Electrochemistry JEE Advanced 2024 Notes

Electrochemistry Deals With 

• The study of electrochemical cells that convert chemical energy into electrical energy or vice versa.

• Oxidation-reduction responses that either produce or use electrical energy and electrochemical responses take place in cells. 

• Each cell has two electrodes, operators through which electrons enter or leave the cell.

Let's understand what Electrochemical cells are!!

Electrochemical Cells:

An electrochemical cell is a device that can induce electrical energy from the chemical responses being in it, or use the electrical energy supplied to it to grease chemical responses in it. This device is able to convert chemical energy into electrical energy, or vice versa.

• A common illustration of an electrochemical cell is a standard 1.5- volt cell which is used to power numerous electrical appliances similar to television remotes and timepieces.

• Electrochemical cells work on oxidation-reduction reactions. Here is a table where you can get an overview of the process that happens in the oxidation and reduction processes of electrochemical cells.

The electrochemical process involves the following effects:

• Either oxidation or reduction occurs at each electrode.

• Electrons flow through an external captain.

• Ions flow in the electrolyte result.

There are Two Types of Electrochemical Cells: Voltaic Cells and Electrolytic Cells.

1. A voltaic cell derives its energy from spontaneous redox reactions. This cell consists of two different electrodes immersed in an electrolytic solution. One electrode undergoes oxidation other undergoes reduction. The chemical reactions that occur at electrodes encourage the production of electrical energy.

2. An electrolytic cell is an electrochemical cell in which the energy from an external power source is used to drive a normally non-spontaneous reaction, i.e. apply a reverse voltage to a voltaic cell.

Below is an illustration that states how electrolytic cell differs from voltaic cell:

Electric current is considered a flow of electrons. The substance which allows the electric current to pass through it is called a conductor.

There are two types of conductors, one is an Electronic conductor and another is an Electrolytic conductor.

The below table explains well the difference between Electronic and Electrolytic conductors.

Now, Let's understand the working of different kinds of cells:

Daniel Cell:

• An example of a galvanic cell is the Daniel cell.

• It consists of zinc and copper half-cells.

• This cell converts the chemical energy liberated during the redox reaction.

$Z{{n}_{(s)}}+C{{u}^{2+}}_{(aq)}\to Z{{n}^{2+}}_{(aq)}+C{{u}_{(s)}}$ 

• Electron flow from Cu to Zn

• Zinc is deposited at the zinc electrode and Copper dissolves at the copper electrode.

Galvanic Cell:

• A Galvanic cell and a voltaic cell are the same thing. There are other cells, however, similar to an electrolytic cell.

• A Galvanic/ voltaic cell converts chemical energy into electrical energy. An electrolytic cell uses electrical energy to drive a non-spontaneous response i.e., converts electrical energy into chemical energy. 

• Example: Zn rod is dipped in zinc sulphate solution

• Zinc undergoes oxidation, and gives electrons, it is electron-rich, so represented by a negative sign. $Zn\to Z{{n}^{2+}}+2{{e}^{-}}$

• $C{{u}^{+2}}$present in the solution undergoes reduction on the Cu rod surface. Cu rod is electron-poor, so it is represented by a positive sign. $C{{u}^{2+}}+2{{e}^{-}}\to Cu$

• Electrons flow in the external circuit from Zn to Cu.

• A salt bridge is used to prevent the accumulation of charges at the electrodes.

• The reaction taking place in the electrochemical cell is$Z{{n}_{(s)}}+CuS{{O}_{4}}_{(aq)}\to ZnS{{O}_{4}}_{(aq)}+C{{u}_{(s)}}$

Electromotive Force of the Cell:

The difference between the two electrode potentials present in the electrochemical cell is the electromotive force of the cell.

Ecell = Standard electrode potential of right-hand electrode-Standard electrode potential of left-hand electrode.

EMF of a Cell

The reduction potential of an electrode is exactly equal in magnitude but opposite in sign to its oxidation potential.

Following are the different conditions of EMF cell that affects the cell reaction:

• If the EMF of the cell is positive, the cell reaction is spontaneous or irreversible. We can draw current from it.

• If the EMF of the cell is negative, the cell reaction is nonspontaneous or reversible. We cannot draw current from it.

• If the EMF of the cell is zero, the cell reaction is at equilibrium.

Standard Hydrogen Electrode:

In electrochemistry, the standard hydrogen electrode (SHE) serves as a reference electrode. It has a platinum electrode with a standard electrode potential of 0 volts that is submerged in a solution of 1 M H2 ions.

When determining the electrode potentials of other half-cells in electrochemical cells, the electrode reaction for the SHE is H₂(g) ⇌ 2H⁺(aq) + 2e⁻.

The below Diagram explains the overview of SHE:

• In the IUPAC system of referring to electrode potentials, standard reducing potentials are simply called standard potentials.

The table below shows how different standard electrode potentials of different elements vary.

Electrochemical series:

• In the electrochemical series, the metals are arranged in increasing order to reduce potential.

• Metals placed above can displace the below metal ions from their solution.

• Oxidation power ∝ reduction potential ∝ $\dfrac{{\text{1}}}{{{\text{Oxidation potential}}}}$

Reducing power ∝ oxidation potential ∝ $\dfrac{{\text{1}}}{{{\text{reduction potential}}}}$

Note: In two half-cell reactions having electrode potentials $E_1^\circ $and $E_2^\circ $ are combined to give a third half-cell reaction having an electrode potential $E_3^\circ$ then,

$\Delta G_3^\circ  = \Delta G_1^\circ  + \Delta G_2^\circ$

$- {n_3}FE_3^\circ  =  - {n_1}FE_1^\circ  - {n_2}FE_2^\circ$

$E_3^\circ  = \dfrac{{{n_1}E_1^\circ  + {n_2}E_2^\circ }}{{{n_3}}}$

If the number of electrons involved is equal to $E_3^\circ  = E_1^\circ  + E_2^\circ $

EMF of Galvanic Cell - Nernst Equation:

The variation of electrode potential and cell potential with a concentration of ions in solutions is given by the Nernst equation which is derived from Vant’s Hoff isotherm equation.

$\Delta G = \Delta G^\circ  + 2.303RT\log Q$

R is the universal gas constant $\left( 8.314J/K/mole \right)$.

T is the temperature in kelvin.

n= number of electrons involved in the reaction

F= Faraday =96500 coulombs

Q= reaction quotient

${{\text{E}}_{\text{cell}}}\text{=E}_{\text{cell}}^{\text{ }\!\!{}^\circ\!\!\text{ }}\text{-}\dfrac{\text{2}\text{.303RT}}{\text{nF}}\text{log}\dfrac{\left[ \text{Product} \right]}{\left[ \text{Reactants} \right]}......\left( 1 \right)$ 

By substituting the value of F, R and T(298 K) in the above equation(1) we get,

${{\text{E}}_{\text{cell}}}\text{=E}_{\text{cell}}^{\text{ }\!\!{}^\circ\!\!\text{ }}\text{-}\dfrac{\text{0}\text{.059}}{\text{n}}\text{log}\dfrac{\left[ \text{Product} \right]}{\left[ \text{Reactants} \right]}$

Let's Solve !!

Practice Problems

1) The standard emf of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25C°. The equilibrium constant of the reaction will be (Given, F = 96500 C mol¯¹, R = 8. 314 JK¯¹ mol¹)

1) 2.0 × 10¹¹

2) 4.0 × 1012

For Cation Electrodes:

The Nernst equation for a single electrode potential $\left( {{E_{{M^{n + }}/M}}} \right)$ at 298 K is $M^{n+}(aq)+ne^-\longrightarrow M(s)$

The equilibrium state EMF is given by:

$M^{n+}(aq)+ne^-\longrightarrow M(s)$

$E={{E}^{o}}-\dfrac{RT}{nF}ln\dfrac{1}{c}$

$E=E{}^\circ -\dfrac{2.303RT}{nF}\log \dfrac{1}{c}$

$\therefore c=\left[ {{M}^{n+}} \right]$

$\left[ {{M}^{n+}} \right]$is the molar concentration of the metal ion .

R is the universal gas constant $\left( 8.314J/K/mole \right)$.

T is the temperature in kelvin.

n is the number of electrons involved in the reaction.

F is the faraday’s constant (96500 C/mole).

The concentration of pure liquid and solid are taken as one so, the expression will be:

(or) $E=E{}^\circ +\dfrac{0.059}{n}\log \left[ {{M}^{n+}} \right]$

Impact of Conductance and Resistance on Electrolytic Solution:

In the context of electrolytic solutions conductance and resistance are important in predicting the flow of current and they relate to Ohm's law as follows:

Ohm’s Law: It states the current (I) flowing through a conductor at a given temperature is proportional to the potential difference (V) and proportional to resistance.

Ohm’s law formula is written as:

V ∝ I

Thus, V = RI where R is a constant called resistance. R depends on the dimensions of the conductor and also on the material of the conductor. Its SI unit is Ohm( Ω)

Resistance(R): It is the opposition to the flow of current offered by the electrolytic solution. Its units are Ohm( Ω.)

The two parallel electrodes of the cell and inversely proportional to the area of cross-section(A) of the electrode i.e.,${{R\alpha }}\dfrac{{\text{l}}}{{\text{A}}}{\text{OR R =}} \rho {\text{.}}\dfrac{{\text{l}}}{{\text{A}}}$

Conductance(G): 

It is the reciprocal of the electrical resistance(R).

$G = \dfrac{1}{R}$

It measures the ease with which the current flows through a conductor.

$G = \dfrac{1}{R}$

$G = \dfrac{1}{\rho } \times \dfrac{1}{{\dfrac{l}{A}}} = k \times \dfrac{1}{{\dfrac{l}{A}}}\because k = G \times \dfrac{l}{A}$ 

Where, G= Conductance.

$\rho$= Specific conductance (Conductivity).

$\dfrac{l}{A}$= Cell Constant 

S.I unit of conductance = Siemens(S).

The conductance or the current conducting capacity of an electrolytic solution can be expressed in terms of

• Specific conductance (k) Kappa

• Molar conductance $\left( {{\Lambda _m}} \right)$

• Equivalent conductance $\left( {{\Lambda }} \right)$

Specific conductance(K): It is reciprocal of specific resistance $\left( \rho  \right)$

$k = \dfrac{1}{\rho }$

$R = \rho .\dfrac{l}{A};\dfrac{1}{G} = \dfrac{1}{k}.\dfrac{l}{A}$

$k = G \times \dfrac{l}{A}$

G= conductance, $\dfrac{l}{A}$=cell constant

Molar conductance $\left( {{\Lambda _m}} \right)$

It is the conducting power of all the ions produced by dissolving 

$\Lambda_{{m}}=\dfrac{k}{molarity}$

$\Lambda_{m}(Scm^2 mol^{−1})=\dfrac{k(Scm^{−1})×1000(cm^3/L)}{molarity (mole/L)}$

Equivalent Conductance:

It is the conducting power of all the ions produced by dissolving 1 gram equivalent of an electrolyte (or) The conductance of a solution containing 1 gram equivalent of the electrolyte placed between two parallel electrodes separated by unit length and an area large enough to contain that 1 mole is called Equivalent conductance.

$\left( {{ \wedge _{eq}}} \right)\left( {oh{m^{ - 1}}c{m^2}e{q^{ - 1}}} \right) = \dfrac{{k\left( {oh{m^{ - 1}}c{m^{ - 1}}} \right) \times 1000}}{{N(eq.wts/L)}}$

$\left( {{ \wedge _{eq}}} \right)\left( {oh{m^{ - 1}}{m^2}e{q^{ - 1}}} \right) = \dfrac{{k\left( {s{m^{ - 1}}} \right) \times {{10}^{ - 3}}}}{{N(eq.wts/L)}}$

Factors Affecting Molar and Equivalent Conductance

• Nature of electrolyte

• Nature of the solvent

• Viscosity of solvent

• Temperature

• Concentration of electrolyte

• Size of the ions produced and their solvation

Effect of Dilution: With both $\left( {{\Lambda _{eq}}} \right)$ as well as $\left( {{\Lambda _m}} \right)$ of both weak strong electrolytes increases.

Specific conductivity (k) decreases with dilution because of the decrease in the number of ions per $c{m^3}$of electrolyte.

Effect of Temperature: The conductivity $\left( {{\Lambda _{eq}}} \right)$ and $\left( {{\Lambda _m}} \right)$ of all electrolytes increases with increase in temperature.

Debye-Huckel-Onsager Equation:

$\wedge_{c}=\wedge_{o}-\left[\dfrac{82.4}{(\mathrm{DT})_{\eta}^{\dfrac{1}{2}}}+\dfrac{8.2 \times 10^{5}}{(\mathrm{DT})^{\dfrac{3}{2}}} \wedge_{\mathrm{o}}\right] \sqrt{C}$

Here,

D= Dielectric constant of water

T = Temperature in Kelvin scale

${ \wedge _c}$ = Equivalent conductance at conc. ‘c’

${ \wedge _o}$ = equivalent conductance at almost zero concentration or infinite dilution.

$\eta$ = Viscosity coefficient of the solvent

${\wedge _c} = { \wedge _o} - b\sqrt C$, where b is constant and depends on the nature of the solvent and temperature.

Let's Solve this:

1) Read the following statements and predict the corresponding law. At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the total equivalent conductance of the electrolyte, irrespective of the nature of the ion.

1) Ostwald's-dilution law

2) Kohl Rausch's law

Electrolysis:

Chemical reactions in an electrolyte are induced by an electric current through a process known as electrolysis. While negative ions travel to the positive electrode (anode) and undergo oxidation, positive ions go to the negative electrode (cathode) and undergo reduction. The electrolyte breaks down into its parts or ions as a result.

These Electrolytes are categorized into two:

1. Non-Electrolyte: The substance that does not conduct electricity either in its molten state or in an aqueous solution is called a non-electrolyte.

2. Electrolyte: A substance that is in the molten state or in the dissolved state containing ions and is functioning as an electrically conducting medium is called an electrolyte.

The table gives an overview of different electrolyte-carrying cell reactions:

Faraday’s Law of Electrolysis:

First Law:

The amount of the substance deposited or electrolysed is directly proportional to the quantity of electricity passed through the electrolyte.

• The weight of the substance deposited on the electrode for 1 coulomb of electricity is called electrochemical equivalent(Z).

• The unit for electrochemical equivalent is gram/coulomb (g/C).

• The electrochemical equivalent depends only on the nature of the electrolyte.

Second Law:

When the same quantity of electricity is passed through different electrolytes connected in series the weights of the substances deposited will be in the ratio of their equivalent weight.

$\dfrac{{{\text{mass of the hydrogen}}}}{{{\text{mass of copper}}}}{\text{ = }}\dfrac{{{\text{Equivalent mass of hydrogen}}}}{{{\text{Equivalent mass of copper}}}}$

• Chemical equivalent weight of an element.

$\left[ \dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}} \right]$Where,

${{W}_{1}}$and ${{W}_{2}}$ are weights deposited of two elements.

${{E}_{1}}$ and ${{E}_{2}}$are the equivalent weight of two element 

• 1 Faraday of electric current or 96500 coulombs of electricity or one mole of electron, when passed through an electrolyte 1G.E.W of a substance will be deposited.

• 1 coulomb deposited

${\text{ = }}\left[ {\dfrac{{{\text{gram equivalent weight}}}}{{{\text{96500}}}}} \right]$

${\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}{\text{ = e (electrochemical equivalent)}}$ 

• 1 Faraday=charge present on $6.023 \times {10^{23}}$ electrons

• 1 coulomb $ = 6.25 \times {10^{18}}$ electrons

Current Efficiency: % of current efficiency $= \dfrac{{{\text{mass actually produced}}}}{{{\text{mass of substance expected}}}}\times 100$

Electrochemical Corrosion:

The process of corrosion may be chemical (or) electrochemical in nature

• The anodic dissolution of a metal under the conditions of corrosion is known as electrochemical corrosion

$M \to {M^{n + }} + n{e^ - }$

Passivity:

The phenomenon of metal reaching a stage of non-reactive state in its reaction with concentrated acids may be called passivity.

• Iron groups of metal and some other transition metals are rendered passive with concentrated acids.

• Passivity of metal can be classified into 

1. Chemical passivity

2. Mechanical passivity

3. Electro-chemical passivity

Chemical Passivity:

• Non-reactivity of metals after initial reaction with concentrated $HN{O_3}$ 

• Passive iron doesn’t dissolve in dill $HN{O_3}$ and iron don’t displace Ag from $AgN{O_3}$

Mechanical Passivity:

In some cases, the dissolution of metal stops due to visible oxide film formation.

• Electrochemical Passivity:

• Metal with more -ve potential functions as an anode in the cell.

• Generally, Fe, Ni, and Co functions are anodes.

• They dissolve as $M \to {M^{n + }} + n{e^ - }$

At a particular stage, the anode stops dissolving due to the formation of an invisible metal oxide film. This phenomenon is called electrochemical passivity.

Importance of Electrochemistry - JEE Advanced

In this chapter, students will learn the advanced concepts of Chemistry related to electrolysis. In this section, you will find out how an electrochemical reaction is conducted in a system using electrodes and electrolytes.

Every reaction will be explained in this chapter using chemical reactions in different phases of electrolysis. It will help you understand the mechanism of an electrochemical reaction, so you will be able to figure out how an expected result is achieved using these methods.

This chapter is very important as quite a few questions related to electrochemical reactions and electrochemical series are expected in JEE Advanced. These revision notes explain how the ions behave in an electrochemical system and the roles of different parts of electrochemical reactions.

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