JEE Chemistry 2026 Question with Solution

Given: The electrochemical cell is $Ag(s) | AgCl(s) | FeCl_2(aq), FeCl_3(aq) | Pt(s)$ at $298 \, \text{K}$.

Find: Which change increases the cell potential $E_{cell}$.

The solution uses the Nernst equation. To increase cell potential, the reaction quotient $Q$ must decrease.

Anode reaction:

$$Ag(s) + Cl^-(aq) \rightarrow AgCl(s) + e^-$$

Cathode reaction:

$$Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$$

Net reaction:

$$Ag(s) + Cl^-(aq) + Fe^{3+}(aq) \rightarrow AgCl(s) + Fe^{2+}(aq)$$

Therefore,

$$E_{cell} = E^\circ_{cell} - \frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$$

So,

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

Now check each statement:

• Decreasing $[Fe^{2+}]$ decreases $Q$, so $E_{cell}$ increases.
• Increasing $[Fe^{3+}]$ decreases $Q$, so $E_{cell}$ increases.
• Increasing $[Cl^-]$ decreases $Q$, so $E_{cell}$ increases.

Hence statements C, D and E are correct.

Therefore, the correct option is C.

Given: The cell potential depends on the reaction quotient.

Find: Which concentration changes make $E_{cell}$ larger.

From the net reaction, the quotient is

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

A smaller $Q$ gives a larger $E_{cell}$ in the Nernst equation.

So the voltage increases when the numerator decreases or the denominator increases:

• decrease $[Fe^{2+}]$,
• increase $[Fe^{3+}]$,
• increase $[Cl^-]$.

Therefore, the correct option is C, corresponding to C, D and E Only.