Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following condition will increase the voltage of the cell represented by the equation \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq})=\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) ?\) (a) increase in the dimension of \(\mathrm{Cu}\) electrode (b) increase in the dimension of Ag electrode (c) increase in the concentration of \(\mathrm{Cu}^{2+}\) ions (d) increase in the concentration of \(\mathrm{Ag}^{+}\)ions

Step-by-Step Solution

Answer

Increasing \( \mathrm{Ag}^{+} \) concentration increases cell voltage.

Step 1: Identify the Cell Reaction

The given reaction is \(\mathrm{Cu}(\mathrm{s}) + 2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq}) + 2 \mathrm{Ag}(\mathrm{s})\). This is an example of a redox reaction involving the oxidation of copper and the reduction of silver ions.

Step 2: Understand Cell Voltage Influence

According to Nernst equation, the cell voltage is affected by the concentration of ions involved in the reaction: \( E = E^0 - \frac{RT}{nF}\ln{Q} \), where \( Q \) is the reaction quotient \( \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^{+}]^2} \). Changes in concentrations of ions directly impact \( Q \) and thus the cell voltage.

Step 3: Examine Each Condition

(a) Increasing the dimension of the \( \mathrm{Cu} \) electrode does not change \( Q \) as it does not affect concentration.(b) Increasing the dimension of the \( \mathrm{Ag} \) electrode has no effect on \( Q \) because it affects neither \( [\mathrm{Cu}^{2+}] \) nor \( [\mathrm{Ag}^{+}] \).(c) Increasing \( [\mathrm{Cu}^{2+}] \) increases \( Q \), leading to a decrease in voltage.(d) Increasing \( [\mathrm{Ag}^{+}] \) adds to the denominator, decreasing \( Q \), hence, increasing the cell voltage.

Step 4: Conclude With the Correct Condition

Out of all the given options, only increasing the concentration of \( \mathrm{Ag}^{+} \) ions decreases the reaction quotient \( Q \) and thus increases the voltage of the cell according to the Nernst equation.

Key Concepts

Redox Reaction

A redox reaction, or reduction-oxidation reaction, is a chemical reaction where the oxidation states of atoms are altered. This involves the transfer of electrons from one molecule to another. In the context of the given reaction, copper (\(\mathrm{Cu}\)) undergoes oxidation, meaning it loses electrons and becomes copper ions (\(\mathrm{Cu}^{2+}\)). Meanwhile, silver ions (\(\mathrm{Ag}^{+}\)) gain electrons through a reduction process and become solid silver.

• Oxidation: Loss of electrons (e.g., \(\mathrm{Cu \rightarrow Cu^{2+} + 2e^-}\))
• Reduction: Gain of electrons (e.g., \(\mathrm{2Ag^{+} + 2e^- \rightarrow 2Ag}\))

Redox reactions are crucial in understanding how electrochemical cells, like the one in the exercise, operate. They facilitate the flow of electrons from one electrode to another, generating an electric current.

Electrode Concentration

Electrode concentration refers to the molarity of the ions that are involved in the electrode reaction within an electrochemical cell. These concentrations affect the reaction quotient (\(Q\)), which is a critical part of the Nernst equation. The concentration of ions influences how close the cell reaction is to equilibrium.In our exercise:

• A higher concentration of \(\mathrm{Cu^{2+}}\) ions increases \(Q\) because it is in the numerator, making the cell reaction less favorable and reducing the cell voltage.
• Conversely, a higher concentration of \(\mathrm{Ag^{+}}\) ions adds to the denominator of \(Q\), which makes the overall quotient smaller and thus the cell voltage higher.

Thus, any alteration in these concentrations will affect the cell voltage, highlighting the importance of electrode concentration in electrochemical cells.

Cell Voltage

Cell voltage, also known as electromotive force (EMF), is the measure of the energy potential that drives electrons through a circuit, which is generated by an electrochemical cell. The Nernst equation provides a means to calculate this voltage under non-standard conditions and is expressed as:\[E = E^0 - \frac{RT}{nF}\ln{Q}\]Where:

• \(E^0\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged in the reaction.
• \(F\) is Faraday's constant.
• \(Q\) is the reaction quotient.

In the exercise question, altering concentrations changes \(Q\), which in turn affects \(E\). Increasing the \(\mathrm{Ag^{+}}\) concentration decreases \(Q\), resulting in an increase in cell voltage, illustrating how sensitive cell voltage is to changes in ion concentration.