Problem 4 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

The balanced equations are: (a) Sn(s) + 2H^{+}(aq) → Sn^{2+}(aq) + H_{2}(g) (b) 14H^{+} + Cr_{2}O_{7}^{2-} + 6Fe^{2+} → 2Cr^{3+} + 6Fe^{3+} + 7H_{2}O.

Step 1: Assign Oxidation Numbers

Identify the oxidation states of each element in the reactants and products for both equations to determine which elements are oxidized and reduced. (a) Sn(s) has 0, and Sn^{2+}(aq) has +2. H^{+}(aq) and H_{2}(g) have +1 and 0. (b) Cr2O7^{2-} has Cr in +6, and Cr^{3+} has Cr in +3. Fe^{2+} and Fe^{3+} have +2 and +3.

Step 2: Write Half-Reactions

For each reaction, separate the process into oxidation and reduction half-reactions. (a) Oxidation: Sn(s) → Sn^{2+}(aq) Reduction: 2H^{+}(aq) → H_{2}(g) (b) Oxidation: Fe^{2+} → Fe^{3+} Reduction: Cr_{2}O_{7}^{2-} → Cr^{3+}

Step 3: Balance Atoms Other than O and H

Make sure all elements other than oxygen and hydrogen are balanced in the half-reactions. (a) Sn(s) → Sn^{2+}(aq) is balanced for Sn. (b) 6Fe^{2+} → 6Fe^{3+} is balanced for Fe. Cr_{2}O_{7}^{2-} → 2Cr^{3+} is balanced for Cr.

Step 4: Balance Oxygen Atoms

Balance the oxygen atoms by adding water molecules to the side lacking oxygen. (b) Add 7 H_{2}O to the right side: Cr_{2}O_{7}^{2-} + 6e^- → 2Cr^{3+} + 7H_{2}O.

Step 5: Balance Hydrogen Atoms

Add H^+ ions to balance hydrogen atoms in the reactions. (a) Added in Reduction: 2H^{+}(aq) + 2e^- → H_{2}(g) is already balanced. (b) Add 14H^+ to the left for balance: 14H^{+} + Cr_{2}O_{7}^{2-} → 2Cr^{3+} + 7H_{2}O.

Step 6: Balance Charges by Adding Electrons

Add electrons to balance the charge on both sides of each half-reaction. (a) Sn(s) → Sn^{2+}(aq) + 2e^- (b) 6Fe^{2+} - 6e^- → 6Fe^{3+} 14H^{+} + Cr_{2}O_{7}^{2-} + 6e^- → 2Cr^{3+} + 7H_{2}O

Step 7: Combine Half-Reactions

Adjust coefficients if needed and combine the half-reactions, ensuring electrons cancel out. (a) Sn(s) + 2H^{+}(aq) → Sn^{2+}(aq) + H_{2}(g) (b) Combine and balance: 14H^{+} + Cr_{2}O_{7}^{2-} + 6Fe^{2+} → 2Cr^{3+} + 6Fe^{3+} + 7H_{2}O

Step 8: Verify Balancing

Confirm that the atoms and charges are balanced in the overall equation for accuracy. Both reactions are balanced with respect to mass and charge.

Key Concepts

Oxidation Numbers

In the context of redox reactions, oxidation numbers serve as a useful tool for keeping track of electron transfers. An oxidation number is an imaginary charge that an atom would have if all bonds to atoms of different elements were entirely ionic.
For example, in the given exercise:

• In part (a), tin (Sn) in its elemental form has an oxidation number of 0. When it's converted to \( \text{Sn}^{2+} \), its oxidation number increases to +2, indicating a loss of electrons, thus being oxidized.
• The hydrogen ion \( \text{H}^+ \) has an oxidation number of +1, while \( \text{H}_2 \) has an oxidation number of 0, indicating the reduction of hydrogen.
• In part (b), chromium in \( \text{Cr}_2\text{O}_7^{2-} \) has an oxidation number of +6, while \( \text{Cr}^{3+} \) in the products has +3. The change shows the reduction of chromium.

By assigning these numbers, we identify which species are undergoing oxidation and which are undergoing reduction.

Half-Reactions

Half-reactions are an essential part of understanding redox reactions. They allow us to separate the oxidation process from the reduction process.
In the exercise:

• For part (a), the oxidation half-reaction is \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) \), where tin loses electrons.
• The reduction half-reaction is \( 2 \text{H}^{+}(aq) + 2e^- \rightarrow \text{H}_2(g) \), where hydrogen ions gain electrons.
• In part (b), the oxidation half-reaction is \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \)+, indicating the loss of electrons.
• The reduction half-reaction is \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \), showing the gain of electrons.

Each of these half-reactions must be balanced for both mass and charge separately before combining them into a full redox equation.

Charge Balancing

Balancing charges is a critical step in ensuring a redox equation is correctly balanced. This involves adding electrons to balance the charge on both sides of each half-reaction.
In the worked solution:

• In part (a), the oxidation half-reaction involves the equation \( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- \). Electrons are added to account for the charge difference.
• The reduction half-reaction already balances hydrogen without needing additional electrons since it is \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \).
• For part (b), the electrons from the oxidation half-reaction are balanced with the reduction by \( 6\text{Fe}^{2+} - 6e^- \rightarrow 6\text{Fe}^{3+} \).
• The reduction half-reaction balances charges with 6 electrons, initially accounted: \( 14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \).

This aligns both sides so they mirror each other in terms of charge.

Step-by-Step Balancing

Following a step-by-step method ensures a systematic approach to balancing redox reactions. Here’s how you can proceed:

1. **Identify Oxidation and Reduction** Recognize the elements that undergo changes in oxidation state by assigning oxidation numbers.

• This helps pinpoint what is oxidized and what is reduced.

2. **Write Half-Reactions** Separate the equation into oxidation and reduction half-reactions for clarity.

• This approach simplifies balancing individual components.

3. **Balance Atoms Other than O and H** Ensure other elements are balanced in each half-reaction.

• Non-oxygen and non-hydrogen atoms are balanced first.

4. **Balance Oxygen with Water** Add \( \text{H}_2\text{O} \) molecules to the side lacking oxygen.

• This ensures oxygens are balanced appropriately.

5. **Balance Hydrogen with Protons** Use \( \text{H}^{+} \) ions to balance hydrogen.

• In acidic solutions, hydrogen balancing is crucial.

6. **Balance Charges with Electrons** Ensure both sides of the half-reactions have equal charges by adding electrons.

• This makes the reactions realistic concerning charge neutrality.

7. **Combine and Simplify** Adjust coefficients if necessary and merge the half-reactions.

• Ensure electrons cancel out for an overall balanced equation.

By following these steps, you can methodically tackle complex redox equations, ensuring both mass and charge are balanced.