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Test: Thermodynamics of Electrochemistry


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24 Questions MCQ Test Chemistry for JEE | Test: Thermodynamics of Electrochemistry

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Test: Thermodynamics of Electrochemistry - Question 1

Only One Option Correct Type

This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

Q.

Temperature coefficient of EMF of a cell in terms of entropy change is

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 1




 is called temperature coefficient of EMF of a cell. Thus,
temperature coefficient of

Test: Thermodynamics of Electrochemistry - Question 2

The standard reduction potential at 298 K of the reaction, 

2H2O + 2e-  H2 + 2OH is 0.8277 V

Thus,thermodynamic equilibrium constant for the reaction. 

2H2 H3O+ + OH- at 298 K is  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 2



Test: Thermodynamics of Electrochemistry - Question 3

Given ,  

Thus ,equilibrium constant for the reaction in terms of log k is

2Fe3+ + 3I-  2Fe2+ + I-3   

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 3


Test: Thermodynamics of Electrochemistry - Question 4

Given ,

Thus ,(log Keq) for the reaction   Cu2+ +In2+  Cu+ + In3+ is   

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 4




Test: Thermodynamics of Electrochemistry - Question 5

An excess of liquid mercury is added to an acidified solution of 1.0 x 10-3 M Fe3+ .Thus  is if 5% of Fe3+ remains at equilibrium at 298 K

  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 5

When equilibrium is set up



Test: Thermodynamics of Electrochemistry - Question 6

EMF of the following cell is 0.2905 V

Zn/Zn2+ (a = 0.1M)|| Fe2+ (a = 0.01M)| Fe

 The equilibrium constant for the cell reaction is   

[IIT JEE 2004]

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 6



Test: Thermodynamics of Electrochemistry - Question 7

The half-cell reactions for rusting of iron are

 

ΔG° (in kJ) for the reaction is     

[IIT JEE 2005]

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 7

In rusting of iron, Fe2+ is formed. Thus 

Test: Thermodynamics of Electrochemistry - Question 8

The Gibbs free energy for the decomposition of Al2O3 at 800 K is as follows :

2Al2O→ 4Al + 3O2 , ΔrG = 2898kJ mol-1

The potential difference needed for electrolytic reduction of Al2O3 is at least  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 8


n= electrons exchanged = 12 in the given reaction 

Test: Thermodynamics of Electrochemistry - Question 9

Given ,  

The value of standard electrode potential for the half-reaction is

Fe3+(aq) + e- → Fe2+(aq) 

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 9


Since, different number of electrons are involved hence

Test: Thermodynamics of Electrochemistry - Question 10

For the reaction, 2H2(g) + O2(g) → 2H2O(l), E0cell = 1.23 V at 298 K 

and  ΔH0(H2O) = - 285.8 kJ mol-1 Thus, ΔS° (standard entropy change ) is  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 10


n = 4 (four electrons are involved) as

Test: Thermodynamics of Electrochemistry - Question 11

For the reaction ,

Thus , for the reaction 

  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 11

Electron involved are different, hence

Signs are taken as per required reaction

III is obtained as (I) - (II)

Test: Thermodynamics of Electrochemistry - Question 12

Consider the following equations for a cell reaction. 

A+B  C+ D, E0 = x volt, Keq = K1

2A +2B  2C+ 2D, E0 = y volt, Keq = K2

Then,   

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 12

When a chemical reaction is m ultiplied/divided EMF of the changed equation remains constant but equilibrium constant is raised to power of that change.

Test: Thermodynamics of Electrochemistry - Question 13

Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always true?

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 13

For a spontaneous reaction,

   ΔG < 0

Also,  ΔG = -nFEcell
∴    -ve = -nFEcell
∴ Ecell > 0

Also, Ecell = 

if equilibrium is reached, Ecell = 0

and Q = Keq

∴   


To make       

Ecell >0

Ecell > Q

Test: Thermodynamics of Electrochemistry - Question 14

For a (Ag-Zn) button cell ,the net reaction is 

Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

ΔG0f(Ag2O) = -11.21kJmol-1,  ΔGf(ZnO) = - 318.3 kJ mol-1

Hence, E°cell of the button cell is  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 14

ΔG0f(element, as Zn, Ag) = 0

Thus, ΔG(Button cell) 

=ΔG(ZnO) - ΔGf (Ag2O)

=-318.30 - (-11.21)

=-307.09 kJ mol-1

 ΔG0 = -nFE0cell        (n=2)

∴   

Test: Thermodynamics of Electrochemistry - Question 15

For the reaction ,

Thus  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 15

for the given reaction

E0cell  = 2.73 V

n(electrons exchanged) = 12

(4 Al → Al3+ + 12e-)

F=96500 C mol-1

ΔG0 (element) = 0

∴ ΔG(reaction) = -nFE0cell

= -12 x 96500 x 2.73

=3161340 J = 3161.34 kJ

thus,

Test: Thermodynamics of Electrochemistry - Question 16

Matching List Type

Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

Q.

The standard reduction potential data at 298 K is given below:

Match E° of a redox pair in Column I with the values given in Column II and select the corect answer using the codes given below:

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 16

In all cases, we use





Equal number of electrons are involved. 



Test: Thermodynamics of Electrochemistry - Question 17

The standard potential of the following cell is 0.23 V at 288 K and 0.21 V at 308 K. 

                                       

Match the parameters in Column I with their values in Column II and select the answer from the codes given below the list.

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 17

(i) Temperature coefficient of EMF 








Test: Thermodynamics of Electrochemistry - Question 18

Comprehension Type

This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)  

Passage I

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several process such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is

M(s) | M+ (aq, 0.05M), || M+(aq) 1M|M(s) | Ecell | = 77mV

Q.

For the above cell,

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 18




Test: Thermodynamics of Electrochemistry - Question 19

Passage I

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several process such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is

M(s) | M+ (aq, 0.05M), || M+(aq) 1M|M(s) | Ecell | = 77mV

Q.

If 0.05 M solution of M+ is replaced by 0.0025 M solution of M+,then |Ecell | would be  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 19




Test: Thermodynamics of Electrochemistry - Question 20

Passage II 

Given,

ΔG0f(AgCl) = -109kJmol-1, ΔG0f(Cl-) = -129kJmol-1

ΔG0f(Ag+) = 77kJmol-1,

Thus E°cell of the cell reaction is  

Ag+(aq) + Cl-(aq) → AgCl(s) is  

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 20




Test: Thermodynamics of Electrochemistry - Question 21

Passage II 

Given,

ΔG0f(AgCl) = -109kJmol-1, ΔG0f(Cl-) = -129kJmol-1

ΔG0f(Ag+) = 77kJmol-1,

Q.

Ksp of AgCl is thus,

Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 21


For equilibrium, AgCl(s)  Ag+  + Cl-    E0 = -0.59V


-0.59 = 0.0591logKsp

∴   log Ksp = -10

∴ Ksp = 10-10

*Answer can only contain numeric values
Test: Thermodynamics of Electrochemistry - Question 22

One Integer Value Correct Type

This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

A platinum electrode is immersed in a solution containing 0.1 M Fe2+ and 0.1 M Fe3+.It iscoupled with SHE.Concentration Fe3+ of increased to 0.1 M without change in [Fe2+], then the change in EMF (in centivolt) is 


Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 22



*Answer can only contain numeric values
Test: Thermodynamics of Electrochemistry - Question 23

Using Cr2O72- aqueous, solution  
E0red = 1.33V and ΔG0 = -770.07 kJmol-1

What is the valency of the ion formed after reduction?  


Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 23


*Answer can only contain numeric values
Test: Thermodynamics of Electrochemistry - Question 24

  Equilibriumconstant of the cell reaction,

Cu + 2Fe3+  2Fe2+ + cu2+ is y x 1014

 What is the value of y? 


Detailed Solution for Test: Thermodynamics of Electrochemistry - Question 24



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