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.

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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 $\dpi{100} (\rho )$

$\dpi{100} \\G=\frac{1}{R}=\frac{1}{\rho }\left ( \frac{a}{l} \right )\\k=G\left ( \frac{l}{a} \right )$

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.

$\dpi{100} \\ \Lambda _{eq}=K\times V\\$

4. Nernst equation

aA+bB $\dpi{100} \rightarrow$ cC+dD

$\dpi{100} E_{cell} = E_{cell}^{o}-\frac{0.0591}{n}log\frac{\left [ C \right ]^{c}\left [ D \right ]^{d}}{\left [ A \right ]^{a}\left [ B \right ]^{b}}$

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

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.

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
$E^{o}cell = E^{o} Mg|Mg^{2+}$

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:-

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

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.

Answer :

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

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.

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

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.

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 $K_c = 9.57\times10^7$

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.

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

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.

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.

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- .

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:-

Answer :

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

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.

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:-

Question 3.2 Given the standard electrode potentials,

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.

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 :-

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

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

Answer :

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

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 $K = 3.13\times10^34$

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

Answer :

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

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 $k\ \approx \ 3.2$

(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.

(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.

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.

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

Answer :

The given reaction is obtained from :-

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 $= - 213.04\ KJ$

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.

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:-

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

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 = $0.219\ cm^{-1}$

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) $\dpi{100} 1\ mol\ of\ Al^{3+}\ to\ Al\ ?$

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) $\dpi{100} MnO_{4}^{-}\ to\ Mn^{2+}\ ?$

Answer :

The given reaction can be written as:-

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

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

$\dpi{100} = 5(96500)$

$= 482500\ C$

(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 $= 1F$

$= 96500\ C$

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.

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.

(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.

(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.

(iii) $\dpi{100} Fe^{3+}_{aq}\: and\ Br^{-}_{aq}$

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.

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.

(v) 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.

(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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### 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 ?

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.

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.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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 help you.

Good luck with your studies!

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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) Option 2) Option 3) Option 4)

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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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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