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Redox Equivalent Concept & Titration formulas

Master Redox Equivalent Concept & Titration through 22 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

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Redox Equivalent Concept & Titration, every formula

22 formulas, typeset and free. Print it, or keep it open beside your practice.

Oxidation
$loss of electrons (increase in oxidation number)$
Q1MCQOxidation
Oxidation is defined as:
- A
  loss of electrons
- B
  gain of electrons
- C
  gain of protons
- D
  loss of neutrons
Reduction
$gain of electrons (decrease in oxidation number)$
Q1MCQReduction
Reduction is defined as:
- A
  gain of electrons
- B
  loss of electrons
- C
  loss of protons
- D
  gain of neutrons
n-factor (acid)
$n = number of replaceable H^{+}$
n-factor (base)
$n = number of replaceable O H^{-}$
n-factor (redox)
$n = change in oxidation number$
Equivalent weight
$E = \frac{molar mass}{n -factor}$
Gram equivalents
$eq = \frac{mass}{E} = n \times moles$
Normality
$N = \frac{gram equivalents}{volume (L)}$
Q1MCQNormality
Normality is:
- A
  $\dfrac{\text{gram equivalents}}{\text{volume (L)}}$
- B
  $\dfrac{\text{moles}}{\text{volume}}$
- C
  $\dfrac{\text{mass}}{\text{volume}}$
- D
  $M V$
Normality–molarity
$N = M \times n -factor$
Law of equivalents
$N_{1} V_{1} = N_{2} V_{2}$
at end point
Equivalents balance
$eq of oxidant = eq of reductant$
Q1MCQEquivalents balance
At the end point of a titration:
- A
  equivalents of acid $=$ equivalents of base
- B
  moles of acid $=$ moles of base
- C
  masses are equal
- D
  volumes are equal
Percentage purity
$% = \frac{mass of pure substance}{mass of sample} \times 100$
Molarity of a mixture
$M = \frac{N _{1} V _{1} + N _{2} V _{2}}{V _{1} + V _{2}}$
Q1MCQMixture normality
When two solutions are mixed, the resulting normality is:
- A
  $\dfrac{N_1V_1+N_2V_2}{V_1+V_2}$
- B
  $N_1+N_2$
- C
  $\dfrac{N_1}{N_2}$
- D
  $N_1 V_1$
Volume-strength of H₂O₂
$V.S. = 5.6 \times N$
Hardness (ppm CaCO₃)
$ppm = \frac{mass of CaCO _{3}}{mass of water} \times 1 0^{6}$
Q1MCQHardness
Water hardness is expressed as ppm of:
- A
  $CaCO_3$
- B
  $NaCl$
- C
  $H_2O$
- D
  $O_2$
Oxidation number of O
$- 2 (usually), - 1 (peroxides)$
Q1MCQOx. no. of O
The usual oxidation number of oxygen is:
- A
  $-2$
- B
  $+2$
- C
  $-1$
- D
  $0$
Oxidation number of H
$+ 1 (usually), - 1 (hydrides)$
Q1MCQOx. no. of H
The oxidation number of hydrogen in metal hydrides is:
- A
  $-1$
- B
  $+1$
- C
  $0$
- D
  $+2$
Sum of oxidation numbers
$neutral molecule = 0, ion = charge$
Q1MCQSum of ox. no.
The sum of oxidation numbers in a neutral molecule is:
- A
  $0$
- B
  $+1$
- C
  $-1$
- D
  equal to the number of atoms
Disproportionation
$same element both oxidised and reduced$
Q1MCQDisproportionation
In a disproportionation reaction, the same element is:
- A
  both oxidised and reduced
- B
  only oxidised
- C
  only reduced
- D
  unchanged
Equivalents from Faraday
$eq = \frac{I t}{96500}$
Q1MCQFaraday equivalents
The number of equivalents deposited by charge $It$ is:
- A
  $\dfrac{It}{96500}$
- B
  $96500\,It$
- C
  $\dfrac{96500}{It}$
- D
  $It$
Back titration principle
$eq_{e x cess} = eq_{a dd e d} - eq_{r e a c t e d}$
Q1MCQBack titration
In a back titration, the equivalents of unreacted reagent equal:
- A
  added $-$ reacted
- B
  added $+$ reacted
- C
  reacted only
- D
  zero
Mole–equivalent relation
$equivalents = n -factor \times moles$
Q1MCQMole–equivalent
The number of equivalents equals:
- A
  $n\text{-factor}\times\text{moles}$
- B
  $\dfrac{\text{moles}}{n\text{-factor}}$
- C
  $\text{moles}$
- D
  $\dfrac{n\text{-factor}}{\text{moles}}$

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Redox Equivalent Concept & Titration formulas

Redox Equivalent Concept & Titration, every formula

Oxidation

Reduction

n-factor (acid)

n-factor (base)

n-factor (redox)

Equivalent weight

Gram equivalents

Normality

Normality–molarity

Law of equivalents

Equivalents balance

Percentage purity

Molarity of a mixture

Volume-strength of H₂O₂

Hardness (ppm CaCO₃)

Oxidation number of O

Oxidation number of H

Sum of oxidation numbers

Disproportionation

Equivalents from Faraday

Back titration principle

Mole–equivalent relation

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