JEE Chemistry Electrochemistry MCQs Set 18

Answer: (c) increase the E and shift the equilibrium to the right

Explanation:
1. Adding H\(_2\)SO\(_4\) raises [H\(^+\)].
2. \( E = E^\circ - \frac{0.0591}{2}\log\frac{[Zn^{2+}]p_{H_2}}{[H^+]^2} \): a larger [H\(^+\)] lowers Q, so E increases.
3. Q becomes smaller than K, so the reaction shifts to the right.

Teacher's Note:
1. Increasing a reactant concentration raises the emf of a galvanic cell.

Answer: (d) 108.0 g

Explanation:
1. 5600 mL of O\(_2\) at STP is 0.25 mol, which is 8 g, i.e. 1 gram equivalent of oxygen.
2. By Faraday's second law, the same charge deposits 1 gram equivalent of silver, 108 g.

Teacher's Note:
1. The same charge liberates equal numbers of equivalents of different substances.

Answer: (a) 4.008%

Explanation:
1. \( \alpha = \frac{\Lambda_m}{\Lambda^\infty_m} = \frac{9.54}{238} = 0.04008 \).
2. The degree of ionisation is 4.008%.

Teacher's Note:
1. For a weak electrolyte, \( \alpha = \Lambda_m / \Lambda^\infty_m \).

Answer: (a) F\(_2\) and I\(^-\)

Explanation:
1. The highest \( E^\circ \) belongs to F\(_2\)/F\(^-\), so F\(_2\) is the strongest oxidising agent.
2. The lowest \( E^\circ \) belongs to I\(_2\)/I\(^-\), so I\(^-\) is the strongest reducing agent.

Teacher's Note:
1. Oxidising ability of the halogens falls from F\(_2\) to I\(_2\); reducing ability of the halide ions rises.

Answer: (c) Z > X > Y

Explanation:
1. Reducing power increases as the reduction potential decreases.
2. The potentials are Z (\(-3.0\)), X (\(-1.2\)), Y (+0.5), so Z > X > Y.

Teacher's Note:
1. More negative reduction potential means a stronger reducing agent.

Answer: (c) I\(^-\) will be oxidised to I\(_2\)

Explanation:
1. Fe\(^{3+}\)/Fe\(^{2+}\) has the higher reduction potential, so Fe\(^{3+}\) is the stronger oxidising agent and oxidises I\(^-\).
2. The reaction 2Fe\(^{3+}\) + 2I\(^-\) \(\rightarrow\) 2Fe\(^{2+}\) + I\(_2\) has \( E^\circ = 0.77 - 0.536 = +0.23 \) V.
3. Iodine cannot oxidise Fe\(^{2+}\).

Teacher's Note:
1. A species oxidises another whose couple has a lower reduction potential.

Answer: (a) increase in ionic mobility of the ions

Explanation:
1. A strong electrolyte is already completely ionised at every concentration, so the number of ions does not increase.
2. On dilution the inter-ionic attraction falls, so the ions move more freely and the equivalent conductance rises.

Teacher's Note:
1. For weak electrolytes the rise on dilution is mainly due to more ionisation.

Answer: (b) More voltage is required to reduce H\(^+\) at Hg than at Pt.

Explanation:
1. Mercury has a very high hydrogen overpotential, so H\(^+\) (or water) is not reduced at an Hg cathode.
2. Na\(^+\) is discharged instead and dissolves in mercury as amalgam.

Teacher's Note:
1. Overpotential decides which species is discharged at an electrode, as in the mercury-cathode chlor-alkali cell.

Answer: (a) energised electrons moving to the other end

Explanation:
1. Metals have free (delocalised) electrons.
2. The electrons near the heated end gain kinetic energy and carry it through the metal, so thermal conduction occurs.

Teacher's Note:
1. Free electrons make metals good conductors of both heat and electricity.

Answer: (b) Mg\(^{2+}\)

Explanation:
1. The strongest oxidising agent has the highest (least negative) reduction potential.
2. Mg\(^{2+}\)/Mg at \(-2.37\) V is the highest of the four.

Teacher's Note:
1. Higher \( E^\circ \) means a stronger tendency to be reduced.

Answer: (b) 63.5 g

Explanation:
1. Cu\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Cu, so 2 F deposits 1 mol of copper.
2. Mass = 63.5 g.

Teacher's Note:
1. Mass deposited = (charge in F / electrons per ion) \(\times\) molar mass.

Answer: (b) 20 F

Explanation:
1. In Mn\(_3\)O\(_4\) the total oxidation number of 3 Mn is +8.
2. In 3 MnO\(_4^{2-}\) it is 3 \(\times\) (+6) = +18.
3. Each Mn\(_3\)O\(_4\) loses 18 \(-\) 8 = 10 electrons, so 2 mol need 20 F.

Teacher's Note:
1. Use the total change in oxidation number per formula unit, not per atom.

Answer: (c) 75 s

Explanation:
1. Initial moles = 0.25 \(\times\) 0.20 = 0.05; final moles = 0.25 \(\times\) 0.10 = 0.025, so 0.025 mol is reduced.
2. Charge = 0.025 \(\times\) 3 F = 0.075 \(\times\) 96500 C.
3. \( t = \frac{Q}{I} = \frac{0.075 \times 96500}{96.5} = 75 \) s.

Teacher's Note:
1. Each Cr\(^{3+}\) needs 3 electrons, so charge = 3F \(\times\) moles reduced.

Answer: (c) 3

Explanation:
1. Charge = 3.20 \(\times\) 2400 = 7680 C, i.e. 0.0796 mol of electrons.
2. Moles of In = \( \frac{3.05}{114.8} = 0.02657 \).
3. Electrons per In = \( \frac{0.0796}{0.02657} = 3.0 \), so the oxidation state is +3.

Teacher's Note:
1. Oxidation state = moles of electrons / moles of metal deposited.

Answer: (d) 6

Explanation:
1. Charge = 1.10 \(\times\) 2400 = 2640 C, i.e. 0.02736 mol of electrons.
2. Moles of W = \( \frac{0.838}{184} = 0.004554 \).
3. Electrons per W = \( \frac{0.02736}{0.004554} = 6.0 \), so the charge on tungsten is +6.

Teacher's Note:
1. Use the same ratio of electrons to atoms as in the indium problem.

Answer: (c) 3

Explanation:
1. For I\(_2\) + 2e\(^-\) \(\rightarrow\) 2I\(^-\), \( E = 0.535 - \frac{0.0591}{2}\log(0.1)^2 = 0.535 + 0.0591 = 0.5941 \) V.
2. The hydrogen electrode is the anode with \( E_{red} = -0.0591\,\text{pH} \).
3. \( E_{cell} = 0.5941 + 0.0591\,\text{pH} = 0.7717 \), so pH = 3.0.

Teacher's Note:
1. The more dilute the iodide, the higher the I\(_2\)/I\(^-\) potential.

Answer: (c) 39

Explanation:
1. Cell constant \( G^* = \kappa R = 0.013 \times 300 = 3.9 \) cm\(^{-1}\).
2. Multiplying by 10 gives 39.

Teacher's Note:
1. Cell constant = conductivity \(\times\) resistance.

Answer: (b) 50

Explanation:
1. The charge on the cation is +3, so one mole of AB\(_3\) contains 3 equivalents.
2. \( \Lambda_{eq} = \frac{\Lambda_m}{3} = \frac{150}{3} = 50 \) S cm\(^2\) eq\(^{-1}\).

Teacher's Note:
1. Equivalent conductivity = molar conductivity divided by the total positive charge per formula unit.

Answer: (c) 100

Explanation:
1. \( \kappa = \frac{G^*}{R} = \frac{2.0}{200} = 0.01 \) S cm\(^{-1}\).
2. \( \Lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{0.01 \times 1000}{0.1} = 100 \) S cm\(^2\) eq\(^{-1}\).

Teacher's Note:
1. The factor 1000 converts the volume to cm\(^3\) per litre.

Answer: (a) 55

Explanation:
1. Calomel is the cathode with \( E_{red} = +0.28 \) V.
2. Anode: H\(_2\) \(\rightarrow\) 2H\(^+\) + 2e\(^-\), \( E_{ox} = 0 - \frac{0.0591}{2}\log\frac{[H^+]^2}{p_{H_2}} = 0.0591\,\text{pH} + \frac{0.0591}{2}\log 0.1 \).
3. \( 0.5755 = 0.28 + 0.0591\,\text{pH} - 0.02955 \), so pH = 5.5 and 10 \(\times\) pH = 55.

Teacher's Note:
1. Write the anode potential from the Nernst equation carefully, including the 0.1 bar pressure term.