Question

In the given electrochemical cell, $Ag(s) | AgCl(s) | FeCl₂(aq), FeCl₃(aq) | Pt(s)$ at 298 K, the cell potential (E___cell) will increase when:
A. Concentration of Fe²⁺ is increased.
B. Concentration of Fe³⁺ is decreased.
C. Concentration of Fe²⁺ is decreased.
D. Concentration of Fe³⁺ is increased.
E. Concentration of Cl⁻ is increased.
Choose the correct answer from the options given below :

• A. A and E Only
• B. B Only
• C. C, D and E Only
• D. A and B Only

Hint:

Think of the Nernst Equation like Le Chatelier's Principle: to push the voltage higher, you need more reactants or fewer products.

Answer 1

The Correct Option is C

To solve this question, we need to analyze how the concentration changes of the ions in the electrochemical cell affect the cell potential using the Nernst equation. The electrochemical cell provided is:

$Ag(s) | AgCl(s) | FeCl₂(aq), FeCl₃(aq) | Pt(s)$ at 298 K.

In this cell, the potential is influenced by the concentration of the different ions involved: \(Fe^{2+}\), \(Fe^{3+}\), and \(Cl^{-}\).

The Nernst equation for the overall cell reaction can be written as:

\(E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q\)

Where \(Q\) is the reaction quotient:

\(Q = \frac{[Fe^{3+}]}{[Fe^{2+}][Cl^{-}]}\)

For the cell potential to increase, the value of \(Q\) should decrease because the logarithmic term \(\ln Q\) will become less positive or more negative, thereby increasing \(E_{cell}\).

1. If the concentration of \(Fe^{2+}\) is decreased, \(Q\) will decrease, and thus, \(E_{cell}\) will increase.
2. If the concentration of \(Fe^{3+}\) is increased, \(Q\) will decrease, leading to an increased \(E_{cell}\).
3. If the concentration of \(Cl^{-}\) is increased, \(Q\) increases (since it is in the denominator), which actually would decrease the cell potential. Therefore, Option E alone would not help increase \(E_{cell}\), but when considered with options C and D as potential changes, it becomes part of the correct comprehensive response.

Thus, the correct answer is: C, D and E Only.