Question

Balance each of the following half-reactions. a. ${\mathrm{N}}_2(g)\to {\mathrm{N}}^{3-}(s)$ b. ${\mathrm{O}}_2^{2-}(aq)\to {\mathrm{O}}_2(g)$ c. $\mathrm{Zn}(s)\to {\mathrm{Zn}}^{2+}(aq)$ d. ${\mathrm{F}}_2(g)\to {\mathrm{F}}^{-}(aq)$

Short answer

a. ${\mathrm{N}}_2(g)\to 2{\mathrm{N}}^{3-}(s)+6e^{-}$ b. $2e^{-}+{\mathrm{O}}_2^{2-}(aq)\to {\mathrm{O}}_2(g)$ c. $\mathrm{Zn}(s)+2e^{-}\to {\mathrm{Zn}}^{2+}(aq)$ d. ${\mathrm{F}}_2(g)\to 2{\mathrm{F}}^{-}(aq)+2e^{-}$

Step-by-step solution

Identify the change in oxidation state

The oxidation state of N changes from 0 in N2 to -3 in N^(3-). To balance this, we need to add electrons to the right side of the reaction.

Balance the electrons

Since 2 nitrogen atoms in N2 are reduced to 2 atoms of N^(3-), a total of 6 electrons (3 for each nitrogen atom) are gained. Thus, we need to add 6e- on the right side of the reaction: ${\mathrm{N}}_2(g)\to 2{\mathrm{N}}^{3-}(s)+6e^{-}$ b. ${\mathrm{O}}_2^{2-}(aq)\to {\mathrm{O}}_2(g)$

Identify the change in oxidation state

The oxidation state of O changes from -1 in O2^(2-) to 0 in O2. To balance this, we need to add electrons to the left side of the reaction.

Balance the electrons

Since 2 oxygen atoms in O2^(2-) are oxidized to 2 atoms of O2, a total of 2 electrons (1 for each oxygen atom) are lost. Thus, we need to add 2e- on the left side of the reaction: $2e^{-}+{\mathrm{O}}_2^{2-}(aq)\to {\mathrm{O}}_2(g)$ c. $\mathrm{Zn}(s)\to {\mathrm{Zn}}^{2+}(aq)$

Identify the change in oxidation state

The oxidation state of Zn changes from 0 in Zn to +2 in Zn^(2+). To balance this, we need to add electrons to the left side of the reaction.

Balance the electrons

Since the Zn atom is oxidized to Zn^(2+), a total of 2 electrons are lost. Thus, we need to add 2e- on the left side of the reaction: $\mathrm{Zn}(s)+2e^{-}\to {\mathrm{Zn}}^{2+}(aq)$ d. ${\mathrm{F}}_2(g)\to {\mathrm{F}}^{-}(aq)$

Identify the change in oxidation state

The oxidation state of F changes from 0 in F2 to -1 in F^(-). To balance this, we need to add electrons to the right side of the reaction.

Balance the electrons

Since 2 fluoride atoms in F2 are reduced to 2 atoms of F^(-), a total of 2 electrons (1 for each fluorine atom) are gained. Thus, we need to add 2e- on the right side of the reaction: ${\mathrm{F}}_2(g)\to 2{\mathrm{F}}^{-}(aq)+2e^{-}$

Key concepts

Oxidation State

When studying chemistry, you'll often hear about an atom's 'oxidation state'—a very important concept to grasp, especially when delving into redox reactions. An oxidation state, also known as oxidation number, is a value that represents how many electrons an atom can lose, gain, or share when it forms chemical compounds or ions. It kind of gives us a 'balance sheet' of electrons for each atom.

Now, let's simplify this: in any given element in its natural state, the oxidation number is zero. For example, oxygen gas (O₂) or zinc metal (Zn). When these elements undergo chemical changes, their oxidation states can change. In compounds, we usually assign oxidation states according to some rules: for instance, hydrogen is typically +1 (but -1 when paired with metals in hydrides), and oxygen is usually -2 (except in peroxides or when it bonds with fluorine).

To make sense of electronic 'give and take' during chemical reactions, identifying changes in oxidation states is crucial. It tells us which atoms got 'richer' in electrons (reduction) and which got 'poorer' (oxidation). Let's look at one of the textbook problems we solved. Nitrogen (N) in N₂ has an oxidation state of 0, but in N^(3-) it changes to -3, indicating it has gained 3 electrons. Therefore, keeping track of oxidation states helps to balance half-reactions accurately by showing us how many electrons are on the move.

Electron Transfer in Redox Reactions

Redox reactions are like a dance of electrons between atoms. One partner 'gives' electrons and the other 'takes' them - this could be thought of as the chemical equivalent of passing a baton in a relay race.

In a redox reaction, the substance that loses electrons is said to be oxidized, and its oxidation state increases. On contrary, the substance that gains electrons is reduced and its oxidation state decreases. This electron shuffle is the heart of balancing half-reactions. It's not just about counting atoms to make sure you have the same number of each element on both sides of the equation; you also need to make sure the charges balance out by offsetting any changes in oxidation state with electrons.

For instance, when zinc metal becomes Zn^(2+), it loses two electrons—these electrons don't just disappear; they have to be accounted for in the balanced equation. Therefore, in the balanced half-reaction, we'd denote the loss of those two electrons: $\mathrm{Zn}(s)+2e^{-}\to {\mathrm{Zn}}^{2+}(aq)$. It is this balance of loss and gain of electrons that drives chemical reactions and the principles behind batteries, metal corrosion, and even how our bodies metabolize food.

Oxidation-Reduction (Redox) Reactions

At the core of many chemical processes lies the oxidation-reduction, or redox, reactions. These reactions are all about the flow of electrons between species. In a redox reaction, two halves come together to form a whole: the oxidation half-reaction, where electrons are lost, and the reduction half-reaction, where electrons are gained.

Typically, we balance redox reactions by separating them into these two halves. This helps us to focus on the changes that occur to each reactant independently before combining them back together. Balancing the number of electrons lost in the oxidation half-reaction with those gained in the reduction half-reaction ensures the overall charge is conserved. And remember, conservation of charge is just as important as conservation of mass in chemistry!

Take fluorine gas (F₂) becoming fluoride ions (F^-), for example. Here, fluorine is reduced, as it gains electrons: ${\mathrm{F}}_2(g)\to 2{\mathrm{F}}^{-}(aq)+2e^{-}$. Each half-reaction is balanced separately, and when you put them side by side, you make sure that the electrons that vanished from one species appear on the other end because in chemistry, just like in real life, what goes out must come in somewhere else. This interconnectedness of oxidation and reduction makes redox reactions especially fascinating and vital to understanding everything from energy production to cellular respiration.