Electrochemistry

The branch of chemistry that studies the interrelationship between chemical reactions and electricity. It primarily studies converting electrical energy into chemical energy and vice versa through redox reactions. Some of the fundamental topics, like the electrolytic and Galvanic Cells, where chemical reactions are driven by electric currents, are discussed in this chapter. The Nernst Equation is one of the important concepts discussed in this chapter, which explains the relationship between temperature, potential of the cell, concentration of reactants, and products. The electrode potential explains how electrons move between the cathode and the anode.

This Story also Contains

1. Important Topics of Electrochemistry
2. Important Concepts and Laws:
3. Importance of Electrochemistry class 12:
4. How to prepare for Electrochemistry
5. Prescribed Books

Electrochemistry is everywhere around us, from electric vehicles to electroplating. The lithium-ion battery used in electric vehicles stores and discharges energy based on electrochemical reactions. Hydrogen-powered buses generate electricity by the conversion of Hydrogen and Oxygen into water. Electrochemistry is widely used in industrial processes like electroplating, which is used to protect material from corrosion and rust is carried out through electrochemical processes.

Important Topics of Electrochemistry

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It focuses on the study of redox reactions, where the transfer of electrons generates or requires electrical energy, forming the basis of batteries, electrolysis, and fuel cells.

Galvanic Cells

Galvanic cells (or voltaic cells) are devices that convert chemical energy into electrical energy. Galvanic cells consist of two electrodes in separate half-cells connected by a conductive solution and a salt bridge, with electrons flowing from the anode to the cathode.

Salt Bridge

A salt bridge is a device used in galvanic cells to maintain electrical neutrality by allowing the flow of ions between the two half-cells. Salt bridge prevents charge buildup and sustains the cell's operation.

Emf of a Cell

The electromotive force (emf) of a cell is the potential difference between its two electrodes when no current flows. Emf of a cell is a measure of the cell's ability to drive electrons through an external circuit and is calculated using electrode potentials.

Standard Hydrogen Electrode (SHE)

The Standard Hydrogen Electrode is a reference electrode with a defined potential of 0 V. It consists of a platinum electrode in contact with H₂ gas and it is used to determine standard electrode potentials of other half-cells.

Electrochemical Series

The electrochemical series ranks elements based on their standard electrode potentials. It helps predict the feasibility of redox reactions, with elements higher in the series acting as better oxidizing agents and those lower as reducing agents.

Standard electrode potential

Standard electrode potential measures the half cell's tendency to gain or lose electrons when connected to the hydrogen electrode under standard conditions.

Standard electrode potential is expressed in Volts. It is represented by $E^{\circ}$.

• Tendency to gain electrons is represented by positive $E^{\circ}$
• Tendency to lose electrons is represented by $E^{\circ}$.

The potential of the Standard Hydrogen electrode is 0.00V.

Nernst Equation

The Nernst equation provides for the cell potential to its standard potential, temperature, and reactant/product concentrations. Nernst equation is crucial for understanding how EMF changes with varying conditions.

The Nernst equation is represented by:

$E=E^{\circ}-\frac{R T}{n F} \ln Q$

Concentration Cell

A concentration cell is a type of galvanic cell where the electrodes are identical but are in solutions of different concentrations. The concentration cell generates EMF due to the concentration gradient, driving ions to equalize the concentrations.

Specific Conductivity

Specific conductivity is the measure of a solution's ability to conduct electric current per unit length. Specific conductivity is expressed in Siemens per meter. It is represented by $\kappa$

The formula of Specific conductivity is:

$\kappa=\frac{1}{R} \times \frac{l}{A}$

Molar Conductivity

It is defined as the ability of an electrolytic solution to conduct electricity when one mole of electrolyte is dissolved and placed between electrodes that are 1 cm apart. Molar conductivity is represented by $\Lambda_m$ and is expressed in $\mathrm{S} \cdot \mathrm{cm}^2 \cdot \mathrm{~mol}^{-1}$.

The formula to calculate molar conductivity is:

$\Lambda_m=\frac{\kappa \times 1000}{C}$

Molar Conductivity for Strong Electrolytes:

Strong electrolytes like $\mathrm{NaCl}, \mathrm{HCl}, \mathrm{KNO}_3$ dissociate completely in solution, which produces a high concentration of ions.

Molar Conductivity for Weak Electrolytes:

Weak electrolytes like $\mathrm{NH}_4 \mathrm{OH}$ dissociates partially in solution producing fewer ions.

Kohlrausch's Law

Kohlrausch’s law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its ions. It is useful in determining dissociation constants and ion conductivities.

Faraday’s Laws of Electrolysis

Faraday gave two laws of Electrolysis

• Faraday’s first law
• Faraday's second law

Faraday's first law:

It states that the mass of substance deposited at the electrode during electrolysis is directly proportional to the electric charge passed through the electrolyte.

It is given by the formula:

$m=Z \times Q$

Second law of electrolysis:

The mass of substance deposited on different electrolytes when the same quantity of electric charge is passed through them is directly proportional to their chemical equivalents.

It is given by the formula:

$\frac{m_1}{m_2}=\frac{E_1}{E_2}$

Gibbs Free Energy of Reaction

The Gibbs free energy change (ΔG) for an electrochemical reaction is related to the cell emf by the equation ΔG=−nFE, where n is the number of electrons and F is Faraday's constant. Negative ΔG indicates a spontaneous reaction.

Types of Batteries - Primary and Secondary Cells

Primary cells, like dry cells, are non-rechargeable and provide energy until the reactants are exhausted. Secondary cells, like lead-acid and lithium-ion batteries, are rechargeable, allowing reversible chemical reactions to restore their charge.

Important Concepts and Laws:

ELECTROCHEMISTRY
Electrochemistry is the branch of science that deals with the transformation of chemical energy into electrical energy and vice versa or it deals with the relationship between electrical and chemical energy produced in a redox reaction.

Galvanic Cell (or Voltanic Cell)
Consider the following redox reaction:

$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq})$

In the above reaction, Zn displaces copper ions (Cu²⁺) from aqueous solution. This reaction can be achieved very easily in practice. Put a Zn rod into a solution of CuSO₄ (containing Cu²⁺ ions). It is observed that the blue color of the CuSO₄ solution disappears after some time. In this situation, Zn loses 2 electrons per atom, and Cu²⁺ ions in the solution accept them. Cu²⁺ ions from the solution in this manner are deposited in the form of solid Cu and Zn goes into the solution as Zn²⁺ (colorless). The reaction can well be understood in terms of two half-reactions:

Oxidation: & $\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$
Reduction: & $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{s})$

Now, we can make the same reaction take place even if the copper ions and zinc rod are not in direct contact. If we put the Cu²⁺ ions and Zn rod in two separate containers and connect the two by a conducting metallic wire and introduce an inverted U shape instrument (called as salt-bridge), then electrons will still be transferred through the connecting wires. The electrons from the Zn rod travel to Cu²⁺ ions through the connecting wires and the same reaction takes place. This flow of electrons through the wire generates electricity.

The function of a Salt Bridge

• A salt bridge acts as an electrical contact between the two half-cells.
• It prevents the mechanical flow of the solution but it provides a free path for the migration of ions to maintain an electric current through the electrolyte solution. It prevents the accumulation of excess charges.
• A salt bridge helps maintain the charge balance in the two half-cells.
• A salt bridge minimizes/eliminates the liquid junction potential.
• Liquid Junction Potential: The unequal rates of migration of the cations and anions across a liquid-liquid junction give rise to a potential difference across the junction. This potential difference across the liquid-liquid junction is called liquid junction
  potential.

Importance of Electrochemistry class 12:

Chemistry has a large syllabus that must be covered in order for a student to qualify for the class 12 board exams. It is significant not just in terms of the board test, but also in terms of the JEE and NEET examinations. Physical chemistry, organic chemistry, and inorganic chemistry are the three major areas that make up the chemistry curriculum.

Electrochemistry is a branch of physical chemistry that is used to store energy electrochemically. It deals with the qualities of materials that make them appropriate for electrodes, different types of electrolytes that help with charge transport, and establishing a relationship between observable and quantifiable electrical potential properties for chemical reactions that occur inside the cell. Given the growing relevance of energy storage systems, electrochemistry becomes increasingly important in understanding how electrodes, separators, chemicals, and other materials are chosen, as well as the computation of the available electric potential due to these processes.

How to prepare for Electrochemistry

• This chapter is one of the most important in Physical chemistry.
• For preparing this chapter, there is no need of any pre-requisite chapter. Thus prepare it freshly and actively.
• All these basic concepts and rules life EMF of a cell, flow of current, redox reaction, mechanism of galvanic cell and electrochemical cell etc. must be understood well . Thus prepare this chapter carefully. Try to solve numerical problems from NCERT as many such questions are asked in the examination.
• In the nutshell, it can be said that although this chapter is not very long, it is a very simple and straightforward one. So always say a "Big YES" to this chapter.

Prescribed Books

For this chapter, first, the NCERT book is best for initial level preparation as well as for board exams. Now, after this, if you want to prepare for competitive exams like JEE and NEET, then these are the best books for you - O.P Tandon. Meanwhile, in the preparation, you must continuously give the mock tests for the depth of knowledge. Our platform will help you with the variety of questions for deeper knowledge with the help of videos, articles, and mock tests.

Some Solved Problems

Example.1 Given :

$\begin{aligned} & \mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+}+; E^0=+1.81 \mathrm{~V} \\ & \mathrm{~Pb}^{4+}+2 e^{-} \rightarrow \mathrm{Pb}^{2+}+; E^0=+1.67 \mathrm{~V} \\ & C e^{4+}+e^{-} \rightarrow C e^{3+}+; E^0=+1.61 \mathrm{~V} \\ & B i^{3+}+3 e^{-} \rightarrow B i ; E^0=+0.20 \mathrm{~V}\end{aligned}$

The oxidizing power of the species will increase in the order:

1) (correct)$\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$

2)$\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}$

3)$\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}$

4)$\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}$

Solution

The greater the standard reduction potential, the greater its oxidizing power.

$\therefore$ The correct sequence will be :

$\mathrm{Co}^{3+}>\mathrm{Pb}^{4+}>\mathrm{Ce}^{4+}>\mathrm{Bi}^{3+}$

Hence, the answer is the option (1).

Example.2 Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given that

$\mathrm{E}_{\mathrm{Ag}+\mid \mathrm{Ag}}^0=\mathrm{x} \mathrm{V}$

$\mathrm{E}_{\mathrm{Fe}^{2+} \mid \mathrm{Fe}}^0=\mathrm{y} \mathrm{V}$

$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}}^0=\mathrm{z} \mathrm{V}$

1) x - z

2) x - y

3) (correct) x + 2y - 3z

4) x + y - z

Solution

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given,

$\mathrm{E}_{\mathrm{Ag}+\mid \mathrm{Ag}}^0=\mathrm{x} \mathrm{V}$

$\mathrm{E}_{\mathrm{Fe}^{2+\mid} \mid \mathrm{Fe}}^0=\mathrm{y} \mathrm{V}$

$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}}^0=\mathrm{z} \mathrm{V}$

The given equation can be represented in the form of a cell as

$\mathrm{Fe}^{2+}\left|\mathrm{Fe}^{+3} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$

Standard EMF of given cell reaction
$
\mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0-\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0
$

It is evident that in order to find the above cell potential, we need to find the electrode potential of the $\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}$ couple.

In order to calculate the value of $\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0$, we need to use the given values of $\mathrm{E}_{\mathrm{Fe}^{2+} \mid \mathrm{Fe}}^0$ and $\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}}^0$.

Now,
$
\mathrm{Fe}^{+2}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \quad \mathrm{E}^0=\mathrm{y}, \Delta \mathrm{G}^0=-2 \mathrm{Fy}
$
$
\mathrm{Fe}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}^0=\mathrm{z}, \Delta \mathrm{G}^0=-3 \mathrm{Fz}
$

Now, subtracting (ii) from (iii), we have

$\mathrm{Fe}^{+3}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2} ; \Delta \mathrm{G}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

Thus we can write

$-1 \times \mathrm{F} \times \mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

$\mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}^{2+}}^0=3 \mathrm{z}-2 \mathrm{y}$

Putting this in equation (i), we have

$\mathrm{E}_{\text {cell }}^0=x-(3 z-2 y)=x+2 y-3 z$

Hence, the answer is the option (3).

Example. 3 The cell,

$ Zn\left|Zn^{2+}(1 M) \| Cu^{2+}(1 M)\right| Cu\left(E_{\text {cell }}^{\circ}=1.10 \mathrm{~V}\right)$

was allowed to be completely discharged at 298 K.The relative concentration of $\mathrm{Zn}^{2+}$ to $\mathrm{Cu}^{2+}$

$\left(\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}\right)$is

1) $9.65 \times 10^4$

2) $\operatorname{antilog}(24.08)$

3) 37.3

4) (correct) $10^{37.3}$

Solution

$\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$

From the Nernst equation, we can write

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$

When the cell is completely discharged, $E_{\text {cell }}=0$

$0=1.1-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$

or $\log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{2 \times 1.1}{0.059}$ or, $\log \frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}=37.3$

or $\frac{Z n^{2+}}{C u^{2+}}=10^{37.3}$

Hence, the answer is the option (4).

Example 4: The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70 \mathrm{~S} \mathrm{~cm} \mathrm{~mol}^1$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its degree of dissociation is given by $\mathrm{x} \times 10^{-1}$. The value of $x$ is _______ (Nearest integer)

Solution:

$\begin{aligned}
& \lambda_{\mathrm{m}}^{\prime \prime} \text { of } \mathrm{NH}_4 \mathrm{Cl}=185 \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}=185 \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4}=185-70=115 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
& \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{\circ}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{OH}^{-}} \\
& =115+170 \\
& \left(\lambda_{\mathrm{m}}^0\right)_{\mathrm{NH}_4 \mathrm{OH}}=285 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
& \text { degree of dissociation }=\frac{\left(\lambda_{\mathrm{m}}\right)_{\mathrm{NH}_4 \mathrm{OH}}}{\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}} \\
& =\frac{85.5}{285} \\
& =0.3 \\
& =3 \times 10^{-1}
\end{aligned}$

Hence, the answer is 3.

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