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12th Standard English Medium Chemistry Subject Electro Chemistry Book Back 1 Mark Questions with Solution Part - I

Published on: 01/06/2021

QB365 provides detailed and simple solution for every Book back Questions in class 12 Chemistry Subject. It will helps to get more idea about question pattern in every book back questions with solution.

Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Std: 12th Standard - Chemistry

chapter: - Book back Questions

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Questions:

Multiple Choice Question

5 x 1 = 5

During electrolysis of molten sodium chloride, the time required to produce 0.1mole of chlorine gas using a current of 3A is _____.

(a)

55 minutes

(b)

107.2 minutes

(c)

220 minutes

(d)

330 minutes

A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is_______. (atomic mass of Ca is 40g / mol and 1F = 96500C).

(a)

(b)

(c)

(d)

How many faradays of electricity are required for the following reaction to occur MnO₄^-→ Mn²⁺

(a)

(b)

(c)

(d)

Faraday constant is defined as_______.

(a)

charge carried by 1 electron

(b)

charge carried by one mole of electrons

(c)

charge required to deposit one mole of substance

(d)

charge carried by 6.22 ×10¹⁰ electrons

Electrolyte	KCl	KNO₃	HCl	NaOAC	NaCl
Λ_- (Scm² mol^-1)	149.9	145.0	426.2	91.0	126.5

Calculate Λ^o_HoAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25^oC_______.

(a)

517.2

(b)

552.7

(c)

390.7

(d)

217.5

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Answers Key:

Multiple Choice Question

5 x 1 = 5

mass of 1 mole of CI₂ gas = 71
∴ mass of 0.1 mole of Cl₂ gas = 7.1 g mol^-1
m = Zlt
t = m/ZI
\(= \frac{7.1}{\frac{71}{2 \times 96500} \times 3}\) (2Cl^- ➝ Cl₂+ 2e^-)
\(= \frac{2 \times 96500 \times 7.1}{71 \times 3}\)
= 6433.33s = 107.2 min

mg = ZIt
\(= \frac{40 \times 3.86 \times 2500}{2 \times 96500} = 2g\)
(∵ t = 41 min 40 sec = 2500 seconds & \(Z = \frac{m}{n \times 96500} = \frac{40}{x \times 96500})\)

7MnO₄^- + 5e^- → Mn²⁺ + 4H₂O
5 moles of electrons i.e., 5F charge is required.

IF = 96500 C = charge of one mole of e^- charge of 6.022 x 10^-23 electron

(Λ^o_HoAC) = [(Λ^o)_HCl + Λ^o)_NaOAC ] - (Λ^o)_NaCl
= (426.2 + 91) - (126.5)
= 390.7