Balance Redox Reaction Using Oxidation Number Method

❓ Question

Balance the following equation:

$$K_2Cr_2O_7 + HCl \rightarrow KCl + CrCl_3 + H_2O + Cl_2$$

using ion-electron method.

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🖼 Question Image

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✍️ Short Explanation

This is a redox balancing question based on:

👉 Oxidation number change
👉 Ion-electron method
👉 Acidic medium balancing.

Main idea:

$$Cr_2O_7^{2-}$$

acts as oxidizing agent and

$$Cl^-$$

gets oxidized to

$$Cl_2$$

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🔷 Step 1 — Identify Oxidation & Reduction 💯

In:

$$K_2Cr_2O_7$$

Chromium oxidation state:

$$Cr=+6$$

In:

$$CrCl_3$$

Chromium becomes:

$$Cr=+3$$

So chromium is reduced.

Now chlorine:

$$Cl^- \rightarrow Cl_2$$

Oxidation state:

$$-1 \rightarrow 0$$

So chloride is oxidized.

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🔷 Step 2 — Reduction Half Reaction

Dichromate reduction in acidic medium:

$$Cr_2O_7^{2-}+14H^+ +6e^- \rightarrow 2Cr^{3+}+7H_2O$$

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🔷 Step 3 — Oxidation Half Reaction

Chloride oxidation:

$$2Cl^- \rightarrow Cl_2+2e^-$$

Multiply by 3 to equalize electrons:

$$6Cl^- \rightarrow 3Cl_2+6e^-$$

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🔷 Step 4 — Add Half Reactions

Adding both reactions:

$$Cr_2O_7^{2-}+14H^+ +6Cl^- \rightarrow 2Cr^{3+}+7H_2O+3Cl_2$$

Now convert ions into molecular form using:

$$K^+ \text{ and } Cl^-$$

from HCl.

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🔷 Step 5 — Final Balanced Equation

$$\boxed{ K_2Cr_2O_7+14HCl \rightarrow 2KCl+2CrCl_3+7H_2O+3Cl_2 }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Oxygen directly balance kar dena

❌ Electron balancing skip kar dena

❌ Acidic medium mein:

$$H^+, H_2O$$

use karna bhool jaana

Remember:

Ion-electron method:

$$\text{Oxidation + Reduction separately}$$

always easiest approach.

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✅ Final Answer

$$\boxed{ K_2Cr_2O_7+14HCl \rightarrow 2KCl+2CrCl_3+7H_2O+3Cl_2 }$$

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📚 Related Topics

• First Order Reaction Rate Constant Using Pressure
• First Order Reaction Using Pressure Data
• First Order Reaction via Pressure Data