Chapter 14. Oxidation-Reduction Reactions
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2. Oxidation-Reduction Reactions
3. Oxidizing Agents and Reducing Agents
4. Balancing Chemical Equations
1. Oxidation Number
⑴ Definition: The hypothetical charge that an element would have if it lost all of its shared electron pairs, assuming the element with the higher electronegativity had taken them.
⑵ Oxidation number of each atom = charge of the atom + valence electrons of the atom - number of bonds - number of unshared electrons
① Fractional oxidation numbers represent average oxidation states
⑶ In nature, the oxidation numbers of atoms are standardized: Priority is given to the order ① ＞ ··· ＞ ⑦
① The sum of oxidation numbers of atoms composing a neutral compound is 0.
② The sum of oxidation numbers of atoms composing an ion is equal to the charge of the ion.
③ The oxidation number of fluorine is always -1.
④ Alkali metals are +1, alkaline earth metals are +2, aluminum is +3.
⑤ Hydrogen is +1.
○ Exception: In LiH, the oxidation number of H is -1.
⑥ Oxygen is -2.
○ Exception 1. In OF2, the oxidation number of O is +2.
○ Exception 2. H2O2 (-1), O2- (-1/2), O3- (-1/3).
⑦ Halogen elements excluding fluorine are -1.
○ Exception: When combined with oxygen (e.g., OCl2), it is +1.
2. Oxidation-Reduction Reactions
⑴ Oxidation: The loss of high-energy electrons by atoms, molecules, ions, etc.
① Expression 1. An increase in oxidation number
② Expression 2. Gaining oxygen
○ Before reaction: The oxidation number of each O atom in O2 is 0.
○ After reaction: The oxidation number of O in molecules becomes -2, and other species experience an increase in oxidation number.
③ Expression 3. Losing hydrogen
○ Before reaction: Generally, the oxidation number of H in a molecule is +1.
○ After reaction: H2 with oxidation numbers of 0 for each atom is produced, causing other species to experience an increase in oxidation number.
④ Expression 4. A reaction that loses energy: Due to the loss of high-energy electrons.
⑵ Reduction: The gain of high-energy electrons by atoms, molecules, ions, etc.
① Expression 1. A decrease in oxidation number
② Expression 2. Losing oxygen
○ Before reaction: Generally, the oxidation number of O in a molecule is -2.
○ After reaction: O2 with oxidation numbers of 0 for each atom is produced, causing other species to experience a decrease in oxidation number.
③ Expression 3. Gaining hydrogen
○ Before reaction: The oxidation number of each H atom in H2 is 0.
○ After reaction: The oxidation number of H in molecules becomes +1, and other species experience a decrease in oxidation number.
④ Expression 4. A reaction that gains energy: Due to the gain of high-energy electrons.
⑶ Simultaneity: Oxidation and reduction always occur simultaneously, and the change in oxidation number is equal to the number of transferred electrons.
⑷ Oxidation-reduction reactions involve the direct movement of electrons, unlike acid-base reactions.
① Electron Activity (electron activity): Concentration of electrons.
3. Oxidizing Agents and Reducing Agents
⑴ Definition
① Oxidizing agent: A substance that is reduced itself while oxidizing another substance.
② Reducing agent: A substance that is oxidized itself while reducing another substance.
⑵ Relative Strength of Oxidizing Agents
① The stronger the tendency to gain electrons (Group 17), the stronger the oxidizing agent.
② Compounds containing metals or nonmetals with high oxidation numbers
○ Example: In K2Cr2O7, the oxidation number of Cr is +6, making it a strong oxidizing agent.
③ If an element can have multiple oxidation numbers, the greater the oxidation number, the stronger the oxidizing agent.
○ KMnO4 > MnO2 > Mn2O3 > MnCl2
○ KMnO4: Since the oxidation number of Mn is +7, it oxidizes other substances to achieve a lower oxidation state.
④ If the oxidation state is lower than usual, the substance can function as an oxidizing agent.
○ Hydrogen peroxide: The oxidation number of O in H2O2 is -1, so it oxidizes other substances to achieve -2.
⑶ Relative Strength of Reducing Agents
① The stronger the tendency to lose electrons (lower left of the periodic table), the stronger the reducing agent.
② Compounds containing metals or nonmetals with low oxidation numbers
③ If an element can have multiple oxidation numbers, the lower the oxidation number, the stronger the reducing agent.
○ Example: H2S (-2) > S (0) > SO2 (+4) > SO3 (+6)
⑷ Relative Nature of Oxidizing Agents and Reducing Agents: Depending on the reaction, substances can act as either oxidizing agents or reducing agents.
① Example 1: Disproportionation reaction: A → A’ + A”, where A’ and A” have different oxidation states.
○ 4SO2 + 3H2O → S2O32- + 2HSO4- + 4H+
4. Balancing Chemical Equations
⑴ Oxidation Number Method
① Definition: Utilizing the fact that the increase in oxidation number is equal to the decrease in oxidation number.
② 1st. Write the skeletal equation
MnO4-(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
② 2nd. Calculate the change in oxidation number (x) for the oxidation skeletal equation
Oxidation number of carbon indicated by [ · ]: H2C2O4(aq) [+6] → 2CO2(g) [+8]
x = +2
③ 3rd. Calculate the change in oxidation number (y) for the reduction skeletal equation
Oxidation number of manganese indicated by [ · ]: MnO4-(aq) [+7] → Mn2+(aq) [+2]
y = -5
④ 4th. Multiply the oxidation and reduction skeletal equations by appropriate factors to make the change in oxidation number for both equations equal to the least common multiple of x and y (10), then add them together
2MnO4-(aq) + 5H2C2O4(aq) → 2Mn2+(aq) + 10CO2(g)
⑤ 5th. In an acidic environment, balance the number of O atoms with H2O and the number of H atoms with H+
○ 5th - 1st.
2MnO4-(aq) + 5H2C2O4(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(ℓ)
2MnO4-(aq) + 5H2C2O4(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(ℓ)
○ 5th - 2nd. Review: Check if charge balance is maintained
○ 5th - 3rd. Charge balance is naturally maintained by the process in ④.
⑥ 6th. In a basic environment, balance the number of O atoms with H2O and the number of H atoms with (H2O - OH-)
○ 6th - 1st.
2MnO4-(aq) + 5H2C2O4(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(ℓ)
2MnO4-(aq) + 5H2C2O4(aq) + 6H2O(ℓ) - 6OH-(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(ℓ)
2MnO4-(aq) + 5H2C2O4(aq) → 2Mn2+(aq) + 10CO2(g) + 2H2O(ℓ) + 6OH-(aq)
○ 6th - 2nd. Review: Check if charge balance is maintained
○ 6th - 3rd. Charge balance is naturally maintained by the process in ④.
⑵ Ion-Electron Method: Obtain the oxidation half-equation and the reduction half-equation, then add the two half-equations together.
Input: 2018.12.28 13:28