Chemca Formula Sheet - Redox Reactions

CHEMCA

EXAM MASTER FORMULA SHEET

Redox Reactions

Core Fundamentals for JEE Main, Advanced & NEET

1. Definitions of Redox

Oxidation

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

Reduction

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

2. Oxidation Number (O.N.) Rules

• Free State: O.N. is always \(0\) (e.g., \(O_2, P_4, S_8, Na\)).

• Fluorine: Always \(-1\) in its compounds.

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

• Hydrogen: \(+1\) with non-metals, \(-1\) with active metals (hydrides).

Sum of O.N. in a neutral molecule = \(0\) | Sum in ion = Charge on the ion

3. Balancing Methods

Ion-Electron Method:

Split into half-reactions, balance atoms, balance charge with \(e^-\), multiply to equalize electrons, and add.

Oxidation Number Method:

Calculate change in O.N. per atom, cross-multiply to equalize the total change, then balance remaining atoms.

Balancing H and O:

Acidic Medium: Add \(H_2O\) to balance \(O\), then add \(H^+\) to balance \(H\).

Basic Medium: Balance as acidic, then add \(OH^-\) to both sides to neutralize \(H^+\).

4. n-factor for Redox Species

General formula for Redox n-factor:

\[ n\text{-factor} = |\text{Change in O.N. per molecule}| \]

Total change = (change per atom) \(\times\) (no. of atoms in the formula)

Reagent	Medium	Reaction Product	n-factor
\(KMnO_4\)	Acidic	\(Mn^{2+}\)	5
\(KMnO_4\)	Neutral / Weak Basic	\(MnO_2\)	3
\(KMnO_4\)	Strong Basic	\(MnO_4^{2-}\)	1
\(K_2Cr_2O_7\)	Acidic	\(2Cr^{3+}\)	6

5. Titrations & Equivalents

Iodometry:

\(Analytic \xrightarrow{KI} I_2 \text{ liberated}\)

\(I_2 + Na_2S_2O_3 \xrightarrow{Starch} \text{End Point}\)

Key: Equivalents of analyte = Eq. of \(I_2\) = Eq. of Hypo

Disproportionation:

Same species is oxidized and reduced.

Effective n-factor:

\[ \frac{1}{n_{net}} = \frac{1}{n_1} + \frac{1}{n_2} \implies n_{net} = \frac{n_1 \cdot n_2}{n_1 + n_2} \]

Master Law:

At the equivalence point: \[ N_1 V_1 = N_2 V_2 \] \[ (\text{Molarity}_1 \times n_1) \times V_1 = (\text{Molarity}_2 \times n_2) \times V_2 \]

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