Advance Stoichiometry Based on Redox Reactions

ChemistryRedox ReactionsFor NEET aspirants

Redox stoichiometry turns a redox reaction into numbers: how much oxidant reacts with how much reductant, and what volume of titrant reaches the end point. It has two routes, the mole method (from the balanced equation) and the equivalent method (from n-factors, with no balancing). This page builds n-factor, equivalent weight, normality and the law of equivalence, then applies them to permanganate and dichromate titrations, iodometry, back titration and the volume strength of $H_{2} O_{2}$ . Redox stoichiometry numericals appear every year in JEE Main and as one-step problems in NEET.

On this page1Two routes2n-factor3Equivalents and normality4Law of equivalence5Redox titrations6Iodometry and iodimetry7Back titration8Volume strength of

H_{2} O_{2}

Key Formulas - Quick Reference

★ Must learnn-factor (redox) $=$ electrons lost or gained per formula unit $=$ (change in ON) $\times$ (atoms that change); add if two elements change.
★ Must learnEquivalent weight $E = \frac{M}{n}$ ; gram equivalents $= \frac{w}{E} =$ moles $\times n$ .
★ Must learnNormality $N = n \times M$ (molarity); milliequivalents $=$ $N \times V$ (mL).
★ Must learnLaw of equivalence: meq of oxidant $=$ meq of reductant, so $N_{1} V_{1} = N_{2} V_{2}$ , or $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$ .
Disproportionation: $n = \frac{n _{1} n _{2}}{n _{1} + n _{2}}$ , with $n_{1}$ , $n_{2}$ the electrons per mole in the two directions.
Iodometry: eq of oxidant $=$ eq of $I_{2}$ liberated $=$ eq of $N a_{2} S_{2} O_{3}$ ( $n = 1$ for $S_{2} O_{3}^{2 -}$ → $S_{4} O_{6}^{2 -}$ ).
Back titration: meq(sample) $=$ meq(excess reagent added) $-$ meq(back-titrant).
★ Must learnVolume strength of $H_{2} O_{2}$ $= 11.2 \times M = 5.6 \times N$ (litres of $O_{2}$ at STP from 1 L).

1. Two Routes to Redox Stoichiometry

Every redox calculation asks the same thing: how many moles of electrons does the reductant give, and how much oxidant is needed to take them? You can answer it with the mole method (balance, then use the mole ratio) or the equivalent method (count electrons per formula unit and equate equivalents). Both always agree.

Figure 1: The mole method needs the balanced equation; the equivalent method needs only the n-factors (5 for

KMn O_{4}

, 1 for

F e^{2 +}

). Both give 20 mL.

Mole method

Needs the balanced equation. Best when the products matter or one reagent is limiting, as in $4N H_{3}$ + $5 O_{2}$ → $4NO$ + $6 H_{2} O$ .

Equivalent method

Needs only n-factors. Fastest for titrations, mixtures, back titration and multistep problems; no balancing at all.

2. n-Factor

The n-factor of a species in a redox reaction is the number of electrons it loses or gains per formula unit, that is, per mole. Find it as (change in oxidation number per atom)

\times

(number of atoms that change).

The n-factor is a property of the reaction, not of the compound. Permanganate gives three different values in three media, and thiosulphate gives 1 with iodine but 8 with chlorine.

Figure 2: The n-factor belongs to the reaction, not the compound:

KMn O_{4}

is 5, 3 or 1 by medium, and thiosulphate is 1 with

I_{2}

but 8 with

C l_{2}

2.1 Two elements changing, and disproportionation

If two parts of a formula change, add their electrons: ferrous oxalate loses 1 from Fe and 2 from oxalate, so $n = 3$ . For a species that both gains and loses electrons, combine the two directions: $n = \frac{n _{1} n _{2}}{n _{1} + n _{2}}$ .

Figure 3: When two parts of one formula change, add their electrons (

Fe C_{2} O_{4}

n = 3

). For a disproportionating species,

n = \frac{n _{1} n _{2}}{n _{1} + n _{2}}

(

C l_{2}

\frac{5}{3}

n-factor in acid-base and salt reactions (used with the same equivalent rules): acids, the number of

H^{+}

replaced (basicity:

H_{2} S O_{4}

H_{3} P O_{4}

H_{3} P O_{3}

H_{3} P O_{2}

1); bases, the

O H^{-}

replaced (

Ca (OH)_{2}

2); salts, the total charge on the cations (

A l_{2} (S O_{4})_{3}

6).

Exam Trick 5-3-1, then 6.

KMn O_{4}

takes 5 electrons in acid, 3 in neutral, 1 in strong alkali;

K_{2} C r_{2} O_{7}

takes 6. Thiosulphate gives just 1 with iodine. These five numbers solve most titration questions.

JEE Advanced One compound, two n-factors. Oxalic acid is 2 as an acid and 2 as a reductant.

KH C_{2} O_{4}

is 1 as an acid but 2 as a reductant. Potassium tetroxalate,

KH C_{2} O_{4} \cdot H_{2} C_{2} O_{4} \cdot 2 H_{2} O

, has 3 replaceable H (n = 3 with NaOH) and two oxalate units (n = 4 with

KMn O_{4}

). So equal moles of it need volumes of 0.1 M NaOH and 0.1 M

KMn O_{4}

in the ratio

3 : \frac{4}{5}

. Always ask which reaction the question is about.

Key idea

n-factor = electrons per formula unit for this reaction. Change the medium or partner and the n-factor can change.

3. Equivalents, Equivalent Weight and Normality

Equivalent weight

E = M / n

is the mass that gives or takes one mole of electrons. One equivalent of any oxidant reacts exactly with one equivalent of any reductant. Normality

N

is the number of gram equivalents per litre:

N = n \times

molarity.

Figure 4: Every quantity in a redox problem turns into equivalents, and one equivalent means one mole of electrons transferred. Equal equivalents of oxidant and reductant always react.

Reagent (reaction)	$M$ / g mol $^{- 1}$	n-factor	$E$ / g
$KMn O_{4}$ (acid, → $M n^{2 +}$ )	158	5	31.6
$KMn O_{4}$ (neutral, → $Mn O_{2}$ )	158	3	52.7
$K_{2} C r_{2} O_{7}$ (acid, → $C r^{3 +}$ )	294	6	49
$FeS O_{4} \cdot 7 H_{2} O$ (→ $F e^{3 +}$ )	278	1	278
Mohr's salt, $FeS O_{4} \cdot (N H_{4})_{2} S O_{4} \cdot 6 H_{2} O$	392	1	392
Oxalic acid, $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ (→ $C O_{2}$ )	126	2	63
$H_{2} O_{2}$ (either role)	34	2	17
$I_{2}$ (→ $I^{-}$ )	254	2	127
$N a_{2} S_{2} O_{3} \cdot 5 H_{2} O$ (with $I_{2}$ )	248	1	248

4. The Law of Equivalence

In any reaction the equivalents of all reacting species are equal. For a titration, meq of oxidant

=

meq of reductant.

N_{1} V_{1} = N_{2} V_{2} or n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}

The law needs no balanced equation, works for unbalanced and multistep changes, and simply adds up for mixtures: meq of a mixture is the sum of the meq of its parts. If 20 mL of 0.05 M oxalic acid needs 20 mL of $KMn O_{4}$ in acid, then $5 \times M \times 20 = 2 \times 0.05 \times 20$ and $M = 0.02$ M, the same as from the balanced equation

2Mn O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} 2M n^{2 +} + 10C O_{2} + 8 H_{2} O

Exam Trick Stay in milli-units. meq

= N \times

= n \times M \times

mL, and mmol

= M \times

mL. You never convert to litres, and a mass in grams becomes meq as

1000 w / E

Figure 5: Five moves solve every redox titration. The branch handles back titration; the most common error is a wrong n-factor at the second step.

Key idea

Equal equivalents react, not equal moles. Moles react in the ratio of the inverse n-factors: 1

Mn O_{4}^{-}

(n = 5) per 5

F e^{2 +}

(n = 1).

Quick Recall: tap to check

n-factor of

Fe C_{2} O_{4}

with acidified

KMn O_{4}

3 (Fe 1 + oxalate 2);

E = 144/3 = 48

Normality of 0.1 M

K_{2} C r_{2} O_{7}

in acid?

0.1 \times 6 = 0.6

Equivalent weight of

KMn O_{4}

in neutral medium?

158/3 = 52.7

5. Redox Titrations

Just as acid-base titrations use a pH indicator, redox titrations find the strength of an oxidant or reductant with a redox-sensitive end point. Three situations are common.

5.1 Permanganate: a self-indicator

Intensely purple $Mn O_{4}^{-}$ is decolourised as long as a reductant ( $F e^{2 +}$ or $C_{2} O_{4}^{2 -}$ ) remains, because $M n^{2 +}$ is almost colourless. The first lasting pale pink marks the end point; the eye detects it at only about $1 0^{- 6}$ mol L $^{- 1}$ of $Mn O_{4}^{-}$ , so the overshoot beyond the equivalence point (where oxidant and reductant are equal in mole stoichiometry) is negligible. Oxalate titrations are run warm (about 60 °C) because the reaction is slow in the cold.

Mn O_{4}^{-} + 5F e^{2 +} + 8 H^{+} M n^{2 +} + 5F e^{3 +} + 4 H_{2} O

Figure 6:

KMn O_{4}

is a self-indicator: purple while it is used up, pale pink at the first drop in excess (about

1 0^{- 6}

mol L

^{- 1}

). The medium is dilute

H_{2} S O_{4}

5.2 Dichromate with an indicator

Orange $C r_{2} O_{7}^{2 -}$ gives green $C r^{3 +}$ , a change too weak to see clearly, so it is not a self-indicator. A redox indicator, diphenylamine, is oxidised by the first drop of excess dichromate after the equivalence point and turns an intense blue (blue-violet).

C r_{2} O_{7}^{2 -} + 6F e^{2 +} + 14 H^{+} 2C r^{3 +} + 6F e^{3 +} + 7 H_{2} O

5.3 Choosing the acid

The acid must not react with the titrant. Dilute $H_{2} S O_{4}$ is used for permanganate. HCl is avoided because $Mn O_{4}^{-}$ ( $E^{\circ} = 1.51$ V) oxidises $C l^{-}$ ( $1.36$ V) to $C l_{2}$ , wasting titrant; dichromate ( $1.33$ V) does not, so it can be used in HCl. $HN O_{3}$ is avoided with both because it is an oxidant itself.

Figure 7: An acid must not react with the titrant.

Mn O_{4}^{-}

(1.51 V) oxidises

C l^{-}

(1.36 V), so HCl wastes permanganate;

C r_{2} O_{7}^{2 -}

(1.33 V) does not.

6. Iodometry and Iodimetry

Iodine reactions give a third way to see the end point. Iodine forms an intense blue complex with starch, and it reacts quickly and cleanly with thiosulphate. Iodine is barely soluble in water, but it stays dissolved in KI solution as $K I_{3}$ ( $I_{3}^{-}$ ).

I_{2} + 2 S_{2} O_{3}^{2 -} 2 I^{-} + S_{4} O_{6}^{2 -}

In iodometry an oxidant that can oxidise $I^{-}$ , such as $C u^{2 +}$ , is treated with excess KI. The liberated iodine is titrated with standard thiosulphate. Starch is added near the end (when the solution turns pale yellow), and the blue colour vanishes when the last iodine is used up.

2C u^{2 +} + 4 I^{-} C u_{2} I_{2} + I_{2}

Figure 8: In iodometry iodine is only a messenger: the oxidant, the liberated

I_{2}

and the thiosulphate have equal equivalents. In iodimetry a standard

I_{2}

solution titrates a reductant directly.

Iodometry (indirect)

Estimates an oxidant. Oxidant + excess KI → $I_{2}$ ; titrate $I_{2}$ with $N a_{2} S_{2} O_{3}$ . End point: blue → colourless.

Iodimetry (direct)

Estimates a reductant with a standard $I_{2}$ solution. End point: the first permanent blue with starch.

Key idea

In iodometry, iodine is only a messenger: eq of oxidant = eq of $I_{2}$ = eq of thiosulphate, with

n = 1

for each

C u^{2 +}

and each

S_{2} O_{3}^{2 -}

Quick Recall: tap to check

Why is

KMn O_{4}

not titrated in HCl?

Mn O_{4}^{-}

(1.51 V) oxidises

C l^{-}

(1.36 V) to

C l_{2}

, so extra titrant is used.

Why is starch added only near the end in iodometry?

At high iodine concentration the starch-iodine complex forms too firmly and the end point becomes sluggish.

Indicator for dichromate titration of

F e^{2 +}

Diphenylamine, which turns intense blue just after the equivalence point.

7. Back Titration

When the sample reacts slowly, is insoluble or cannot be titrated directly (for example $Mn O_{2}$ in pyrolusite ore), add a known excess of a reagent, let it react completely, and titrate the left-over excess.

meq(sample) = meq(reagent added) - meq(back-titrant)

Mn O_{2} + C_{2} O_{4}^{2 -} + 4 H^{+} M n^{2 +} + 2C O_{2} + 2 H_{2} O

Figure 9: In a back titration a known excess reagent is added and the left-over part is titrated. Here

10 - 2 = 8

meq of

Mn O_{2}

, i.e. 4 mmol, since

n = 2

for

Mn O_{2}

8. Volume Strength of Hydrogen Peroxide

$H_{2} O_{2}$ bottles are labelled in volumes. " $x$ volume" means that 1 L of the solution gives $x$ L of $O_{2}$ at STP when it decomposes:

2 H_{2} O_{2} 2 H_{2} O + O_{2}

Two moles of $H_{2} O_{2}$ give one mole of $O_{2}$ , 22.4 L at STP. So 1 L of an $M$ molar solution gives $\frac{M}{2} \times 22.4 = 11.2 M$ litres. With $n = 2$ , $N = 2 M$ :

volume strength = 11.2 \times M = 5.6 \times N, strength (g L^{- 1}) = 34 \times M

Figure 10: "20 volume" means 1 L gives 20 L of

O_{2}

at STP. With

V = 11.2 M = 5.6 N

, it is 1.79 M, 3.57 N and 6.07% (w/v).

$H_{2} O_{2}$ is usually estimated by titration with acidified $KMn O_{4}$ , in which it acts as a reductant ( $n = 2$ ):

2Mn O_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} 2M n^{2 +} + 5 O_{2} + 8 H_{2} O

Exam Trick 11.2 and 5.6. Volume strength

= 11.2 M = 5.6 N

. Quick checks: 10 V ≈ 3%, 20 V ≈ 6%, 100 V ≈ 30% (w/v). Exams use 22.4 L for STP unless they say otherwise; with 22.7 L (1 bar) the factor becomes 11.35.

Key idea

x

volume" =

x

L of

O_{2}

at STP per litre, so $M = x /11.2$ and

N = x /5.6

Quick Recall: tap to check

Molarity of "11.2 volume"

H_{2} O_{2}

1 M (

11.2/11.2

); normality 2 N.

In a back titration, which meq equals the sample?

Total meq of reagent added minus meq of the back-titrant.

n-factor of

H_{2} O_{2}

with acidified

KMn O_{4}

2: O goes from

- 1

to 0 in

O_{2}

8.1 The whole concept at a glance

Figure 11: The whole concept on one page. Revise from the centre outwards.

9. Solved Examples

Solved Example 1

What volume of 0.1 M

KMn O_{4}

(acidic) is needed to oxidise 1.52 g of

FeS O_{4}

(

M = 152

Solution:

meq of $FeS O_{4}$ $= \frac{1000 \times 1.52}{152/1} = 10$ . Normality of $KMn O_{4}$ $= 5 \times 0.1 = 0.5$ N.

$0.5 \times V = 10$ , so $V = 20$ mL. (Mole check: 10 mmol $F e^{2 +}$ need $10/5 = 2$ mmol $Mn O_{4}^{-}$ , i.e. $2/0.1 = 20$ mL.)

Solved Example 2

The equivalent weight of

K_{2} C r_{2} O_{7}

(

M = 294

) in acidic medium is
(A) 294
(B) 147
(C) 49
(D) 98

Solution:

Answer: (C). Each Cr falls from $+ 6$ to $+ 3$ , and there are two Cr: $n = 6$ . $E = 294/6 = 49$ .

Solved Example 3

20 mL of 0.05 M oxalic acid needs 20 mL of a

KMn O_{4}

solution in dilute

H_{2} S O_{4}

. Find the molarity of

KMn O_{4}

Solution:

$n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$ : $5 \times M \times 20 = 2 \times 0.05 \times 20$ , so $M = 0.02$ M (normality 0.1 N).

Solved Example 4

How many grams of Mohr's salt (

M = 392

) react with 25 mL of 0.1 N

K_{2} C r_{2} O_{7}

in acid?

Solution:

meq of dichromate $= 0.1 \times 25 = 2.5$ . Mohr's salt has $n = 1$ ( $F e^{2 +}$ → $F e^{3 +}$ ), so $E = 392$ .

Mass $= 2.5 \times 1 0^{- 3} \times 392 = 0.98$ g.

Solved Example 5

A 0.5 g sample of a copper ore is dissolved. The

C u^{2 +}

liberates iodine from excess KI, and the iodine needs 25 mL of 0.1 M

N a_{2} S_{2} O_{3}

. Find the percentage of copper (

M = 63.5

Solution:

Iodometry: meq $C u^{2 +}$ $=$ meq $I_{2}$ $=$ meq thiosulphate $= 1 \times 0.1 \times 25 = 2.5$ . Copper has $n = 1$ ( $C u^{2 +}$ → Cu(I)), so 2.5 mmol Cu.

Mass of Cu $= 2.5 \times 63.5 = 158.75$ mg; percentage $= \frac{0.15875}{0.5} \times 100 = 31.75%$ .

Solved Example 6

A sample of

H_{2} O_{2}

is labelled "20 volume". Its normality is
(A) 1.79 N
(B) 3.57 N
(C) 5.6 N
(D) 0.89 N

Solution:

Answer: (B). $N = \frac{20}{5.6} = 3.57$ N (molarity $20/11.2 = 1.79$ M; strength $1.79 \times 34 = 60.7$ g L $^{- 1}$ , about 6.1% w/v).

Solved Example 7

25 mL of an

H_{2} O_{2}

solution needs 20 mL of 0.04 M

KMn O_{4}

in acid. Find the molarity and the volume strength of the

H_{2} O_{2}

Solution:

meq $KMn O_{4}$ $= 5 \times 0.04 \times 20 = 4$ = meq $H_{2} O_{2}$ . With $n = 2$ , mmol $H_{2} O_{2}$ $= 2$ , so $M = 2/25 = 0.08$ M.

Volume strength $= 11.2 \times 0.08 = 0.896$ volume.

Solved Example 8

0.5 g of pyrolusite ore is warmed with 50 mL of 0.1 M oxalic acid in dilute

H_{2} S O_{4}

. The excess oxalic acid needs 20 mL of 0.02 M

KMn O_{4}

. Find the percentage of

Mn O_{2}

(

M = 86.9

Solution:

meq oxalic acid added $= 2 \times 0.1 \times 50 = 10$ . meq $KMn O_{4}$ (back) $= 5 \times 0.02 \times 20 = 2$ . So meq $Mn O_{2}$ $= 10 - 2 = 8$ .

$Mn O_{2}$ has $n = 2$ (Mn $+ 4 \to + 2$ ): 4 mmol, mass $= 4 \times 86.9 = 347.8$ mg. Percentage $= \frac{0.3478}{0.5} \times 100 = 69.6%$ .

Solved Example 9

Find the equivalent weight of

C l_{2}

(

M = 71

) in

3C l_{2}

6NaOH

→

5NaCl

NaCl O_{3}

3 H_{2} O

Solution:

Per $C l_{2}$ : to 2 $C l^{-}$ it gains $n_{1} = 2$ ; to 2 $Cl O_{3}^{-}$ it would lose $n_{2} = 10$ .

n = \frac{2 \times 10}{2 + 10} = \frac{5}{3}

Check: 3 $C l_{2}$ transfer 5 electrons, i.e. $\frac{5}{3}$ per $C l_{2}$ .

$E = 71 \div \frac{5}{3} = 42.6$ .

Solved Example 10

1 mmol of potassium tetroxalate,

KH C_{2} O_{4} \cdot H_{2} C_{2} O_{4} \cdot 2 H_{2} O

, is titrated separately with 0.1 M NaOH and with 0.1 M

KMn O_{4}

in acid. Find each volume.

Solution:

As an acid it has 3 replaceable H: meq $= 3$ , so $V_{NaOH} = 3/0.1 = 30$ mL.

As a reductant its two oxalate units give $4 e^{-}$ : meq $= 4$ , and $5 \times 0.1 \times V = 4$ gives $V_{KMn O_{4}} = 8$ mL.

Practice Questions

Show with three examples that a reductant in excess gives a lower oxidation state product and an oxidant in excess gives a higher one.Answer: C + $O_{2}$ : excess C gives CO, excess $O_{2}$ gives $C O_{2}$ . $P_{4}$ + $C l_{2}$ : excess $P_{4}$ gives $PC l_{3}$ , excess $C l_{2}$ gives $PC l_{5}$ . Na + $O_{2}$ : excess Na gives $N a_{2} O$ , excess $O_{2}$ gives $N a_{2} O_{2}$ .
Why does thiosulphate react differently with iodine ( $2 S_{2} O_{3}^{2 -}$ + $I_{2}$ → $S_{4} O_{6}^{2 -}$ + $2 I^{-}$ ) and bromine ( $S_{2} O_{3}^{2 -}$ + $4B r_{2}$ + $5 H_{2} O$ → $2S O_{4}^{2 -}$ + $8B r^{-}$ + $10 H^{+}$ )?Answer: $B r_{2}$ is a stronger oxidant than $I_{2}$ : it takes S to $+ 6$ ( $n = 8$ per $S_{2} O_{3}^{2 -}$ ), while $I_{2}$ only reaches $S_{4} O_{6}^{2 -}$ (S $+ 2.5$ , $n = 1$ ).
In the Ostwald process, $4N H_{3}$ + $5 O_{2}$ → $4NO$ + $6 H_{2} O$ . What is the maximum mass of NO from 10.00 g $N H_{3}$ and 20.00 g $O_{2}$ ?Answer: $N H_{3}$ 0.587 mol needs 0.734 mol $O_{2}$ , but only 0.625 mol is present, so $O_{2}$ is limiting: NO $= 0.625 \times \frac{4}{5} = 0.500$ mol $= 15.0$ g.
Find the equivalent weight of $Fe C_{2} O_{4}$ ( $M = 144$ ) in its titration with acidified $KMn O_{4}$ .Answer: $n = 3$ , $E = 48$ .
What volume of 0.1 N $K_{2} C r_{2} O_{7}$ oxidises 50 mL of 0.15 N $FeS O_{4}$ ?Answer: $0.1 V = 0.15 \times 50$ , $V = 75$ mL.
Find the volume strength of 1.5 N $H_{2} O_{2}$ .Answer: $5.6 \times 1.5 = 8.4$ volume.
What is the normality of 0.1 M $KMn O_{4}$ used in neutral medium, and how many electrons does each $Mn O_{4}^{-}$ take?Answer: 3 electrons ( $Mn O_{2}$ forms), so 0.3 N.

Common Mistakes to Avoid

Watch out

Writing $M_{1} V_{1} = M_{2} V_{2}$ for a redox titration. Use $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$ (or normalities).
Taking $n = 5$ for $KMn O_{4}$ in every medium (it is 3 in neutral and 1 in strongly alkaline solution).
Taking $n = 2$ for thiosulphate with iodine (it is 1 per $S_{2} O_{3}^{2 -}$ ) or $n = 1$ for $H_{2} O_{2}$ (it is 2).
Treating n-factor as a fixed property of a compound: $KH C_{2} O_{4}$ is 1 as an acid but 2 as a reductant.
Titrating permanganate in HCl or $HN O_{3}$ instead of dilute $H_{2} S O_{4}$ .
In a back titration, equating the sample to the back-titrant instead of to (added $-$ back-titrated).
Mixing units: meq $= N \times$ mL, but gram equivalents $= N \times$ L.
Adding starch at the start of an iodometric titration rather than near the end point.

Frequently Asked Questions

What is n-factor in redox reactions?

The n-factor is the number of electrons lost or gained by one formula unit of a species in a given reaction. It equals the change in oxidation number per atom times the number of atoms that change. KMnO4 has n equal to 5 in acid, 3 in neutral and 1 in strongly alkaline medium.

What is the law of equivalence?

The law of equivalence says that in any reaction the equivalents of all reacting species are equal. In a redox titration the milliequivalents of oxidant equal those of reductant, so N1V1 = N2V2, or n1M1V1 = n2M2V2. It works without a balanced equation and for mixtures.

Why is the equivalent weight of KMnO4 different in acidic, neutral and basic media?

Equivalent weight is molar mass divided by n-factor, and the product of permanganate depends on the medium. It gains 5 electrons in acid (Mn2+), 3 in neutral (MnO2) and 1 in strong alkali (MnO4 2-), so its equivalent weight is 31.6, 52.7 or 158.

What is the difference between iodometry and iodimetry?

Iodometry is indirect: an oxidant liberates iodine from excess potassium iodide and the iodine is titrated with sodium thiosulphate. Iodimetry is direct: a standard iodine solution titrates a reductant such as sulphite or thiosulphate. Starch shows the end point in both.

Why is KMnO4 called a self-indicator?

Permanganate is intensely purple, while its reduction product Mn2+ is almost colourless. As long as reductant remains each drop is decolourised; the first drop in excess gives a permanent pale pink at about 0.000001 mol per litre, so no separate indicator is needed.

What does 20 volume hydrogen peroxide mean?

It means one litre of the solution gives 20 litres of oxygen at STP on decomposition. Since 2 moles of H2O2 give 22.4 litres of O2, volume strength equals 11.2 times molarity, so 20 volume H2O2 is 1.79 M, 3.57 N and about 6 percent weight by volume.

Which redox stoichiometry questions are asked in NEET?

NEET asks one-step problems: molarity or normality from titration volumes, equivalent weight of KMnO4 or K2Cr2O7, volume strength of H2O2, and the mole ratio of oxidant to reductant from a balanced equation, plus NCERT facts on KMnO4 as a self-indicator and starch in iodometric titrations.

How do JEE Main and JEE Advanced test redox titrations?

JEE uses n-factors that change with the medium, back titration of ores such as pyrolusite, iodometric estimation of copper, mixtures titrated by two reagents, compounds like tetroxalate with different acid and redox n-factors, disproportionation n-factors and volume strength conversions, often as integer-answer questions.

Previous year questions on Advance Stoichiometry Based on Redox Reactions

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

Ready to master Redox Reactions?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test