JEE Chemistry Electrochemistry MCQs Set 05

Answer: (a) 0.354 V

Explanation:
1. \( E^\circ=\dfrac{0.0591}{n}\log K_c=\dfrac{0.0591}{2}\times12 \).
2. \( E^\circ=0.354 \) V.

Teacher's Note:
1. n = 2 for this reaction.
2. Use E° = (0.0591/n) log K.

Answer: (c) 0.52 V

Explanation:
1. \( \Delta G_1=-2(0.94)F \) for OCl- + 2e giving Cl-, and \( \Delta G_2=-(-1.36)F=+1.36F \) for Cl- giving 1/2 Cl2 + e.
2. Sum is -1.88F + 1.36F = -0.52F for a one electron change.
3. \( E^\circ=0.52 \) V.

Teacher's Note:
1. Add free energies, not potentials, when electron counts differ.
2. Cl-/Cl2 value here is an oxidation potential.

Answer: (c) 0.521 V

Explanation:
1. \( \Delta G^\circ \) for Cu2+ + 2e giving Cu is -2(0.337)F and for Cu2+ + e giving Cu+ is -0.153F.
2. Cu+ + e giving Cu: \( -E F=-0.674F+0.153F \), so \( E=0.521 \) V.

Teacher's Note:
1. Combine through ΔG° because the number of electrons differs.
2. 0.674 minus 0.153 equals 0.521.

Answer: (b) -0.177 V

Explanation:
1. E = -0.059 x pH = -0.059 x 3.
2. E = -0.177 V.

Teacher's Note:
1. The potential decreases with pH.
2. Each pH unit is 59 mV.

Answer: (c) 0.295 V

Explanation:
1. This is a concentration cell with \( E=0.0591\log\dfrac{[H^+]_{cathode}}{[H^+]_{anode}} \).
2. \( E=0.0591\log\dfrac{10^{-3}}{10^{-8}}=0.0591\times5=0.295 \) V.

Teacher's Note:
1. Electrode with higher [H+] is the cathode.
2. n = 1 for a hydrogen concentration cell.

Answer: (c) Increases by 0.0295 V

Explanation:
1. \( E=E^\circ-\dfrac{0.059}{2}\log\dfrac{[Cu^{2+}]}{[Ag^+]^2} \).
2. Q changes by 10/100 = 1/10, so \( \log Q \) falls by 1 and E rises by 0.0295 V.

Teacher's Note:
1. Count the powers: Ag+ is squared.
2. Q decreases, so E increases.

Answer: (a) 1.0

Explanation:
1. 112 mL of H2 is 0.005 mol, which needs 0.01 F = 965 C.
2. I = 965 C / 965 s = 1.0 A.

Teacher's Note:
1. 1 mol H2 needs 2 F.
2. 22.4 L at STP per mole.

Answer: (d) 3 : 2

Explanation:
1. Mass deposited is proportional to equivalent weight, M/n.
2. Fe2+: M/2 and Fe3+: M/3, so the ratio is 1/2 : 1/3 = 3 : 2.

Teacher's Note:
1. Faraday's second law.
2. More electrons per atom means less mass.

Answer: (c) 50

Explanation:
1. Charge = 9.65 x 600 = 5790 C = 0.06 F.
2. Equivalent weight = 3.0 / 0.06 = 50.

Teacher's Note:
1. Eq. wt = mass deposited per faraday.
2. Convert minutes to seconds.

Answer: (b) 0.17 g

Explanation:
1. Ni2+: E = 59/2 = 29.5 and Cr3+: E = 52/3 = 17.33.
2. Mass of Cr = 0.3 x 17.33/29.5 = 0.176 g, about 0.17 g.

Teacher's Note:
1. Masses are in the ratio of equivalent weights.
2. Cr3+ needs three electrons.

Answer: (c) \( 1.93\times10^5 \) C

Explanation:
1. \( 2H_2O\to O_2+4H^++4e^- \), so 2 mol of electrons per mole of water.
2. Charge = 2 x 96500 = \( 1.93\times10^5 \) C.

Teacher's Note:
1. Four electrons for two water molecules.
2. That is 2 F per mole of water.

Answer: (c) 6.35 g

Explanation:
1. 2.24 L of H2 is 0.1 mol, needing 0.2 F.
2. 0.2 F deposits 0.1 mol of Cu, which is 6.35 g.

Teacher's Note:
1. Cu2+ needs 2 F per mole.
2. The concentrations given are not needed.

Answer: (a) +4

Explanation:
1. Moles of metal = 2.977/106.4 = 0.02798.
2. Charge per mole = 10800/0.02798 = 3.86 x 10^5 C, about 4 F, so the charge is +4.

Teacher's Note:
1. Charge per mole divided by 96500 gives the charge number.
2. The ratio is close to 4.

Answer: (a) 1 : 8

Explanation:
1. H2 (E = 1) is liberated at the cathode and O2 (E = 8) at the anode.
2. The mass ratio equals the equivalent weight ratio, 1 : 8.

Teacher's Note:
1. Equivalent weights: H = 1, O = 8.
2. Same charge liberates equal equivalents.

Answer: (b) Anode

Explanation:
1. CaH2 is an ionic hydride containing H- ions.
2. H- is oxidised to H2 at the anode.

Teacher's Note:
1. Anions move to the anode.
2. Ionic hydrides give H2 at the anode.

Answer: (c) 1.0787 g

Explanation:
1. Charge = 965 C = 0.01 F.
2. Mass = 0.01 x 107.87 = 1.0787 g.

Teacher's Note:
1. Ag+ needs one electron.
2. 1 F deposits one mole of Ag.

Answer: (c) 1.5

Explanation:
1. 13.5 g is 0.5 mol of Al.
2. Al3+ needs 3 F per mole, so 1.5 F.

Teacher's Note:
1. Moles x charge number gives faradays.
2. Al = 27 g/mol.

Answer: (d) H2, O2

Explanation:
1. Water is reduced more easily than Na+, giving H2 at the cathode.
2. Water is oxidised more easily than sulphate, giving O2 at the anode.

Teacher's Note:
1. Na+ and SO4(2-) are not discharged.
2. Net change is electrolysis of water.

Answer: (b) Silver nitrate

Explanation:
1. One mole of metal needs one mole of electrons only for a monovalent ion such as Ag+.
2. Au3+, Ca2+ and Cu2+ need 3, 2 and 2 F.

Teacher's Note:
1. Charge on ion decides faradays per mole.
2. Silver is monovalent.

Answer: (a) 0.5 mol

Explanation:
1. Ba2+ needs 2 electrons per atom.
2. 1 F deposits 0.5 mol of Ba.

Teacher's Note:
1. In aqueous solution H2 would be liberated instead.
2. Molten state gives the metal.

Answer: (b) 18 kg

Explanation:
1. 40 kg of Ca is 1000 mol, needing 2000 F.
2. 2000 F gives 2000/3 mol of Al = 666.7 mol x 27 g = 18 kg.

Teacher's Note:
1. Same charge: compare equivalents.
2. Ca2+ takes 2 F per mole and Al3+ takes 3 F.