NEET and AIPMT NEET Chemistry Electrochemistry MCQ Questions Solved


An electrochemical cell is shown below Pt, H2(1 atm)| HCI (0.1 M)CH3COOH (0.1 M)|

H2(1 atm), Pt The EMF of the cell will not be zero, because

(a) EMF depends on molarities of acids used

(b) pH of 0.1 M HCl and 0.1 M CH3COOH is not same

(c) the temperature is constant

(d) acids used in two compartments are different

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Electrode Potential / Introduction

(b) The EMF of the cell will not be zero because concentration of H+ ions in two electrolytic solutions is different. Mean HCl is strong acid where, acetic acid is weak acid and gives different pH.

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Saturated solution of KNO3 is used to make 'salt-bridge' because:

(a) velocity of K+ is greater than that of NO3-

(b) velocity of NO3- is greater than that of K+

(c) Velocities of both K+ and NO3- are nearly the same

(d) KNO3 is highly soluble in water

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(c) The salt bridge possesses, the electrolyte having nearly same ionic mobilities of its cation and anion.

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A current is passed through two voltameters connected in series. The first voltmeter connected in series. The first voltmeter contains XSO4(aq) while the second voltmeter contains Y2SO4(aq). The relative  atomic masses of X and Y are in the ratio of 2:1. The ration of the mass of X liberated to the mass of Y liberated is: 

(a) 1:1

(b) 1:2

(c) 2:1

(d) none of these

(a) Equal eqivalents of each are liberated.

      Eq. of X=Eq. of Y

                m12M2=m2M1                           m1=m2

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The mass of silver(eq. mass = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is:

(a) 54 g

(b) 108 g

(c) 5.4 g

(d) none of these

Concept Videos :-

#1 | Introduction to Electrolysis
#2 | Types of Electrolysis
#3 | Faraday's Laws of Electrolysis
#4 | Problems on Laws of Electrolysis

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Faraday’s Law of Electrolysis

(a) Eq. of Ag = Eq. of H2;

         m108=5600×222400×1      mAg = 54 g

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A silver cup is plated with silver by passing 965 coulomb of electricity. The amount of Ag deposited is:

(a) 1.08 g

(b) 1.0002 g

(c) 9.89 g

(d) 107.89 g

 

Concept Videos :-

#1 | Introduction to Electrolysis
#2 | Types of Electrolysis
#3 | Faraday's Laws of Electrolysis
#4 | Problems on Laws of Electrolysis

Concept Questions :-

Faraday’s Law of Electrolysis

(a) m = 108x965/96500 = 1.08g

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At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 Ω-1 cmmol-1 and at infinite dilution its molar conductance is 238 Ω-1 cmmol-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

(a) 2.080 %

(b) 20.800 %

(c) 4.008 %

(d) 40.800 %

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Specific and molar Conductance/Conductivity

(C) Given, molar conductance at 0.1 M concentration,

λc = 9.54Ω-1 cmmol-1

Molar conductance at infinite dilution, 

λc= 238 Ω-1 cmmol-1.

We know that,

degree of ionisation, 

α= λc/λc

=(9.54/238) x 100 = 4.008%

 

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Which is the correct representation for Nernst equation ?

(a) ERP = ERP + 0.059nlogoxidantreductant

(b) EOP = EOP - 0.059nlogoxidantreductant

(c) EOP = EOP + 0.059nlogreductantoxidant

(d) All of the above

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#6 | Nernst Equation
#7 | Problem Set 1 on Nernst Equation
#8 | Problem Set 2 on Nernst Equation

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Nernst Equation

(d) All are same.

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When a copper wire is immersed in a solution of AgNO3, the colour of solution becomes blue because copper:

(a) forms a soluble complex with AgNO3

(b) is oxidised to Cu2+

(c) is reduced to Cu2-

(d) splits up into atoic form and dissolves

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Electrochemical series

(b) Cu is above Ag in electrochemical series and thus,

      Cu + 2Ag+  Cu2+ + 2Ag reaction occurs.

      

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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The specific conductance of a 0.1N KCl solution at 23°C is 0.012 Ω-1cm-1. The resistance of cell containing the solution at the same temperature was found to be 55 Ω. The cell constant will be 

(a) 0.142 cm-1

(b) 0.66 cm-1

(c) 0.918 cm-1

(d) 1.12cm-1

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Concept Questions :-

Specific and molar Conductance/Conductivity

(b) Specific conductivity, k=0.012Ω-1cm-1

              k=1resistance×1aG=1R1a=cell constant1a=55 × 0.012 = 0.66cm-1

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Given below are the half-cell reactions,

Mn2+ + 2e-  

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Electrode Potential / Introduction

(a) 

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