Electrochemistry

Electrochemistry is the study of generating electricity from the energy produced during spontaneous chemical reactions, as well as the application of electrical energy to non-spontaneous chemical changes. This topic is important from both a theoretical and a practical point of view. Electrochemical processes are used to make a variety of metals, sodium hydroxide, chlorine, fluorine, and other compounds. Batteries and fuel cells are widely utilized in a variety of equipment and systems to convert chemical energy to electrical energy. Electrochemical reactions can be both energy-efficient and environmentally friendly.

Electrochemistry is a broad and interdisciplinary field of study. We will only explore several of the most significant fundamental components of it in this Unit. Class 12 electrochemistry notes aim to put the concepts behind chemical processes in context. Students will learn that the ionic conductor is an important part of cells by reading the notes from chapter 3 of Chemistry class 12 thoroughly.

Important Concepts and Laws:

ELECTROCHEMISTRY
Electrochemistry is the branch of the 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 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.

NCERT Solutions Subject-wise

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.

Some Solved Problems

Example.1 Given :

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

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).

NCERT Exemplar Solutions Subject-wise

• NCERT exemplar solutions for class 12 Physics.
• NCERT exemplar solutions for class 12 Chemistry.
• NCERT exemplar solutions for class 12 Mathematics.
• NCERT exemplar solutions for class 12 Biology.