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Multiple Choice Questions

1. Electrolysis of dilute aqueous $\text{NaCl}$ solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of $\text{H}_2$ gas at the cathode is ($1\text{ Faraday} = 96500\text{ C}$) [JEE 2008 P-II]

(a) $9.65 \times 10^4\text{ sec}$

(b) $19.3 \times 10^4\text{ sec}$

(d) $38.6 \times 10^4\text{ sec}$

Answer: (b)

Explanation: $2\text{H}_2\text{O} + 2e^- \longrightarrow \text{H}_2 + 2\text{OH}^-$

To liberate 0.01 mole of $\text{H}_2$, 0.02 Faraday charge is required.

$$Q = i \times t \implies 10 \times 10^{-3} \times t = 0.02 \times 96500\text{ C}$$

$$\therefore t = \frac{0.02 \times 96500}{10^{-2}} = 19.30 \times 10^4\text{ sec}$$

2. Among the electrolytes $\text{Na}_2\text{SO}_4$, $\text{CaCl}_2$, $\text{Al}_2(\text{SO}_4)_3$ and $\text{NH}_4\text{Cl}$, the most effective coagulating agent for $\text{Sb}_2\text{S}_3$ sol is: [JEE 2009 P-I]

(a) $\text{Na}_2\text{SO}_4$

(b) $\text{CaCl}_2$

(d) $\text{NH}_4\text{Cl}$

Answer: (c)

Explanation: As $\text{Sb}_2\text{S}_3$ is a negative sol, $\text{Al}_2(\text{SO}_4)_3$ will be the most effective coagulant due to higher charge density on $\text{Al}^{3+}$ in accordance with Hardy-Schulze rule.

Order of effectiveness of cations: $\text{Al}^{3+} > \text{Ca}^{2+} > \text{Na}^+ > \text{NH}_4^+$

3. $\text{AgNO}_3(aq.)$ was added to an aqueous $\text{KCl}$ solution gradually and the conductivity of the solution was measured. The plot of conductance ($\Lambda$) versus the volume of $\text{AgNO}_3$ is: [JEE 2011 P-I]

(a)
Conductance plot a

(b)
Conductance plot b

(d)
Conductance plot d

Answer: (d)

Explanation: When $\text{AgNO}_3$ is added to $\text{KCl}$, $\text{AgCl}$ precipitates out while $\text{K}^+$ and $\text{NO}_3^-$ remain in solution. The mobility of $\text{NO}_3^-$ is nearly equal to that of $\text{Cl}^-$, so the conductance remains almost constant until the end point. After the end point, addition of excess $\text{AgNO}_3$ increases the number of ions ($\text{Ag}^+$ and $\text{NO}_3^-$), causing conductance to increase sharply.

4. Consider the following cell reaction: $2\text{Fe}(s) + \text{O}_2(g) + 4\text{H}^+(aq) \longrightarrow 2\text{Fe}^{2+}(aq) + 2\text{H}_2\text{O}(l)$, $E^\circ = 1.67\text{ V}$. At $[\text{Fe}^{2+}] = 10^{-3}\text{ M}$, $P(\text{O}_2) = 0.1\text{ atm}$ and $\text{pH} = 3$ at $25^\circ\text{C}$, the cell potential is: [JEE 2011 P-II]

(a) 1.47 V

(b) 1.77 V

(d) 1.57 V

Answer: (d)

Explanation: $n = 4$ (no. of moles of electron involved)

From Nernst's equation,

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

$$\because \text{pH} = 3 \implies [\text{H}^+] = 10^{-3}\text{ M}$$

$$E_{\text{cell}} = 1.67 - \frac{0.0591}{4} \log \left( \frac{[\text{Fe}^{2+}]^2}{P(\text{O}_2) [\text{H}^+]^4} \right) = 1.67 - \frac{0.0591}{4} \log \left( \frac{(10^{-3})^2}{0.1 \times (10^{-3})^4} \right)$$

$$= 1.67 - 0.106 = 1.57\text{ V}$$

5. Given: $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33\text{ V}$; $E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51\text{ V}$; $E^\circ_{\text{Cl}/\text{Cl}^-} = 1.36\text{ V}$; $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\text{ V}$. Based on the data given above, strongest oxidizing agent will be: [JEE Main 2013]

(a) $\text{Cl}^-$

(b) $\text{Cr}^{3+}$

(d) $\text{MnO}_4^-$

Answer: (d)

Explanation: $\text{MnO}_4^-$ has the highest standard reduction potential ($E^\circ = 1.51\text{ V}$), so it is the strongest oxidising agent.

6. In a galvanic cell, the salt bridge [JEE Adv. 2014 P-I]

(a) does not participate chemically in the cell reaction.

(b) stops the diffusion of ions from one electrode to another.

(d) ensures mixing of the two electrolytic solutions

Answer: (a)

Explanation: The salt bridge maintains electrical neutrality and completes the inner circuit without participating chemically in the cell reaction.

7. Two Faraday of electricity is passed through a solution of $\text{CuSO}_4$. The mass of copper deposited at the cathode is: [JEE Main 2015]

(a) 0 g

(b) 63.5 g

(d) 127 g

Answer: (b)

Explanation: $\text{Cu}^{2+} + 2e^- \longrightarrow \text{Cu}(s)$

2 moles of electrons (2 Faraday) deposit 1 mole of $\text{Cu} = 63.5\text{ g}$.

8. For the following electrochemical cell at 298 K, $\text{Pt}(s) \mid \text{H}_2(g, 1\text{ bar}) \mid \text{H}^+(aq, 1\text{ M}) \mid\mid \text{M}^{4+}(aq), \text{M}^{2+}(aq) \mid \text{Pt}(s)$, $E_{\text{cell}} = 0.092\text{ V}$ when $\frac{[\text{M}^{2+}(aq)]}{[\text{M}^{4+}(aq)]} = 10^x$. Given: $E^\circ_{\text{M}^{4+}/\text{M}^{2+}} = 0.151\text{ V}$; $\frac{RT}{F} \times 2.303 = 0.059\text{ V}$. The value of x is: [JEE Adv. 2016-II]

(a) –2

(b) –1

(d) 2

Answer: (d)

Explanation:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left( \frac{[\text{M}^{2+}][\text{H}^+]^2}{[\text{M}^{4+}] P(\text{H}_2)} \right)$$

$$0.092 = 0.151 - \frac{0.059}{2} \log(10^x)$$

$$0.092 = 0.151 - \frac{0.059}{2} x \implies x = 2$$

9. Given $E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36\text{ V}$, $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\text{ V}$, $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33\text{ V}$, $E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51\text{ V}$. Among the following, the strongest reducing agent is : [JEE Main 2017]

(a) $\text{Cl}^-$

(b) $\text{Cr}$

(d) $\text{Cr}^{3+}$

Answer: (b)

Explanation:

(i) $E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51\text{ V}$

(ii) $E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36\text{ V}$

(iii) $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33\text{ V}$

(iv) $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\text{ V}$

Since $\text{Cr}^{3+}/\text{Cr}$ is having least reduction potential (most negative), $\text{Cr}$ is the best reducing agent.

10. For the following cell, $\text{Zn}(s) \mid \text{ZnSO}_4(aq) \mid\mid \text{CuSO}_4(aq) \mid \text{Cu}(s)$ when the concentration of $\text{Zn}^{2+}$ is 10 times the concentration of $\text{Cu}^{2+}$, the expression for $\Delta G$ (in $\text{J mol}^{-1}$) is [$F$ is Faraday constant; $R$ is gas constant; $T$ is temperature; $E^\circ(\text{cell}) = 1.1\text{ V}$] [JEE Adv. 2017 P-II]

(a) $2.303 RT - 2.2 F$

(b) $1.1 F$

(d) $2.303 RT + 1.1 F$

Answer: (a)

Explanation: $\text{Zn}(s) \mid \text{ZnSO}_4(aq) \mid\mid \text{CuSO}_4(aq) \mid \text{Cu}(s)$

$$\Delta G = \Delta G^\circ + RT \ln Q$$

$$\Delta G = \Delta G^\circ + 2.303 RT \log Q$$

Given $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10$

$$\Delta G^\circ = -nF E^\circ_{\text{cell}} = -2 \times F \times 1.1 = -2.2 F$$

$$\therefore \Delta G = -2.2 F + 2.303 RT \log(10) = 2.303 RT - 2.2 F$$

11. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of $\text{B} = 10.8\text{u}$) [JEE Main 2018]

(a) 1.6 hours

(b) 6.4 hours

(d) 3.2 hours

Answer: (d)

Explanation: $\text{B}_2\text{H}_6 + 3\text{O}_2 \longrightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$

27.66 g of $\text{B}_2\text{H}_6 = 1\text{ mole of B}_2\text{H}_6$ which requires 3 moles of oxygen ($\text{O}_2$) for complete burning.

$$6\text{H}_2\text{O} \longrightarrow 6\text{H}_2 + 3\text{O}_2 \text{ (On electrolysis)}$$

Number of Faradays $= 12 = \text{Amount of charge in Faradays}$

$$12 \times 96500 = i \times t \implies 12 \times 96500 = 100 \times t$$

$$t = \frac{12 \times 96500}{100}\text{ seconds} = \frac{12 \times 96500}{100 \times 3600}\text{ hours} \implies t = 3.2\text{ hours}$$

12. Given that: $E^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.23\text{ V}$, $E^\circ_{\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}} = 2.05\text{ V}$, $E^\circ_{\text{Br}_2/\text{Br}^-} = +1.09\text{ V}$, $E^\circ_{\text{Au}^{3+}/\text{Au}} = +1.4\text{ V}$. The strongest oxidising agent is [JEE Main 2019]

(a) $\text{O}_2$

(b) $\text{Br}_2$

(d) $\text{Au}^{3+}$

Answer: (c)

Explanation: For strongest oxidising agent, standard reduction potential should be highest ($\text{S}_2\text{O}_8^{2-}$ has $E^\circ = 2.05\text{ V}$).

13. Calculate the standard cell potential in (V) of the cell in which following reaction takes place: $\text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \longrightarrow \text{Fe}^{3+}(aq) + \text{Ag}(s)$. Given that $E^\circ_{\text{Ag}^+/\text{Ag}} = x\text{V}$, $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = y\text{V}$, $E^\circ_{\text{Fe}^{3+}/\text{Fe}} = z\text{V}$. [JEE Main 2019]

(a) $x + 2y - 3z$

(b) $x - z$

(d) $x + y - z$

Answer: (a)

Explanation: $\text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \longrightarrow \text{Fe}^{3+}(aq) + \text{Ag}(s)$

Anode: $\text{Fe}^{2+}(aq) \longrightarrow \text{Fe}^{3+}(aq) + e^-$; $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = mV$

Cathode: $\text{Ag}^+(aq) + e^- \longrightarrow \text{Ag}(s)$; $E^\circ_{\text{Ag}^+/\text{Ag}} = x\text{V}$

$\implies E^\circ_{\text{cell}} = (x + m)\text{V}$

To find $m$:

(1) $\text{Fe}^{2+} + 2e^- \longrightarrow \text{Fe}$; $E^\circ = y\text{V} \implies \Delta G^\circ_1 = -2Fy$

(2) $\text{Fe}^{3+} + 3e^- \longrightarrow \text{Fe}$; $E^\circ = z\text{V} \implies \Delta G^\circ_2 = -3Fz$

(3) $\text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + e^-$; $\Delta G^\circ_3 = -1Fm$

$$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2 = -2Fy - (-3Fz) = -Fm \implies m = 2y - 3z$$

$$\implies E^\circ_{\text{cell}} = (x + 2y - 3z)\text{V}$$

14. The anodic half cell of lead acid battery is recharged using electricity of 0.05 Faraday. The amount of $\text{PbSO}_4$ electrolysed in g during the process in: (Molar mass of $\text{PbSO}_4 = 303\text{ g mol}^{-1}$) [JEE Main 2019]

(a) 22.8

(b) 15.2

(d) 11.4

Answer: (b)

Explanation:

(i) $\text{PbSO}_4(s) + 2\text{OH}^- \longrightarrow \text{PbO}_2 + \text{H}_2\text{SO}_4 + 2e^-$ (0.05 F $\rightarrow$ 0.05/2 mole)

(ii) $\text{PbSO}_4 + 2e^- + 2\text{H}^+ \longrightarrow \text{Pb}(s) + \text{H}_2\text{SO}_4$ (0.05 F $\rightarrow$ 0.05/2 mole)

Total $n(\text{PbSO}_4) = 0.05\text{ mole}$

$$m_{\text{PbSO}_4} = 0.05 \times 303 = 15.2\text{ gm}$$

15. If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant ($K$) for the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \longrightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$ at 300 K is approximately. ($R = 8\text{ J K}^{-1}\text{ mol}^{-1}$, $F = 96000\text{ C mol}^{-1}$) [JEE Main 2019]

(a) $e^{160}$

(b) $e^{320}$

(d) $e^{-80}$

Answer: (a)

Explanation: $\Delta G^\circ = -RT \ln K = -nFE^\circ_{\text{cell}}$

$$\ln K = \frac{nFE^\circ}{RT} = \frac{2 \times 96000 \times 2}{8 \times 300} = 160 \implies K = e^{160}$$

16. Consider the following reduction processes:

$\text{Zn}^{2+} + 2e^- \longrightarrow \text{Zn}(s)$; $E^\circ = -0.76\text{ V}$

$\text{Ca}^{2+} + 2e^- \longrightarrow \text{Ca}(s)$; $E^\circ = -2.87\text{ V}$

$\text{Mg}^{2+} + 2e^- \longrightarrow \text{Mg}(s)$; $E^\circ = -2.36\text{ V}$

$\text{Ni}^{2+} + 2e^- \longrightarrow \text{Ni}(s)$; $E^\circ = -0.25\text{ V}$

The reducing power of the metals increases in the order: [JEE Main 2019]

(a) $\text{Ca} < \text{Zn} < \text{Mg} < \text{Ni}$

(b) $\text{Ni} < \text{Zn} < \text{Mg} < \text{Ca}$

(d) $\text{Ca} < \text{Mg} < \text{Zn} < \text{Ni}$

Answer: (b)

Explanation: Higher the oxidation potential (or lower/more negative the reduction potential), better will be the reducing power. Order: $\text{Ni} < \text{Zn} < \text{Mg} < \text{Ca}$.

17. In the cell $\text{Pt}(s) \mid \text{H}_2(g, 1\text{ bar} \mid \text{HCl}(aq) \mid \text{AgCl}(s) \mid \text{Ag}(s) \mid \text{Pt}(s)$, the cell potential is 0.92 V when a $10^{-6}\text{ molal HCl}$ solution is used. The standard electrode potential of $(\text{AgCl}/\text{Ag}, \text{Cl}^-)$ electrode is: {given, $\frac{2.303RT}{F} = 0.06\text{V}$ at 298 K} [JEE Main 2019]

(a) 0.20 V

(b) 0.76 V

(d) 0.94 V

Answer: (a)

Explanation: For $10^{-6}\text{ mol HCl}$

Anode: $\frac{1}{2}\text{H}_2(g) \longrightarrow \text{H}^+ + e^-$

Cathode: $e^- + \text{AgCl}(s) \longrightarrow \text{Ag}(s) + \text{Cl}^-(aq)$

Overall: $\frac{1}{2}\text{H}_2(g) + \text{AgCl}(s) \longrightarrow \text{H}^+(aq) + \text{Ag}(s) + \text{Cl}^-(aq)$

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{1} \log ([\text{H}^+][\text{Cl}^-])$$

$$0.92 = E^\circ_{\text{AgCl}/\text{Ag}, \text{Cl}^-} - 0.06 \log (10^{-6} \times 10^{-6})$$

$$0.92 = E^\circ_{\text{AgCl}/\text{Ag}, \text{Cl}^-} - 0.06 \times (-12)$$

$$0.92 = E^\circ + 0.72 \implies E^\circ_{\text{AgCl}/\text{Ag}, \text{Cl}^-} = 0.92 - 0.72 = 0.20\text{ V}$$

18. The electrolytes usually used in the electroplating of gold and silver, respectively, are: [JEE Main 2019]

(a) $[\text{Au}(\text{OH})_4]^-$ and $[\text{Ag}(\text{OH})_2]^-$

(b) $[\text{Au}(\text{CN})_2]^-$ and $[\text{AgCl}_2]^-$

(d) $[\text{Au}(\text{CN})_2]^-$ and $[\text{Ag}(\text{CN})_2]^-$

Answer: (d)

Explanation: Cyanide complexes $[\text{Au}(\text{CN})_2]^-$ and $[\text{Ag}(\text{CN})_2]^-$ are used as electrolytes in electroplating of gold and silver respectively.

19. For the cell $\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \mid\mid \text{M}^{x+}(aq) \mid \text{M}(s)$, different half cells and their standard electrode potentials are given below:

$\text{M}^{x+}(aq)/\text{M}(s)$	$\text{Au}^{3+}(aq)/\text{Au}(s)$	$\text{Au}^+(aq)/\text{Au}(s)$	$\text{Fe}^{3+}(aq)/\text{Fe}^{2+}(aq)$	$\text{Fe}^{2+}(aq)/\text{Fe}(s)$
$E^\circ_{\text{M}^{x+}/\text{M}}$ (V)	1.40	0.80	0.77	–0.44

If $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$, which cathode will give a maximum value of $E^\circ_{\text{cell}}$ per electron transferred? [JEE Main 2019]

(a) $\text{Fe}^{3+} / \text{Fe}^{2+}$

(b) $\text{Ag}^+ / \text{Ag}$

(d) $\text{Fe}^{2+} / \text{Fe}$

Answer: (b)

Explanation: Calculating $E^\circ_{\text{cell}}$ per electron transferred gives maximum value for $\text{Ag}^+/\text{Ag}$ cathode system.

20. Given the equilibrium constant: $K_c$ of the reaction: $\text{Cu}(s) + 2\text{Ag}^+(aq) \longrightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$ is $10 \times 10^{15}$, calculate the $E^\circ_{\text{cell}}$ of this reaction at 298 K $\left[\text{at 298 K }, \frac{2.303 RT}{F} = 0.059\text{V}\right]$ [JEE Main 2019]

(a) 0.04736 V

(b) 0.4736 V

(d) 0.04736 mV

Answer: (b)

Explanation: At equilibrium, $E_{\text{cell}} = 0$

$$E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_c = \frac{0.059}{2} \log(10^{16}) = \frac{0.059}{2} \times 16 = 0.472\text{ V} \approx 0.4736\text{ V}$$

21. Which of the following plot is correct regarding its conductance? [JEE Adv. 2019 P-I]

(a)
Plot a

(b)
Plot b

(d)
Plot d

Answer: (b)

Explanation: Plot (b) correctly shows the variation of molar conductivity with concentration for the given electrolyte systems.

Matrix Match/Column Match

22. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in Column I. The variation in conductivity of these reactions shown in Column II. [JEE Adv. 2013 P-II]

Column I	Column II
(A) $(\text{C}_2\text{H}_5)_3\text{N} (\text{X}) + \text{CH}_3\text{COOH} (\text{Y})$	1. Conductivity decreases and then increases
(B) $\text{KI} (0.1\text{M}) (\text{X}) + \text{AgNO}_3 (0.01\text{M}) (\text{Y})$	2. Conductivity decreases and then does not change much
(C) $\text{CH}_3\text{COOH} (\text{X}) + \text{KOH} (\text{Y})$	3. Conductivity increases and then does not change much
(D) $\text{NaOH} (\text{X}) + \text{HI} (\text{Y})$	4. Conductivity does not change much and then increases

(a) A $\rightarrow$ 2; B $\rightarrow$ 3; C $\rightarrow$ 1; D $\rightarrow$ 4

(b) A $\rightarrow$ 1; B $\rightarrow$ 2; C $\rightarrow$ 3; D $\rightarrow$ 4

(d) A $\rightarrow$ 3; B $\rightarrow$ 4; C $\rightarrow$ 2; D $\rightarrow$ 1

Answer: (d)

Explanation: A $\rightarrow$ 3; B $\rightarrow$ 4; C $\rightarrow$ 2; D $\rightarrow$ 1

(A) Initially conductivity increases due to ion formation ($(\text{C}_2\text{H}_5)_3\text{NH}^+ \text{CH}_3\text{COO}^-$), after that it becomes practically constant because X alone cannot form ions. Hence (3) is the correct match.

(B) Number of ions in the solution remains constant until all the $\text{AgNO}_3$ is precipitated as $\text{AgI}$. Thereafter conductance increases due to increase in number of ions. Hence (4) is the correct match.

(C) Initially conductance decreases due to the decrease in the mobility/number of highly conducting ions, thereafter it slowly changes due to excess weak acid. Hence (2) is the correct match.

(D) Initially it decreases due to decrease in $\text{H}^+$ ions and then increases due to increase in $\text{OH}^-$ ions. Hence (1) is the correct match.

Paragraph

Paragraph for Question Nos. 23 to 25

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of $\text{NaCl}$ is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: $\text{Na}=23, \text{Hg}=200, 1\text{ Faraday}= 96500\text{ coulombs}$) [JEE 2007 P- I]

23. The total number of moles of chlorine gas evolved is

(a) 0.5

(b) 1.0

(d) 3.0

Answer: (b)

Explanation: At anode: $2\text{Cl}^- \longrightarrow \text{Cl}_2 + 2e^-$. Moles of $\text{NaCl}$ in 500 mL $= 4.0 \times 0.5 = 2.0\text{ moles}$. Since $2\text{ moles}$ of $\text{Cl}^-$ form $1.0\text{ mole}$ of $\text{Cl}_2$, 1.0 mole of $\text{Cl}_2$ evolves.

24. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is:

(a) 200

(b) 225

(d) 446

Answer: (d)

Explanation: $\text{Na}^+ + e^- + \text{Hg} \longrightarrow \text{Na-Hg}$ (amalgam). 2 moles of $\text{Na-Hg}$ formed at cathode. Mass $= 2 \times (23 + 200) = 2 \times 223 = 446\text{ g}$.

25. The total charge (coulombs) required for complete electrolysis is:

(a) 24125

(b) 48250

(d) 193000

Answer: (d)

Explanation: 2 moles of electrons (2 Faraday) are required. Charge $= 2 \times 96500 = 193000\text{ C}$.

Paragraph for Question Nos. 26 to 27

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is:

$$\text{M}(s) \mid \text{M}^+(aq, 0.05\text{molar}) \mid\mid \text{M}^+(aq, 1\text{molar}) \mid \text{M}(s)$$

For the above electrolytic cell the magnitude of the cell potential $\mid E_{\text{cell}} \mid = 70\text{ mV}$. [JEE 2010 P-I]

26. For the above cell

(a) $E_{\text{cell}} < 0; \Delta G > 0$

(b) $E_{\text{cell}} > 0; \Delta G < 0$

(a) $E_{\text{cell}} < 0; \Delta G^\circ > 0$

(d) $E_{\text{cell}} > 0; \Delta G^\circ > 0$

Answer: (b)

Explanation: According to Nernst equation, $E_{\text{cell}} = -\frac{RT}{F} \ln \left( \frac{0.05}{1} \right) > 0$. Hence, $E_{\text{cell}} > 0$ and $\Delta G < 0$ for the feasibility of the reaction.

27. If the 0.05 molar solution of $\text{M}^+$ is replaced by 0.0025 molar solution, then the magnitude of the cell potential would be

(a) 35 mV

(b) 70 mV

(d) 700 mV

Answer: (c)

Explanation: From Nernst equation, $E_1 = -\frac{RT}{F} \ln(0.05) = 70\text{ mV}$.

For $0.0025\text{ M} = (0.05)^2\text{ M}$:

$$E_2 = -\frac{RT}{F} \ln((0.05)^2) = 2 \times \left( -\frac{RT}{F} \ln(0.05) \right) = 2 \times 70\text{ mV} = 140\text{ mV}$$

Integer and Subjective

28. The molar conductivity of a solution of a weak acid $\text{HX}$ (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid $\text{HY}$ (0.10 M). If $\lambda^\circ_{\text{X}^-} \approx \lambda^\circ_{\text{Y}^-}$, the difference in their $\text{p}K_a$ values, $\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY})$, is (consider degree of ionisation of both acids to be $\ll 1$) [JEE Adv. 2015 P-II]

Answer: 3

Explanation: $\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-$, $K_a(\text{HX}) = \frac{[\text{H}^+][\text{X}^-]}{[\text{HX}]}$

$\text{HY} \rightleftharpoons \text{H}^+ + \text{Y}^-$, $K_a(\text{HY}) = \frac{[\text{H}^+][\text{Y}^-]}{[\text{HY}]}$

For weak electrolytes, $K_a = C \alpha^2 = C \left( \frac{\Lambda_m}{\Lambda_m^\circ} \right)^2$

$$\frac{K_a(\text{HX})}{K_a(\text{HY})} = \frac{C_1}{C_2} \times \left( \frac{\Lambda_{m1}}{\Lambda_{m2}} \right)^2 = \frac{0.01}{0.1} \times \left( \frac{1}{10} \right)^2 = 0.1 \times \frac{1}{100} = 10^{-3}$$

$$\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}) = -\log \left( \frac{K_a(\text{HX})}{K_a(\text{HY})} \right) = -\log(10^{-3}) = 3$$

29. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinised Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of $1\text{ cm}^2$. The conductance of this solution was found to be $5 \times 10^{-7}\text{ S}$. The pH of the solution is 4. The value of limiting molar conductivity ($\Lambda_m^\circ$) of this weak monobasic acid in aqueous solution is $Z \times 10^2\text{ S cm}^2\text{ mol}^{-1}$. The value of Z is: [JEE Adv. 2017 P-I]

Answer: 6

Explanation: $\kappa = G \times \frac{l}{A} = 5 \times 10^{-7}\text{ S} \times \frac{120\text{ cm}}{1\text{ cm}^2} = 6 \times 10^{-5}\text{ S cm}^{-1}$

$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-5} \times 1000}{0.0015} = 40\text{ S cm}^2\text{ mol}^{-1}$$

$\text{pH} = 4 \implies [\text{H}^+] = 1.0 \times 10^{-4}\text{ M}$

$$\alpha = \frac{[\text{H}^+]}{C} = \frac{10^{-4}}{0.0015} = \frac{1}{15}$$

$$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \implies \Lambda_m^\circ = \frac{\Lambda_m}{\alpha} = \frac{40}{1/15} = 600 = 6 \times 10^2\text{ S cm}^2\text{ mol}^{-1}$$

$$\therefore Z = 6$$

30. For the electrochemical cell, $\text{Mg}(s) \mid \text{Mg}^{2+}(aq, 1\text{M}) \mid\mid \text{Cu}^{2+}(aq, 1\text{M}) \mid \text{Cu}(s)$ the standard emf of the cell is 2.70 V at 300 K. When the concentration of $\text{Mg}^{2+}$ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is_____. (Given, $\frac{F}{R} = 11500\text{ K V}^{-1}$, where $F$ is the Faraday constant and $R$ is the gas constant, $\ln(10) = 2.30$) [JEE Adv. 2018 P-I]

Answer: 10

Explanation: Cell reaction: $\text{Mg}(s) + \text{Cu}^{2+}(1\text{M}) \longrightarrow \text{Cu}(s) + \text{Mg}^{2+}(x\text{M})$

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln x$$

$$2.67 = 2.70 - \frac{300}{2 \times 11500} \ln x$$

$$0.03 = \frac{300}{23000} \ln x = \frac{3}{230} \ln x$$

$$\ln x = \frac{0.03 \times 230}{3} = 2.30 = \ln(10) \implies x = 10$$

31. Consider an electrochemical cell: $\text{A}(s) \mid \text{A}^{n+}(aq, 2\text{ M}) \mid\mid \text{B}^{2n+}(aq, 1\text{ M}) \mid \text{B}(s)$. The value of $\Delta H^\circ$ for the cell reaction is twice that of $\Delta G^\circ$ at 300 K. If the emf of the cell is zero, the $\Delta S^\circ$ (in $\text{J K}^{-1}\text{ mol}^{-1}$) of the cell reaction per mole of B formed at 300 K is: (Given: $\ln(2) = 0.7$, $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$) [JEE Adv. 2018 P-II]

Answer: -11.62

Explanation: Cell reaction: $2\text{A} + \text{B}^{2n+} \longrightarrow 2\text{A}^{n+} + \text{B}$

Given $\Delta H^\circ = 2 \Delta G^\circ$ and $E_{\text{cell}} = 0$

$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 2 \Delta G^\circ - T \Delta S^\circ \implies \Delta G^\circ = T \Delta S^\circ \implies \Delta S^\circ = \frac{\Delta G^\circ}{T}$$

Since $E_{\text{cell}} = 0 \implies E^\circ_{\text{cell}} = \frac{RT}{2nF} \ln Q = \frac{RT}{2nF} \ln \left( \frac{[\text{A}^{n+}]^2}{[\text{B}^{2n+}]} \right) = \frac{RT}{2nF} \ln \left( \frac{2^2}{1} \right) = \frac{RT}{2nF} \ln 4$

$$\Delta G^\circ = -2nF E^\circ_{\text{cell}} = -RT \ln 4 = -2 RT \ln 2$$

$$\Delta S^\circ = \frac{\Delta G^\circ}{T} = -2 R \ln 2 = -2 \times 8.3 \times 0.7 = -11.62\text{ J K}^{-1}\text{ mol}^{-1}$$