Redox

Long, Extensive, and In-Depth Notes on Redox’ing

To understand which reactions are redox reactions, let’s first understand why redox is named redox.

OK let’s get straight into it!!

Oxidation and Reduction: The Basics

Oxidation is defined as the loss of electrons, while reduction is defined as the gain of electrons

Hence… we can formulate this acronym to aid us in remembering this:

Here are an example to get you started:

\[Zn_{(s)}\space+\space2H^{+}_{(aq)}\space->H_{2(g)}\space+\space Zn^{2+}_{(aq)}\]

This is a single displacement reaction, where the hydrogen ion that is bonded to the anion is displaced by the zinc metal forming hydrogen gas. This is a redox reaction because there is a transfer of electrons from the zinc to the hydrogen, which we can see by the change in charge.

\[Zn_{(s)}->Zn^{2+}_{(aq)}\space+\space2e^{-}\] \[2e^{-}\space+\space2H^{+}_{(aq)}\space->H_{2(g)}\]

Splitting the first example into two half equation, we get these two equations. Half equations are equations that show each reactant, oxidizing or reducing, and the movement of electrons. We can see that the electrons are present on the right side for the first equation, and the left side for the second equation. Electrons that are present as products means that electrons are lost, meaning this half equation represents the oxidation of the zinc. Electrons that are present as reactants means that electrons are gained, meaning this half equation represents the reduction of the hydrogen.

The zinc is getting oxidized, while the hydrogen is getting reduced. Well that means that the hydrogen is oxidizing the zinc, and the zinc is reducing the hydrogen. We can give names to this idea:

Therefore, we can call zinc in this reaction the reducing agent (or the reductant), and likewise the hydrogen the oxidizing agent (or oxidant). Pretty confuzzling huh? Take some time to wiggle your brain cells around this part to continue. It’ll help with the whole redox topic.

ALRIGHT! I’ll give you some practice ones…

Q1. What are the half equations for this reaction?

\[Zn_{(s)}\space+\space Ag^{+}_{(aq)}\space->Ag_{(s)}\space+\space Zn^{2+}_{(aq)}\]

Answer: See bottom of page

Q2. Half reactions? What is getting oxidized, and reduced in this reaction? Which reactant is the oxidizing agent and which is the reducing agent?

\[F_{2(g)}\space+\space2Br^{-}_{(aq)}\space->Br_{2(l)}\space+\space 2F^{-}_{(aq)}\]

Answer: See bottom of page

Oxidation Numbers

To make assigning oxidation and reduction names to molecules easier to recognise, we can give each of the constituents a positive/negative number representing the oxidation state. For example:

Combustion of Methane

The blue numbers are the oxidation numbers. Can you see any patterns regarding these numbers?

Oxidation Numbers: The Rules

  1. All Elements have an oxidation number of 0
    • E.g. Hydrogen Gas has an oxidation number of 0
  2. All compounds’ oxidation numbers add to a total of 0
    • E.g. Sulfuric Acid’s hydrogens, sulphur, and oxygen’s oxidation number will add up to 0
  3. For monoatomic ions, the charge is the oxidation number
    • E.g. Sodium Ions have a charge of 1+, and thus have an oxidation number of +1 when in solution or when part of a compound.
  4. Combined Oxygen have an oxidation number of -2
    • E.g. Water
    • Exception!
      • When part of a peroxide, it has an oxidation number of -1
  5. Combined Hydrogen have an oxidation number of +1
    • E.g. Hydrochloric Acid
    • Exception!
      • When part of a hydride, it has an oxidation number of -1
  6. Polyatomic Ions have an oxidation number of its charge
    • E.g. Nitrate Ion has a 1- charge
      • N has +5 oxidation number
      • O3 has -2 oxidation number per oxygen so a total of -6
      • Total of -1

Using these oxidation numbers, we can make it easier to construct half equations.

Constructing Half Equations

In acidic conditions, we can use this step by step procedure to guarantee success.

  1. Balance species other than Oxygen & Hydrogen
  2. Balance Oxygen’s with Water molecules
  3. Balance Hydrogen’s with Hydrogen ions
  4. Balance charges with electrons

For example:

$MnO_{4}^{-}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space->Mn^{2+}\space$

$MnO_{4}^{-}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space->Mn^{2+}\space+\space 4H_{2}O$

$MnO_{4}^{-}\space+\space8H^{+}\space\space\space\space\space\space\space\space\space\space\space\space\space->Mn^{2+}\space+\space 4H_{2}O$

$MnO_{4}^{-}\space+\space8H^{+}\space+\space5e^{-}->Mn^{2+}\space+\space 4H_{2}O$

Since there are 4 oxygens on the reactant side, we can balance this out by adding 4 water molecules onto the product side. Next, to balance the hydrogens we just added, we can add hydrogen ions onto the reactant side, in this case 8. Finally, we need to balance the charges. On the product side, there is a total of 2+ charge due to the manganese ion. This means we need a 2+ charge on the reactant side. There is a 8+ charge from the hydrogen ions, and a 1- charge from the permanganate ion, resulting in a 7+ charge. To get it to the target of 2+, we add 5 electrons which are negatively charged to successfully balance out the charges.

Combining Half Equations

Once we have two half equations, we can combine them into one big equation. This equation can be non-ionic, meaning if there is no change in state symbols, we would not emit them as spectator ions.

To combine half equations, first we would balance the electrons and then combine. The coefficients should be balanced when combined.

For Example:

Answers

Question One.

\[Zn_{(s)}->Zn^{2+}_{(aq)}\space+\space2e^{-}\]
\[e^{-}\space+\space Ag^{+}_{(aq)}\space->Ag_{(s)}\]

Question Two.

\[F_{2(g)}->2F^{-}_{(aq)}\space+\space2e^{-}\]
\[2e^{-}\space+\space 2Br^{-}_{(aq)}\space->Br_{2(l)}\]

Published under Creative Commons Attribution-Share Alike 3.0 - Go Home