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NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

Edited By Irshad Anwar | Updated on Mar 06, 2023 02:45 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry- Electrochemistry is the study of the interconversion of chemical energy and electrical energy. The devices where these conversions take place are known as cells. NCERT solutions for Class 12 Chemistry chapter 3 Electrochemistry deal with questions based on mainly electrochemical and galvanic cells and also on 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. therefor electrochemistry class 12 NCERT solutions become very important to get in-depth understanding of concepts.

Also Read,

Important points and formulas of NCERT Class 12 Chemistry Chapter 3 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

Topics and Sub-topics of NCERT Electrochemistry Class 12 Chemistry Chapter 3 -

3.1 Electrochemical Cells

3.2 Galvanic Cells

3.3 Nernst Equation

3.4 Conductance of Electrolytic Solutions

3.5 Electrolytic Cells and Electrolysis

3.6 Batteries

3.7 Fuel Cells

3.8 Corrosion

Find Solutions of NCERT Class 12 Chemistry Chapter 3 Electrochemistry

Solutions to In-Text Questions Ex 3.1 to 3.15

Question 3.1 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.

1643888463274 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 3.2 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 3.3 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.

Question 3.4 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.

1643888510951

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 3.5 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 3.6 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

Question 3.7 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 3.8 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 3.8 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}

Question 3.10 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 3.11 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 3.12 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- .

Question 3.13 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 3.14 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 3.15 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.

Question 3.1 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:-

Mg > Al > Zn > Fe > Cu


Question 3.2 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 3.3(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 3.3(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 3.3(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 3.4(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:-

1646201684726

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 3.4(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 :-

1643888607928

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 3.5(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 3.5(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 3.5(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 3.5(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 3.6 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 :-

1643888647683

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 3.7 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 3.8 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 3.9 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 3.11 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 3.12(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 3.12(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 3.12(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 3.13(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 3.13(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 3.14(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 3.14(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 3.15 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 3.16 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 3.17(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 3.17(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 3.17(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 3.17(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 3.17(v) Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(v) 1643868790759Answer :

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 3.18(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 3.18(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 3.18(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.

H^+\ +\ e^-\ =\ \frac{1}{2}H_2_{(g)}

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 3.18(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.

Cu^{2+}\ +\ 2e^-\ =\ Cu_{(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 Electrochemistry NCERT Solutions

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This chapter of Class 12 NCERT solutions is the third chapter of NCERT Class 12 Chemistry book. Electrochemistry basically deals with concepts related to an electrochemical cell, electrolytic cell, Nernst equation, emf of a cell, Faraday's law, Battery, fuel cell, corrosion etc. NCERT Solutions for Class 12 Chemistry Chapter 3 does not have many linkages with the previous chapter however it deals with some terms mentioned in chapter 2 like concentration, molarity etc.

Class 12 NCERT solutions is for chapter Electrochemistry is a good source to cover it in a comprehensive manner. Students can score decent marks in this chapter as questions are mostly based on Faraday's law and if the concept is clear, questions can be handled easily. Apart from NCERT, students can refer to class notes for Chemistry Class 12 Chapter 3 to revise and score well in the final board examination as well as competitive exams.

Class 12 NCERT solutions provide questions that cover almost all the topics of the chapter. Hence it becomes inevitable to at least solve them once before the final examination. Class 12 Chemistry Chapter 3 Electrochemistry is a must to read for students as it has decent weightage in all the exams ranging from Board to NEET and JEE.

Ch 3 Chemistry Class 12 solutions will take 10-12 hours to complete if all the concepts of 11th class related to concentration of solutions, Anode, cathode, oxidation and reduction are well-read. Students can easily score more than 90 per cent marks in this chapter as limited concepts are there and the difficulty level of questions is not that high.


NCERT Exemplar Class 12 Solutions

Features of Electrochemistry Class 12 Solutions

Electrochemistry is an important chapter for both CBSE Board exam as well as competitive exams like JEE, NEET, BITSAT, VITEE and KVPY, etc. It carries 5 marks in the CBSE board exams hence learning the concepts of this chapter is very important to get good marks. In this chapter, there are 18 exercise questions. The step-by-step NCERT solutions for class 12 chemistry chapter 3 Electrochemistry are prepared by subject experts which not only help you to clear your doubts but also help you to improve your writing skills.

The NCERT solutions provided here are completely free of cost and you can also download them for offline usage. Please scroll down to get NCERT solutions for class 12 chemistry chapter 3 electrochemistry. By referring to the NCERT solutions for class 12 , students can understand all the important concepts and practice questions well enough before their examination.

NCERT Solutions for Class 12 Chemistry

Chapter 1
Chapter 2
Chapter 3
Electrochemistry
Chapter 4
Chemical Kinetics
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16

NCERT Solutions for Class 12 Subject wise

Benefits of CBSE NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

Hope you have understood well with the help of the free solutions provided here. After completing the solutions of NCERT class 12 chemistry chapter 3 Electrochemistry, students will be able to describe an electrochemical cell, to differentiate between electrolytic and galvanic cells, to apply Nernst equation for calculating the emf of galvanic cell, also will be able to derive relation between standard potential of the cell, Gibbs energy of cell reaction and its equilibrium constant, etc. Keep working, keep improving and also keep enjoying.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. What is the weightage of NCERT class 12 Chemistry chapter 3 in NEET?

This chapter holds weightage of 2% for NEET exam. The weightage given is average based on the previous year papers. To practice questions on the chapter refer to NCERT exemplar problems and NCERT book exercises.

2. What is the weightage of NCERT class 12 Chemistry chapter 3 in JEE Main?

4 marks, that is 1 question can be expected from Electrochemistry for JEE Main exam. Follow previous year papers and NCERT books for good score in the exam.

3. What is the weightage of NCERT class 12 Chemistry chapter 3 in CBSE board exam ?

Around 6 marks questions are asked from the NCERT syllabus of Class 12 chapter 3 Electrochemistry.

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 this chapter?
  • Type of cells
  • Electrode potential.
  • Standard electrode potential.
  • Anode.
  • Cathode.
  • Cell potential.
  • Cell electromotive force (emf)
  • SHE (Standard Hydrogen Electrode)

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer
Sajal Trivedi 29 January,2024
i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer
Upasana Barman 24 August,2022
i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Read Complete Answer
Mayankjha.student2007 23 August,2022
if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer
Nuthalapati Vyshnavi 21 August,2022
I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer
Muskan Dhalwal 17 August,2022
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Popular CBSE Class 12th Questions

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
Read More

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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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