Redox Reactions Study Notes

Redox Reactions

Entry Task Review

The entry task questions cover fundamental concepts related to redox reactions:

1. OIL RIG: This is an acronym that stands for "Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)."

2. **Oxidation Numbers in $Cr<em>2O</em>7^{2-}$

   • Oxygen in a compound typically has an oxidation number of -2.
   • The sum of oxidation numbers should equal the charge of the polyatomic ion.
   • In $Cr<em>2O</em>7^{2-}$, the oxidation number of Cr is +6 and O is -2.

3. Reaction Classification: The example reaction
   $2I^- \rightarrow I_2 + 2e^-$
   is classified as oxidation because iodine loses electrons.

4. Balancing Redox Half Equation: Balancing the half-equation $Br<em>2 \rightarrow Br^-$. The balanced equation is: $Br</em>2 + 2e^- \rightarrow 2Br^-$

5. Identifying the Oxidant: In the reaction
   $2I^- + (something) \rightarrow I_2 + (something else)$
   Iodine is oxidized.

Acknowledgement of Country

The lesson acknowledges the Wurundjeri people of the Kulin Nation as the traditional owners of the land and pays respect to their elders.

Redox Reactions: Study Design Dot Points

• Redox reactions involve simultaneous oxidation and reduction processes.
• Oxidation numbers are used to identify reducing agents, oxidizing agents, and conjugate redox pairs.

Drawing Ions Activity – MgO (aq)

This activity involves determining the correct representation of ions in Magnesium Oxide (MgO).

Recap: Predicting Ionic Charges

• Group 1 Metals: Lose 1 electron to form +1 ions.
• Group 2 Metals: Lose 2 electrons to form +2 ions.
• Transition Metals: Exhibit a range of electron donations (refer to Roman numerals for specific charges).
• Group 13 Elements: Lose 3 electrons to form +3 ions.
• Group 15 Elements: Gain 3 electrons to form -3 ions.
• Group 16 Elements: Gain 2 electrons to form -2 ions.
• Group 17 Elements: Gain 1 electron to form -1 ions.
• Group 18 Elements (Noble Gases): Generally do not form ions due to their full outer electron shells.
• Group 14 Elements: Requires a lot of energy to lose or receive 4 electrons – not ideal

Recap: Polyatomic Ions

A polyatomic ion is a group of tightly bound atoms that behave as a single unit with an overall charge. Common examples include:

• Ammonium: $NH_4^+$
• Hydroxide: $OH^-$
• Phosphate: $PO_4^{3-}$
• Hydrogen Carbonate (Bicarbonate): $HCO_3^-$
• Carbonate: $CO_3^{2-}$
• Nitrate: $NO_3^-$
• Sulfate: $SO_4^{2-}$
• Permanganate: $MnO_4^-$
• Dichromate: $Cr<em>2O</em>7^{2-}$

These ions can be found in the data book (pp. 6-7).

Recap: What is Redox?

Redox reactions involve the transfer of electrons from one chemical species to another.

Recap: Reduction & Oxidation

• Oxidation: Loss of electrons (increase in oxidation number).
• Reduction: Gain of electrons (decrease in oxidation number).

Recap: Reductants & Oxidants

• Reducing Agents (Reductants): Reduce other species and are themselves oxidized in the process.
• Oxidizing Agents (Oxidants): Oxidize other species and are themselves reduced in the process.

Reductants & Oxidants: Examples

1. $Ca(s) + Cl<em>2(g) \rightarrow CaCl</em>2(s)$

   • Calcium (Ca) loses electrons and is oxidized; it is the reducing agent.
   • Chlorine ($Cl_2$) gains electrons and is reduced; it is the oxidizing agent.

2. $2AgBr(s) \rightarrow 2Ag(s) + Br_2(g)$

   • $Ag^+$ cations gain electrons to form solid silver (Ag), undergoing reduction. Silver ions are the oxidizing agent.
   • $Br^-$ ions lose electrons to form $Br_2$ gas, undergoing oxidation. Bromide ions are the reducing agent.

Conjugate Redox Pairs

A conjugate redox pair consists of an electron donor and its corresponding electron acceptor, typically written with a forward slash (/).

Examples:

• $Li(s) \rightarrow Li^+(aq) + e^- ; Li(s) / Li^+(aq)$
• $Br<em>2(l) + 2e^- \rightarrow 2Br^-(aq) ; Br</em>2(l) / Br^-(aq)$
• $2KCl (aq) \rightarrow 2K (s) + Cl_2(g) ; K^+(aq) / K(s)$
• $Cl^-(aq) / Cl_2(g)$

Oxidation Numbers (ON)

Oxidation numbers help identify which species are oxidized or reduced.

• ON decreases: Reduction
• ON increases: Oxidation

Rules for Assigning Oxidation Numbers:

• Elements: ON = 0 (e.g., $Cl<em>2$, Mg, C, K, $P</em>4$).
• Ions: ON = charge of the ion (e.g., $Na^+$, $Cl^-$).
• Covalent Compounds: The most electronegative element has the negative ON, as if it were an ionic substance.
• Hydrogen Compounds: ON(H) = +1, except in metal hydrides (e.g., LiH) where ON(H) = -1.
• Oxygen Compounds: ON(O) = -2, except in $H<em>2O</em>2$ and $BaO_2$ where ON(O) = -1.

Examples:

• NaCl: Na (+1), Cl (-1)
• $NH_4^+$: N (-3), H (+1)
• HCl: H (+1), Cl (-1)
• $H_2O$: H (+1), O (-2)
• LiH: Li (+1), H (-1)
• $H<em>2O</em>2$: H (+1), O (-1)

Oxidation Numbers: Examples

Example 1: $CuO(s) + H<em>2(g) \rightarrow Cu(s) + H</em>2O(l)$

1. Assign oxidation numbers:
   • CuO: O = -2, so Cu = +2
   • $H_2$: H = 0 (elemental form)
   • Cu: Cu = 0 (elemental form)
   • $H_2O$: O = -2, so H = +1
2. Identify the element undergoing reduction:
   • Copper's oxidation number decreases from +2 to 0, so CuO is undergoing reduction.

Example 2: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(l)$

1. Assign oxidation numbers:
   • $CH_4$: H = +1, so C = -4
   • $O_2$: O = 0 (elemental form)
   • $CO_2$: O = -2, so C = +4
   • $H_2O$: O = -2, so H = +1
2. Identify the element undergoing reduction:
   • Oxygen's oxidation number decreases from 0 to -2, so $O_2$ is undergoing reduction.

Quality Vote Sample

$C<em>6H</em>{12}O<em>6(aq) + 6O</em>2(g) \rightarrow 6CO<em>2(g) + 6H</em>2O(l)$

Glucose is oxidized during cellular respiration. Evidence:

• Carbon (in glucose) oxidation number increases from 0 to +4.