Question

Balance these net ionic equations for redox reactions. a. \(A u^{3+}(a q)+I^{-(a q)} \rightarrow \operatorname{Au}(s)+I_{2}(s)\) b. \(C e^{4+}(a q)+S n^{2+}(a q) \rightarrow C e^{3+}(a q)+S n^{4+}(a q)\)

Short answer

The balanced net ionic equations for the redox reactions are: a. \(2Au^{3+}(aq) + 6I^{-}(aq) \rightarrow 2Au(s) + 3I_{2}(s)\) b. \(2Ce^{4+}(aq) + Sn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Sn^{4+}(aq)\)

Step-by-step solution

Identify the half-reactions

a. In the first reaction, Au(III) is reduced to Au(0), while I⁻ is oxidized to I₂. Reduction half-reaction: \(Au^{3+} (aq) \rightarrow Au(s)\) Oxidation half-reaction: \(2I^{-}(aq) \rightarrow I_{2}(s)\) b. In the second reaction, Ce(IV) is reduced to Ce(III), while Sn(II) is oxidized to Sn(IV). Reduction half-reaction: \(Ce^{4+}(aq) \rightarrow Ce^{3+}(aq)\) Oxidation half-reaction: \(Sn^{2+}(aq) \rightarrow Sn^{4+}(aq)\)

Balance atoms in half-reactions

a. - Reduction half-reaction is already balanced for Au atoms. - Oxidation half-reaction is already balanced for I atoms. b. - Reduction half-reaction is already balanced for Ce atoms. - Oxidation half-reaction is already balanced for Sn atoms.

Balance charges in half-reactions

a. - Reduction half-reaction: To balance the charges, add 3 electrons to the left side. \(Au^{3+}(aq) + 3e^{-} \rightarrow Au(s)\) - Oxidation half-reaction: To balance the charges, add 2 electrons to the right side. \(2I^{-}(aq) \rightarrow I_{2}(s) + 2e^{-}\) b. - Reduction half-reaction: To balance the charges, add 1 electron to the left side. \(Ce^{4+}(aq) + e^{-} \rightarrow Ce^{3+}(aq)\) - Oxidation half-reaction: To balance the charges, add 2 electrons to the right side. \(Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) + 2e^{-}\)

Balance half-reactions and combine them

a. We must multiply the reduction half-reaction by 2 and oxidation half-reaction by 3 to get equal number of electrons on both sides. Reduction half-reaction: \(2Au^{3+}(aq) + 6e^{-} \rightarrow 2Au(s)\) Oxidation half-reaction: \(6I^{-}(aq) \rightarrow 3I_{2}(s) + 6e^{-}\) Adding the two half-reactions to get the balanced redox equation: \(2Au^{3+}(aq) + 6I^{-}(aq) \rightarrow 2Au(s) + 3I_{2}(s)\) b. To balance the reaction and have the same number of electrons in both half-reactions, we need to multiply the reduction half-reaction by 2. Reduction half-reaction: \(2Ce^{4+}(aq) + 2e^{-} \rightarrow 2Ce^{3+}(aq)\) Oxidation half-reaction: \(Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) + 2e^{-}\) Now, combining the two half-reactions to get the balanced redox equation: \(2Ce^{4+}(aq) + Sn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Sn^{4+}(aq)\) Finally, we have balanced both the given net ionic equations for redox reactions.

Key concepts

Oxidation Half-Reaction

Understanding oxidation half-reactions in redox processes focuses on the element that loses electrons. In any redox equation, one reactant will lose electrons, which is the essence of oxidation. For example, in the provided exercise of the redox reaction involving iodine:

• The iodide ion \(I^-\) is oxidized to iodine gas \(I_2\).
• Here, \(I^-\) loses electrons to form \(I_2\), taking part in the oxidation half-reaction: \[2I^-(aq) \rightarrow I_2(s) + 2e^-\]

To fully comprehend, remember that oxidation involves an increase in oxidation state due to the loss of electrons. This can be observed in the formula, where iodide ions (with an oxidation state of -1) transform into elemental iodine (0 oxidation state), releasing electrons. This release balances out elsewhere in the reaction thanks to the simultaneous reduction that occurs.

Reduction Half-Reaction

Reduction half-reactions are vital for balancing redox reactions and involve the gaining of electrons by a substance. This process decreases the oxidation state of an element within a compound. Consider the example from the exercise:

• Gold ions, \(Au^{3+}\), are reduced to metallic gold, \(Au(s)\).
• This can be represented in the reduction half-reaction: \[Au^{3+}(aq) + 3e^- \rightarrow Au(s)\]

In this reaction, the oxidation state of gold changes from +3 to 0 as it gains electrons, showcasing the reduction process. Reduction is paired with oxidation in every redox reaction to ensure conservation of charge and mass. Here, electrons are added to the left side to balance the charges in the reaction.

Charge Balancing in Reactions

Charge balancing is crucial for correctly forming and understanding redox equations. Both half-reactions need to have their charges balanced in order to form a complete redox reaction. In the exercise solutions:

• The reduction half-reaction involving \(Au^{3+}\) adds electrons to the left side to counteract the positive charge.
• Similarly, in oxidation, electrons are added to the right side to balance out negative charges lost by iodide ions forming \(I_2\).

For each example provided:

• The charges on each side of the half-reactions must equate, allowing them to combine into a full balanced reaction.
• For instance, multiplying half-reactions to equalize electrons ensures the overall build leads to a balanced redox equation.

This careful balancing verifies that both mass and charge remain constant during chemical reactions, allowing for the coherent progression from reactants to products.