20. ⇒ (JEE Main 2016 (Online) 10th April Morning Slot )

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :

(A) 2.24 L

(B) 4.48 L

(C) 6.72 L

(D) 8.96 L

Explanation

Correct answer is D

The reaction is

$2 C H_{3} C O O K + 2 H_{2} O \overset{E l e c t r o l y s i s}{⟶} C H_{3} - C H_{3} + 2 C O_{2} + H_{2} + 2 K O H$

At the anode (Oxidation):

JEE Main 2016 (Online) 10th April Morning Slot Chemistry - Electrochemistry Question 106 English Explanation

At the cathode (Reduction):

$2 H_{2} O \overset{+ 2 e^{-}}{⟶} 2 O H^{-} + 2 \overset{∙}{H}$

$2 \overset{∙}{H} \overset{}{⟶} H_{2}$

Total number of moles of gases

= moles of C₂H₆ + moles of CO₂ + moles of H₂

$n = \frac{0.2}{2} + \frac{0.2}{1} + \frac{0.2}{2} = 0.4$

$V = \frac{n R T}{p}$

$= \frac{(0.4 \times 0.0821 \times 273)}{1} = 8.96$ L

21. ⇒ (JEE Main 2015 (Offline) )

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)

(A) 63.5 g

(B) 2 g

(C) 127 g

(D) 0 g

Explanation

Correct answer is A

$C u^{2 +} + 2 e^{-} \overset{}{⟶} C u$

$2 F i . e . 2 \times 96500 C$ deposit $C u = 1$ mol $= 63.5$ $g$

22. ⇒ (AIEEE 2005)

Aluminium oxide may be electrolysed at 1000^oC to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al³⁺ + 3e^- $\to$ Al^o To prepare 5.12 kg of aluminium metal by this method would require

A. 5.49 $\times$ 10⁷ C of electricity

B. 1.83 $\times$ 10⁷ C of electricity

C. 5.49 $\times$ 10⁴ C of electricity

D. 5.49 $\times$ 10¹ C of electricity

Correct Answer is Option (A)

$1$ mole of $e^{-} = 1 F = 96500 C$

$27 g$ of $A l$ is deposited by $3 \times 96500 C$

$5120$ $g$ of $A l$ will be deposited by

$= \frac{3 \times 96500 \times 5120}{27}$

$= 5.49 \times 10^{7} C$

23. ⇒ (AIEEE 2003)

When during electrolysis of a solution of AgNO₃, 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be :

A. 10.8 g

B. 21.6 g

C. 108 g

D. 1.08 g

Correct Answer is Option (A)

When $96500$ coulomb of electricity is passed through the electroplating bath the amount of Ag deposited $= 108 g$

$∴$ when $9650$ coulomb of electricity is passed deposited Ag.

$= \frac{108}{96500} \times 9650 = 10.8 g$

24. ⇒ (AIEEE 2002)

When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are :

A. cathode = pure zinc, anode = pure copper

B. cathode = impure sample, anode = pure copper

C. cathode = impure zinc, anode = impure sample

D. cathode = pure copper, anode = impure sample

Correct Answer is Option (D)

Pure metal always deposits at cathode.

⇒Electrochemistry

Topic 2 : Electrolysis and Types of Electrolysis 3