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\(. \mathrm{Ag}\left|\mathrm{Ag}^{+}(\mathrm{IM}) \| \mathrm{Ag}^{+}(2 \mathrm{M})\right| \mathrm{Ag}\) 1 L solution 1 L solution \(0.5 \mathrm{~F}\) of electricity in the LHS (anode) the \(1 \mathrm{~F}\) 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

Short Answer

Expert verified
The EMF decreases after electrolysis.

Step by step solution

01

Determine Initial Cell Setup

Initially, the cell is represented as \( \text{Ag} | \text{Ag}^+(1 \text{M}) \| \text{Ag}^+ (2 \text{M}) | \text{Ag} \). It is an electrochemical cell with different concentrations of \( \text{Ag}^+ \) ions at the anode and cathode.
02

Analyze Electrolysis Process

During electrolysis, electricity is passed through both cell compartments. \(0.5\text{ F}\) is passed in the anode side and \(1\text{ F}\) in the cathode side. This indicates that silver ions at the cathode side are reduced to metallic silver and at the anode side silver dissolves to increase the concentration of \( \text{Ag}^+ \) ions. Hence, silver ions decrease in the cathode and increase in the anode.
03

Compare Ion Concentration Changes

After the completion of electrolysis, the concentrations change such that the concentration of \( \text{Ag}^+ \) ions in the cathode decreases and in the anode increases as explained. This modifies the concentration gradient of \( \text{Ag}^+ \) ions between the two compartments.
04

Apply Nernst Equation to Determine EMF Change

The Nernst equation determines the EMF of a concentration cell: \[ E = E^0 - \frac{RT}{nF} \ln \frac{\text{[Ag}^+\text{]}_{anode}}{\text{[Ag}^+\text{]}_{cathode}} \]. After electrolysis, the concentration ratio \( \frac{\text{[Ag}^+]_{anode}}{\text{[Ag}^+]_{cathode}} \) increases, thus leading to a decrease in EMF according to the Nernst equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
The Nernst Equation is a pivotal formula in electrochemistry, used to calculate the electromotive force (EMF) of a concentration cell. It allows us to understand how the voltage of a cell changes with varying ion concentrations. The equation is expressed as:\[E = E^0 - \frac{RT}{nF} \ln \frac{\text{[Ag}^+\text{]}_{anode}}{\text{[Ag}^+\text{]}_{cathode}}\]Let's break this down:
  • \(E^0\) is the standard electrode potential, representing the voltage under standard conditions.
  • \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday’s constant.
  • \( \frac{\text{[Ag}^+\text{]}_{anode}}{\text{[Ag}^+\text{]}_{cathode}} \) is the ratio of ion concentrations between the anode and the cathode.
During electrolysis, the concentration of silver ions increases at the anode and decreases at the cathode. This adjustment in ion concentrations enhances the concentration ratio, directly influencing the cell's EMF. As the equation shows a natural logarithm, even small changes in concentration can significantly impact the voltage. Thus, after electrolysis, the cell experiences a decrease in EMF since the concentration gradient is intensified.
Electrolysis
Electrolysis involves passing an electric current through an electrolyte to induce a chemical change. In electrochemical cells, this process usually transforms ions into their elemental form:
  • At the anode, where oxidation occurs, ions lose electrons. In our exercise, silver ( \(\text{Ag}\)) dissolves to increase the concentration of \(\text{Ag}^+\) ions.
  • At the cathode, where reduction happens, ions gain electrons. Here, silver ions reduce to form metallic silver, thus depleting \(\text{Ag}^+\) ions.
The amount of electricity passed affects these reactions. In our scenario, the anode receives \(0.5\text{ F}\) and the cathode \(1\text{ F}\). Faraday's constant \((\text{F})\) is essential as it communicates the relationship between charge and substance converted. The result is a shift in ion concentrations, displaying the transformative power of electrolysis in modifying the component makeup of the cell solutions.
Electrode Concentration
Electrode concentration is a crucial factor in electrochemical cells, and it refers to the concentration of ions that are available at the electrodes in the cell compartments. In a concentration cell, it is particularly significant because the EMF arises primarily from the concentration gradient:
  • When there is a higher concentration of ions in one compartment compared to the other, a potential difference is established. This motivates ions to move across the cell to balance the difference.
  • In our exercise, initially, the anode housed a \(1\text{ M}\) concentration of \(\text{Ag}^+\) ions, while the cathode had \(2\text{ M}\).
  • Applying electrolysis changes these concentrations: the anode's increases, the cathode's decreases. Thus, modifying the potential difference or EMF of the cell.
This concentration change alters the voltage according to the Nernst Equation. The cell's ability to perform work, like creating electricity, heavily depends on maintaining this concentration gradient. When electrolysis alters ion concentrations, it adjusts the potential energy and thus the EMF of the cell as observed in the original exercise.

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Most popular questions from this chapter

\(\mathrm{K}_{s}\) of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10 .}\) If the ionic conductances of \(\mathrm{Ba}^{+}\)and \(\mathrm{SO}_{4}^{2-\text { ions are }} 64\) and \(80 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively, then its specific conductance is (a) \(1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (b) \(144 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (c) \(1.44 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (d) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

Which is/are correct statement about salt bridge? (a) Ions of salt bridge discharge at electrode (b) Ions of salt bridge do not discharge at electrode (c) Velocity of ions of salt bridge are almost equal (d) Salt bridge complete the electric circuit.

The standard emf of a cell, involving one electron change is found to be \(0.591 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.\) \(\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(1.0 \times 10^{30}\) (b) \(1.0 \times 10^{1}\) (c) \(1.0 \times 10^{5}\) (d) \(1.0 \times 10^{10}\)

A solution of \(\mathrm{CuSO}_{4}\) is electrolyzed for 7 minutes with a current of \(0.6 \mathrm{~A}\). The amount of electricity passed equal to (a) \(26 \mathrm{C}\) (b) \(4.2 \mathrm{C}\) (c) \(2.6 \times 10^{-4} \mathrm{~F}\) (d) \(2.6 \times 10^{-3} \mathrm{~F}\)

The standard reduction potentials of \(\mathrm{Ag}, \mathrm{Cu}, \mathrm{Co}\) and \(\mathrm{Zn}\) are \(0.799,0.337,-0.277\) and \(-0.762 \mathrm{~V}\) respectively. Which of the following cells will have maximum cell emf? (a) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{IM}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (b) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{lM})\right| \mathrm{Ag}\) (c) \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{IM})\right| \mathrm{Ag}\) (d) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Co}^{2+}(\mathrm{IM})\right| \mathrm{Co}\)

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