The value of \(E^{\circ}\) at \(298 \mathrm{~K}\) for the given reaction is \(1.5\) \(\mathrm{V}\) find the \(\Delta \mathrm{G}\) value here? \(\mathrm{Cr}+3 \mathrm{Ag}^{+}(0.1 \mathrm{M}) \rightarrow 3 \mathrm{Ag}^{\prime}+\mathrm{Cr}^{3+}\) (a) \(-422.83 \mathrm{KJ}\) (b) \(-211.415 \mathrm{KJ}\) (c) \(+211.415 \mathrm{KJ}\) (d) \(+422.83 \mathrm{KJ}\)

The concept of standard cell potential, denoted as \(E^{\circ}\), is fundamental to electrochemistry. It represents the voltage difference between two half-cells in an electrochemical cell when all components are in their standard states, which means solutes are at 1 M concentration, gases are at 1 atm pressure, and the temperature is usually at 25°C (298 K).

In the context of the given exercise, we focus on the cell potential as it relates to the Gibbs free energy, \(\Delta G\), of the reaction. The standard cell potential can indicate whether a reaction is spontaneous. If \(E^{\circ}\) is positive, the reaction tends to occur spontaneously, giving a negative \(\Delta G\), and vice versa. The equation \(\Delta G = -nFE^{\circ}\) forms the bridge between these two thermodynamic quantities.

The answer to the textbook problem regarding the Gibbs free energy change of a reaction, given its standard cell potential, demonstrates the practical use of this relationship. By understanding the standard cell potential, students can predict the feasibility of reactions and calculate the maximum work that can be harnessed from electrochemical processes.

Electrochemical reactions are at the heart of batteries, corrosion, and many industrial processes. These reactions involve the transfer of electrons between species, leading to chemical changes. They are composed of two parts: oxidation, where a species loses electrons, and reduction, where a species gains electrons. Together, these make up redox reactions.

The exercise provided showcases a typical electrochemical reaction where chromium (Cr) is oxidized, and silver ions (Ag+) are reduced. It's important to understand that the movement and transfer of electrons facilitate the conversion of chemical energy to electrical energy, or vice versa, depending on the system's context. By identifying the number of electrons transferred in the reaction, as shown in the exercise, and by knowing the concept of cell potential, students can grasp how these reactions are essential for calculating energy changes in an electrochemical cell.

Faraday's constant, symbolized by \(F\), is a critical factor in electrochemistry calculations. It represents the total electric charge carried by one mole of electrons, approximately 96485 coulombs/mol. Named after the scientist Michael Faraday, this constant helps quantify the relationship between the amount of substance altered at an electrode and the electric charge passing through the electrode.

In the step-by-step solution of our exercise, \(F\) is used to convert the product of the number of moles of electrons and the cell potential into energy units (joules or kilojoules). Understanding Faraday's constant is vital in solving various electrochemical problems, as it not only enables the calculation of Gibbs free energy changes but also assists in determining the amount of substance produced or consumed during electrolysis, based on the electric current passed through the system.