Redox Reactions

Classical Definition:

• Oxidation: Addition of O / electronegative element OR removal of H / electropositive element
• Reduction: Addition of H / electropositive element OR removal of O / electronegative element

Electronic Concept:

• Oxidation: Loss of electrons (LEO)
• Reduction: Gain of electrons (GER)
• Mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Redox Reaction: Oxidation and reduction occur simultaneously; total electrons transferred remain equal.

Oxidation Number (ON): Hypothetical charge on an atom if all bonds were ionic (most EN atom assigned all electrons).

Rules to Determine ON:

Identifying Oxidizing and Reducing Agents:

• Oxidizing agent (OA): Causes oxidation; itself gets reduced; gains electrons; ON decreases
• Reducing agent (RA): Causes reduction; itself gets oxidized; loses electrons; ON increases

Common OAs: $KMnO_4, K_2Cr_2O_7, HNO_3$ (concentrated), $H_2O_2$, $F_2, Cl_2$, conc. $H_2SO_4$, $O_3$. Common RAs: $H_2, C, CO, H_2S$, metals like Na, Al, $SO_2$, $H_2O_2$ (in some), $Fe^{2+}$.

Stock Notation: Roman numeral in parentheses indicates ON. e.g., Fe(II), Fe(III), Mn(VII) in KMnO₄.

Two methods:

1. Oxidation Number (ON) method
2. Ion-electron (Half-reaction) method

Ion-electron Method (Acidic Medium):

Step 1: Write skeletal equation with separate half-reactions (oxidation and reduction) Step 2: Balance atoms other than O and H Step 3: Balance O with $H_2O$ Step 4: Balance H with $H^+$ Step 5: Balance charges by adding electrons Step 6: Multiply half-reactions to equalize electrons; add

Ion-electron Method (Basic Medium): Same first 6 steps, then: Step 7: Add OH⁻ to both sides equal to number of H⁺ Step 8: Combine $H^+ + OH^-$ on same side as $H_2O$ Step 9: Cancel water molecules

Worked Example: Balance: $MnO_4^- + Fe^{2+} \to Mn^{2+} + Fe^{3+}$ (acidic).

Oxidation: $Fe^{2+} \to Fe^{3+} + e^-$ Reduction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ (balance O with $4H_2O$, H with $8H^+$; charge: left $-1+8 = +7$, plus 5e⁻ gives $+2$ to match right)

Multiply oxidation by 5: $5Fe^{2+} \to 5Fe^{3+} + 5e^-$ $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$

Add: $MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$. ✓

ON Method (for whole equation):

1. Identify atoms changing ON
2. Calculate change in ON (oxidation, reduction)
3. Multiply coefficients to balance electrons gained/lost
4. Balance other atoms (especially O and H using H₂O and H⁺/OH⁻)

Equivalent Weight ($E$): Mass of substance combining with or displacing $1$ g of H, $8$ g of O, or $35.5$ g of Cl.

For Acids: $E = M/\text{basicity}$ (number of replaceable H⁺). e.g., $H_2SO_4$: $E = 98/2 = 49$.

For Bases: $E = M/\text{acidity}$ (number of OH⁻). e.g., NaOH: $E = 40/1 = 40$.

For Salts: $E = M/(\text{total +ve charge or -ve charge})$. e.g., $Al_2(SO_4)_3$: $E = 342/6 = 57$.

For Oxidizing/Reducing Agents: $$E = \frac{M}{\text{change in oxidation number per molecule}}$$

Normality (N): Equivalents/L. $N = M \times n$, where $n = $ n-factor.

Equivalent Concept: Equivalents of OA = Equivalents of RA in reaction. So in titration: $$N_1V_1 = N_2V_2$$

Volume Strength of $H_2O_2$: Volume of O₂ released at STP per volume of $H_2O_2$ solution. $$\text{Volume strength} = 5.6 \times N = 11.2 \times M$$

Disproportionation Reaction: Same element is both oxidized and reduced in the same reaction. Requires the element to have an intermediate ON, with higher and lower states accessible.

Examples:

• $2H_2O_2 \to 2H_2O + O_2$ (O: $-1 \to -2$ and $-1 \to 0$)
• $3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$ (Mn: $+6 \to +7$ and $+6 \to +4$)
• $Cl_2 + 2NaOH \to NaCl + NaClO + H_2O$ (Cl: $0 \to -1$ and $0 \to +1$)
• $3Cl_2 + 6NaOH \to 5NaCl + NaClO_3 + 3H_2O$ (hot, concentrated)
• $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$ (P: $0 \to -3$ and $0 \to +1$)
• $4KClO_3 \to 3KClO_4 + KCl$ (heated; Cl: $+5 \to +7$ and $+5 \to -1$)
• $Cu^+ \to Cu^{2+} + Cu$ in aqueous solution

Comproportionation (Synproportionation): Two species with same element in different ON combine to form a single product with intermediate ON. Reverse of disproportionation.

Examples:

• $IO_3^- + 5I^- + 6H^+ \to 3I_2 + 3H_2O$ (I: $+5$ and $-1 \to 0$)
• $H_2S + SO_2 \to 3S + 2H_2O$ (S: $-2$ and $+4 \to 0$)
• $Fe + 2Fe^{3+} \to 3Fe^{2+}$ (Fe: $0$ and $+3 \to +2$)

Conditions for Disproportionation:

• Element must have at least three oxidation states
• Intermediate state should be present

Why some elements don't disproportionate:

• F: cannot, since $-1$ is most stable (no positive ON commonly)
• Pure metals in zero state to negative state — typically not possible

Auto-redox: Disproportionation can be called auto-oxidation-reduction in some texts; same concept.

Decomposition Redox Reactions: Single substance breaks down into two with different ON.

• $2H_2O \to 2H_2 + O_2$ (electrolysis)
• $2NaN_3 \to 2Na + 3N_2$ (airbag chemistry)
• $2KClO_3 \to 2KCl + 3O_2$ (catalyst MnO₂)
• $2HgO \to 2Hg + O_2$
• $NH_4NO_3 \to N_2O + 2H_2O$