Problem 4 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in acid solution. (a) \(\mathrm{Sn}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})\) (c) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{CH}_{2} \mathrm{O}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{HCO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\)

Step-by-Step Solution

Answer

(a) \( \mathrm{Sn} + 2\mathrm{H}^+ \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_2 \); (b) \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} + 6\mathrm{Fe}^{2+} + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{Fe}^{3+} + 7\mathrm{H}_2\mathrm{O} \); (c) \( \mathrm{MnO}_2 + 4\mathrm{H}^+ + 2\mathrm{Cl}^- \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O} \); (d) \( \mathrm{CH}_2 \mathrm{O} + \mathrm{H}_2\mathrm{O} + 2\mathrm{Ag}^+ \rightarrow \mathrm{HCO}_2\mathrm{H} + 2\mathrm{H}^+ + 2\mathrm{Ag} \).

Step 1: Identify and Separate Half Reactions (a)

For the reaction \( \mathrm{Sn} + \mathrm{H}^+ \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_2 \):- Oxidation half-reaction: \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} \).- Reduction half-reaction: \( \mathrm{H}^+ \rightarrow \mathrm{H}_2 \).

Step 2: Balance Electrons in Half Reactions (a)

For the oxidation reaction \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \) and the reduction reaction \( 2 \mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2 \). The electrons lost and gained are already balanced with 2 electrons each.

Step 3: Combine Balanced Half Reactions (a)

The overall balanced equation is \( \mathrm{Sn} + 2 \mathrm{H}^+ \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_2 \).

Step 4: Identify and Separate Half Reactions (b)

For the reaction \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} + \mathrm{Fe}^{2+} \rightarrow \mathrm{Cr}^{3+} + \mathrm{Fe}^{3+} \):- Oxidation half-reaction: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} \).- Reduction half-reaction: \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+} \).

Step 5: Balance Electrons in Half Reactions (b)

1. Balance the Cr reaction: \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \).2. Balance the Fe reaction: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^- \), apply a coefficient of 6 to balance electrons, \( 6\mathrm{Fe}^{2+} \rightarrow 6\mathrm{Fe}^{3+} + 6e^- \).

Step 6: Combine Balanced Half Reactions (b)

Combine the two balanced half-reactions: \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} + 6\mathrm{Fe}^{2+} + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{Fe}^{3+} + 7\mathrm{H}_2\mathrm{O} \).

Step 7: Identify and Separate Half Reactions (c)

For the reaction \( \mathrm{MnO}_2 + \mathrm{Cl}^- \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_2 \):- Oxidation half-reaction: \( \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 \).- Reduction half-reaction: \( \mathrm{MnO}_2 \rightarrow \mathrm{Mn}^{2+} \).

Step 8: Balance Electrons in Half Reactions (c)

1. Balance MnO2: \( \mathrm{MnO}_2 + 4\mathrm{H}^+ + 2e^- \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \).2. Balance Cl: \( 2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2e^- \).

Step 9: Combine Balanced Half Reactions (c)

Combine the two balanced half-reactions: \( \mathrm{MnO}_2 + 4\mathrm{H}^+ + 2\mathrm{Cl}^- \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O} \).

Step 10: Identify and Separate Half Reactions (d)

For the reaction \( \mathrm{CH}_2 \mathrm{O} + \mathrm{Ag}^+ \rightarrow \mathrm{HCO}_2 \mathrm{H} + \mathrm{Ag} \):- Oxidation half-reaction: \( \mathrm{CH}_2 \mathrm{O} \rightarrow \mathrm{HCO}_2 \mathrm{H} \).- Reduction half-reaction: \( \mathrm{Ag}^+ \rightarrow \mathrm{Ag} \).

Step 11: Balance Electrons in Half Reactions (d)

1. Oxidation: \( \mathrm{CH}_2 \mathrm{O} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCO}_2\mathrm{H} + 2\mathrm{H}^+ + 2e^- \).2. Reduction: \( \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag} \), apply a coefficient of 2 to balance electrons, \( 2\mathrm{Ag}^+ + 2e^- \rightarrow 2\mathrm{Ag} \).

Step 12: Combine Balanced Half Reactions (d)

Combine the two balanced half-reactions: \( \mathrm{CH}_2 \mathrm{O} + \mathrm{H}_2\mathrm{O} + 2\mathrm{Ag}^+ \rightarrow \mathrm{HCO}_2\mathrm{H} + 2\mathrm{H}^+ + 2\mathrm{Ag} \).

Key Concepts

Understanding Half-Reactions in Redox Chemistry

In redox reactions, understanding half-reactions is crucial. A half-reaction divides the overall reaction into two parts—oxidation and reduction. This separation allows us to see exactly how electrons are transferred.

• Oxidation Half-Reaction: In this part, a substance loses electrons. For example, in the reaction \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} \), tin is oxidized as it loses two electrons.
• Reduction Half-Reaction: Here, a substance gains electrons. The reaction \( \mathrm{H}^+ \rightarrow \mathrm{H}_2 \) shows hydrogen gaining electrons to form hydrogen gas.

Separating these half-reactions makes it easier to identify electron transfer and balance the redox equation, ensuring both the atom number and charge are balanced.

Mastering Electron Balancing

Electron balancing is the heart of balancing redox reactions. It ensures that the electrons lost in oxidation equal those gained in reduction, making the chemical reaction accurate and stable.

• Identify Electrons: Check the oxidation states to count how many electrons are transferred. For instance, in \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} \), tin loses 2 electrons.
• Balancing Electrons: Adjust the number of electrons in each half-reaction, so they are equal. Using the tin and hydrogen reaction, the electrons are balanced as \( 2 \mathrm{e}^- \) are consistent in both half-reactions.
• Using Coefficients: Sometimes, coefficients are applied to multiply the substances to ensure same number of electrons. For example, \( 6\mathrm{Fe}^{2+} \rightarrow 6\mathrm{Fe}^{3+} + 6\mathrm{e}^- \) balances with another reaction involving 6 electrons.

Correct electron balancing ensures that mass and charge are conserved in a chemical reaction.

Delving into Oxidation and Reduction

Oxidation and reduction are fundamental in redox reactions, characterized by the transfer of electrons.

• Oxidation: This process involves the loss of electrons. When a metal like tin becomes \( \mathrm{Sn}^{2+} \), it undergoes oxidation by losing electrons, increasing its oxidation number.
• Reduction: The gain of electrons takes place in reduction, where oxidation numbers decrease. A typical example is \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+} \), where chromium is reduced.

Together, oxidation and reduction maintain balance in redox reactions. The electrons lost in oxidation are always matched with the electrons gained in reduction, facilitating complete and balanced chemical reactions. These essential processes are key to understanding various chemical and biological systems.