Question

Balance these ionic redox equations by any method. a. \(\mathrm{Mg}+\mathrm{Fe}^{3+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}\) b. \(\mathrm{ClO}_{3}^{-}+\mathrm{SO}_{2} \rightarrow \mathrm{Cl}^{-}+\mathrm{SO}_{4}^{2-}(\text { in acid solution })\)

Short answer

The balanced ionic redox equations are: \\ a) \( 3\mathrm{Mg} + 2\mathrm{Fe}^{3+} \rightarrow 3\mathrm{Mg}^{2+} + 2\mathrm{Fe} \) \\ b) \( 3\mathrm{SO}_2 + 3\mathrm{ClO}_{3}^- + 6\mathrm{H}^+ \rightarrow 3\mathrm{SO}_{4}^{2-} + 3\mathrm{Cl}^- + 9\mathrm{H}_2\mathrm{O} \)

Step-by-step solution

Identify the oxidation states

For the first reaction (Mg + Fe³⁺ → Mg²⁺ + Fe), let's assign oxidation states to each of the elements: Mg: 0 (in its elemental form) \\ Fe³⁺: +3 \\ Mg²⁺: +2 \\ Fe: 0 (in its elemental form)

Write the half-reactions

From the oxidation states, we can now write the half-reactions: Oxidation half-reaction: Mg → Mg²⁺ + 2e⁻ \\ Reduction half-reaction: Fe³⁺ + 3e⁻ → Fe

Balance the electron transfer

In order to balance the electron transfer, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2: Oxidation: (3) x (Mg → Mg²⁺ + 2e⁻) \\ Reduction: (2) x (Fe³⁺ + 3e⁻ → Fe)

Combine the balanced half-reactions

Now, we can combine the balanced half-reactions and cancel out the electrons: 3Mg + 2Fe³⁺ → 3Mg²⁺ + 2Fe #b. Chlorate and Sulfur dioxide reaction#

Identify the oxidation states

For the second reaction, assign the oxidation states in acid solution: Cl in ClO₃⁻: +5 \\ O in SO₂: -2 \\ Cl⁻: -1 \\ S in SO₄²⁻: +6

Write the half-reactions

From the oxidation states, write the half-reactions: Oxidation half-reaction: SO₂ → SO₄²⁻ \\ Reduction half-reaction: ClO₃⁻ → Cl⁻

Balance the half-reactions including H and OH⁻ ions in acid solution

Balance the half-reactions by adding appropriate numbers of H₂O, H⁺, and e⁻ (as needed) to each side: Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ \\ Reduction: ClO₃⁻ + 6e⁻ + 6H⁺ → Cl⁻ + 3H₂O

Balance the electron transfer

In order to balance the electron transfer, multiply the oxidation half-reaction by 3 and leave the reduction half-reaction as is: Oxidation: (3) x (SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻) \\ Reduction: ClO₃⁻ + 6e⁻ + 6H⁺ → Cl⁻ + 3H₂O

Combine the balanced half-reactions

Combine the balanced half-reactions and cancel out the electrons: 3SO₂ + 3ClO₃⁻ + 6H⁺ → 3SO₄²⁻ + 3Cl⁻ + 9H₂O The complete balanced ionic redox equations are: \\ a) 3Mg + 2Fe³⁺ → 3Mg²⁺ + 2Fe \\ b) 3SO₂ + 3ClO₃⁻ + 6H⁺ → 3SO₄²⁻ + 3Cl⁻ + 9H₂O

Key concepts

Oxidation and Reduction

Understanding oxidation and reduction is crucial for grasping the concept of redox reactions. In the realm of chemistry, these two processes are always linked—where one substance loses electrons (oxidation), another gains electrons (reduction).

Oxidation is the process which involves the loss of electrons. When magnesium (Mg) turns into a magnesium ion (Mg²⁺), it loses two electrons, signifying it has been oxidized. Reduction, on the contrary, entails gaining electrons. Iron (III) ion (Fe³⁺) gains three electrons to revert back to metallic iron (Fe), demonstrating a reduction.

In contextualizing the concept:

• Oxidation: Mg → Mg²⁺ + 2e⁻ (loses 2 electrons)
• Reduction: Fe³⁺ + 3e⁻ → Fe (gains 3 electrons)

During a redox process, the number of electrons lost in oxidation must equal the number of electrons gained in reduction, ensuring the law of conservation of charge is upheld.

Half-Reaction Method

The half-reaction method is fundamental for balancing redox reactions. It is a systematic approach that breaks down the overall redox reaction into two separate half-reactions—one for oxidation and one for reduction. Each half-reaction is balanced individually for mass and charge, before combining them back together for the final balanced equation.

For instance:

Balancing Magnesium and Iron Reaction

• Write two half-reactions for oxidation and reduction.
• Balance the atoms other than oxygen and hydrogen first.
• Balance oxygen atoms by adding H2O, hydrogen atoms by adding H⁺ (in acidic solutions), and charge by adding electrons.
• Multiply each half-reaction by appropriate coefficients to balance the electrons transferred.
• Add the half-reactions together and simplify to remove any species that appear on both sides of the equation.

The use of coefficients ensures that the same number of electrons is transferred in both half-reactions, effectively balancing the entire reaction.

Oxidation States

Oxidation states, also known as oxidation numbers, provide insight into the degree of oxidation or reduction of an atom within a molecule or ion. They are hypothetical charges that an atom would have if all bonds were ionic, with no covalent component.

In the provided example, oxidation states allow us to determine which elements have been oxidized or reduced:

• Mg goes from 0 to +2, indicating oxidation.
• Fe³⁺ goes from +3 to 0, indicating reduction.

Knowing the oxidation states simplifies writing the half-reactions, as they directly show the loss or gain of electrons. The standard rules for assigning oxidation states include:

• An element in its standard state has an oxidation state of 0.
• The sum of oxidation states for a neutral compound is 0.
• In ions, the sum of oxidation states equals the charge of the ion.
• Typically, oxygen has an oxidation state of -2, and hydrogen has +1 in compounds (except metal hydrides where it's -1).

These rules assist in determining the necessary changes to balance the equation based on the atoms' oxidation states.