JEE Chemistry 2025 Question with Solution

Given: Standard reduction potentials are $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V}$ and $E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \, \text{V}$. The solution states that after forming the cell, the emf is obtained using the Nernst equation and gives the final value $0.40 \, \text{V}$.

Find: The emf of the electrochemical cell at $298 \, \text{K}$.

Using the standard potentials:

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

the solution then applies the Nernst equation in the form:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$

It uses:

$$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]}$$

and substitutes:

$$E_{\text{cell}} = 0.46 \, \text{V} - 0.0591 \log \left(\frac{1.5}{0.2}\right)$$

Now,

$$\log \left(\frac{1.5}{0.2}\right) = \log(7.5) \approx 0.875$$

So,

$$E_{\text{cell}} = 0.46 - 0.0591 \times 0.875$$ $$E_{\text{cell}} = 0.46 - 0.0518 = 0.4082 \, \text{V}$$

Therefore, the emf of the cell is $0.40 \, \text{V}$. Hence the numerical answer is 0.4.

Note: The solution does not consistently show the concentration changes after electrolysis, but it explicitly concludes the final emf as $0.40 \, \text{V}$, which is taken as the answer.

Given: the solution provides the standard potentials and a stepwise Nernst-equation approach.

Find: The final emf after the two electrolytic cells are connected.

From the extracted working:

1. Standard cell potential is written as

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

2. Nernst equation is used:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$

3. The extracted solution takes

$$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]}$$

4. Substitution shown in the working is

$$E_{\text{cell}} = 0.46 \, \text{V} - 0.0591 \log \left(\frac{1.5}{0.2}\right)$$

5. Then

$$\log(7.5) \approx 0.875$$

6. Hence

$$E_{\text{cell}} = 0.46 - 0.0591 \times 0.875 = 0.4082 \, \text{V}$$

So the emf is approximately $0.40 \, \text{V}$.

Therefore, the required numerical value is 0.4.