NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

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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 Question (FAQs)

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

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

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

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

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

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

4. Where can I find complete solutions of NCERT Class 12 Chemistry?

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

5. What are the important topics of this chapter?

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

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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

hello,

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

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

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

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

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

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

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

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

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Read Complete Answer

khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

Read Complete Answer

Parvati Solanki 02 June,2022

View All

Popular CBSE Class 12th Questions

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

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

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 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

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

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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

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

In the reaction,

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

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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

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

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

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

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

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

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Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

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Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

Also Read: Career as Nurse

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Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

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Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

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Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

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Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

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Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

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Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

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Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

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Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

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Visual Communication Designer

Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

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Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

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Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

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Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

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Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

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Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

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Advertising Manager

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

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Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

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Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

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Product Manager

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Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

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Team Lead

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Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

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Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

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Procurement Manager

The procurement Manager is also known as Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

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Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

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Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

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Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

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Product Manager

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ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

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.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of creating, designing and developing applications using .NET languages such as VB and C#.

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Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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DevOps Architect

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

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Cloud Solution Architect

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

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