Balance K₂Cr₂O₇ + HCl Using Oxidation Number Method | Redox Reaction | JEE Chemistry Doubtify

 

🔥 Balance Redox Reaction Using Oxidation Number Method | JEE Chemistry | Doubtify

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⚛️ Question :

Balance the following equation by the oxidation number method:
K₂Cr₂O₇ + HCl → KCl + CrCl₃ + H₂O + Cl₂

🖼️ Question Image:

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📘 Solution:

This is a redox reaction where potassium dichromate reacts with hydrochloric acid to produce potassium chloride, chromium(III) chloride, water, and chlorine gas.

Let’s balance it step-by-step using the oxidation number method:

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🔬 Step 1: Assign oxidation numbers

Reactants:

• K₂Cr₂O₇: Cr has oxidation number +6

• HCl: Cl has oxidation number –1

Products:

• CrCl₃: Cr is +3

• Cl₂: Cl is 0

So here’s what’s happening:

• Cr is getting reduced: +6 → +3 (Reduction)

• Cl is getting oxidized: –1 → 0 (Oxidation)

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🔁 Step 2: Write the changes in oxidation numbers

• Chromium (Cr): 2 atoms × change of 3 = total change = 6

• Chlorine (Cl): x atoms × change of 1 = total change = x

So to balance the changes:

6 electrons lost = 6 electrons gained
⇒ x = 6 Cl⁻ ions oxidized

So we need 6 Cl⁻ to be oxidized, which means 3 Cl₂ molecules will be formed (since each Cl₂ has 2 atoms).

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🧪 Step 3: Write the partial balanced redox equation

Unbalanced skeletal:
K₂Cr₂O₇ + HCl → KCl + CrCl₃ + H₂O + Cl₂

Now, we use the changes in oxidation number to balance the atoms.

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⚖️ Step 4: Balance atoms one by one

1. Cr atoms:
   K₂Cr₂O₇ → 2 CrCl₃
   (Already 2 Cr atoms on both sides)

2. Cl atoms:
   From KCl, CrCl₃, and Cl₂

   • 2 KCl → 2 Cl

   • 2 CrCl₃ → 6 Cl

   • 3 Cl₂ → 6 Cl
     ⇒ Total = 2 + 6 + 6 = 14 Cl
     So, we need 14 HCl on reactant side.

3. K atoms:
   2 KCl ⇒ 2 K → So we need 1 K₂Cr₂O₇

4. H and O atoms:
   From 14 HCl → 14 H
   7 H₂O → 14 H and 7 O
   ⇒ Balance with 7 H₂O

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✅ Final Balanced Equation:

K₂Cr₂O₇ + 14 HCl → 2 KCl + 2 CrCl₃ + 7 H₂O + 3 Cl₂

🧠 Solution Image:

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🎯 Why This Question is Important

• It's a classic example of balancing redox reactions using oxidation number method, a very important topic in JEE Mains and Advanced.

• Helps build strong conceptual clarity on oxidation, reduction, and electron transfer.

• Strengthens understanding of stoichiometry, valency, and chemical reaction mechanisms.

• Often seen in both CBSE board exams and competitive entrance tests.

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🧠 Pro Tip:

Always identify elements undergoing oxidation and reduction first. Match total increase and decrease in oxidation numbers to keep electron balance perfect.

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🎥 Video Solution:

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