Chemca Formula Sheet - Redox Reactions

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EXAM MASTER FORMULA SHEET

Redox Reactions

Ultimate Revision for JEE Main, Advanced & NEET

1. Definitions of Redox

Oxidation & Reducing Agent

• • Addition of O / Electronegative element
• • Removal of H / Electropositive element
• • Loss of Electrons (De-electronation)
• • Increase in Oxidation Number

Reducing Agent (Reductant): The species that undergoes oxidation (reduces others).

Reduction & Oxidizing Agent

• • Addition of H / Electropositive element
• • Removal of O / Electronegative element
• • Gain of Electrons (Electronation)
• • Decrease in Oxidation Number

Oxidizing Agent (Oxidant): The species that undergoes reduction (oxidizes others).

2. Oxidation Number (O.N.) Rules & Anomalies

General Rules:

• Free State / Allotropes: O.N. is always $0$ (e.g., $O_2, P_4, S_8, Fe$).

• Fluorine: Always $-1$ in all compounds.

• Oxygen: Usually $-2$.
Exceptions: Peroxides ($-1$), Superoxides ($-1/2$), $OF_2$ ($+2$), $O_2F_2$ ($+1$).

• Hydrogen: $+1$ with non-metals, $-1$ with active metals (metal hydrides like $NaH$).

• Alkali Metals (G-1): Always $+1$.

• Alkaline Earth (G-2): Always $+2$.

Sum of O.N. in a neutral molecule = $0$   |   Sum in a polyatomic ion = Charge on the ion

High-Yield Paradoxical Oxidation Numbers:

Tribromooctaoxide ($Br_3O_8$)

Average O.N. of Br = $+16/3$.

Structure reveals individual O.N.: $+6, +4, +6$.

Tetrathionate Ion ($S_4O_6^{2-}$)

Average O.N. of S = $+2.5$.

Structure reveals two central S are $0$, two terminal S are $+5$.

Carbon Suboxide ($C_3O_2$)

Average O.N. of C = $+4/3$.

Structure ($O=C=C=C=O$): Central C is $0$, terminal C are $+2$.

Caro's Acid ($H_2SO_5$) & Marshall's ($H_2S_2O_8$)

Calculated O.N. of S = $+8$ (Impossible!). Max is $+6$.

Reason: Presence of peroxide linkages ($-O-O-$). True O.N. of S is $+6$.

3. Balancing Methods

Ion-Electron Method (Half-Reaction)

1. Split reaction into Oxidation and Reduction halves.
2. Balance atoms other than O and H.
3. Balance O and H (see rules below).
4. Balance charge by adding electrons ($e^-$).
5. Multiply halves by integers to equalize $e^-$.
6. Add the two half-reactions together.

Oxidation Number Method

1. Identify atoms undergoing O.N. change.
2. Calculate total increase and decrease in O.N.
3. Cross-multiply to equalize the increase/decrease in O.N.
4. Balance all other atoms except O and H.
5. Balance O and H.

Rule for Balancing H and O:

Acidic Medium:
Add $H_2O$ to the side deficient in Oxygen to balance O.
Add $H^+$ to the opposite side to balance H.

Basic Medium:
Balance exactly like acidic medium first. Then, add $OH^-$ equal to the number of $H^+$ on both sides to neutralize $H^+$ into water.

4. n-factor (Valency Factor) Calculation

General formula for Redox n-factor:

\[ n\text{-factor} = |\text{Change in O.N. per atom}| \times (\text{Number of atoms}) \]

Calculated per mole of the molecule reacting.

Reagent	Medium / Condition	Reaction Product	n-factor
$KMnO_4$ ($Mn$ is +7)	Acidic ($H^+$)	$Mn^{2+}$ (+2)	5
	Neutral / Weak Basic	$MnO_2$ (+4)	3
	Strong Basic ($OH^-$)	$MnO_4^{2-}$ (+6)	1
$K_2Cr_2O_7$ ($Cr$ is +6)	Acidic	$2Cr^{3+}$ (+3)	6
Oxalate ($C_2O_4^{2-}$)	Oxidation	$2CO_2$	2
Thiosulfate ($S_2O_3^{2-}$)	With weak oxidant ($I_2$)	$S_4O_6^{2-}$ (Tetrathionate)	1
Ferrous Oxalate ($FeC_2O_4$)	Complete Oxidation	$Fe^{3+} + 2CO_2$	3

5. Titrations & Equivalents Concept

Law of Chemical Equivalence (Master Law):

In any redox reaction, substances react in equal number of equivalents.

\[ \text{Equivalents of OA} = \text{Equivalents of RA} \] \[ N_1 V_1 = N_2 V_2 \]

Formulas for Equivalents ($Eq$):
$Eq = \text{Moles} \times n\text{-factor} = \text{Normality (N)} \times \text{Volume (L)}$
$Eq = \text{Molarity (M)} \times \text{Volume (L)} \times n\text{-factor} = \frac{\text{Weight}}{\text{Equivalent Weight (E)}}$

Iodometry vs Iodimetry:

Iodometry (Indirect):
Analyte (OA) reacts with excess KI to liberate $I_2$. The liberated $I_2$ is titrated with Hypo ($Na_2S_2O_3$).
Eq. of Analyte = Eq. of liberated $I_2$ = Eq. of Hypo

Iodimetry (Direct):
Direct titration of a reducing agent with a standard solution of $I_2$.

Disproportionation Reactions:

A reaction where the same element is simultaneously oxidized and reduced (e.g., $H_2O_2 \to H_2O + O_2$).

Effective n-factor calculation:

\[ \frac{1}{n_{net}} = \frac{1}{n_{ox}} + \frac{1}{n_{red}} \implies n_{net} = \frac{n_{ox} \cdot n_{red}}{n_{ox} + n_{red}} \]

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