Electrochemistry — NEET UG practice

1. Concept: For a strong electrolyte, the Debye-Hückel-Onsager equation describes how molar conductivity varies with concentration:

$$\Lambda_{\mathrm{m}} = \Lambda_{\mathrm{m}}^{\circ} - A\sqrt{c}$$

where $\Lambda_{\mathrm{m}}^{\circ}$ is the limiting molar conductivity and $A$ is the given slope (with a negative sign already built into the equation).

2. Finding $\sqrt{c}$: since $c = 0.01\ \mathrm{M}$,

$$\sqrt{c} = \sqrt{0.01} = 0.1$$

3. Substituting the known values $\Lambda_{\mathrm{m}} = 145.0$ and slope $= -90.0$:

$$145.0 = \Lambda_{\mathrm{m}}^{\circ} - 90.0 \times 0.1$$ $$\Lambda_{\mathrm{m}}^{\circ} = 145.0 + 9.0 = 154.0\ \mathrm{S\ cm^2\ mol^{-1}}$$

4. Using Kohlrausch's law of independent migration of ions: the limiting molar conductivity of a salt is the sum of the limiting ionic conductivities of its ions.

$$\lambda_{\mathrm{m(XY)}}^{\circ} = \lambda_{\mathrm{X}^{+}}^{\circ} + \lambda_{\mathrm{Y}^{-}}^{\circ}$$ $$154.0 = 74.0 + \lambda_{\mathrm{Y}^{-}}^{\circ}$$ $$\lambda_{\mathrm{Y}^{-}}^{\circ} = 80.0\ \mathrm{S\ cm^2\ mol^{-1}}$$

Hence, the answer is Option B: 80.0.

1. Understanding the setup: When $\mathrm{CuSO}_4$ is electrolysed, $\mathrm{Cu}^{2+}$ ions move to the cathode and gain electrons to form copper metal:

$$\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu(s)}$$

Since 2 electrons are needed per copper atom, $n = 2$.

2. Applying Faraday's law: the mass deposited depends on the molar mass, the current, and the time, according to:

$$W = \frac{M \times I \times t}{n \times F}$$

3. Converting time to seconds: $t = 10 \times 60 = 600\ \mathrm{s}$, and using $I = 1.5\ \mathrm{A}$, $M = 63\ \mathrm{g/mol}$:

$$W = \frac{63 \times 1.5 \times 600}{2 \times 96487}$$

4. Calculating the result:

$$W = \frac{56700}{192974} \approx 0.2938\ \mathrm{g}$$

Hence, the answer is Option B: 0.2938 g.

1. Writing the electrode reaction: at this electrode, hydrogen gas is oxidised to hydrogen ions:

$$\mathrm{H}_2(\mathrm{~g}) \rightarrow 2\mathrm{H}^{+}(\mathrm{aq}) + 2e^-$$

Since 2 electrons are transferred, $n = 2$.

2. Applying the Nernst equation: this equation adjusts the standard potential $E^\circ$ for the actual concentrations and pressures present:

$$E = E^\circ - \frac{2.303RT}{nF}\log\frac{[\mathrm{H}^+]^2}{P_{\mathrm{H}_2}}$$

3. Substituting the known values: $E^\circ = 0$, $[\mathrm{H}^+] = 0.02\ \mathrm{M}$, $P_{\mathrm{H}_2} = 2\ \mathrm{atm}$, and $\frac{2.303RT}{F} = 0.059$:

$$E = 0 - \frac{0.059}{2}\log\frac{(0.02)^2}{2}$$

4. Simplifying the term inside the logarithm:

$$\frac{(0.02)^2}{2} = \frac{0.0004}{2} = 2\times10^{-4}$$ $$E = -0.0295 \log(2\times10^{-4})$$

5. Splitting the logarithm using $\log(a\times b) = \log a + \log b$, with $\log 2 = 0.3010$ and $\log 10^{-4} = -4$:

$$E = -0.0295(0.3010 - 4) = -0.0295(-3.699)$$ $$E = 0.109\ \mathrm{V}$$

Hence, the answer is Option D: 0.109 V.