# NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

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. CBSE 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. This chapter will also acknowledge you to various types of batteries and their benefits. The chapter is important for both theoretical and practical purposes.

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 solutions of NCERT class 12 chemistry chapter 3 electrochemistry.

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

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

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

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

## 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 | MgSO4 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
$E^{o}cell = E^{o} Mg|Mg^{2+}$

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:-

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

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 :-   F2,  Cl2,  Br2,   Ag+1  etc.

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

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.

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                                                $K_c = 9.57\times10^7$

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.

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 $\lambda ^{0}(HCOO^{-})=54.6\,\: S\, cm^{2}\, \: mol^{-1}$

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

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.

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.

It is clear from the given reaction that reduction of 1 mol of Cr2O72- 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 Cr2O72- .

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

For recharging PbSO4 is converted into Pb at the anode and into PbO2 at the cathode.

The chemical reactions are as follows:-

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

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.

## 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.

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?

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.

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.

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:

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

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                                             $K = 3.13\times10^34$

### 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.

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                                              $k\ \approx \ 3.2$

### Question 3.5(i)     Write the Nernst equation and emf of the following cells at 298 K:

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

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

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:

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

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.

The given reaction is obtained from :-

So the Eocell  can be obtained directly.

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

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

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.

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.

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 $0.146\times10^{-3}\ S cm^{-1}$.

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?

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

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)  $1\ mol \ of\ Cu^{2+}\ to\ Cu?$

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

The given reaction can be written as:-

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

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

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

$= 482500\ C$

### Question 3.13(i)     How much electricity in terms of Faraday is required to produce

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

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 $AI_{2}O_{3}$ ?

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   $H_{2}O\ to\ O_{2}$ ?

According to question the equation of oxidation will be :-

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

Thus,      for oxidation of   O2- ,  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  $FeO\ to\ Fe_{2}O_{3}$ ?

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?

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?

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

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

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

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:

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

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)$Br_{2}_{(aq)}\ and\ Fe^{2+}_{aq}$

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.

For the given solution :

At cathode :-    Reaction with greater E0 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.

For the given solution :

At cathode :-    Reaction with greater E0 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 O2 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.

For the given solution :

At cathode :-    Reaction with greater E0 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, H2 gas will be generated at cathode and O2 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.

For the given solution :

At cathode :-    Reaction with greater E0 will take place.

$Cu^{2+}\ +\ 2e^-\ =\ Cu_{(s)}$

At anode :-

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

Hence, Cu will get deposited at cathode and Cl2 will be produced from anode.

## NCERT Solutions for Class 12 Chemistry

 Chapter 1 NCERT solutions for class 12 chemistry chapter 1 The Solid State Chapter 2 CBSE NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 NCERT solutions for class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry Chapter 6 NCERT solutions for class 12 chemistry chapter 6 General Principles and Processes of isolation of elements Chapter 7 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements Chapter 9 NCERT solutions for class 12 chemistry chapter 9 Coordination compounds Chapter 10 Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols, and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT class 12 chemistry chapter 15 Polymers Chapter 16 NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

## NCERT Solutions for Class 12 Subject wise

 Solutions of NCERT class 12 biology NCERT solutions for class 12 maths CBSE NCERT solutions for class 12 chemistry Solutions of NCERT 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. Happy learning!!!