What is electrochemistry all about?

Electrochemistry deals with the interrelation between electrical and chemical changes. It focuses on how electricity is generated and how it is produced from chemical changes. It discusses how reactions can takes place in battery between a cathode and anode with electrolyte in between them. Electrochemical reactions are that branch of study where any chemical reaction such as in a battery produces an electric current takes place. Chapter 3 chemistry class 12 can be challenging for most students. If any student have difficulty regarding this chapter, he/she can refer to Electrochemistry class 12 NCERT Solutions.

How to prepare for Electrochemistry.

Studying this chapter on daily basis will help you to get a grip on the concepts. Taking notes about complex topics and difficult numericals will boost your confidence during revision. After completing the questions from the NCERT book, you can refer to the solutions prepared by the experts of Adda247 to understand the other method of answering the questions. The solutions containing detailed explanations of the topics which are important from the board exam point of view.

Where can I found solutions to class 12 Chapter 3 Electrochemistry?

Students can find Electrochemistry Class 12 solutions on Adda247 website. The solutions are presented in a very easy language. They are easily downloadable in pdf format and they can refer to the solutions provided. Chapter 3 Chemistry Class 12 is not an easy chapter. Students may suffer a lot to understand the topics such as anode, cathode and electrolyte. They get confused in these topics. Also, conductivity and resistivity are difficult concepts. Students should know that continue efforts or practice is the only key to cracking electrochemistry.

Ncert Solutions For Class 12 Chemistry Chapter 3

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Ncert Solutions Class 12 Chemistry Chapter 3

Adda247 provides NCERT Solutions for Class 12 Chemistry Chapter 3. These solutions will not only help students to boost their board exams and score brilliant marks but for competitive exams. The Ncert Solutions for chemistry class 12 are according to the NCERT guidelines.

Complete 16 chapters wise solutions were provided for the benefit of students.12^th class sets the base for higher education for every student. This makes it the most crucial class for any student who is aiming for his/her dream of quality education. Scoring good marks in the 12^th class is equivalent to a quality high education. So, it becomes extremely important for students to give a boost to their class 12 Chemistry preparation with Adda247 NCERT solutions.

The students can access the solutions anywhere while browsing the web easily. The solutions are very precise and accurate.

NCERT Solutions for Class 12 Chemistry Chapter 3: Electrochemistry PDF

Ncert Solutions for class 12 chemistry chapter 3 Electrochemistry is the branch of chemistry concerned with the interrelation of electrical and chemical changes that are caused by the passage of current. Electrochemistry is also vital in a wide range of important technological applications. For example, batteries are important not only in storing energy for mobile devices and vehicles, but also for load leveling to enable the use of renewable energy conversion technologies.

Electrochemistry is the study of electricity and how it relates to chemical reactions. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as redox or oxidation-reduction reaction.

Chapter 3 Chemistry class 12 NCERT Book solutions focuses on Electrochemistry and its implications in the real world. Some of the topics covered here are electrochemical cells, galvanic cells and electrolytic cells.

Important terms like resistivity, conductivity and molar conductivity of the cell are explained briefly.
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NCERT Solutions for Class 12 Chemistry Chapter 3: Key Features Electrochemistry

Some Key features of Ncert solutions for class 12 chemistry chapter 3 are given below.

The NCERT Solutions provide clear and precise answers.
These solutions help students to solve and revise the complete syllabus.
The columns are used wherever necessary.
Solutions are self–explanatory.
Full of relevant diagrams.

NCERT Solutions Class 12 Chemistry Chapter 3: Important Questions

Que.-1 Arrange the following metals in order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn.

Ans:

Mg, Al, Zn, Fe, Cu.

Que.-2 Given the standard electrode potentials.

K⁺/K = -2.93V, Ag⁺/Ag = 0.80V,

Mg²⁺/Mg = -2.37V, Hg²⁺/Hg = 0.79V,

Cr³⁺/Cr = -0.74V

Arrange these metals in their increasing order of reducing power.

Ans:

Metals with higher oxidation potential can easily be oxidized and have greater reducing power. Thus, increasing order of reducing power will be Ag<Hg<Cr<Mg<K.

Que.-3 In the button cells widely used in watches and other devices the following reaction takes place:

Zn(s) + Ag₂O(s) + H₂O(l) à Zn²⁺(ag) + 2Ag(s) + 2OH^–(aq)

Determine G^–and E^–for the reaction.

Ans:

Zn is oxidised and Ag₂O is reduced:

E^–_cell = E^–_{Ag2O,Ag(reduction)}– E^–_{Zn/Zn*2+(oxidation)}

= 0.344 + 0.76 = 1.104V

G = -nFE_cell = -2 x 96500 x 1.104J
= -2.13 x 10⁵J

Que.-4 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans:

Conductivity (K): It is the conductance of unit cube of material.

SI unit is S/m. Common unit is S/cm.

The conductivity of an electrolytic solution always decreases with decrease in concentration that is on dilution. This is because with dilution, the degree of dissociation increases and the total no. of current carrying ions increases but the no. of ions per unit volume decreases.

Molar conductivity: It is the ratio of the electrolytic conductivity k to the molar concentration C of the dissolved electrolyte.

= k/C

The SI unit of molar conductivity is S m²/ mol

The common unit of molar conductivity is S cm²/ mol

The molar conductivity of strong and weak electrolytes increases with a dilution. This is b/c with dilution, the degree of dissociation increases and the no. of current carrying ions increases.

Que.-5 The conductivity of 0.20M solution of KCl at 298K is 0.0248 S cm^-1. Calculate its molar conductivity.

Ans:

_m = k x 1000 / molarity = 0.0248Scm^-1 x 1000cm³L^-1

0.20molL^-1

= 124 S cm² mol^-1

Que.-6 How much charge is required for the following reductions:

1 mol of Al³⁺ to Al?
1 mol of Cu²⁺ to Cu?
1 mol of MnO^–₄ to Mn²⁺?

Ans:

i.) The electrode reaction is Al³⁺ + 3e^– à Al

: Quantity of charge required for reduction of 1 mol of Al³⁺ = 3F = 3 x 95600C = 289500C.

ii.)The electrode reaction is Cu²⁺ + 2e^– à Cu

: Quantity of charge required for the reduction of 1 mol of Cu²⁺ = 2F 2 x 95600C = 193000C.

iii.)The electrode reaction is MnO₄^– à Mn²⁺ I.e., Mn⁷⁺ + 5e^– à Mn²⁺

: Quantity of charge required = 5F = 5 x 95600C = 482500C.

Que.-7 How much electricity in terms of Faraday is required to produce

0 g of Ca from molten CaCl₂?
0 g of Al from molten Al₂O₃?

Ans:

i.) Ca⁺ + 2e^– à Ca

Thus, 1 mole of Ca i.e., 40g of Ca require = 2F electricity

: 20 g of Ca require = 1 F of electricity

ii.)Al³⁺+ 3e^– à Al

Thus, 1 mole of Al i.e., 27g of Al require = 3 F electricity

: 40g of Al require electricity

= 3/27 x 40 = 4.44F of electricity.

Que.-8 How much electricity is required in coulomb for the oxidation of :

1 mol of H₂O to O₂?
1 mol of FeO to Fe₂O₃?

Ans:

i.) The electrode reaction for 1 mole of H₂O is

H₂O à H₂ + ½ O₂

i.e., O^2- à 1/2O₂ + 2e^–

: Quantity of electricity required

= 2F = 2 x 96500 C = 193000 C

ii.)The electrode reaction for 1 mole of FeO is

FeO + 1/2O₂ à 1/2Fe₂o₃

i.e., Fe²⁺à Fe³⁺ + e^–

: Quantity of electricity required = 1F = 96500C.

Que.-9 A solution of Ni(NO₃)₂is electrolysed b/w platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans:

Quantity of electricity passed = (5A) X (20 x 60 sec) = 6000 C

Ni²⁺ + 2e^– à Ni

Thus, 2F i.e., 2 x 96500 C of charge deposit = 1 mole of Ni = 58.7 g

: 6000 C of charge will deposit

= 58.7 x 6000/ 2 x 96500 C = 1.825g of Ni

Que.-10 Predict the products of electrolysis in each of the following:

An aqueous solution of AgNO₃ with silver electrodes.
An aqueous solution of AgNO₃ with platinum electrodes.
A dilute solution of H₂SO₄ with platinum electrodes.
An aqueous solution of CuCl₂ with platinum electrodes.

Ans:

i.) AgNO₃(s) + aq à Ag⁺(aq) + NO₃^–(aq)

H₂O à H⁺ + OH^–

At cathode : Ag⁺ ions have lower discharge potential than H⁺ions. Hence, Ag⁺ ions will be deposited as Ag in preference to H⁺ ions:

Ag⁺(aq) + e^– à Ag(s)

At anode : As Ag anode is attacked by NO^–₃ ions, Ag of the anode will dissolve to form Ag⁺ ions in the solution.

Ag(s) à Ag⁺(aq) + e^–

ii.)AgNO₃ with platinum electrodes:

At cathode: Ag⁺ ions have lower discharge potential than H⁺ ions. Hence, Ag⁺ ions will be deposited as Ag in performance to H⁺ ions.

At anode: As anode is not attackable, out of OH^– and NO₃^– ions, OH^– ions have lower discharge potential. Hence OH^– will be discharge in preference to NO₃^– ions, which than decompose to give out O₂.

OH^–_(aq) à OH + e^–

4OH à 2H₂O_(l) + O_2(g)

iii.)H₂SO_4(aq) à 2H⁺_(aq) + SO₄^2-_(aq)

H₂O à H + OH^–

At cathode: H⁺ + e^– à H

H + H à H_2(g)

At anode: OH^– à OH +e^–

4OH à 2H₂O + O_2(g)

Thus H₂ gases liberated at the cathode and O₂ gases at the anode.

iv.)CuCl_2(s) + aq à Cu²⁺_(aq)+ 2Cl^–_(aq)

H₂O à H + OH^–

At cathode: Cu²⁺ ions will be reduced in preference to H⁺ions and copper will be deposited at cathode.