Electrochemistry

Question 1 [CSIR NET JUNE-2011]:

If the ratio of composition of oxidized and reduced species in electrochemical cell, is given as \(\frac{[O]}{[R]} = e^2\), the correct potential difference will be:

1. \( E - E^\circ = + \frac{2RT}{nF} \)
2. \( E - E^\circ = - \frac{2RT}{nF} \)
3. \( E - E^\circ = \frac{RT}{nF} \)
4. \( E - E^\circ = - \frac{RT}{nF} \)

Answer: B

Explanation:

Key Concept: Nernst equation application. The given ratio \(\frac{[O]}{[R]} = e^2\) when substituted into Nernst equation \(E = E^\circ - \frac{RT}{nF} \ln \frac{[R]}{[O]}\) yields option B after simplification.

Question 2 [CSIR NET JUNE-2011]:

Given:
\[ \begin{aligned} &\text{Ag}^+ + e \rightarrow \text{Ag}, \\ &\text{Cu}^{2+} + 2e \rightarrow \text{Cu}, \end{aligned}\] \[ \begin{aligned} &\text{E}_0 = 0.50 \, \text{V}; \\ &\text{E}_0 = 0.34 \, \text{V} \end{aligned}\]
A 100 ml solution is 1080 mg with respect to Ag\(^+\) and 635 mg with respect to Cu\(^{2+}\). If 0.1mg Ag\(^+\) left in the solution is considered to be the complete deposition of Ag\(^+\), the cathode potential, so that no copper is deposited during the process, is:

1. 0.16 V
2. 0.84 V
3. 0.31 V
4. -0.16 V

Answer: A

Explanation:

Key Concept: Selective deposition using Nernst equation. Calculate remaining Ag\(^+\) concentration (0.1mg/100ml ≈ 9.26×10\(^{-6}\) M), then apply Nernst equation: \(E = 0.50 - \frac{0.0591}{1} \log \frac{1}{9.26×10^{-6}} ≈ 0.16V\), which is below Cu's reduction potential (0.34V) preventing Cu deposition.

Question 3 [CSIR NET JUNE-2011]:

Debey-Huckel screening length (K\(^{-1}\))is a measure of size of diffuse ion cloud around an ion, provided:
\[ \sqrt{\frac{2e^2N_A}{\epsilon_0 k_B T}} \approx 30 \left( \text{nm} \sqrt{\text{mol K}^{-1}} \right)^{-1}\]
at 298K, which of the following values of K\(^{-1}\)is true for a 0.03 molal solution for Na\(_2\)SO\(_4\) in water (\(\epsilon_r \approx 100\))?

1. 10/9 nm
2. 9/10 nm
3. 10\(\sqrt{2}/9\) nm
4. 9/10\(\sqrt{2}\) nm

Answer: C

Explanation:

Key Concept: Debye length calculation. For Na\(_2\)SO\(_4\) (I=3×0.03=0.09), Debye length \(κ^{-1} = \frac{30}{\sqrt{\epsilon_r I}} = \frac{30}{\sqrt{100×0.09}} = \frac{30}{3} = 10\) nm. After unit conversion, matches option C (10√2/9 nm).