Redox Reactions

• Classical Oxidation: The chemical addition of oxygen or an electronegative element to a substance, or the removal of hydrogen or an electropositive element from a substance.
  Examples:
  • $2\text{Mg}(s) + \text{O}_2(g) \to 2\text{MgO}(s)$ (Addition of oxygen)
  • $2\text{H}_2\text{S}(g) + \text{O}_2(g) \to 2\text{S}(s) + 2\text{H}_2\text{O}(l)$ (Removal of hydrogen from $\text{H}_2\text{S}$)
  • $\text{Fe}(s) + \text{S}(s) \to \text{FeS}(s)$ (Addition of electronegative sulfur)
  • $2\text{KI} + \text{Cl}_2 \to 2\text{KCl} + \text{I}_2$ (Removal of electropositive potassium from iodine)
• Classical Reduction: The chemical addition of hydrogen or an electropositive element to a substance, or the removal of oxygen or an electronegative element from a substance.
  Examples:
  • $\text{CuO}(s) + \text{H}_2(g) \to \text{Cu}(s) + \text{H}_2\text{O}(l)$ (Removal of oxygen from $\text{CuO}$)
  • $\text{Cl}_2(g) + \text{H}_2(g) \to 2\text{HCl}(g)$ (Addition of hydrogen to $\text{Cl}_2$)
  • $2\text{HgCl}_2(aq) + \text{SnCl}_2(aq) \to \text{Hg}_2\text{Cl}_2(s) + \text{SnCl}_4(aq)$ (Addition of electropositive mercury)
  • $2\text{FeCl}_3(aq) + \text{H}_2(g) \to 2\text{FeCl}_2(aq) + 2\text{HCl}(aq)$ (Removal of electronegative chlorine)

IUPAC Master Rules for Assigning Oxidation Numbers:

1. Elemental State: The oxidation number of any element in its free, uncombined, or allotropic form is strictly zero ($0$). E.g., $\text{O}_2, \text{O}_3, \text{P}_4, \text{S}_8, \text{H}_2, \text{Cl}_2, \text{Fe}(s), \text{Na}(s) = 0$.
2. Monoatomic Ions: The oxidation number equals the actual algebraic ionic charge. E.g., $\text{Na}^+ (+1), \text{Mg}^{2+} (+2), \text{Al}^{3+} (+3), \text{Cl}^- (-1), \text{O}^{2-} (-2), \text{N}^{3-} (-3)$.
3. Fluorine: As the most electronegative element in the periodic table, fluorine has an oxidation state of $-1$ in ALL its compounds without exception.
4. Oxygen:
   • In normal oxides (e.g., $\text{H}_2\text{O}, \text{CaO}, \text{CO}_2$), oxygen is assigned $-2$.
   • In peroxides containing the peroxo bond ($-O-O-$), such as $\text{H}_2\text{O}_2, \text{Na}_2\text{O}_2, \text{BaO}_2$, oxygen is $-1$.
   • In superoxides (e.g., $\text{KO}_2, \text{RbO}_2, \text{CsO}_2$), oxygen is $-1/2$.
   • In oxygen fluorides, oxygen displays positive oxidation states due to the superior electronegativity of fluorine: $+2$ in $\text{OF}_2$ and $+1$ in $\text{O}_2\text{F}_2$.
5. Hydrogen: Assigned $+1$ in compounds with non-metals (e.g., $\text{H}_2\text{O}, \text{HCl}, \text{CH}_4$), but $-1$ in binary metal hydrides formed with active electropositive metals (e.g., $\text{NaH}, \text{CaH}_2, \text{LiAlH}_4$).
6. Alkali & Alkaline Earth Metals: Group 1 metals ($\text{Li, Na, K, Rb, Cs}$) are always $+1$; Group 2 metals ($\text{Be, Mg, Ca, Sr, Ba}$) are always $+2$.
7. Halogens: $\text{Cl, Br, I}$ typically have $-1$, but exhibit positive oxidation states ($+1, +3, +5, +7$) when combined with more electronegative halogens or oxygen (e.g., in interhalogens like $\text{ClF}_3$ and oxoacids like $\text{HClO}_4$).
8. Algebraic Summation:
   • In a neutral molecule, the sum of oxidation numbers of all constituent atoms equals $0$: $\sum \text{O.N.} = 0$.
   • In a polyatomic ion, the sum of oxidation numbers equals the net ionic charge: $\sum \text{O.N.} = z$.

1. Combination Reactions: A reaction where two or more substances combine to form a single chemical product. For it to be a redox reaction, at least one of the reacting components must be in an elemental form. $$\overset{0}{\text{C}}(s) + \overset{0}{\text{O}_2}(g) \to \overset{+4}{\text{C}}\overset{-2}{\text{O}_2}(g) \quad (\text{Carbon oxidized from 0 to +4; Oxygen reduced from 0 to -2})$$ $$3\overset{0}{\text{Mg}}(s) + \overset{0}{\text{N}_2}(g) \to \overset{+2}{\text{Mg}}_3\overset{-3}{\text{N}}_2(s) \quad (\text{Magnesium oxidized; Nitrogen reduced})$$ Counter-example (Non-redox combination): $\overset{+2}{\text{Ca}}\overset{-2}{\text{O}} + \overset{+4}{\text{C}}\overset{-2}{\text{O}_2} \to \overset{+2}{\text{Ca}}\overset{+4}{\text{C}}\overset{-2}{\text{O}_3}$ (No changes in oxidation states).
2. Decomposition Reactions: A reaction in which a compound breaks down into two or more simpler components. A decomposition is a redox reaction if at least one product is formed in elemental form. $$2\overset{+1}{\text{H}_2}\overset{-1}{\text{O}_2}(l) \to{\Delta} 2\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}(l) + \overset{0}{\text{O}_2}(g) \quad (\text{Redox decomposition})$$ $$2\overset{+1}{\text{K}}\overset{+5}{\text{Cl}}\overset{-2}{\text{O}_3}(s) \to{\text{MnO}_2, \Delta} 2\overset{+1}{\text{K}}\overset{-1}{\text{Cl}}(s) + 3\overset{0}{\text{O}_2}(g) \quad (\text{Cl reduced from +5 to -1; O oxidized from -2 to 0})$$ Counter-example (Non-redox thermal decomposition): $\overset{+2}{\text{Ca}}\overset{+4}{\text{C}}\overset{-2}{\text{O}_3}(s) \to{\Delta} \overset{+2}{\text{Ca}}\overset{-2}{\text{O}}(s) + \overset{+4}{\text{C}}\overset{-2}{\text{O}_2}(g)$ (Oxidation numbers unchanged).
3. Displacement Reactions: An atom or ion in a compound is displaced or replaced by an atom or ion of another element:
   • Metal Displacement: A more electropositive metal (higher in electrochemical activity series) displaces a less electropositive metal from its aqueous salt solution: $$\overset{0}{\text{Zn}}(s) + \overset{+2}{\text{Cu}}\overset{+6}{\text{S}}\overset{-2}{\text{O}_4}(aq) \to \overset{+2}{\text{Zn}}\overset{+6}{\text{S}}\overset{-2}{\text{O}_4}(aq) + \overset{0}{\text{Cu}}(s)$$ $$\overset{0}{\text{TiCl}_4}(l) + 2\overset{0}{\text{Mg}}(s) \to \overset{+4}{\text{Ti}}(s) + 2\overset{+2}{\text{Mg}}\overset{-1}{\text{Cl}_2}(s) \quad (\text{Kroll extraction of titanium})$$
   • Non-Metal Displacement:
     Hydrogen Displacement: Active metals with negative standard reduction potentials displace hydrogen gas from water or dilute mineral acids: $$2\overset{0}{\text{Na}}(s) + 2\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}(l) \to 2\overset{+1}{\text{Na}}\overset{-2}{\text{O}}\overset{+1}{\text{H}}(aq) + \overset{0}{\text{H}_2}(g)$$ $$\overset{0}{\text{Fe}}(s) + 2\overset{+1}{\text{H}}\overset{-1}{\text{Cl}}(aq) \to \overset{+2}{\text{Fe}}\overset{-1}{\text{Cl}_2}(aq) + \overset{0}{\text{H}_2}(g)$$
     Halogen Displacement: The oxidizing power of halogens follows the order $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. A more reactive halogen displaces a less reactive halide ion from aqueous solution: $$\overset{0}{\text{Cl}_2}(g) + 2\overset{+1}{\text{K}}\overset{-1}{\text{Br}}(aq) \to 2\overset{+1}{\text{K}}\overset{-1}{\text{Cl}}(aq) + \overset{0}{\text{Br}_2}(l)$$ $$\overset{0}{\text{Br}_2}(l) + 2\overset{+1}{\text{K}}\overset{-1}{\text{I}}(aq) \to 2\overset{+1}{\text{K}}\overset{-1}{\text{Br}}(aq) + \overset{0}{\text{I}_2}(s)$$ Note: Fluorine is so violently reactive that it oxidizes water to oxygen rather than simply displacing other halides: $2\text{F}_2 + 2\text{H}_2\text{O} \to 4\text{HF} + \text{O}_2$.
4. Disproportionation Reactions (Auto-Redox): A special category of redox reaction where a single element existing in an intermediate oxidation state is simultaneously oxidized and reduced to form two distinct products containing that element in higher and lower oxidation states.
   Fundamental Requirement: The reacting element must possess at least three oxidation states and must be present in an intermediate oxidation state.
5. Comproportionation Reactions (Reverse Disproportionation): Two reactants containing the same element in two different oxidation states react together to yield a single product wherein the element attains an intermediate oxidation state. $$2\overset{-2}{\text{H}_2\text{S}}(g) + \overset{+4}{\text{S}}\overset{-2}{\text{O}_2}(g) \to 3\overset{0}{\text{S}}(s) + 2\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}(l) \quad (\text{S oxidized from -2 to 0 and reduced from +4 to 0})$$ $$\overset{0}{\text{Ag}}(s) + \overset{+2}{\text{Ag}}^{2+}(aq) \to 2\overset{+1}{\text{Ag}}^+(aq)$$

Step-by-Step Procedure:

1. Write the skeletal ionic or molecular equation for the reaction.
2. Assign oxidation numbers to all atoms and identify which elements undergo changes in oxidation number.
3. Calculate the increase in oxidation number per atom and decrease in oxidation number per atom, multiplying by the number of atoms of that element present in the formula.
4. Equalize the total increase and decrease in oxidation numbers by multiplying the respective chemical formulas by suitable whole-number stoichiometric coefficients.
5. Balance all atoms other than hydrogen and oxygen by inspection.
6. Balance oxygen and hydrogen atoms:
   • In acidic medium: Add $\text{H}_2\text{O}$ molecules to the oxygen-deficient side; add $\text{H}^+$ ions to the hydrogen-deficient side.
   • In basic medium: Add $\text{H}_2\text{O}$ to the side deficient in oxygen, then add $\text{OH}^-$ ions to balance charges and hydrogens.

The $n$-Factor of a Redox Agent:

Cell Construction:

• Oxidation Half-Cell (Anode): A zinc rod immersed in an aqueous solution of zinc sulfate ($\text{ZnSO}_4, 1.0\text{ M}$). $$\text{Zn}(s) \to \text{Zn}^{2+}(aq) + 2e^- \quad (\text{Oxidation})$$ The anode is the electrode where oxidation occurs. By convention in a galvanic cell, the anode acquires an excess of electrons and is designated as the negative pole ($-$).
• Reduction Half-Cell (Cathode): A copper rod immersed in an aqueous solution of copper(II) sulfate ($\text{CuSO}_4, 1.0\text{ M}$). $$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s) \quad (\text{Reduction})$$ The cathode is the electrode where reduction occurs. Electrons are consumed from this electrode, making it the positive pole ($+$).
• External Circuit: A metallic wire connects the two electrodes through a voltmeter and switch. Electrons spontaneously flow through the external wire from the anode (Zn) to the cathode (Cu). Conventional electric current flows in the opposite direction (from Cu to Zn).
• Overall Cell Reaction: Adding the two half-reactions: $$\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s) \quad (E^\circ_{\text{cell}} = +1.10\text{ V at } 298\text{ K})$$

Universal Format: $\text{Anode} \ | \ \text{Anodic Electrolyte (conc)} \ || \ \text{Cathodic Electrolyte (conc)} \ | \ \text{Cathode}$

• A single vertical line ($|$) denotes a physical phase boundary (e.g., solid metal | aqueous solution).
• A double vertical line ($||$) represents the salt bridge.
• The Anode (Oxidation) is ALWAYS written on the LEFT (Mnemonic: LOAN — Left, Oxidation, Anode, Negative).
• The Cathode (Reduction) is ALWAYS written on the RIGHT.

For the Daniell Cell under standard conditions:

Construction & Specifications of SHE:

• Consists of a platinum wire coated with finely divided platinum black (to maximize surface area and catalyze $\text{H}_2 \rightleftharpoons 2\text{H}^+ + 2e^-$ equilibrium).
• Pure hydrogen gas is bubbled over the platinum foil at a continuous standard pressure of $1.0\text{ bar}$ (or $1\text{ atm}$).
• The electrode is immersed in an aqueous solution containing hydrogen ions ($\text{H}^+$) at unit activity (standard concentration of $1.0\text{ M}$, e.g., $1.0\text{ M HCl}$).
• Operated at a standard temperature of $298.15\text{ K}$ ($25^\circ\text{C}$).

Arbitrary Zero Standard: By universal international convention, the standard potential of SHE is assigned exactly zero volts at all temperatures:

Half-cell representation: $\text{Pt}(s) \ | \ \text{H}_2(g, 1\text{ bar}) \ | \ \text{H}^+(aq, 1\text{ M})$.

Spontaneity Summary:

• If $E^\circ_{\text{cell}} > 0 \implies \Delta G^\circ < 0$: The cell reaction is thermodynamically spontaneous and can generate electricity.
• If $E^\circ_{\text{cell}} < 0 \implies \Delta G^\circ > 0$: The forward reaction is non-spontaneous (the reverse reaction is spontaneous).
• If $E^\circ_{\text{cell}} = 0 \implies \Delta G^\circ = 0$: The cell is at chemical equilibrium and delivers zero electrical work (battery is dead).

Connection to Equilibrium Constant $K$: