Equivalent weight of H2SO4 in the reaction: 2KMnO4+3H2SO4 + 10HCl→ 2MnSO4 + K2SO4 +5Cl2+8H2O, is (a)M/2 (b) M (c)3M/10 (d)3M/5

1. Concept of Equivalent Weight in Redox Medium

For any species,
$\text{Equivalent weight} = \frac{\text{Molar mass (M)}}{\text{n-factor}}$

• n-factor in redox = total electrons transferred per molecule of that substance as represented in the balanced reaction.
• If the substance itself is not oxidised or reduced (like $H_2SO_4$ here) we look at the reaction stoichiometry: "How many electrons flow while 1 mole of this substance is used?"

2. Electron Accounting for the Given Reaction

$2\,KMnO_4 + 3\,H_2SO_4 + 10\,HCl \;\longrightarrow\; 2\,MnSO_4 + K_2SO_4 + 5\,Cl_2 + 8\,H_2O$

Step-wise oxidation–reduction changes:

1. Manganese:
   $\text{Mn}^{+7}$ in $KMnO_4 \;\rightarrow\; \text{Mn}^{+2}$ in $MnSO_4$
   Electrons gained per Mn: $7-2 = 5$
   For 2 Mn: $5 \times 2 = 10$ electrons gained.

2. Chlorine:
   $\text{Cl}^{-1}$ in $HCl \;\rightarrow\; \text{Cl}_2^0$
   Electrons lost per Cl atom: $(-1) - 0 = -1$ (i.e. 1 e⁻ lost)
   Each $Cl_2$ uses 2 Cl atoms ⇒ 2 e⁻ lost per $Cl_2$.
   For 5 $Cl_2$: $2 \times 5 = 10$ electrons lost.

The 10 e⁻ lost = 10 e⁻ gained → balanced.

3. Relating $H_2SO_4$ to the Electron Transfer

• 3 moles of $H_2SO_4$ are present when 10 electrons move.
• Therefore, electrons per mole of $H_2SO_4$:

$n = \frac{10\;\text{e⁻}}{3\;\text{mol}} = \frac{10}{3}\;\text{e⁻ mol}^{-1}$

4. Calculating Equivalent Weight

$\text{Equivalent weight} = \frac{M}{n} = \frac{M}{\tfrac{10}{3}} = \frac{3M}{10}$

Hence, the correct option is (c) 3M⁄10.

Imagine this like sharing sweets at a party 🍬

1. Big Picture – In this chemical "party" we have three main guests:

   • Purple KMnO₄ (a strong oxidising agent)
   • Colourless HCl (which will lose electrons and turn into greenish Cl₂ gas)
   • Strong acid H₂SO₄ (it just makes the place acidic and supplies sulphate ions – it doesn’t change colour itself).

2. Passing electrons = exchanging sweets – Whenever something gains or loses electrons we count how many "sweets" (electrons) move around.

3. How many sweets move?

   • 2 molecules of KMnO₄ each need 5 sweets (electrons) to become Mn²⁺. That’s 10 sweets gained.
   • 10 molecules of HCl together give away 10 sweets lost when turning into Cl₂.

4. What about H₂SO₄?

   • H₂SO₄ does not give or take sweets. It only stands there and lets the others trade.
   • But in the final equation we can see that 3 molecules of H₂SO₄ are present while 10 sweets are being traded.

5. Equivalent idea – Ask: "How much H₂SO₄ corresponds to 1 sweet (1 electron) traded?"

   • If 3 molecules correspond to 10 sweets, then 1 molecule corresponds to 10⁄3 sweets.

6. So its n-factor (sweets per molecule) is 10⁄3 and the equivalent weight (weight for 1 sweet) becomes 3M⁄10 (where M is the molar mass of H₂SO₄).

👉 Option (c) 3M⁄10 is correct.

Step-by-Step Solution

1. Write the balanced equation

$2\,KMnO_4 + 3\,H_2SO_4 + 10\,HCl \;\longrightarrow\; 2\,MnSO_4 + K_2SO_4 + 5\,Cl_2 + 8\,H_2O$

2. Calculate total electrons transferred

• Mn change: $+7 \rightarrow +2$ ⇒ $5$ e⁻ gained per Mn.

• For $2$ Mn: $2 \times 5 = 10$ e⁻ gained.

• Cl change: $-1 \rightarrow 0$ ⇒ $1$ e⁻ lost per Cl.

• $5\,Cl_2$ uses $10$ Cl atoms ⇒ $10$ e⁻ lost.

Electrons gained = electrons lost = 10. ✔️

3. Determine n-factor of $H_2SO_4$

• 3 moles of $H_2SO_4$ accompany the transfer of 10 e⁻.

• Electrons per mole of $H_2SO_4$:

$n = \frac{10}{3}$

4. Equivalent weight

$\text{Equivalent weight} = \frac{\text{Molar mass, }M}{n} = \frac{M}{\tfrac{10}{3}} = \frac{3M}{10}$

5. Final answer

Option (c) $\displaystyle\frac{3M}{10}$