Question

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)\) (acidic solution) (b) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q)\) (acidic solution) (c) \(\mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}{ }^{+}(\mathrm{aq})\) (acidic solution) (e) \(\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g)\) (basic solution) (f) \(\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{N}_{2}(\mathrm{~g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{~g})\) (basic solution)

Short answer

(a) \(\mathrm{Sn^{2+}(aq) \longrightarrow Sn^{4+}(aq) + 2e^{-}}\) (oxidation) (b) \(\mathrm{TiO_{2}(s) + 4H^{+} \longrightarrow Ti^{2+}(aq) + 2H_{2}O(l) + 2e^{-}}\) (reduction) (c) \(\mathrm{ClO_{3}^{-}(aq) + 6e^{-} \longrightarrow Cl^{-}(aq) + 3H_{2}O(l)}\) (reduction) (d) \(\mathrm{N_{2}(g) + 6H_{2}O(l) + 6H^{+} \longrightarrow 2NH_{4}^{+}(aq)}\) (reduction) (e) \(\mathrm{4OH^{-}(aq) \longrightarrow 2H_{2}O(l) + O_{2}(g)}\) (oxidation) (f) \(\mathrm{SO_{3}^{2-}(aq) + H_{-2}O(l) + 2OH^{-} \longrightarrow SO_{4}^{2-}(aq) + 2OH^{-}(aq)}\) (oxidation) (g) \(\mathrm{N_{2}(g) \longrightarrow 2NH_{3}(g)}\) (reduction)

Step-by-step solution

Balance the atoms

In this case, the Sn atoms are already balanced.

Balance the charges

Since the Sn is going from 2+ to 4+, we add 2 electrons to the product side to balance the charges: \(\mathrm{Sn^{2+}(aq) \longrightarrow Sn^{4+}(aq) + 2e^{-}}\)

Determine if oxidation or reduction

The oxidation state of Sn increases from +2 to +4, so this is an oxidation half-reaction. (b)

Balance the atoms

Balance the Ti atoms by adding 1 Ti to the product side: \(\mathrm{TiO_{2}(s) \longrightarrow Ti^{2+}(aq)}\)

Balance the charges

Since the Ti is going from 4+ in TiO2 to 2+, we add 2 electrons to the product side to balance the charges: \(\mathrm{TiO_{2}(s) + 4H^{+} \longrightarrow Ti^{2+}(aq) + 2H_{2}O(l) + 2e^{-}}\)

Determine if oxidation or reduction

The oxidation state of Ti decreases from +4 to +2, so this is a reduction half-reaction. (c)

Balance the atoms

Balance the Cl atoms by adding 1 Cl to the product side: \(\mathrm{ClO_{3}^{-}(aq) \longrightarrow Cl^{-}(aq)}\)

Balance the charges

Since the Cl is going from +5 in ClO3- to -1, we add 6 electrons to the reactant side to balance the charges: \(\mathrm{ClO_{3}^{-}(aq) + 6e^{-} \longrightarrow Cl^{-}(aq) + 3H_{2}O(l)}\)

Determine if oxidation or reduction

The oxidation state of Cl decreases from +5 to -1, so this is a reduction half-reaction. (d)

Balance the atoms

Balance the N atoms by adding 2 NH4+ to the product side, and balance the H atoms by adding 6 H2O to the reactant side: \(\mathrm{N_{2}(g) + 6H_{2}O(l) \longrightarrow 2NH_{4}^{+}(aq)}\)

Balance the charges

Add 6 H+ ions to the reactant side, since there are now 2 net positive charges on the product side: \(\mathrm{N_{2}(g) + 6H_{2}O(l) + 6H^{+} \longrightarrow 2NH_{4}^{+}(aq)}\)

Determine if oxidation or reduction

The oxidation state of N decreases from 0 to -3, so this is a reduction half-reaction. (e)

Balance the atoms

Balance the O atoms by first adding 2 OH- ions to the reactant side, and then adding \(H_2O\) molecules to the product side: \(\mathrm{4OH^{-}(aq) \longrightarrow 2H_{2}O(l) + O_{2}(g)}\)

Balance the charges

Since charges are balanced, there's no need to add electrons.

Determine if oxidation or reduction

The oxidation state of O increases from -2 to 0, so this is an oxidation half-reaction. (f)

Balance the atoms

Since S and O atoms are already balanced, there's no need to add new atoms.

Balance the charges

Add 2 H2O molecules to the product side and 4 OH- ions to the reactant side: \(\mathrm{SO_{3}^{2-}(aq) + H_{2}O(l) + 2OH^{-}(aq) \longrightarrow SO_{4}^{2-}(aq) + 2OH^{-}(aq)}\)

Determine if oxidation or reduction

The oxidation state of S increases from +4 to +6, so this is an oxidation half-reaction. (g)

Balance the atoms

Balance the N atoms by adding 2 NH3 molecules to the product side: \(\mathrm{N_{2}(g) \longrightarrow 2NH_{3}(g)}\)

Balance the charges

Since charges are balanced, there's no need to add electrons.

Determine if oxidation or reduction

The oxidation state of N decreases from 0 to -3, so this is a reduction half-reaction.

Key concepts

Oxidation

Oxidation involves the loss of electrons in a chemical reaction. It's a key concept in redox reactions, where one substance transfers electrons to another. In a redox process:

• The substance that loses electrons is said to undergo oxidation.
• The oxidation state of this substance increases.

In the exercise provided, Sn going from a +2 to a +4 oxidation state by losing two electrons (\[\mathrm{Sn^{2+}(aq) \longrightarrow Sn^{4+}(aq) + 2e^{-}}\]) is an classic example of an oxidation half-reaction. This increase in charge indicates that electrons are being lost, which is the hallmark of oxidation.

Reduction

Reduction is the gain of electrons by a substance in a chemical reaction. It is the companion process to oxidation in redox reactions. In such processes:

• The substance that gains electrons undergoes reduction.
• The oxidation state of this substance decreases.

An instance from the exercise would be Ti reducing from a +4 state to a +2 state by gaining electrons (\[\mathrm{TiO_{2}(s) + 4H^{+} \longrightarrow Ti^{2+}(aq) + 2H_{2}O(l) + 2e^{-}}\]). This decrease in charge is a clear indicator of reduction. Other examples include reduction of chlorine in \[\mathrm{ClO_{3}^{-}(aq) \longrightarrow Cl^{-}(aq)}\], which gains electrons as its oxidation state moves from +5 to -1.

Half-Reactions

Half-reactions provide a clearer picture of redox processes by dividing them into two parts: oxidation and reduction. Each half-reaction shows either the loss or gain of electrons. This separation helps in:

• Balancing redox reactions by focusing individually on oxidation and reduction.
• Understanding the flow of electrons in the reaction.

By examining half-reactions, such as those in the exercise, you can identify how atoms change their oxidation states and how electrons are exchanged between reacting species. In redox chemistry, recognizing half-reactions helps you comprehend which component of the reaction is being oxidized and which is being reduced.

Balancing Chemical Equations

Balancing chemical equations is crucial to accurately reflect the law of conservation of mass in chemical reactions. For redox reactions, this involves:

• Balancing the number of atoms for each element in the reactants and products.
• Ensuring that the total charge is the same on both sides of the equation.

A well-balanced redox equation shows the exact stoichiometry of the substances involved. This means you account for all atoms and charges as seen in the steps provided for the exercise's half-reactions. To balance a redox reaction, you typically start by balancing the elements involved and then adjust the charge using electrons, as demonstrated by adding electrons to either the reactant or product side to equalize the charges. This method ensures all aspects of the conservation laws are respected in chemistry.