Redox Reactions

OIL RIG: Mnemonic for remembering the definitions of oxidation and reduction. It stands for:
- Oxidation is Loss of electrons
- Reduction is Gain of electrons
Oxidation:
- Definition: The loss of electrons, leading to an increase in the oxidation state or charge of the atom.
Reduction:
- Definition: The gain of electrons, leading to a reduction in the oxidation state or charge of the atom.
Oxidation Number:
- Definition: The assigned charge on an atom based on the assumption of complete transfer of electrons in a compound.

Apparent Charge: The oxidation number reflects the change in electric charge due to the loss or gain of electrons.

Neutral Compounds: The sum of all oxidation numbers in a neutral compound must equal zero.
Polyatomic Ions: For polyatomic ions, the sum of oxidation numbers must equal the overall charge on the ion.
Hydrogen: Has an oxidation number of +1 unless combined with a metal, where it is -1.
Fluorine: Always has an oxidation number of -1.
Oxygen: Typically has an oxidation number of -2 in compounds.
Binary Compounds: The more electronegative element is assigned the oxidation number equal to the charge it would exhibit as an ion.
Standard State Elements: Elements in their standard state (uncombined form) have an oxidation number of 0.

Assign oxidation states to the following compounds:
- a. CO₂
- b. SF₆
- c. NO₃
Note: Non-integer oxidation states can occur, as seen in Fe₃O₄, where the oxidation state can be calculated as follows: All oxidation states sum to -8 for the 4 oxygen atoms, divided across 3 iron ions gives: Fe
ightarrow rac{8}{3}^+.

Reaction: When powdered aluminum (Al) is mixed with iodine (I₂) and water is added, the reaction produces a large amount of energy, resulting in flames and purple smoke from excess iodine.
- Chemical Equation:
  2 ext{Al}(s) + 3 ext{I}2(s) ightarrow 2 ext{AlI}3(s)
- Identify: Determine which atoms are oxidized and which are reduced.

Metallurgy: The process of producing metals from ores always involves redox reactions.
- Reactions:
1. Conversion of lead sulfide (PbS) to lead oxide (PbO):
  2 ext{PbS}(s) + 3 ext{O}2(g) ightarrow 2 ext{PbO}(s) + 2 ext{SO}2(g)
2. Reduction process with carbon monoxide:
  ext{PbO}(s) + ext{CO}(g)
  ightarrow ext{Pb}(s) + ext{CO}_2(g)
- Identify: Determine which atoms are oxidized and reduced in each reaction.

Divide the overall reaction into oxidation and reduction half-reactions (using OIL RIG).
Balance all elements in each half-reaction except hydrogen and oxygen.
Balance oxygen atoms by adding H₂O to the appropriate side of each half-reaction.
Balance hydrogen atoms by adding H⁺ to the appropriate side of each half-reaction.
Balance the charge of each half-reaction by adding electrons (e⁻) where needed.
To make the number of moles of electrons equal in both half-reactions, multiply half-reactions as necessary. This allows for the cancellation of electrons when half-reactions are combined.
Cancel out any common species in both half-reactions.
If the reaction is in a basic medium, neutralize any excess H⁺ by adding the same number of OH⁻ to each side of the balanced equation; this forms H₂O and can simplify the equation further.
Finally, check the equation to ensure that it is balanced in terms of both mass and charge.

Balance the following equation using the half-reaction method (in acidic conditions):
ext{MnO}4^-(aq) + ext{I}^-(aq) ightarrow ext{Mn}^{2+}(aq) + ext{I}2(aq)

Potassium Dichromate (K₂Cr₂O₇): A bright orange compound that can be reduced to blue-violet Cr³⁺ ions when reacted with ethyl alcohol: ext{C}2 ext{H}5 ext{OH} + ext{K}2 ext{Cr}2 ext{O}7 ightarrow ext{Cr}^{3+}(aq) + ext{CO}2(g) + ext{H}_2 ext{O}(l)
- Balance this reaction using the half-reaction method.

Extraction of Silver: Silver can be found as nuggets or mixed with ores. An aqueous solution containing cyanide ion is used to extract silver through the following reaction: ext{Ag}(s) + ext{CN}^-(aq) + ext{O}2(g) ightarrow ext{Ag(CN)}2^-(aq)
- Balance this and identify the half-reactions involved in the process.