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Test: Electrolysis & Faraday's Law (November 12) - NEET MCQ


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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Electrolysis & Faraday's Law (November 12)

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Test: Electrolysis & Faraday's Law (November 12) - Question 1

Only One Option Correct Type

This section contains 16 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

Q.

During the electrolysis of aqueous Zn(NO3)2 solution

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 1


In preference to reduction of Zn2+ at cathode, O2- (aq) is oxidised to O2 at anode.

Test: Electrolysis & Faraday's Law (November 12) - Question 2

Refining of impure metal is done by electrolysis using impure metal as anode.Select the correct statement about this refining.

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 2

When impure metal (M) is anode and pure metal is made cathode, then 

Thus, impure metal from the anode is deposited at the cathode.  

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Test: Electrolysis & Faraday's Law (November 12) - Question 3

In the electrosynthesis,potassium manganate (VII) is converted to manganese(IV) dioxide. By passage of 1F of electrolysis ,one mole of potassium manganate(VII) will form manganese dioxide.

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 3

Manganate (VII) is MnO4- with ON of Mn = + 7
1 mole required 3F of electricity thus, 1 F will reduce only (1/3) mole MnO4- t o MnO2.
Thus, MnO2 formed = 0.33 mol.

Test: Electrolysis & Faraday's Law (November 12) - Question 4

1 Faraday of electricity is passed through the solution containing 1 mole each of AgNO3, CuSO4, AlCland SiCl4. Elements are discharged at the cathode.Number of moles of Ag, Cu,Al and Si formed will be in the ratio of  

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 4

(d)

Test: Electrolysis & Faraday's Law (November 12) - Question 5

100 mL of a buffer of 1 M NH3(aq) and 1 M NH4+(aq) are placed in two volatic cells separately.A current of 1.5 A is passed through both cells for 20 min. If electrolysis of water only takes place

2H2O +O2 + 4e- → 4OH- (RHS

2H2O  → 4H+. + O2 + 4e- (LHS)

then pH of the 

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 5


In RHS, [NH3] increases hence, pOH decreases
pH = 14 - pOH
Hence, pH of RHS increases.
In LHS, [NH4+] increases, hence pOH increases and pH decreases.

Test: Electrolysis & Faraday's Law (November 12) - Question 6

What product are formed during the electrolysis of a concentrated aqueous solution of sodium chloride using an electrolytic cell in which electrodes are separated by a porous pot?

I. Cl2(g)
II. NaOH(aq)
III. H2(g)
IV. NaClO(aq)
V. NaClO3(aq) 

Select the correct choice.  

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 6

Cl2(g) is formed at anode and NaOH and H2 are formed at cathode. Since, cathode and anode are separated. Hence. there is no reaction between the products formed at the cathode and anode. 

Note: but these reaction do not occur

Note: NaCl is neutral pH = 7 initially. After electrolysis , NaOH is formed

Hence, solution become, basic pH >7

*Multiple options can be correct
Test: Electrolysis & Faraday's Law (November 12) - Question 7

Select the correct statement(s) about electrolysis of aqueous CuSO4 solution.

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 7

(a) 

Oxidation takes place at anode - correct

2H2O(l) →4H+ (aq) + O2(g) + 4e-

(b) Reduction take place at cathode - correct

since, E0cell < 0, in electrolysis.

Cell reaction is spontaneous.

(c) Copper is deposited at the cathode, hence the meted to be electroplated (say Fe), should be made cathode - correct (d) Since. Cu2+(aq)- (blue) is reduced Cu(s), hence blue colour fades - correct. 
 

*Multiple options can be correct
Test: Electrolysis & Faraday's Law (November 12) - Question 8

Which pair of electrolysis could be distinguished by the products of electrolysis using inert electrodes?

Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 8

a(Cu So4 and CuCl2 changes to Cu and solution becomes colourless

(b) KCl changes to Cl2 (yellow)
KI changes to I2 (voilet)

Thus, distinguished.

(c) AgNo3 No change

CuSo4 (blue) changes to colourless

(d) CuBr2 (blue) both changes to colourles.

NiBr2 (Blue)

*Answer can only contain numeric values
Test: Electrolysis & Faraday's Law (November 12) - Question 9

An electrochemical cell was based on the following reaction:

Mn(OH)2(s) + H2O2(aq) → MnO2(s) + 2H2O (e)

During the opeartion of this for 1 min, 0.135 g of MnO2 was produced. What is the average electric current (in ampere ) produced by the cell?  


Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 9

*Answer can only contain numeric values
Test: Electrolysis & Faraday's Law (November 12) - Question 10

A fully charged battery contains 500 mL of 5.00 M H2SO4. What is the concentration of H2SO4 in the battery after 6.0 A of current is drawn from the battery for 13.40 h?


Detailed Solution for Test: Electrolysis & Faraday's Law (November 12) - Question 10

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