Problem 3 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in acid solution. $$\text { (a) } \mathrm{Ag}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{Ag}^{+}(\mathrm{aq})$$ $$\begin{aligned} &\text { (b) } \mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \rightarrow\\\ &&\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned}$$ (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}-(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Cr}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) \(2\mathrm{Ag(s)} + \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} \rightarrow 2\mathrm{Ag^+(aq)} + \mathrm{NO_2(g)} + \mathrm{H_2O(l)}\). (b) \(2\mathrm{MnO_4^-(aq)} + 5\mathrm{HSO_3^-(aq)} + 6\mathrm{H^+(aq)} \rightarrow 2\mathrm{Mn^{2+}(aq)} + 5\mathrm{SO_4^{2-}(aq)} + 3\mathrm{H_2O(l)}\). (c) \(4\mathrm{Zn(s)} + 2\mathrm{NO_3^-(aq)} + 10\mathrm{H^+} \rightarrow 4\mathrm{Zn^{2+(aq)}} + \mathrm{N_2O(g)} + 5\mathrm{H_2O(l)}\). (d) \( \mathrm{Cr(s)} + \mathrm{NO_3^-(aq)} + 4\mathrm{H^+} \rightarrow \mathrm{Cr^{3+(aq)} } + \mathrm{NO(g)} + 2\mathrm{H_2O(l)} \).

Step 1: Separate into Half-Reactions (a)

The redox reaction given involves Ag and NO₃⁻. Separate it into oxidation and reduction half-reactions. **Oxidation half-reaction:** \[ \mathrm{Ag(s)} \rightarrow \mathrm{Ag^+(aq)} + \mathrm{e^-} \] **Reduction half-reaction:** \[ \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} + \mathrm{e^-} \rightarrow \mathrm{NO_2(g)} + \mathrm{H_2O(l)} \]

Step 2: Balance Atoms and Charge (a)

Each half-reaction should have a balance of atoms and charge. The silver half-reaction is balanced. The nitrogen half-reaction needs 2 electrons.**Balanced oxidation half-reaction:** \[ \mathrm{Ag(s)} \rightarrow \mathrm{Ag^+(aq)} + \mathrm{e^-} \] **Balanced reduction half-reaction with electrons:** \[ \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{NO_2(g)} + \mathrm{H_2O(l)} \] Now, you need to balance the number of electrons by multiplying the oxidation half-reaction by 2.

Step 3: Combine Half-Reactions (a)

After balancing the electrons, combine the two half-reactions:\[2\mathrm{Ag(s)} + \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} + 2\mathrm{e^-} \rightarrow 2\mathrm{Ag^+(aq)} + \mathrm{NO_2(g)} + \mathrm{H_2O(l)} + 2\mathrm{e^-} \]Cancel out the electrons:\[2\mathrm{Ag(s)} + \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} \rightarrow 2\mathrm{Ag^+(aq)} + \mathrm{NO_2(g)} + \mathrm{H_2O(l)} \]

Step 4: Balance Equations (b)

For the permanganate and sulfite reaction, separate into half-reactions:**Oxidation half-reaction:** \[ \mathrm{HSO_3^-(aq)} \rightarrow \mathrm{SO_4^{2-}(aq)} + \mathrm{H^+(aq)} + 2\mathrm{e^-} \] **Reduction half-reaction:** \[ \mathrm{MnO_4^-(aq)} + 8\mathrm{H^+(aq)} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}(aq)} + 4\mathrm{H_2O(l)} \]Balance atoms, protons, and electrons.

Step 5: Multiply and Combine (b)

Balance electrons by multiplying the oxidation half-reaction by 5 and the reduction by 2:\[5(\mathrm{HSO_3^-(aq)} \rightarrow \mathrm{SO_4^{2-}(aq)} + \mathrm{H^+(aq)} + 2\mathrm{e^-})\]\[2(\mathrm{MnO_4^-(aq)} + 8\mathrm{H^+(aq)} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}(aq)} + 4\mathrm{H_2O(l)})\]Combine and cancel electrons:\[2\mathrm{MnO_4^-(aq)} + 5\mathrm{HSO_3^-(aq)} + 6\mathrm{H^+(aq)} \rightarrow 2\mathrm{Mn^{2+}(aq)} + 5\mathrm{SO_4^{2-}(aq)} + 3\mathrm{H_2O(l)}\]

Step 6: Separate into Half-Reactions (c)

Separate the given redox reaction into half-reactions:**Oxidation half-reaction:**\[ \mathrm{Zn(s)} \rightarrow \mathrm{Zn^{2+(aq)}} + 2\mathrm{e^-} \]**Reduction half-reaction:**\[ \mathrm{2NO_3^-(aq)} + 10\mathrm{H^+} + 8\mathrm{e^-} \rightarrow \mathrm{N_2O(g)} + 5\mathrm{H_2O(l)} \]

Step 7: Balance Electrons and Combine (c)

To balance, multiply the entire oxidation reaction by 4 to match the 8 electrons required for the reduction:\[4(\mathrm{Zn(s)} \rightarrow \mathrm{Zn^{2+(aq)}} + 2\mathrm{e^-})\]Combine the half-reactions:\[4\mathrm{Zn(s)} + 2\mathrm{NO_3^-(aq)} + 10\mathrm{H^+} \rightarrow 4\mathrm{Zn^{2+(aq)}} + \mathrm{N_2O(g)} + 5\mathrm{H_2O(l)} \]

Step 8: Separate into Half-Reactions (d)

The reaction involves chromium and nitrate ions. Separate into half-reactions:**Oxidation half-reaction:**\[ \mathrm{Cr(s)} \rightarrow \mathrm{Cr^{3+(aq)} } + 3\mathrm{e^-} \]**Reduction half-reaction:**\[ \mathrm{NO_3^-(aq)} + 4\mathrm{H^+} + 3\mathrm{e^-} \rightarrow \mathrm{NO(g)} + 2\mathrm{H_2O(l)} \]

Step 9: Balance and Combine Equations (d)

As the electrons are already balanced, combine these half-reactions directly:\[ \mathrm{Cr(s)} + \mathrm{NO_3^-(aq)} + 4\mathrm{H^+} \rightarrow \mathrm{Cr^{3+(aq)} } + \mathrm{NO(g)} + 2\mathrm{H_2O(l)} \]

Key Concepts

Half-Reactions

In redox chemistry, a common technique used to simplify complex reactions is to split them into half-reactions. This approach helps in identifying and balancing the oxidation and reduction processes separately.
Each redox reaction involves two half-reactions — one half-reaction for oxidation and another for reduction. For example, in the reaction where silver (\(\mathrm{Ag}\)) reacts with nitrate (\(\mathrm{NO_3^-}\)), we have:

• **Oxidation Half-Reaction:** Here, \( \mathrm{Ag(s)} \) gets converted to \( \mathrm{Ag^+(aq)} \) by losing an electron:
\[ \mathrm{Ag(s)} \rightarrow \mathrm{Ag^+(aq)} + \mathrm{e^-} \]

• **Reduction Half-Reaction:** Nitrate ions gain electrons to form nitrogen dioxide:
\[ \mathrm{NO_3^-(aq)} + 2\mathrm{H^+} + \mathrm{e^-} \rightarrow \mathrm{NO_2(g)} + \mathrm{H_2O(l)} \]

This splitting is a fundamental step in diagnosing the electron flow and further facilitates balancing the equation. By isolating the two processes, it becomes easier to ensure all atoms and charges are balanced.

Electron Balance

Balancing electrons is a crucial step in redox reactions. In a redox reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction to maintain charge neutrality.
For example, consider the half-reactions involving the conversion of \( \mathrm{MnO_4^-} \) to \( \mathrm{Mn^{2+}} \) and \( \mathrm{HSO_3^-} \) to \( \mathrm{SO_4^{2-}} \):

• In the oxidation half-reaction, two electrons are lost.
• In the reduction half-reaction, five electrons are gained per permanganate ion.

To balance them, multiply the whole oxidation reaction by 5 and the reduction by 2 prior to combining them. This method ensures each element and charge are perfectly balanced across the resultant equation. The combined balanced equation is achieved by canceling out equal numbers of electrons from both sides.

Oxidation-Reduction

Understanding oxidation and reduction is key to comprehending redox reactions. Oxidation refers to the loss of electrons, while reduction denotes the gain of electrons. These two processes occur simultaneously and cannot happen independently.
Consider the reaction involving zinc (\(\mathrm{Zn}\)) and nitrate ions (\(\mathrm{NO_3^-}\)):

• **Oxidation:** Zinc metal loses electrons converting to zinc ions: \[ \mathrm{Zn(s)} \rightarrow \mathrm{Zn^{2+(aq)}} + 2\mathrm{e^-} \]
• **Reduction:** Simultaneously, nitrate ions gain electrons to form \( \mathrm{N_2O} \): \[ \mathrm{2NO_3^-(aq)} + 10\mathrm{H^+} + 8\mathrm{e^-} \rightarrow \mathrm{N_2O(g)} + 5\mathrm{H_2O(l)} \]

Recognizing and balancing these processes help keep track of electron transfers and assure the overall charge is neutralized.

Acidic Solution Reactions

Many redox reactions occur in acidic solutions which necessitate the presence of hydrogen ions (\(\mathrm{H^+}\)) to balance hydrogen and oxygen atoms. In acidic conditions, water molecules and \(\mathrm{H^+}\) ions are used critically to complement the balancing of the half-reactions.
For instance, in the reaction involving chromium (\(\mathrm{Cr}\)) and nitrate ions:

• The reduction of nitrate to nitrogen monoxide (\(\mathrm{NO}\)) necessitates the inclusion of \(\mathrm{H^+}\) ions to produce water:
• \[ \mathrm{NO_3^-(aq)} + 4\mathrm{H^+} + 3\mathrm{e^-} \rightarrow \mathrm{NO(g)} + 2\mathrm{H_2O(l)} \]

• This not only aligns the stoichiometry of hydrogen and oxygen atoms but ensures that the reaction maintains electrical neutrality.

Using acidic solutions hence plays a pivotal role in achieving a balanced redox equation.