CBSE Class 12 Physics Chapter 3 – Electrochemistry Important Question Answer

Very Short Type Question – 1 Mark

1. Define the term – standard electrode potential?
Ans. When the concentration of all the species involved in a half-cell is unity, then the electrode potential is called standard electrode potential.
2. What is electromotive force of a cell? 
Ans. Electromotive force of a cell is also called the cell potential. It is the difference between the electrode potentials of the cathode and anode.

3. Can an electrochemical cell act as electrolytic cell? How?
Ans. Yes, An electrochemical cell can be converted into electrolytic cell by applying an external opposite potential greater than its own electrical potential.

4. Can you store AgCl solution in Zinc pot?
Ans. No. We can’t store AgCl solution in Zinc pot because standard electrode potential of Zinc is less than silver.
5. Single electrode potential cannot be determined. Why? 
Ans. A single half cell does not exist independently as reduction and oxidation occur simultaneously therefore single electrode potential cannot be measured.
6. What is SHE? What is its electrode potential?
Ans. SHE stands for standard Hydrogen electrode. By convention, its electrode potential is taken as 0 (zero).
7. What does the positive value of standard electrode potential indicate?
Ans. The positive value of standard electrode potential indicates that the element gets reduced more easily than ions and its reduced form is more stable than Hydrogen gas.

8. What is an electrochemical series? How does it predict the feasibility of a certain redox reaction?

Ans. The arrangement of metals and ions in increasing order of their electrode potential values is known as electrochemical series.
The reduction half reaction for which the reduction potential is lower than the other will act as anode and one with greater value will act as cathode. Reverse reaction will not occur.
9. Give some uses of electrochemical cells?
Ans. Electrochemical cells are used for determining the
a) pH of solutions
b) solubility product and equilibrium constant
c) in potentiometric titrations
10. A cell is represented by notation –
 Calculate e.m.f of the cell if and?          
Ans. 

=0.80V – (+0.34V)
=+0.46V
11. What would happen if Nickel spatula is used to stir a solution of?
 ,?                                                             
Ans. From the reduction potential values, it is indicated that Nickel (more negative value) is more reactive than copper and will, then displace copper from 
.
12. State the factors that affect the value of electrode potential?                                   
Ans. Factors affecting electrode potential values are –
a) Concentration of electrolyte
b)  Temperature.
13. Write Nernst equation for a Daniel cell?     
Ans. Daniel cell:

Nernst equation – at 298 K
 
14. Define the term specific resistance and give its SI unit
Ans. The specific resistance of a substance is its resistance when it is one meter long and its area of cross Section is one m² . Its SI unit is (ohm meter)
15. Give the unit of conductance?
Ans. The SI unit of conductance is Siemens, denoted by the symbol, S & is equal to.
16. What do you understand by the term- conductivity?

Ans. Conductivity of a material in is its conductance when it is 1m long and its area of cross – section is. It is represented by .
17. What do you understand by strong and weak electrolytes?

Ans. An electrolyte that ionises completely in solution is a strong electrolyte eg. NaCl, etc and an electrolyte that ionizes partially in solution is weak electrolyte eg .
18. State kohlrausch’s Law?
Ans. Kohlrausch Law of independent migration of ion states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
19. What is meant by Faraday’s constant?
Ans. Faraday’s constant is the quantity of electricity carried by one mole of electrons.
1 F = 96500 C/mol
20. How many faradays are needed to reduce 3g mole of to Cu metal?
Ans. 
Two faradays are needed to reduce 1g mole 6 Faradays will be needed to reduce 3g mole of.
21. Give the reaction taking place in lead storage battery when it is on charging?
Ans. When the lead storage battery is on charging –
 
22. A Leclanche cell is also called dry cell. Why?
Ans. Leclanche cell consists of zinc anode (container) and carbon cathode. The electrolyte is a moist paste of and carbon black. Because there is no free liquid in the cell, it is called dry cell.
23. Why is the voltage of a mercury cell constant during its working?
Ans. As all the products and reactants are either in solid or liquid state, their concentration does not change with the use of the cell.
24. What are fuel cells?
Ans. A fuel cell is a galvanic cell for converting the energy of a fuel directly into electrical energy without use of a heat engine.
25. What do you understand by corrosion?
Ans. Corrosion is an electrochemical phenomenon in which metal gets decomposed in the presence of air and water and forms compounds like oxides, sulphates, carbonates, sulphides etc.
26. Name two metals that can be used for cathodic protection of iron?
Ans. Names of the metals are –  Zinc and Magnesium.
27.  Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn
Ans. The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu
28.  Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Ans. Methane and methanol can be used as fuels in fuel cells.
29.  Suggest a list of metals that are extracted electrolytically.
Ans. Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

Short type question answers – 2 Marks

1. How is standard electrode potential of a cell related to :-
1) Equilibrium constant?
2) Gibbs free energy change.
Ans. (i) Standard electrode potential and equilibrium constant

Where = standard electrode potential of cell
R = Gas constant
T = temperature in Kelvin
n = no. of electrons.
F = Faraday’s constant and
Kc = Equilibrium constant
(ii) Standard electrode potential and Gibbs free energy change-

Where= Change in Gibbs’ free energy
n =  No. of electrons
F =  Faraday’s Constant
=  Standard electrode Potential of cell.

2. What is the half cell potential for  electrode in which.
        
Ans. 
According to Nernst Equation –
   log 
=   0.771 V – 
=   0.771 V – 0.0197 V
=   +0.7513V

3. Calculate pH of following half cell. Pt , , if its electrode potential is 0.03V.       
Ans. 
The cell reaction is –

According to Nearest Equation

0.03V  =  0 + 

PH =      =  5.07 V

4. What are the factors on which conductivity of an electrolyte depend?
Ans. The conductivity of an electrolyte depends upon
i)  The nature of electrolyte
ii)  Size of the ions produced
iii)  Nature of solvent and its viscosity.
iv) Concentration of electrolyte.
v)  Temperature

5. How is molar conductance related to conductivity of an electrolyte ?
Ans. Molar conductance, m is related to conductively by the relation.
m = 
Where = conductivity in s/m.
C = concentration in 

6. Write an expression relating cell constant and conductivity?
Ans. Cell constant and conductivity are related by the expression-
 =   where   G = Cell constant
 = conductivity
R = Resistance.

7. The conductivity of an aqueous solution of NaCl in a cell is  the resistance offered by this cell is 247.8. Calculate the cell constant?
Ans. Specific conductivity = 
Or cell constant = 
= 92 
= 22797.6 

8. The molar conductivity of 0.1M CH₃COOH solution is. What is the conductivity and resistivity of the solution?
Ans. 
= 
 = 
= 
= 0.00046 s/cm
Resistivity = 
= .
9. The conductivity of metals decreases while that of electrolytes increases with increases in temperature. Why?
Ans. With increase in temperature, the K.E. of metal cation increases and obstructs the free flow of electrons decreasing the conducts of metal while in case of electrolytes, increased temperature increases the mobility of ions this increases the conductance of ions.

10. The measured resistance of a cell containing solution of KCl at  was 1005  calculate
 (a) Specific conductance and
 (b) Molar conductance of the solution.  Cell Constant = 
Ans. 
.

11. How is Limiting molar conductivity related to
 i)  degree of ionization and  
 ii) dissociation constant
Ans. Relation between limiting molar conductance and degree of dissociation –
      where = degree of dissociation
molar conductance molar
Limiting molar conductance
Relation between dissociation constant and limiting molar conductance –
where c = concentration

12. In fig. (1), identify the nature  of electrolyte A& B. In which case it is not possible to obtain value of limiting molar conductance?   

Ans. A = strong Electrolyte
B = weak Electrolyte
In case of B , it is not possible to get an exact value of limiting molar conductance.

13. At 298 K , the molar conductivities at infinite dilution of , NaOH and NaCl are 129.8 , 217.4 and respectively .It molar conductivity of 0.01M solution is,calculate the degree of dissociation of at this dilution?
Ans. 
= 129.8 +217.4 – 108.9    
Degree of dissociation,  = 
= 0.039  or  3.9 %.

14. State Faraday’s Laws of electrolysis?
Ans. Faraday’s Laws of electrolysis
First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
Second Law: The amount of different substances liberated by the same quantity of electricity passing through the electrolytic solution is proportional to their chemical equivalent weights.

15. How many g of chlorine can be produced by the electrolysis of molten NaCl with a current of 1 amp. for 15 min?
Ans. Q = It    
The reaction is 
 2mol  1mol  2mol

1mol of 

900 C  will produce    .

16. How many electrons flow when a current of 5 amps is passed through a solution for 193 sec. Given f = 96500 C.?
Ans. Q = It   
96500C = 1 mol of electrons
965 C = 
electrons.

17. There are two possible reactions for cathode in the electrolysis of aqueous : E = -0.76v,E = – 0.83v Which one will take place ?
Ans. has higher reduction potential (-0.76v) Than and therefore is reduced to Zn preferentially at cathode.

18. What do you mean by primary and secondary battery?
Ans. In the primary batteries, the reaction occurs only once and after the use over a period of time battery becomes dead and cannot be reused again. A secondary battery , after used, can be recharged by passing current through it in the opposite direction so that it can be used again.

19. Name the cell used for low current devices like hearing aids, watches etc. Also give the half cell reactions for such a cell?

Ans. This cell is mercury cell – Half cell reactions are Anodeand those 

20. Rusting of iron is quicker in saline water than in ordinary water. Explain?
Ans. Saline water consists of greater no. of ions than normal water which increases the electrochemical reaction. This increases rate of corrosion.

21. Enlist the factors affecting corrosion?
Ans. Factors affecting corrosion are –
1) Water and air
2) Presence of electrolytes in water.
3) Presence of gases like  .

22.  The conductivity of 0.20 M solution of KCl at 298 K is . Calculate its molar conductivity.
Ans. Given,
K= 0.0248 S 
c = 0.20 M
Therefore, Molar conductivity, 

23.  The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is .
Ans. Given,
Conductivity, 
Resistance, R = 
Therefore, Cell constant = 
= 
= 

24.  In the button cells widely used in watches and other devices the following reaction takes place:  Determine andfor the reaction.
Ans.

Therefore, = 1.104 V
We know that,

= 
= –213043.296 J
= –213.04 kJ

25.  What is the quantity of electricity in coulombs needed to reduce 1 mol of ? Consider the reaction: 
Ans. The given reaction is as follows: 
Therefore, to reduce 1 mole of , the required quantity of electricity will be:
=6 F = 
= 578922 C

26.   Why does the conductivity of a solution decrease with dilution?
Ans. The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

27.   Suggest a way to determine the  value of water.
Ans. Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows:



Hence, by knowing the values of HCl, NaOH, and NaCl, the value of water can be determined.

28.   Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Ans. Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.

This implies that the substances having higher reduction potentials than
+0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are,, and .

Short type question answer – 3 Marks

1.Calculate  for the reaction at  , , R = 8.314 J/K. 
Ans.The half cell reactions are
Anode: 
Cathode: 
Nernst Equation
 
= (- 0.403 – (-0.763) – 
= 0.36V  – 0.0798V = 0.4398V


=  -8488  J mol^-1

2.Calculate Equilibrium constant K for the reaction at = +0.34v.
Ans.From the reaction, n =2

= + 0.34v – (-0.76v) = 1.10V

At 298k , 
Log k_c =
= 

3. For what concentration of will the emf of the given cell be zero at 
 if the concentration of is 0.1 M ?   
Ans.

4.Calculate the standard free energy change for the cell- reaction.
How is it related to the equilibrium constant of the reaction?, 
Ans.

5.How much charge is required for the following reductions:
(i) 1 mol of  to Al.
(ii) 1 mol of to Cu.
(iii) 1 mol of  to .
Ans.(i) 
Therefore, Required charge = 3 F

= 289461 C
(ii) 
Therefore, Required charge = 2 F

= 192974 C
(iii) 
i.e., 
Therefore, Required charge = 5 F

= 482435 C

6. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten .
(ii) 40.0 g of Al from molten .
Ans.(i) According to the question,

Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium =
= 1 F
(ii) According to the question,

Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al = 
= 4.44 F

7. How much electricity is required in coulomb for the oxidation of
(i) 1 mol of  to .
(ii) 1 mol of FeO to .
Ans.(i) According to the question,

Now, we can write:

Electricity required for the oxidation of 1 mol of to = 2 F

= 192974 C
(ii) According to the question,

Electricity required for the oxidation of 1 mol of FeO to = 1 F
= 96487 C

8. A solution of  is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Ans.Given,
Current = 5A
Time = = 1200 s
Therefore, 

= 6000 C
According to the reaction,

Nickel deposited by = 58.71 g
Therefore, nickel deposited by 6000 C 
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.

9. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans.A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide as the cathode, and a 38% solution of sulphuric acid  as an electrolyte.

When the battery is in use, the following cell reactions take place:
At anode: 
At cathode: 
The overall cell reaction is given by,

When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, present at the anode and cathode is converted into and respectively.

Ans.I = 0.5 A
t = 2 hours = 
Thus, Q = It
= = 3600 C
We know that  number of electrons.
Then,  number of electrons
 number of electrons
Hence, number of electrons will flow through the wire.

11.  Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Ans.For hydrogen electrode, , it is given that pH = 10
Therefore,  = 10 – 10M
Now, using Nernst equation:



= – 0.0591 log  = – 0.591 V
12. Calculate the emf of the cell in which the following reaction takes place:

Given that 
Ans.Applying Nernst equation we have:



= 1.05 – 0.02955 log 4   104
= 1.05 – 0.02955 (log 10000 + log 4)
= 1.05 – 0.02955 (4 + 0.6021)
= 0.914 V

13.  How would you determine the standard electrode potential of the systemMg2+| Mg?
Ans.The standard electrode potential of | Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), (1 atm) | (aq) (1M).

A cell, consisting of Mg | (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

Here, for the standard hydrogen electrode is zero.
Therefore, 
=

Long term question answer – 5 Marks

1. Explain construction and working of standard Hydrogen electrode?
Ans. Construction :
SHE consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure Hydrogen gas is bubbled through it. The concentration of both the reduced and oxidized. Forms of Hydrogen is maintained at unity i.e) pressure of gas is 1 bar and concentration of Hydrogen ions in the solution is 1 molar.
Working – The reaction taking place in SHE is At 298 K , the emf of the cell constructed by taking SHE as anode and other half cell as cathode, gives the reduction potential of the other half cell where as for a cell constructed by taking SHE as anode gives the oxidation potential of other half cell as conventionally the electrode potential of SHE is zero.

2.   The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S . Calculate its degree of dissociation and dissociation constant. Given = 349.6 Sand 
Ans. C = 0.025 


= 349.6 + 54.6 = 
Now, degree of dissociation:


= 0.114 (approximately)
Thus, dissociation constant:

3.  Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans. In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by, 
Electrons released at the anodic spot move through the metallic object and go to another spot of the object.
There, in the presence of ions, the electrons reduce oxygen. This spot behaves as the cathode. These H+ ions come either from, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is given by, 
The overall reaction is: 
Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide i.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

4.  Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i)  (ii) 
Calculate the ¸ and equilibrium constant of the reactions
Ans. (i) 

The galvanic cell of the given reaction is depicted as: 
Now, the standard cell potential is 
= 0.40 – (–0.74)
= +0.34 V

In the given equation,
n = 6
F = 96487 C mol – 1
= +0.34 V
Then, 
= –196833.48 CV 
= –196833.48 J 
= –196.83 kJ v
Again, 



= 34.496
Therefore, K = antilog (34.496)

(ii) 

The galvanic cell of the given reaction is depicted as: 
Now, the standard cell potential is 
= 0.80 – 0.77
= 0.03 V
Here, n = 1.
Then, 
= 
= –2894.61 J 
= –2.89 kJ 
Again, 


= 0.5073
Therefore, K = antilog (0.5073)
= 3.2 (approximately)

5.  Write the Nernst equation and emf of the following cells at 298 K:
(i) 
(ii) 
(iii) 
(iv) 
Ans. (i) For the given reaction, the Nernst equation can be given as:



= 2.7 – 0.02955
= 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:


= 0.44–0.02955(–3)
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:


= 
= 0.14–0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:





= –1.09–0.02955(0.0453+7)
= –1.09–0.208
= –1.298 V

6.  Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans. Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol K. If is resistivity, then we can write: 

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.
i.e., 
(Since a = 1, l = 1)
Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length. 
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
Therefore, 
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of with for strong and weak electrolytes is shown in the following plot:

7.  The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100  
Calculate for all concentrations and draw a plot betweenand . Find the value of .
Ans. Given, 
Then,  
Therefore, 


Given,

Then, 
Therefore, 

= 
Given,

Then, 
Therefore, 


Given,

Then, 
Therefore, 

= 
Given,

Then, 
Therefore, 


Now, we have the following data:

	0.0316	0.1	0.1414	0.2236	0.3162
	123.7	118.5	115.8	111.1	106.74

Since the line interrupts at , 

8.  Conductivity of 0.00241 M acetic acid is . Calculate its molar conductivity and if for acetic acid is , what is its dissociation constant?
Ans. Given, 

Then, molar conductivity, 


Again, 
Now, 
= 0.084
Therefore, Dissociation constant, 

9.  Three electrolytic cells A,B,C containing solutions of  and respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Ans. According to the reaction:

i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = 
= 1295.43 C
Given,
Current = 1.5 A
Therefore, Time 
= 863.6 s
= 864 s
= 14.40 min
Again,

i.e., C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = 
= 0.426 g of Cu

i.e., C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit =
= 0.439 g of Zn

10.  Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) and
(ii) and Cu(s)
(iii) and  
(iv) Ag(s) and  
(v) and .
Ans.

Sincefor the overall reaction is positive, the reaction between and is feasible.

Since for the overall reaction is positive, the reaction between and Cu(s) is feasible.

Sincefor the overall reaction is negative, the reaction between and is not feasible.

Since E for the overall reaction is negative, the reaction between Ag(s) and is not feasible.

Sincefor the overall reaction is positive, the reaction between and is feasible.

11.  Predict the products of electrolysis in each of the following:
(i) An aqueous solution of with silver electrodes.
(ii) An aqueous solution of with platinum electrodes.
(iii) A dilute solution of with platinum electrodes.

(iv) An aqueous solution of with platinum electrodes.
Ans. (i) At cathode:
The following reduction reactions compete to take place at the cathode.


The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by ions. Therefore, the silver electrode at the anode dissolves in the solution to form .
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.


The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by ions. Therefore, or ions can be oxidized at the anode. But ions having a lower discharge potential and get preference and decompose to liberate.


(iii) At the cathode, the following reduction reaction occurs to produce gas.

At the anode, the following processes are possible.
……………(1)
……………(2)
For dilute sulphuric acid, reaction (i) is preferred to produce gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.


The reaction with a higher value of takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
Atanode:
The following oxidation reactions are possible at the anode.


At the anode, the reaction with a lower value of is preferred. But due to the over-potential of oxygen, gets oxidized at the anode to produce gas.