Faraday Law and Nernst Equation Combined Concept

 

❓ Concept Question

How do we calculate EMF after:

👉 Electrolysis changes ion concentrations
👉 Two half-cells are connected to form an electrochemical cell?

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🖼 Concept Image

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✍️ Short Concept

This is a combination of:

• Faraday’s laws of electrolysis
• Concentration change
• Nernst equation

Main idea:

👉 First update concentrations after electrolysis
👉 Then calculate EMF.

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🔷 Step 1 — Faraday Rule 💯

$$1 \text{ Faraday} = 1 \text{ mole electrons}$$

Reactions:

$$Ag^+ + e^- \rightarrow Ag$$
$$Cu^{2+} + 2e^- \rightarrow Cu$$

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🔷 Step 2 — $Ag^+$ Cell (0.1 F Passed)

Electrons passed:

$$0.1 \text{ mole } e^-$$

Since:

$$Ag^+ + e^- \rightarrow Ag$$

Ag⁺ consumed:

$$0.1 \text{ mole}$$

Initial moles:

$$0.2M \times 1L = 0.2$$

Remaining:

$$0.2 - 0.1 = 0.1$$

New concentration:

$$[Ag^+] = 0.1M$$

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🔷 Step 3 — $Cu^{2+}$ Cell (1 F Passed)

Electrons passed:

$$1 \text{ mole } e^-$$

Reaction:

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

Cu²⁺ consumed:

$$0.5 \text{ mole}$$

Initial moles:

$$1.5M \times 1L = 1.5$$

Remaining:

$$1.5 - 0.5 = 1$$

New concentration:

$$[Cu^{2+}] = 1M$$

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🔷 Step 4 — Cell Reaction

Ag has higher reduction potential.

So:

Cathode:

$$Ag^+ + e^- \rightarrow Ag$$

Anode:

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

Net reaction:

$$Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$$

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🔷 Step 5 — Final EMF

Standard EMF:

$$E^\circ_{cell} = 0.80 - 0.34$$
$$= 0.46V$$

Using Nernst equation:

$$E = E^\circ_{cell} - \frac{0.06}{n}\log Q$$

Final value:

$$\boxed{0.40V}$$

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✅ Final Takeaway

Electrochemistry combo problems follow:

$$\text{Faraday Law} \rightarrow \text{New Concentration} \rightarrow \text{Nernst Equation}$$

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⭐ Golden JEE Insight

Most students make mistakes in:

❌ Mole calculation after electrolysis

❌ Forgetting electron stoichiometry

Always use:

$$\text{Moles reacted} = \frac{\text{Faraday passed}}{n}$$