Problem 4 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in acid solution. (a) \(\operatorname{Sn}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})\) (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

a) \(\operatorname{Sn} + \mathrm{2H}^{+} \rightarrow \operatorname{Sn}^{2+} + \mathrm{H}_{2}\). 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) \(\mathrm{MnO}_{2} + 2\mathrm{Cl}^{-} + 4\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_{2} + 2\mathrm{H}_{2}\mathrm{O}\). d) \(\mathrm{CH}_{2}\mathrm{O} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{HCOOH} + 2\mathrm{Ag}\).

Step 1: Assign Oxidation States (Part a)

For the equation \(\operatorname{Sn}(\mathrm{s}) + \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g})\), the oxidation state of tin \(\operatorname{Sn}\) changes from 0 to +2, indicating oxidation; while hydrogen changes from +1 in \(\mathrm{H}^{+}\) to 0 in \(\mathrm{H}_{2}\), indicating reduction.

Step 2: Balance Oxidation and Reduction (Part a)

Write the half-reactions. Oxidation: \(\operatorname{Sn} \rightarrow \operatorname{Sn}^{2+} + 2e^-\). Reduction: \(\mathrm{2H}^{+} + 2e^- \rightarrow \mathrm{H}_{2}\). Both half-reactions involve 2 electrons, so they are already balanced in terms of electrons.

Step 3: Combine and Check Atoms (Part a)

Combine the half-reactions: \(\operatorname{Sn} + \mathrm{2H}^{+} \rightarrow \operatorname{Sn}^{2+} + \mathrm{H}_{2}\). Check that all elements have the same quantities on both sides of the equation.

Step 4: Assign Oxidation States (Part b)

For \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(\mathrm{aq}) + \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq}) + \mathrm{Fe}^{3+}(\mathrm{aq})\), the oxidation state of Cr changes from +6 to +3 (reduction), and Fe changes from +2 to +3 (oxidation).

Step 5: Write and Balance Half-Reactions (Part b)

Reduction: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\). Oxidation: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-\).

Step 6: Equalize Electrons and Combine (Part b)

Multiply the oxidation half-reaction by 6 to equalize the electrons: \(6\mathrm{Fe}^{2+} \rightarrow 6\mathrm{Fe}^{3+} + 6e^-\). Combine: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{Fe}^{2+} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{Fe}^{3+} + 7\mathrm{H}_2\mathrm{O}\).

Step 7: Assign Oxidation States (Part c)

For \(\mathrm{MnO}_{2} + \mathrm{Cl}^{-} \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_{2}\), Mn is reduced from +4 to +2, and Cl is oxidized from -1 to 0.

Step 8: Write and Balance Half-Reactions (Part c)

Reduction: \(\mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2e^- \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O}\). Oxidation: \(2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2e^-\).

Step 9: Combine Half-Reactions (Part c)

Combine the half-reactions without modifying, as they already involve two electrons each: \(\mathrm{MnO}_{2} + 2\mathrm{Cl}^{-} + 4\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_{2} + 2\mathrm{H}_{2}\mathrm{O}\).

Step 10: Assign Oxidation States (Part d)

For \(\mathrm{CH}_{2}\mathrm{O} + \mathrm{Ag}^{+} \rightarrow \mathrm{HCOOH} + \mathrm{Ag}\), carbon is oxidized from 0 in formaldehyde to +2 in formic acid, and silver is reduced from +1 to 0.

Step 11: Redox Half-Reactions and Balance (Part d)

Oxidation: \(\mathrm{CH}_{2}\mathrm{O} \rightarrow \mathrm{HCOOH} + 2e^-\). Reduction: \(2\mathrm{Ag}^{+} + 2e^- \rightarrow 2\mathrm{Ag}\). Both are balanced because both transfer two electrons.

Step 12: Final Balancing (Part d)

Combine them: \(\mathrm{CH}_{2}\mathrm{O} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{HCOOH} + 2\mathrm{Ag}\), ensuring all reactants and products are balanced.

Key Concepts

Oxidation States

In a redox reaction, understanding oxidation states is crucial as they help identify what gets oxidized and what gets reduced. An oxidation state is a hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. Here is how you can assign oxidation states:

• Atoms in their elemental form have an oxidation state of zero. For example, in the reaction \(\mathrm{Sn} + \mathrm{H}^{+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_2\), the tin (\(\mathrm{Sn}\)) and the hydrogen molecule (\(\mathrm{H}_2\)) both have an oxidation state of 0 initially.
• For simple ions, the oxidation state is equal to the charge of the ion. For instance, \(\mathrm{H}^{+}\) has an oxidation state of +1, and \(\mathrm{Cl}^-\) has an oxidation state of -1.
• In molecules, the more electronegative atom is assigned negative oxidation states. Considerations such as these allow identification of changes in oxidation states that signify a redox reaction.

Remember, when oxidation states increase, the element is oxidized, and when they decrease, the element is reduced.

Half-Reactions

To effectively balance redox reactions, breaking them into half-reactions can be very helpful. Half-reactions separately show the oxidation process and the reduction process. Here is a simple guide:

• Firstly, write the unbalanced half-equations for oxidation and reduction. For the reaction \(\mathrm{Sn} + \mathrm{H}^{+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_2\), the half-reactions could be:
  • Oxidation: \(\mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^-\)
  • Reduction: \(2\mathrm{H}^{+} + 2e^- \rightarrow \mathrm{H}_2\)
• Next, balance the electrons between the two processes. Multiply the half-reactions by integers if necessary so that the same number of electrons are transferred in both half-reactions.

By clearly detailing each process, it becomes easier to balance the overall redox equation. The half-reactions approach simplifies the account of electron transfer, making it apparent how electrons are gained and lost.

Acidic Solution Balancing

Balancing redox reactions occurring in acidic solutions requires some specific steps. This is necessary because the presence of \(\mathrm{H}^{+}\) ions and sometimes water is integral to these reactions. Follow these steps in balancing in acidic conditions:

• Start by balancing all the elements involved, except for hydrogen and oxygen.
• To balance the oxygen atoms, add \(\mathrm{H}_2\mathrm{O}\) molecules where necessary. For example, in the reaction \(\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{Fe}^{2+} \rightarrow \mathrm{Cr}^{3+} + \mathrm{Fe}^{3+}\), adding water helps balance the oxygen: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\).
• Balance the hydrogen atoms by adding \(\mathrm{H}^{+}\) ions. This aligns with the acidic solution conditions.
• Make sure that the charge is balanced by adjusting the coefficients, ensuring that however many electrons are lost in oxidation, the same number are gained in the reduction.

Following these steps ensures that the reactions will be appropriately balanced in acidic environments, satisfying both mass and charge conservation laws.