NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

Edited By Shivani Poonia | Updated on Jun 05, 2025 05:54 PM IST | #CBSE Class 12th
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CBSE Class 12th  Application Date : 30 May' 2025 - 17 Jun' 2025

Electrochemistry is an important branch of chemistry that explores between electricity and chemical reactions. It focuses on how chemical energy changes into electrical energy, and vice versa. Have you ever wondered how a battery powers your phone or why metals are coated with a layer of gold? The answers to these questions lie in Electrochemistry.

This Story also Contains
  1. NCERT Solutions Of Class 12 Chemistry Chapter 2 Electrochemistry (Intext Questions 2.1 to 2.15)
  2. NCERT Solutions for class 12 Electrochemistry (Exercise Questions)
  3. Class 12 Chemistry NCERT Chapter 2: Higher Order Thinking Skill (HOTS) Questions
  4. Approach to Solve Questions of Chapter 4 Electrochemistry
  5. Topics of NCERT Class 12 Chemistry Chapter 2
  6. What Extra Should Students Study Beyond NCERT for JEE?
  7. Formulas Of NCERT Class 12 Chemistry Electrochemistry
  8. NCERT Solutions for Class 12 Chemistry
  9. NCERT Solutions for Class 12 Subject-wise
NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry
NCERT Solutions for Class 12 Chemistry Chapter 2 - Electrochemistry

NCERT solutions for class 12 Chemistry Chapter 2 Electrochemistry deals with questions based on mainly electrochemical and galvanic cells and also on the Nernst equation in order to calculate electromotive force potential. Electrochemistry Class 12 will also acknowledge you to various types of batteries and their benefits. The chapter is important for both theoretical and practical purposes. Therefore electrochemistry class 12 NCERT solutions become very important to get an in-depth understanding of concepts.

Also Read,

NCERT Solutions Of Class 12 Chemistry Chapter 2 Electrochemistry (Intext Questions 2.1 to 2.15)

Page 36

Question 2.1 How would you determine the standard electrode potential of the system $Mg^{2+} | Mg$ ?
Answer :

To determine the standard electrode potential of the given system we need to use a hydrogen electrode. In the setup, we shall put a hydrogen electrode as cathode and Mg | MgSO 4 as an anode.

$\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right|\left|\mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{H}_2(\mathrm{~g}, \mathrm{I}$ bar $), \mathrm{Pt}_{(\mathrm{s})}$

Now we will measure the emf of the cell. This emf will be the standard electrode potential of the magnesium electrode.

E°cell = E° right – E°left
E°left =0 ( The standard hydrogen electrode is always zero)
Hence
$E^{o}cell = E^{o} Mg|Mg^{2+}$

Question 2.2 Can you store copper sulphate solutions in a zinc pot?

Answer:

The standard electrode potential of Zinc is - 0.76 whereas that of Copper is 0.34. So Zinc will reduce copper into the lower state.

It is known that zinc is more reactive than copper. Thus if we will store copper sulphate solution in zinc pot then zinc will displace copper from its solution.

The following reaction will take place:-

$Zn \:+\:CuSO_{4}\rightarrow ZnSO_{4}\:+\:Cu$

Question 2.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Answer :

The oxidising strength of elements increases as the standard electrode potential increases.

So all the elements having greater standard potential than iron can oxidise it to a higher state.

Few such elements are :- F 2 , Cl 2 , Br 2 , Ag +1 etc.

Page 41

Question 2.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer :

It is given that pH of the solution is 10,i.e., the hydrogen ion concentration in the solution is 10 -10 M.

$\mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2$

By Nernst equation we have :-

$E_{Cell} = E_{cell}^{\circ}\ - \frac{RT}{2F}ln \frac{1}{\left [ H^+ \right ]}$

So, $= 0 - \frac{0.0591}{1}log \frac{1}{\left [ 10^{-10} \right ]}$

or $= -\ 0.591\ V$

So the required potential is - 0.591 V.

Question 2.5 Calculate the emf of the cell in which the following reaction takes place:

$Ni(s)+2Ag^{+} (0.002M)\rightarrow Ni^{2+}(0.160M)+ 2Ag(s)$

Given that $E^{\Theta }_{(cell) }= 1.05 \, V$

Answer :

Here we can directly apply the nernst equation :-

$E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{[Ni^{+2}]}{\left [ Ag^+ \right ]^2}$

Putting the value in this equation :-

$= 1.05\ - \frac{0.0591}{2}log \frac{0.160}{(0.002)^2}$

or $= 1.05\ - 0.02955\ log (4\times10^4)$

or $= 0.914\ V$

Hence the required potential is 0.914 V.

Question 2.6 The cell in which the following reaction occurs:
$2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(s)$ has $E^{0}_{cell}=0.236\, V$ at $298 K.$
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer :

For finding Gibbs free energy we know the relation :-

$\Delta G_r^{\circ} = -\ nFE_{cell}^{\circ}$

$= -\ 2\times96487\times 0.236$

$= -\ 45541.864\ J\ mol^{-1}$

$= -\ 45.54\ KJ\ mol^{-1}$

Now, for equilibrium constant we will use :-

$\Delta G_r^{\circ} = -2.303\ RTlog\ K_c$

So, $logK_c = -\frac{-45.54\times10^3}{2.303\times8.314\times298}$

or $logK_c = 7.981$

or $K_c = 9.57\times10^7$

Page 51

Question 2.7 Why does the conductivity of a solution decrease with dilution?

Answer :

The conductivity of a solution depends upon the number of ions and the distance between them. In the process of dilution, we don't increase the number of ions in the solution instead we increase the distance between them. So the conductivity of the solution decreases due to dilution.

Question 2.8 Suggest a way to determine the $\Lambda ^{0}_{m}$ value of water.

Answer :

We know :

$\Lambda _m=\Lambda _m^{\circ} - A c^{\frac{1}{2}}$

If we draw a straight line between $\Lambda _m$ and $\sqrt c$ , its slope will be -A and the intercept on the y-axis will be $\Lambda _m^{\circ}$ .

In this way, we can obtain the value of limiting molar conductivity.

Question 2.9 The molar conductivity of $0.025 mol L^{-1}$ methanoic acid is $46.0 \, S cm^2\, S\, cm^2 mol ^{-1}$
Calculate its degree of dissociation and dissociation constant. Given $\lambda ^{0}(H+)=349.6 \, S cm^{2} mol ^{-1}$ and $\lambda ^{0}(HCOO^{-})=54.6\,\: S\, cm^{2}\, \: mol^{-1}$

Answer :

We know that :-

$\Lambda _m = \lambda^{\circ}(H^+) + \lambda^{\circ}(HCOO^-)$

$= 349.6 +54.6$

$= 404.2\ Scm^2\ mol^{-1}$

For degree of dissociation, we have :-

$\alpha = \frac{\Lambda _m(HCOOH)}{\Lambda ^{\circ}(HCOOH)}$

or $\alpha = \frac{46.1}{404.2} = 0.114$

For dissociation constant, we have :-

$K_a = \frac{c\alpha ^2}{1-\alpha }$

or $K_a = \frac{0.025\times(0.114)^2}{1-0.114 }$

or $= 3.67\times 10^{-4}\ mol\ L^{-1}$

Page no 54

Question 2.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Answer :

Firstly we will find total charge flown through the wire then we will calculate number of electrons.

We are given :- I = 0.5 A, Time = 2 hours = 7200 seconds.

We have, Q = I.t

= (0.5)7200 = 3600 C.

Now we will convert charge into number of electrons.

We know that $96487\ C = 6.023\times10^{23}\ No.\ of\ electrons$

So toal number of electrons :

$=\frac{3600}{96487}\times 6.023\times10^{23}$

or $=2.25\times 10^{22}$ no. of electrons will flow through wire.

Question 2.11 Suggest a list of metals that are extracted electrolytically.

Answer :

Metals like Na, Mg, Al, etc. are produced on a large scale by electrochemical reduction of their respective cations or by the process electrolysis because there are no suitable reducing agents available for this purpose.

Question 2.12 Consider the reaction: $Cr_{2}O_{7}^{2-}+ 14H^+ + 6e^-\rightarrow 2cr^{3+} + 7H_{2}O$
What is the quantity of electricity in coulombs needed to reduce 1 mol of $Cr_{2}O_{7}^{2-}$ ?

Answer :

It is clear from the given reaction that reduction of 1 mol of Cr 2 O 7 2- will be

= 6 F (as 6 electrons are required to balance the reaction; Charge required = nF)

$= 6\times96500$

$= 579000\ C$

Thus 578922 C charge is required for reduction of 1 mol of Cr 2 O 7 2- .

Page no 58

Question 2.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Answer :

The lead storage battery can be recharged by reversing the direction of current passing through it.

For recharging PbSO 4 is converted into Pb at the anode and into PbO 2 at the cathode.

The chemical reactions are as follows:-

1643888556447

Question 2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Answer :

The two materials are methane and methanol that can be used as fuels in fuel cells.

Question 2.15 Explain how rusting of iron is envisaged as setting up of a electrochemical cell.

Answer :

The chemistry of corrosion is quite complex but it can be understood by considering it as an electrochemical phenomenon. Consider a particular spot on an object where corrosion takes place. At here oxidation takes place and this spot behaves as the anode. The released electrons at anodic spot go through the metal and go to another spot on the metal and reduction of oxygen takes place in the presence of H+. This spot behaves as a cathode with the reaction. In this way this analogy is possible.

NCERT Solutions for class 12 Electrochemistry (Exercise Questions)

Question 2.1 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.

Answer :

The order in which metals displace each other from the solution of their salts can be given with the help of their standard electrode potential. Since magnesium has the least standard electrode potential so it is the most strong reducing agent. So the required order we get is:-

$M g>A l>Z n>F e>C u$

Question 2.2 Given the standard electrode potentials,

$K^{+}/K=-2.93V , Ag^{+}/Ag=0.80V,$

$Hg^{2+}/Hg=0.79V$

$Mg^{2+}/Mg=-2.37V, Cr^{3+}/Cr =-0.74 V$

Arrange these metals in their increasing order of reducing power.

Answer :

Elements with reducing power or reducing agents have least/minimum standard electrode potential i.e., reducing power increases with a decrease in standard electrode potential. So the result obtained is:-

K > Mg > Cr > Hg > Ag

Question 2.3(i) Depict the galvanic cell in which the reaction

$Zn(s)+2Ag^{+}(aq)\rightarrow Zn^{2+}(aq) +2Ag(s))$ takes place.Further show

(i) Which of the electrode is negatively charged?

Answer :

The galvanic cell of the given reaction is depicted below :-

Zn (s) | Zn +2 (aq) || Ag + (aq) | Ag (s)

Clearly Zn electrode is negatively charged.

Question 2.3(ii) Depict the galvanic cell in which the reaction

$Zn(s)+2Ag^{+}(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)$ takes place.

(ii) Further show: The carriers of the current in the cell.

Answer :

The carriers of current in the cell are ions . and Current flows from silver to zinc in the external circuit.

Question 2.3(iii) Depict the galvanic cell in which the reaction $Zn(s)+2Ag^{2+}(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)$

takes place.

(iii) Further show: Individual reaction at each electrode.

Answer :

The reaction taking place at both cathode and anode are shown below :-

(i) Cathode reaction :-

$Ag^+_{(aq)} + e^- \rightarrow Ag_{(s)}$

(ii) Anode reaction :-

$Zn_{(s)}\ \rightarrow Zn^{2+}_{(aq)}\ +\ 2 e^-$

Question 2.4(i) Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

$2Cr(s)+3Cd^{2+}(aq)\rightarrow 2Cr^{3+}(aq)+3Cd$

Calculate the $\Delta _{r}G^{e}$ and equilibrium constant of the reactions.

Answer :

The galvanic cell of the given reaction is shown below:-

$\mathrm{Cr}_{(s)}\left|\mathrm{Cr}_{(\mathrm{aq})}^{3+} \| \mathrm{Cd}_{(\mathrm{aq})}^{2+}\right| \mathrm{Cd}_{(s)}$

The standard electrode potential of Cr and Cd can be found in table of standard electrode potential.

So, we get :

$E^{\circ} = E^{\circ}_R\ -\ E^{\circ} _L$

$= -0.40\ -\ (-0.74)$

$=\ 0.34\ V$

Now

$\Delta G_r^{\circ}\ = -\ nFE_{cell}^{\circ}$

Putting values :

$\Delta G_r^{\circ}\ = -\ 6\times96487\times 0.34$

$= -196.83\ KJ\ mol^{-1}$

Now for finding equlilibrium constant we have :

$log\ k = \frac{-\Delta G _r^{\circ}}{2.303\times R\times T}$

or $log\ k = 34.496$

or $K = 3.13\times10^34$

Question 2.4(ii) Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

$Fe^{2+}(aq)+Ag^{+}(aq)\rightarrow Fe^{3+}(aq)+Ag(s)$

Calculate the $\Delta _{r}G^{e}$ and equilibrium constant of the reactions.

Answer :

The galvanic cell of the given reaction is shown below :-

$\mathrm{Fe}_{(a q)}^{2+}\left|\mathrm{Fe}_{(a q)}^{3+}\right|\left|\mathrm{Ag}_{(a q)}^{+}\right| \mathrm{Ag}_{(s)}$

We can know about the electrode potential of Fe and Ag with the help of table of standard electrode potential.

We have : $E_{(cell)}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ}$

or $= 0.80 - 0.77$

or $= 0.03\ V$

Now consider : $\Delta G_r^{\circ} = -\ nFE_{(cell)}^{\circ}$

or $= -1\times96487\times0.03$

$= -2.89\ KJ\ mol^{-1}$

Now for equilibrium constant :

$log\ K =\ -\frac{\Delta G_r^{\circ}}{2.303\times RT}$

or $=\ -\frac{-2894.61}{2.303\times 8.314\times298}$

or $=\ 0.5073$

Thus $k\ \approx \ 3.2$

Question 2.5(i) Write the Nernst equation and emf of the following cells at 298 K:

(i) $Mg(s)| Mg^{2+}(0.001M)|| Cu^{2+}(0.000.1M)|Cu(s)$

Answer :

The nernst equation gives :

$E_{Cell} = E_{cell}^{\circ}\ - \frac{0.059}{n}log \frac{[Mg^{2+}]}{\left [ Cu^{2+} \right ]}$

This gives,

$= {0.34 - (-2.36)} - \frac{0.059}{2}log \frac{0.001}{ 0.0001}$

or $= 2.7 - 0.02955$

or $= 2.67\ V$

So the emf of the cell is 2.67 V.

Question 2.5(ii) Write the Nernst equation and emf of the following cells at 298 K:

(ii) $Fe(s)|Fe^{2+}(0.001M)||H^{+}(1M)|H_{2}(g)(1 bar)|Pt(s)$

Answer :

The nernst equation for this gives :

$E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{[Fe^{+2}]}{\left [ H^+ \right ]^2}$

This gives : $= 0 - (-0.44)- \frac{0.0591}{2}log \frac{0.001}{1^2}$

or $= 0.44 - 0.02955(-3) = 0.53\ V$

Thus the emf of the given galvanic cell is 0.53 V.

Question 2.5(iii) Write the Nernst equation and emf of the following cells at 298 K:

(iii) $Sn(s)|Sn^{2+}(0.050M)||H^{+}(0.020M)|H_{2}(g)(1 bar)Pt(s)$

Answer :

The nernst equation for this reaction gives :-

$E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{[Sn^{+2}]}{\left [ H^+ \right ]^2}$

Now for emf, just put all the values.

$E_{Cell} =0 - (-0.14) - \frac{0.0591}{2}log \frac{0.050}{0.020^2}$

or $= 0.14 - 0.0295 \times log125$

or $= 0.14 - 0.062 = 0.078\ V$

Thus emf of the cell is 0.078 V.

Question 2.5(iv) Write the Nernst equation and emf of the following cells at 298 K:

$Pt(s)|Br^{-}(0.010M)|Br_{2}(1)||H^{+}(0.030 M)| H_{2}(g)(I bar)Pt(s)$

Answer :

The nernst equation of the given reaction gives :

$E_{Cell} = E_{cell}^{\circ}\ - \frac{0.0591}{n}log \frac{1}{\left[Br^-]^2 [ H^+ \right ]^2}$

or $=(0-1.09)\ - \frac{0.0591}{2}\ log \frac{1}{(0.010)^2 (0.030)^2}$

or $=-1.09\ - 0.02955\times\ log(1.11\times10^7)$

or $=-1.09\ - 0.208 =\ -1.298\ V$

So the required emf of the cell is -1.298 V.

Question 2.6 In the button cells widely used in watches and other devices the following reaction takes place:

$Zn (s)+ Ag_{2}O(s)+H_{2}O(l)\rightarrow Zn^{2+}(aq)+2Ag(s)+2OH^{-}(aq)$

Determine $\Delta_{r}G^{e}$ and $E^{e}$ for the reaction.

Answer :

The given reaction is obtained from :-

$\mathrm{Zn}_{(s)} \longrightarrow \mathrm{Zn}_{(a q)}^{2+}+2 \mathrm{e}^{-} ; E^{\ominus}=0.76 \mathrm{~V}$

$\mathrm{Ag}_2 \mathrm{O}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(t)}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{OH}_{(a q)}^{-} ; E^{\ominus}=0.344 \mathrm{~V}$

So the E o cell can be obtained directly.

$E_{cell}^{\circ} = 0.76 - (-0.344) = 1.104\ V$

Now for free energy calculation, we have :-

$\Delta G_r^{\circ} = -nFE^{\circ}_{cell}$

or $= -2\times96487\times1.04$

or $= - 213043.29\ J$

or $= - 213.04\ KJ$

Question 2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer :

Conductivity(k) or specific conductance of a solution is defined as the inverse of resistivity.

Mathematically, it can be written as:-

$G = \kappa \frac{A}{L}$

In the above equation is $\kappa$ the conductivity of a solution. Thus the definition of conductivity becomes as the conductance of a substance which is 1 cm long and has 1 sq. cm of cross-sectional area.

With dilution conductivity of a solution decreases due to an increase in distance between ions.\

Molar conductivity: - It is defined as the conductivity of a solution per unit concentration

i.e., $\Lambda _M\ =\ \frac{\kappa }{C}$

It is clear from the above mathematical expression of the molar conductivity that, if we dilute the solution or decrease its concentration then molar conductivity increases. This is because, on dilution of a solution, a decrease in is $\kappa$ more than compensated by the increase in its volume.

Question 2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.

Answer :

We know that the molar conductivity of a solution is defined as:-

$\Lambda _M\ = \frac{\kappa }{C}$

Putting the value of conductivity and concentration in the above equation:-

$\Lambda _M\ = \frac{0.0248\times1000 }{0.20} = 124\ Scm^2\ mol^{-1}$

Question 2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 ohm . What is the cell constant if conductivity of 0.001M KCl solution at 298 K is $0.146\times10^{-3}\ S cm^{-1}$ .

Answer:

We are given with conductivity of cell $\kappa =$ $0.146\times10^{-3}\ S cm^{-1}$ and resistance R = 1500 $\Omega$ .

Also, Cell constant = $\kappa\times R$

or = $0.146\times 10^{-3}\times 1500$

or = $0.219\ cm^{-1}$

Question 2.11 Conductivity of 0.00241 M acetic acid is $7.896\times 10^{-5} \: S cm^{-1}$ . Calculate its molar conductivity. If $\Delta _{m}^{0}$ for acetic acid is $390.5 \: S cm^{2} mol^{-1}$ , what is its dissociation constant?

Answer :

Molar conductivity of a solution is given by :-

$\Lambda _M = \frac{\kappa }{C}$

So, $= \frac{7.896\times10^{-5}}{0.00241}\times1000$

or $= 32.76\ Scm^2\ mol^{-1}$

Also, it is given that $\Lambda _m^{\circ}= 390.5\ Scm^2\ mol^{-1}$ .

$\alpha = \frac{\Lambda _m}{\Lambda _m^{\circ}}$

or $\alpha = \frac{32.76}{390.5}$

$\alpha = 0.084$

For dissociation constant we have,

$K_d\ = \frac{c\alpha ^2}{(1-\alpha )}$

so, $= \frac{0.00241\times0.084^2}{(1-0.084 )}$

or $= 1.86\times 10^{-5}\ mol\ L^{-1}$

Question 2.12(i) How much charge is required for the following reductions:

(i) $1\ mol\ of\ Al^{3+}\ to\ Al\ ?$

Answer :

The equation becomes:-

$Al^{+3}\ + 3e^-\ =\ Al$

So required charge is 3F.

Q = n*96500

Q = 3*96500 = 289500 C

Question 2.12(ii) How much charge is required for the following reductions:

(ii) $1\ mol \ of\ Cu^{2+}\ to\ Cu?$

Answer :

The equation can be written as:-

$Cu^{2+}\ +\ 2e^-\ =\ Cu$

Thus charge required is $=\ 2F$

$= 2(96500) = 193000\ C$

Question 2.12(iii) How much charge is required for the following reductions:

(iii) $MnO_{4}^{-}\ to\ Mn^{2+}\ ?$

Answer :

The given reaction can be written as:-

$Mn^{+7}\ +\ 5e^- =\ Mn^{+2}$

Thus charge required in above equation $=\ 5F$

$= 5(96500)$

$= 482500\ C$

Question 2.13(i) How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten $CaCI_{2}$ ?

Answer :

The equation for the question is given by :-

$Ca^{2+}\ +\ 2e^-\ =\ Ca$

In this equation, for 1 mol of Ca, 2F charge is required or we can say that for 40 g of Ca charge required is 2F.

So, for 20 g of Ca charge required will be = F = 96500 C.

Question 2.13(ii) How much electricity in terms of Faraday is required to produce

(ii)40.0 g of AI from molten $AI_{2}O_{3}$ ?

Answer :

The equation for the given question is :-

$Al^{+3}\ + 3e^-\ =\ Al$

Thus for 1 mol of Al, charge required is 3F.

So the required amount of electricity in terms of charge will be :-

$=\ \frac{3}{27}\times40F = 4.44F$

Question 2.14(i) How much electricity is required in coulomb for the oxidation of

(i) 1 mol of $H_{2}O\ to\ O_{2}$ ?

Answer :

According to question the equation of oxidation will be :-

$O^{2-}\ \rightarrow \ \frac{1}{2}O_2\ +\ 2e^-$

Thus, for oxidation of O 2- , 2F charge is required.

$= 2\times96500\ C$

$= 193000\ C$

Question 2.14(ii) How much electricity is required in coulomb for the oxidation of

(ii)1 mol of $FeO\ to\ Fe_{2}O_{3}$ ?

Answer :

The oxidation equation for the given reaction will be :-

$Fe^{+2}\ \rightarrow\ Fe^{+3}\ +\ e^-$

So for oxidation of 1 mol $Fe^{+2}$ charge required $= 1F$

$= 96500\ C$

Question 2.15 A solution of $Ni(No_{3})_{2}$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of Ni is deposited at the cathode?

Answer :

We are given:

I = 5A

and t = 20(60) = 1200 sec.

So total charge = 5(1200) = 6000 C.

The equation for nickel deposition will be:-

$Ni^{+2}\ +\ 2e^-\ \rightarrow\ Ni$

Thus, from 2F charge 58.7 g of nickel deposition takes place.

i.e., $2(96487)\ C \rightarrow 58.7\ g\ Ni$

So for 6000 C charge total nickel deposition will be:-

$= \frac{58.7}{2\times96487}\times6000$

or $= 1.825\ g$

Hence 1.825 g Ni will be deposited in the given conditions.

Question 2.16 Three electrolytic cells A,B,C containing solutions of $ZnSO_{4}$ , $AgNO_{3}$ and $CuSO_{4}$ , 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?

Answer :

Since the cells are connected in series so the current passing through each cell will be equal.(1.5 A)

Now we are given that 1.45 g of silver is deposited. So firstly we will consider the cell containing silver.

$Ag^+\ +\ e^-\ \rightarrow Ag$

Since for deposition of 108 g silver 96487 C charge is required, thus for 1.45 g deposition of silver charge required will be:-

$= \frac{96487}{108}\times1.45$ $= 1295.43\ C$

Now we can find the time taken by 1.5 A current to deposit 1.45 g silver.

$Time\ taken = \frac{1295.43}{1.5} \approx 864\ sec.$

For copper:-

$Cu^{+2}\ + 2e^-\ =\ Cu$

Since 2F charge will deposit 63.5 g of Cu, then deposition by 1295.43 C will be:-

$= \frac{63.5}{2\times96487}\times1295.43$ $= 0.426\ g$

Hence 0.426 g of copper will be deposited.

For zinc:-

$Zn^{+2}\ +\ 2e^-\ \rightarrow\ Zn$

Since 2F charge will deposit 65.4 g of Zn, then deposition by 1295.43 C will be:-

$= \frac{65.4}{2\times96487}\times1295.43$ $= 0.439\ g$

Hence 0.439 g of zinc will be deposited.

Question 2.17(i) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) $Fe^{3+}_{aq}\ and\ I^{-}_{aq}$

Answer :

The concept used here will be that a reaction is feasible only if $E_{cell} ^{\circ}$ is positive.

Anode and cathode reactions will be as follows:-

$Fe^{3+}\ +\ e^-\ =\ Fe^{2+}$ $E^{\circ}\ = 0.77\ V$

$2I^{-}\ =\ I_2\ +\ 2e^-$ $E^{\circ}\ = -0.54\ V$

So $E_{cell} ^{\circ} = 0.77 - 0.54 = 0.23\ V$

So this reaction is feasible.

Question 2.17(ii) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(ii) $Ag^{+}_{aq}\ and\ Cu_{(s)}$

Answer :

A reaction is feasible only if $E_{cell} ^{\circ}$ is positive.

So, anode and cathode reactions will be as follows :-

$(Ag^{+}\ +\ e^-\ =\ Ag)\times 2$ $E^{\circ}\ = 0.80\ V$

$Cu\ =\ Cu^{+2}\ +\ 2e^-$ $E^{\circ}\ = -0.34\ V$

and $E_{cell} ^{\circ} = 0.80 - 0.34 = 0.46\ V$

So this reaction is feasible.

Question 2.17(iii) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(iii) $Fe^{3+}_{aq}\: and\ Br^{-}_{aq}$

Answer :

A reaction is feasible only if $E_{cell} ^{\circ}$ is positive.

So, anode and cathode reactions will be as follows :-

$(Fe^{+3}\ +\ e^-\ =\ Fe^{+2})\times 2$ $E^{\circ}\ = 0.77\ V$

$2Br^-\ =\ Br_2\ +\ 2e^-$ $E^{\circ}\ = -1.09\ V$

and $E_{cell} ^{\circ} = 0.77 - 1.09 = -0.32\ V$

So this reaction is not feasible.

Question 2.17(iv) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

$Ag_{s}\ and\ Fe^{3+}_{aq}$

Answer :

A reaction is feasible only if $E_{cell} ^{\circ}$ is positive.

So, anode and cathode reactions will be as follows:-

$Ag\ =\ Ag^{+}\ +\ e^-$ $E^{\circ}\ = -0.80\ V$

$Fe^{+3}\ +\ e^-\ =\ Fe^{+2}$ $E^{\circ}\ = 0.77\ V$

and $E_{cell} ^{\circ} = -0.80 + 0.77 = -0.03\ V$

So this reaction is not feasible.

Question 2.17(v) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(v) $B r_{2(a q)}$ and $F e_{a q}^{2+}$

Answer :

A reaction is feasible only if $E_{cell} ^{\circ}$ is positive.

So, anode and cathode reactions will be as follows :-

$Br_2\ +\ 2e^-\ =\ 2Br^{-}$ $E^{\circ}\ = 1.09\ V$

$Fe^{2+}\ =\ Fe^{3+}\ +\ e^-$ $E^{\circ}\ = -0.77\ V$

and $E_{cell} ^{\circ} = 1.09 - 0.77 = 0.32\ V$

So this reaction is feasible.

Question 2.18(i) Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.

Answer :

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

$Ag^+\ +\ e^-\ =\ Ag_{(s)}$

At anode :-

$Ag\ +\ NO^{3-}\ =\ AgNO_3\ +\ e^-$

Hence, silver will get deposited at the cathode and it will be getting dissolved at anode.

Question 2.18(ii) Predict the products of electrolysis in each of the following:

(ii)An aqueous solution of $AgNO_{3}$ with platinum electrodes.

Answer :

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

$Ag^+\ +\ e^-\ =\ Ag_{(s)}$

At anode :- Self ionisation will take place due to presence of water.

$H_2O\ \rightarrow\ 2H^+\ + \frac{1}{2}O_2\ +\ 2e^-$

Hence, silver will get deposited at the cathode and O 2 will be produced from anode.

Question 2.18(iii) Predict the products of electrolysis in each of the following:

(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.

Answer :

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

$\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow 1 / 2 \mathrm{H}_2$

At anode :- Self ionisation of water will take place due to presence of platinum electrode.

$H_2O\ \rightarrow\ 2H^+\ + \frac{1}{2}O_2\ +\ 2e^-$

Hence, H 2 gas will be generated at cathode and O 2 will be produced from anode.

Question 2.18(iv) Predict the products of electrolysis in each of the following:

(iv) An aqueous solution of $CuCI_{2}$ with platinum electrodes.

Answer :

For the given solution :

At cathode :- Reaction with greater E 0 will take place.

$\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$

At anode :-

$2Cl^-\ =\ Cl_2\ +\ 2e^-$

Hence, Cu will get deposited at cathode and Cl 2 will be produced from anode.

Class 12 Chemistry NCERT Chapter 2: Higher Order Thinking Skill (HOTS) Questions

Question. The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70 \mathrm{~S} \mathrm{~cm} \mathrm{~mol}^1$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its degree of dissociation is given by $\mathrm{x} \times 10^{-1}$. The value of $x$ is _______ (Nearest integer)

Answer. $\begin{aligned}
& \lambda_{\mathrm{m}}^{\prime \prime} \text { of } \mathrm{NH}_4 \mathrm{Cl}=185 \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}=185 \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4}=185-70=115 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{\circ}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{OH}^{-}} \\
& =115+170 \\
& \left(\lambda_{\mathrm{m}}^0\right)_{\mathrm{NH}_4 \mathrm{OH}}=285 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
& \text { degree of dissociation }=\frac{\left(\lambda_{\mathrm{m}}\right)_{\mathrm{NH}_4 \mathrm{OH}}}{\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}} \\
& =\frac{85.5}{285} \\
& =0.3 \\
& =3 \times 10^{-1}
\end{aligned}$

Hence, the answer is 3.

Approach to Solve Questions of Chapter 4 Electrochemistry

Electrochemistry is the study of electron movement in an oxidation or reduction reaction at a polarized electrode surface. Here is a more detailed approach on how students can approach to solve questions of NCERT class 12 Electrochemistry:

1. Understand the Key Concepts First

Before attempting questions, ensure you're clear on:

  • Redox reactions and balancing
  • Galvanic and Electrolytic cells
  • Electrode potential, standard electrode potential
  • Cell notation and EMF (E°cell)
  • Nernst equation
  • Gibbs free energy and cell potential
  • Conductance and molar conductivity
  • Kohlrausch’s Law
  • Electrolysis and Faraday’s laws
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2. Resources to Use

  • NCERT Textbook & NCERT Exemplar
  • Coaching material or reference books (e.g., OP Tandon, Pradeep)
  • Previous Year Papers (NEET/JEE/Boards)
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3. Tips to solve questions

  • Always write units in numericals
  • For EMF, ensure you use sign of the standard potentials carefully.
  • Practice step-by-step problem solving rather than shortcuts.
  • Make a sheet of formulas for quick revision

Topics of NCERT Class 12 Chemistry Chapter 2

2.1 Electrochemical Cells
2.2 Galvanic Cells
2.3 Nernst Equation
2.4 Conductance of Electrolytic Solutions
2.5 Electrolytic Cells and Electrolysis
2.6 Batteries
2.7 Fuel Cells
2.8 Corrosion

What Extra Should Students Study Beyond NCERT for JEE?

Formulas Of NCERT Class 12 Chemistry Electrochemistry

1. Conductance(G) is the reciprocal of resistance (R) and specific conductance or conductivity(k) is inverse of resistivity $(\rho )$

$\\G=\frac{1}{R}=\frac{1}{\rho }\left ( \frac{a}{l} \right )\\k=G\left ( \frac{l}{a} \right )$

2. l/a is called the cell constant of conductivity cell.

3. Equivalent Conductivity is defined as the conductance of a solution containing 1g of an electrolyte.

$\\ \Lambda _{eq}=K\times V\\$

4. Nernst equation

aA+bB $\rightarrow$ cC+dD

$E_{cell} = E_{cell}^{o}-\frac{0.0591}{n}log\frac{\left [ C \right ]^{c}\left [ D \right ]^{d}}{\left [ A \right ]^{a}\left [ B \right ]^{b}}$

class 12 chemistry electrochemistry ncert solutions

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NCERT Solutions for Class 12 Chemistry

Frequently Asked Questions (FAQs)

1. What are electrochemical cell in class 12 Chapter 2 Electrochemistry NCERT solutions?

An electrochemical cell is a device that converts chemical energy into electrical energy through a redox reaction. It consists of two half-cells connected by a wire and a salt bridge. The two types of electrochemical cells are Galvanic Cells and electrolytic Cells

2. What is oxidation and reduction in electrochemistry?

In an electrochemical cell, oxidation occurs at the anode, In this process, a substance loses electrons. Reduction occurs at the cathode. In this process, a substance gains electrons.

3. What is the difference between a galvanic cell and an electrolytic cell in electrochemistry?

The main difference between a Galvanic cell and an Electrolytic cell is how they work.

  • Galvanic Cell: Generates electrical energy from spontaneous reactions (e.g., batteries). Here energy is produced.

  • Electrolytic Cell: Uses electrical energy to drive non-spontaneous reactions (e.g., electroplating). Here energy is consumed.
4. Where can I find complete solutions of NCERT Class 12 Chemistry?

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry Chapter wise links are provide to get NCERT solutions for Class 12 Chemistry PDF.

5. What are the important topics of Electrochemistry?
  • Type of cells
  • Electrode potential.
  • Standard electrode potential.
  • Anode.
  • Cathode.
  • Cell potential.
  • Cell electromotive force (emf)
  • SHE (Standard Hydrogen Electrode)
6. What is the relationship between cell potential and Gibbs free energy?

The relationship between cell potential (E) and Gibbs free energy $(\Delta G)$ is:

$\Delta G=-n F E$

A positive cell potential ( E ) means the reaction is spontaneous ( $\Delta \mathrm{G}$ is negative).

A negative cell potential means the reaction is non-spontaneous ( $\Delta \mathrm{G}$ is positive).

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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