Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The following electrochemical cell has been set up. \(\mathrm{Pt}(1)\left|\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}(a=1)\right| \mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}(a=1) \mid \mathrm{Pt}(2)\) \(E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V}: E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)=1.61 \mathrm{~V}\) If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?

Step-by-Step Solution

Answer

Current flows from Ce half-cell to Fe half-cell and decreases over time.

Step 1: Identify the Half-Reactions

The electrochemical cell involves two half-reactions. The oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode. We have the following standard electrode potentials: - For the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) couple, \(E^{\circ} = 0.77 \, \text{V}\).- For the \(\text{Ce}^{4+}/\text{Ce}^{3+}\) couple, \(E^{\circ} = 1.61 \, \text{V}\).Fe will be oxidized, and Ce will be reduced based on their standard potentials.

Step 2: Determine the Direction of Electron Flow

Electrons will flow from the more negative potential to the more positive potential. Since the \(E^{\circ}\) of \(\text{Fe}^{3+}/\text{Fe}^{2+}\) (0.77 V) is less than that of \(\text{Ce}^{4+}/\text{Ce}^{3+}\) (1.61 V), electrons flow from the iron half-cell (anode) to the cerium half-cell (cathode). Thus, current, by convention, flows in the reverse direction, from cerium half-cell to the iron half-cell.

Step 3: Predict the Change in Current Over Time

The current in an electrochemical cell is affected by the concentration of the ions involved in the redox reactions. Since both reactions start with activities of 1, and assuming a minimal cell setup without replenishing sources, concentrations of reactants will decrease, leading to reduced current flow. Therefore, the current will decrease over time as the concentration of active ions diminishes.

Key Concepts

Half-Reactions

In an electrochemical cell, the overall reaction is split into two parts, called half-reactions. These half-reactions represent the oxidation process at the anode and the reduction process at the cathode.
For our cell:

• The oxidation half-reaction involves iron (\(\text{Fe}\)) ions: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\).
• The reduction half-reaction involves cerium (\(\text{Ce}\)) ions: \(\text{Ce}^{4+} + \text{e}^- \rightarrow \text{Ce}^{3+}\).

These half-reactions ensure that electrons are transferred from one substance to another, facilitating the electrochemical process. Identifying the species that gain or lose electrons is crucial in understanding how the cell operates and which direction the electrons will flow.

Standard Electrode Potentials

Standard electrode potentials (\(E^{\circ}\)) help us determine the tendency of a species to gain or lose electrons. They are measured under standard conditions and expressed in volts.
In our cell:

• For the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) couple, \(E^{\circ} = 0.77 \, \text{V}\).
• For the \(\text{Ce}^{4+}/\text{Ce}^{3+}\) couple, \(E^{\circ} = 1.61 \, \text{V}\).

A higher standard electrode potential indicates a stronger affinity for electrons, meaning the species is more likely to be reduced. Thus, in our cell, cerium (\(\text{Ce}^{4+}/\text{Ce}^{3+}\)) has a higher potential compared to iron (\(\text{Fe}^{3+}/\text{Fe}^{2+}\)), making it the preferred site for reduction. This comparison of potentials helps us assign the half-reactions correctly and predict electron movement.

Electron Flow Direction

Electrons in an electrochemical cell flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). This flow direction is determined by comparing the standard electrode potentials.
In our example:

• The \(E^{\circ}\) value for iron (\(0.77 \, \text{V}\)) is lower than cerium's (\(1.61 \, \text{V}\)).
• This means electrons will leave the iron half-cell and flow towards the cerium half-cell.

Since electrons move from lower to higher potential, they provide a current in the opposite direction. By convention, this means the current flows from the cerium half-cell to the iron half-cell. Understanding electron flow direction is essential for predicting how the cell will be wired in practical applications.

Current Change Over Time

The current in an electrochemical cell is influenced by the concentration of the ions in solution, directly affecting the reaction rate.
Initially, both cerium and iron ions have an activity of 1, indicating that they are in abundant supply. However, as the cell operates:

• The ions will be consumed in their respective half-reactions.
• Without replenishing these ions, their concentration diminishes over time.

As the concentration decreases, the reaction rate slows, leading to a decrease in current. Therefore, predicting that the current will reduce over time is logical when no additional reactants are introduced. This insight helps in the practical management and design of long-term usage for electrochemical cells.