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.

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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 ?
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

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

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 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 value of water.

Answer :

We know :

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

If we draw a straight line between and _, its slope will be -A and the intercept on the y-axis will be _.

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
Calculate its degree of dissociation and dissociation constant. Given and

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

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:
What is the quantity of electricity in coulombs needed to reduce 1 mol of ?

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

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 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 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 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.,

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 .

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 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 _.

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

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 ?

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 ?

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 ?

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 ?

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 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.,

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 , and , respectively are connected in series. A steady current of $1.5$ amperes was passed through them until 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 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 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 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

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)

Answer :

A reaction is feasible only if 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

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 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

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) 1643868790759 Answer :

A reaction is feasible only if 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

So this reaction is feasible.

Question 3.18(i) Predict the products of electrolysis in each of the following:

(i) An aqueous solution of 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 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 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 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

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

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Biology Solutions

NCERT Exemplar Class 12 Physics 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	The Solid State
Chapter 2	Solutions
Chapter 3	Electrochemistry
Chapter 4	Chemical Kinetics
Chapter 5	Surface chemistry
Chapter 6	General Principles and Processes of isolation of elements
Chapter 7	The P-block elements
Chapter 8	The d and f block elements
Chapter 9	Coordination compounds
Chapter 10	Haloalkanes and Haloarenes
Chapter 11	Alcohols, Phenols, and Ethers
Chapter 12	Aldehydes, Ketones and Carboxylic Acids
Chapter 13	Amines
Chapter 14	Biomolecules
Chapter 15	Polymers
Chapter 16	Chemistry in Everyday life

NCERT Solutions for Class 12 Subject wise

NCERT Solutions for class 12 biology

NCERT solutions for class 12 maths

NCERT solutions for class 12 chemistry

NCERT Solutions for class 12 physics

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 Questions (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)

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

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!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

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.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

Find Solutions of NCERT Class 12 Chemistry Chapter 3 Electrochemistry

Class 12 Chemistry Electrochemistry NCERT Solutions

NCERT Exemplar Class 12 Solutions

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