Nernst Equation with Faraday Electrolysis Concept

 

❓ Question

1 faraday electricity was passed through:

• $Cu^{2+}(1.5M, 1L)/Cu$
  
• $Ag^+(0.2M, 1L)/Ag$
  

electrolytic cells.

After electrolysis, both cells are connected to form an electrochemical cell.

Find the EMF of the cell at 298 K.

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🖼 Question Image

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✍️ Short Concept

This question combines:

👉 Electrolysis
👉 Change in concentration
👉 Nernst equation

First update concentrations after electrolysis, then calculate EMF.

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🔷 Step 1 — New Concentration of $Cu^{2+}$💯

Reaction:

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

1 faraday deposits:

$$\frac{1}{2} \text{ mole } Cu$$

Initial moles:

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

Final moles:

$$1.5 - 0.5 = 1$$

So:

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

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🔷 Step 2 — New Concentration of $Ag^+$

Reaction:

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

0.1 faraday deposits:

$$0.1 \text{ mole } Ag$$

Initial moles:

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

Final moles:

$$0.2 - 0.1 = 0.1$$

So:

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

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🔷 Step 3 — Standard EMF

Cathode:

$$Ag^+ / Ag$$

Anode:

$$Cu^{2+} / Cu$$
$$E^\circ_{cell} = 0.80 - 0.34$$
$$= 0.46V$$

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🔷 Step 4 — Net Cell Reaction

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

Number of electrons:

$$n = 2$$

Reaction quotient:

$$Q = \frac{[Cu^{2+}]}{[Ag^+]^2}$$
$$= \frac{1}{(0.1)^2} = 100$$

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🔷 Step 5 — Apply Nernst Equation

$$E_{cell} = E^\circ_{cell} - \frac{0.06}{n}\log Q$$
$$= 0.46 - \frac{0.06}{2}\log(100)$$
$$= 0.46 - 0.03 \times 2$$
$$= 0.40V$$

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✅ Final Answer

$$\boxed{0.40 \text{ V}}$$

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⭐ Golden JEE Insight

Electrochemistry combo questions usually follow:

$$\text{Electrolysis}\to \text{New concentration}\to \text{Nernst equation}$$

Most mistakes happen in updating concentrations.