gained an electron. This is a reduction process.

• Always determine the initial and final oxidation states of the element undergoing change.
• Use these values to identify whether the process is oxidation (increase) or reduction (decrease).

Identifying oxidation state changes helps lay the groundwork for balancing the electrons and completing the half-reaction seamlessly.

Reduction and Oxidation

Reduction and oxidation occur simultaneously in redox reactions. Reduction involves gaining electrons, while oxidation involves losing them. These processes are crucial in balancing half-reactions as they show how electrons shift between reactants.

For example, in the half-reaction involving nickel, \( \mathrm{Ni}(\mathrm{OH})_2 \rightarrow \mathrm{Ni}_2\mathrm{O}_3 \), nickel is oxidized since its oxidation state increases from +2 to +(3/2). This shows nickel loses electrons. Such changes explain the electron flow:

• Reduction: A gain of electrons resulting in a lower oxidation state.
• Oxidation: A loss of electrons leading to a higher oxidation state.

Grasping these concepts is essential as they dictate how we structure and balance reactions to reflect the movement of electrons correctly.

Balancing Electrons

Balancing electrons ensures that both sides of a half-reaction have equal charge. This is done by adding electrons to one side of the equation to counterbalance any changes in oxidation state. Without this step, a half-reaction would be incomplete.

In acidic solutions, balancing electrons may involve introducing hydrogen ions \(\mathrm{H}^+\) and water molecules \(\mathrm{H}_2\mathrm{O}\) where necessary. For instance, in the half-reaction \( \mathrm{NO}_3^- \rightarrow \mathrm{NO}_2 \), nitrogen reduced its oxidation state by 1, requiring the addition of one electron and corresponding \(\mathrm{H}^+\) ions to balance.

• Always add electrons to the more positive side to equalize charges.
• Adjust the protons (\(\mathrm{H}^+\)) in acidic solutions, or hydroxide ions (\(\mathrm{OH}^-\)) in basic ones, to match the electron count.

This careful act of balancing highlights the need for attention to detail, ensuring that the redox equation reflects both mass and charge conservation.