Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{Ag}\left|\mathrm{Ag}^{+}(\mathrm{IM}) \| \mathrm{Ag}^{+}(2 \mathrm{M})\right| \mathrm{Ag}\) 1 L solution 1 L solution \(0.5 \mathrm{~F}\) of electricity in the LHS (anode) the \(1 F\) electricity in the RHS (cathode) is first passed making them independent electro cells at \(298 \mathrm{~K}\). The emf of the cell after electrolysis will (a) increase (b) decrease (c) not change (d) time is also required

Step-by-Step Solution

Answer

(b) decrease

Step 1: Understanding the Cell Setup

We have a galvanic cell with two compartments, both containing silver nitrate solutions but with different concentrations. The left side (LHS) is the anode with \([\text{Ag}^+] = 1\text{ M}\), and the right side (RHS) is the cathode with \([\text{Ag}^+] = 2\text{ M}\). Silver metal plates are used as electrodes on both sides.

Step 2: Understanding the Effect of Current

When electricity is passed through the cell, it will cause redox reactions at the electrodes. We are given \(0.5\) Faraday on the anode side and \(1\) Faraday on the cathode side. This electricity will cause changes in silver ion concentrations at the anode and cathode.

Step 3: Electrolysis at the Anode

At the anode, when 0.5 Faraday of electricity is passed, silver metal \(\text{Ag}\) will dissolve into the solution as \({\text{Ag}^+}\). The reaction is \[\text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-\] Since 0.5 moles of electrons corresponds to 0.5 moles of \( ext{Ag}^+\) ions formed, the concentration of \(\text{Ag}^+\) in the anode compartment will increase.

Step 4: Electrolysis at the Cathode

At the cathode, with 1 Faraday of electricity, \(\text{Ag}^+\) ions will be reduced to \(\text{Ag}\) metal: \[\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\] This corresponds to a decrease in the \(\text{Ag}^+\) concentration in the cathode compartment as 1 mole of Ag will be deposited.

Step 5: Determining Cell Potential Change

According to the Nernst Equation, cell potential depends on the concentration ratio of ions in the two half-cells. With the anode concentration increasing and the cathode concentration decreasing, the concentration cell potential measured decreases as the concentration gradient between the two half-cells diminishes.

Step 6: Concluding the Effect on emf

The cell potential \(E_{cell}\) after electrolysis will decrease because the ion concentration at the anode increases, and at the cathode decreases, reducing the driving force for electron flow.

Key Concepts

Galvanic Cell

A Galvanic Cell, also known as a voltaic cell, is a device that generates electricity through a spontaneous redox reaction. These cells consist of two different metals connected by a salt bridge or divided by a porous partition to allow ion flow. In our example, the Galvanic Cell is built with silver electrodes and silver nitrate solutions of differing concentrations.
When the cell operates, a redox reaction occurs:

• Anode (Oxidation): Silver metal dissolves into the solution, releasing electrons and forming silver ions (\( ext{Ag}(s) \rightarrow ext{Ag}^+(aq) + e^-\)).
• Cathode (Reduction): Silver ions in solution gain electrons and form silver metal (\( ext{Ag}^+(aq) + e^- \rightarrow ext{Ag}(s)\)).

These reactions cause a flow of electrons from the anode to the cathode, creating an electric current. The potential difference between the two electrodes is called cell potential.

Nernst Equation

The Nernst Equation is crucial in understanding how the cell potential or EMF (electromotive force) changes with the concentration of reactants and products. For our Galvanic Cell, the Nernst Equation is:\[E_{cell} = E^⦵_{cell} - \frac{RT}{nF} \ln \left(\frac{[products]}{[reactants]}\right)\]Where

• \(E^⦵_{cell}\) is the standard cell potential,
• \(R\) is the gas constant, \(T\) the temperature, \(n\) the number of electrons exchanged,
• \(F\) is the Faraday constant.

In our exercise, the concentrations of silver ions change due to the electrolysis occurring at both electrodes, altering the EMF. As the concentration gradient between the anode and cathode diminishes, the cell potential decreases.

Electrolysis

In electrochemistry, electrolysis involves passing electricity through a solution to drive a non-spontaneous chemical reaction. Our electrochemical cell uses electrolysis to change ion concentrations inside the cell. For example:

• At the anode, an input of 0.5 Faraday dissolves silver metal, forming silver ions, thus increasing ion concentration.
• At the cathode, 1 Faraday of electricity reduces silver ions, depositing silver metal and decreasing ion concentration.

Electrolysis overcomes the natural energy barrier to facilitate these reactions, and in this case, results in changes that influence overall cell potential.

Cell Potential

Cell Potential refers to the voltage or electromotive force across the electrodes of a working cell. It depends on the inherent tendency of reactions at the electrodes and the concentration of ions involved. In our Galvanic Cell, the potential difference measures the ability to drive electrons from the anode to the cathode.
Initially, the difference in concentration of silver ions induces the flow of electrons. When electrolysis alters these concentrations, the Cell Potential changes too. The decrease in driving force for electron flow between the electrodes leads to a diminished EMF, meaning the cell potential drops. This is explained by the diminishing concentration gradient that decreases the voltage output of the cell.