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# Equivalent weight - Definition, Examples, Practice problems & FAQs

Equivalent weight - Definition, Examples, Practice problems & FAQs

Imagine going to a friend's house and their mother greets you with an assortment of snacks ranging from  samosa to dhokla, from kachori to pakoda, etc. Some people might just gorge on sandwiches. Others might prefer eating samosa.

What would be the fallout of such behaviour? Each and every person would get their stomach full with any of the delicacies.

In a similar manner, in chemical terms, we use the term “equivalent weight”.

You might ask the purpose of such a term.

During chemical reactions, the tendency of an element to react with other elements can be monitored using this concept.

So let's discuss this in detail.

Equivalent weight

Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine, is known as the equivalent weight of that element.

Example: Find the equivalent weight of Mg for its oxide formation.

From the above equation we can say,

16 grams of oxygen combined with 24 g of magnesium

1 gram of oxygen combine with 2416 g of magnesium

8 gram of oxygen combine with 2416×8=12 g of magnesium

Equivalent weight of magnesium = 12

For elements; Equivalent weight = Molar massvalency

Similarly

For acid; Equivalent weight = Molar massbasicity

For base; Equivalent weight = Molar massacidity

For reducing agents; Equivalent weight = Molar massno of electrons lost per mole of the molecule

For oxidizing agents; Equivalent weight = Molar massno of electrons gained  per mole of the molecule

In general, Equivalent weight = Molar massn-factor

Calculation of n-factor

Case 1: n-factor of acid = number of H+ ions furnished per molecule of an acid

(number of replacable hydrogen present in an acid is called its basicity)

 Acid Basicity (H2SO4) 2 (dibasic acid) (H3PO4) 3 (tribasic acid) (H3PO2) 1 (monobasic acid) (HNO3) 1 (monobasic acid)

Note: 1. H2SO4→2H++SO42-                                      n-factor = 2

2. H2SO4H++HSO4-                                    n-factor = 1

Case 2: n-factor of base = number of OH- ions furnisher per molecules of base

(number of replaceable hydroxide ion present in an acid is called its acidity)

 Base Acidity NaOH 1 (mono-acidic base) Ca(OH)2 2 (di-acidic base) Al(OH)3 3 (tri-acidic base)

Calculation of n-factor for redox reactions

n-factor of reductants and oxidants are calculated on the basis of their change in oxidation state.

Case 1: When only one atom is undergoing a change in oxidation number either by reduction or oxidation.

Example: Oxidation: C2O42-CO2

 Element Initial O.S Final O.S C +3 +4 O -2 -2

We can see, only oxidation state of carbon is changes from +3 to +4

n-factor of C2O42-=2×|4-3|=2

(Note: per molecule 2 Carbon atomss are present in C2O42- ion)

Example: Reduction: Cr2O72-Cr3+ (in acidic medium)

 Element Initial O.S Final O.S Cr +6 +3 O -2 -2

We can see, only oxidation state of chromium is changes from +6 to +3

n-factor of Cr2O72-=2×|6-3|=6

(Note: Per molecule, 2 chromium atom is present in Cr2O72- ion

Case 2: If an atom from a compound oxidised or reduced but appears in two product in same oxidation state  as a result of either only oxidation or only reduction then same rule is applicable

Example: Reduction: MnO4-Mn2++Mn2+ (in acidic medium)

 Element Initial O.S Final O.S Mn +7 +2 & +2 O -2 -2

We can see, only oxidation state of chromium is changes from +6 to +3

n-factor of MnO4-=1×|7-2|=5

Case 3: If an atom from a compound oxidised or reduced and appears in two different product in two different oxidation state as a result of either only oxidation or only reduction.

Example: Reduction: MnO4-Mn2++MnO2

 Element Initial O.S Final O.S Mn +7 +2 for Mn2+ & +4 in MnO2 O -2 -2

We can calculate n-factor only if we know the stoichiometric coefficients of Mn, Mn2+ and MnO2 in the simplest balanced chemical equation.

If a balanced equation is represented like this: xMnO4-yMn2++zMnO2

n-factor of MnO4-={y * |7-2|} + {z * |7-4|}x

where, x=y+z

Case 4: When one atom from a single molecule oxidised or reduced and appear in two different compounds, one is same as initial oxidation state and one in different oxidation state as a result of either only oxidation or only reduction.

Example:Cr2O72-+Cl-Cl-+CrCl3+Cl2

 Element Initial O.S Final O.S Cl -1 -1 & 0 Cr +6 +3

n-factor of Cr2O72-=2×|6-3|=6

Method 1:  Cr2O72-+14Cl-→2Cl-+2CrCl3+3Cl2+7H2O

Apply the law of equivalence

Number of the equivalent of  Cr2O72- = No of the equivalent of Cl-

Moles of Cr2O72-×n-factor of  Cr2O72-=Moles of Cl-×n-factor of Cl-

1×6=14×n-factor of Cl-

n-factor of Cl-=  614=37

Method 2: Balance out the chemical reaction in the simplest stochiometric coefficients Cr2O72-+14Cl-→2Cl-+2CrCl3+3Cl2+7H2O

Out of 14 Cl-only 6 Clhave undergone a change in oxidation state i.e 14Cl- →3Cl2

n-factor of Cl-= 8×|-1-(-1)|+6×|0-(-1)|14=37

Case 5: When two or more atom both are either oxidised or reduced as a result of either only oxidation or only reduction.

FeC2O4Fe3++CO2

 Element Initial O.S Final O.S Fe +2 +3 C +3 +4

n-factor of FeC2O4= 1×|3-2|+  2×|4-3|=3

Case 6: when one atom from a compound is oxidised and another atom is reduced as a result of either only oxidation or only reduction.

KClO3KCl+O2

 Element Initial O.S Final O.S Cl +5 -1 O -2 0

n-factor for oxidation = 3×|0-(-2)|=6

n-factor for reduction = 1×|-1-5|=6

You can consider n-factor by oxidation or reduction.

Case 7: Disproportionation reaction: Single-atom from a compound or molecule is reduced as well as oxidised

Type 1. When the number of electrons lost or gained are the same.

H2O2H2O+12O2

 Element Initial O.S Final O.S O -1 -2 in  H2O & 0 in  O2

Reduction: H2O2H2O

n-factor for oxidation = 2×|-2-(-1)|=2

Oxidation: H2O212O2

n-factor for oxidation = 2×|0-(-1)|=2

Type 2. When the number of electrons lost or gained are different.

12OH-+6Cl2→10Cl-+2ClO3-+6H2O

 Element Initial O.S Final O.S Cl 0 -1 in Cl- & +5 in  ClO3-

Out of 12 Cl (from reactant) 10 Cl acquire -1 charge (Cl-) and 2 Cl acquire +5 charge (Cl-)

Number of electron loss in oxidation = 2×|5-0|=10

Number of electron loss in oxidation = 10×|-1-0|=10

Total change in electron for 6 Cl2 or 12 Cl =10

So, change per molecule of Cl2=106=53