Zn and Nitrate Reaction Ion Electron Method

❓ Question

Balance the equation by ion-electron method:

$$Zn + NO_3^- \rightarrow Zn^{2+} + NH_4^+$$

(in basic medium)

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🖼 Question Image

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✍️ Short Explanation

This is a redox balancing problem in basic medium.

👉 Zinc gets oxidised
👉 Nitrate gets reduced
👉 Use ion-electron method carefully.

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🔷 Step 1 — Identify Oxidation and Reduction 💯

Oxidation half reaction:

$$Zn \rightarrow Zn^{2+}$$

Reduction half reaction:

$$NO_3^- \rightarrow NH_4^+$$

Oxidation states of nitrogen:

$$+5 \rightarrow -3$$

So nitrogen gains:

$$8e^-$$

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🔷 Step 2 — Oxidation Half Reaction

Balance charge:

$$Zn \rightarrow Zn^{2+}+2e^-$$

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🔷 Step 3 — Reduction Half Reaction

Start with:

$$NO_3^- \rightarrow NH_4^+$$

Balance oxygen using water:

$$NO_3^- \rightarrow NH_4^+ +3H_2O$$

Balance hydrogen using:

$$10H^+$$
$$10H^+ +NO_3^- \rightarrow NH_4^+ +3H_2O$$

Balance charge using electrons:

$$8e^- +10H^+ +NO_3^- \rightarrow NH_4^+ +3H_2O$$

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🔷 Step 4 — Convert into Basic Medium

Add:

$$10OH^-$$

on both sides:

$$8e^- +10H^+ +10OH^- +NO_3^- \rightarrow NH_4^+ +3H_2O+10OH^-$$

Since:

$$H^+ +OH^- = H_2O$$
$$8e^- +10H_2O +NO_3^- \rightarrow NH_4^+ +3H_2O+10OH^-$$

Cancel water:

$$8e^- +7H_2O +NO_3^- \rightarrow NH_4^+ +10OH^-$$

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🔷 Step 5 — Equalise Electrons

Oxidation reaction:

$$Zn \rightarrow Zn^{2+}+2e^-$$

Multiply by 4:

$$4Zn \rightarrow 4Zn^{2+}+8e^-$$

Add both half reactions:

$$4Zn +7H_2O +NO_3^- \rightarrow 4Zn^{2+}+NH_4^+ +10OH^-$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Acidic medium pe reaction stop kar dena

❌ $OH^-$ add karna bhool jaana

❌ Nitrogen oxidation state galat nikal dena

❌ Electron balance equal na karna

Remember:

$$\text{Basic medium} \Rightarrow H^+ \text{ must be removed using } OH^-$$

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✅ Final Answer

$$\boxed{ 4Zn +7H_2O +NO_3^- \rightarrow 4Zn^{2+}+NH_4^+ +10OH^- }$$

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📚 Related Topics

• First Order Reaction Rate Constant Using Pressure
• Quantum Numbers and Orbitals Statement Analysis
• Inert Pair Effect in Group 13 Elements