HS Chemistry - Redox Reactions

Ionic Equations & Redox Reactions

Overview of The Page

This page will cover:

• What is an ionic equation?
• What is a net ionic equation?
• What is a redox reaction?

In order to understand redox reactions, we'll first look at ionic equations.

Ionic Equations

Ionic equations are chemical equations that show only the ions that react together and change - that is, the ions that form new substances. The ions that don't are called spectator ions.

Let's take a chemical equation

2 NaH (aq) + Cl₂ → 2NaCl (aq) + H₂ (g)

First, let's split up all the ionic compounds into separate ions.

2Na⁺ (aq) + 2H^- (aq) + Cl₂ (aq) → 2Na⁺ (aq) + 2Cl^- (aq) + H₂ (g)

The molecules on either side have not been split up, as they were molecules and not ionic compounds.

Now let's cancel out the ions that appear unchanged on both sides of the equation. These are the spectator ions; they don't change in the reaction.

2Na⁺ (aq) + 2H^- (aq) + Cl₂ (aq) → 2Na⁺ (aq) + 2Cl^- (aq) + H₂ (g)

The result is called the net ionic equation. It shows that in this equation, 2 Hydrogen anions (Hydride anions) are turning into 1 Hydrogen molecule, and 1 Chlorine molecule turns into 2 Chloride anions.

2H^- (aq) + Cl₂ → 2Cl^- + H₂ (g)

This is the net ionic equation. In this equation, 2 dissolved Hydride ions react with 1 dissolved Chlorine molecules to form 2 dissolved Chloride ions and 1 molecule of Hydrogen gas.

This net ionic equation doesn't include the Sodium cations. In fact, the Sodium ions could be switched out for Lithium ions, or Potassium ions, or any other metal that forms a cation with a charge of +1 $including Transition metals, like Sc^\+^$, and the net ionic equation wouldn't change. The Hydride ions would still react with the Chlorine molecules to form Chloride ions and Hydrogen gas.

There's a few more important things to note in this net ionic equation:

1. Afer they lost their excess electrons to the Chlorine molecules, the $now neutral$ Hydrogen atoms bonded together and formed Hydrogen atoms bonded together and formed Hydrogen molecules $H~2~$.
2. When the Chlorine molecules $Cl~2~$ took the excess electrons, they split up.
3. The Chlorine molecules $on the reactant side$ are dissolved in water, but the Hydrogen molecules $on the product side$ are a gas.

Why does this happen?

1. First, the NaH (from the original equation) ionizes in water, and the Na⁺ ions split apart from the H^- ions. When the Hydride ions lose their extra electron, they are left as neutral Hydrogen atoms. But they still want to complete their outer shell, so when they find another Hydrogen atom, they bond together and form a Hydrogen molecule (H₂).


2. When the Hydride anions come near the dissolved Chlorine molecules, the Chlorine molecules take the extra electron from the Hydrogen. But the Chlorine atoms already have a full shell, because they're in a molecule - they don't have space for more atoms. So, they split up and then take the extra electron. Afterwards, the Chloride ions don't bond back together, because now they each have full shells individually.


3. In the reactants, Chlorine gas is dissolved, but in the products, Hydrogen gas isn't. The reason they are both gases is because the temperature of the reaction is above their boiling point. Chlorine gas is dissolved because it is soluble, while Hydrogen gas isn't dissolved because it is insoluble. This is just something you have to memorize.

This is great, but...

For simple equations like this one, we can tell exactly how the substances ionize $NaH ionizes into Na^\+^ and H^\-^, and Cl~2~ won't ionize$, but what if there are more complex compounds?

Fortunately, there is a set of rules we can use $most of the time$:

• Ionic compounds that dissolve in water will ionize. Almost all ionic compounds will dissolve in water. The major exceptions to this are carbonate compounds, hydroxide compounds, and oxide compounds.
  • There are other exceptions, and exceptions to these exceptions, but they are not needed at this moment. This page will be updated to include those exceptions if/when they are needed.
• Molecules, like O₂, will not ionize.
• Polyatomic ions are not ionized further. For example, if we had H₂SO₄, that would ionize into 2H⁺ and SO₄^2-, but the SO₄^2- wouldn't further ionize.
• Additionally, if we're given the state symbols in the reaction equation:
  • Solids and gases won't ionize, and liquids won't ionize unless they're ionic compounds in the liquid state $like liquid NaCl$.
  • If something is dissolved, it's usually ionized. The exception to this is when it's a molecule that's dissolved $like Cl~2~ dissolved in water. It won't split up into Cl^\+^ and Cl^\-^$.

Before moving on, here's the net ionic equation again:

2H^- (aq) + Cl₂ → 2Cl^- + H₂ (g)

Although there's one equation here, it actually shows two reactions going on.

Redox Reactions

The above equation is a redox reaction, a combination of both a reduction reaction and an oxidation reaction.

Reducation is when an atom gets electrons, while oxidation is when an atom lets go of electrons. Think of it this way: in reduction, an atom's charge is reduced.

So if a redox reaction is a combination of a reduction reaction and an oxidation reaction, then we can split up the redox reaction into those two parts. These will be the reduction half-equation and the oxidation half-equation.

Reduction half-equation: What atom$s$ get$s$ electrons? Chlorine. So the reduction half-equation will show Chlorine getting electrons, as well as how many electrons it gets.

The reduction half-equation is:

Cl₂ (aq) + 2e^- → 2Cl^- OR Cl (aq) + e^- → Cl^- (aq)

Oxidation half-equation: What atom$s$ let$s$ go of excess electrons? Hydrogen. So the oxidation half-equation will show Hydrogen letting go of excess electrons.

The oxidation half-equation is:

2H^- (aq) → 2e^- + H₂ (g) OR H^- (aq) → e^- + H^- (g)

With these half-equations, we can see what happened to the Hydride ions and the Chlorine molecules in the redox reaction. It also makes it easy to see why we got the net ionic equation first; if we attempted to get the half-equations from the original equation, it might've been difficult to determine which elements were getting reduced and oxidized. Once we get the net ionic equation, though, we can tell which elements are getting reduced and oxidized.

It is important to note that not all chemical reactions with ionic compounds are redox reactions.

Let's take an example:

CaCl₂ (aq) + MgF₂ (aq) → CaF₂ (aq) + MgCl₂ (aq)

Split up the parts that will ionize:

Ca²⁺ (aq) + 2Cl^- (aq) + Mg²⁺ (aq) + 2F^- (aq) → Ca²⁺ (aq) + 2F^- (aq) + Mg²⁺ (aq) + Cl^- (aq)

And cross out all the like terms:

Ca²⁺ (aq) + 2Cl^- (aq) + Mg²⁺ (aq) + 2F^- (aq) → Ca²⁺ (aq) + 2F^- (aq) + Mg²⁺ (aq) + Cl^- (aq)

There are no terms left in the net ionic equation, and thus there is no redox reaction. The ions don't change, they simply re-arrange themselves in different compounds. Thus, no element undergoes reduction or oxidation, and therefore there is no redox reaction.

Similarly, sometimes there'll also be ionic equations with a net ionic equation that aren't a redox reaction.

Practice

For the equation SrO $aq$ + 2Na $s$ → Na₂O $aq$ + Sr $s$:

• What is the net ionic equation? Sr²⁺ (aq) + O^2- (aq) → Sr (s) + O (aq) Sr²⁺ (aq) + 2Na (s) → 2Na⁺ (aq) + Sr (s) 2Na (s) + O^2- (aq) → 2Na⁺ (aq) + O (aq)
  Show Answer & Explanation

• What is the oxidation half-equation? Sr²⁺ + 2e^- → Sr Na → Na⁺ + e^- O^2- → O + 2e^- Na⁺ + e^- → Na Sr → Sr²⁺ + 2e^-
  Show Answer & Explanation

• What is the reduction half-equation? Sr → Sr²⁺ + 2e^- Na⁺ + e^- → Na O^2- → O + 2e^- Sr²⁺ + 2e^- → Sr O + 2e^- → O^2-
  Show Answer & Explanation