The overall cell reaction is governed by the half reactions:
At Ag/AgCl electrode:
$$ \text{AgCl(s)} + e^- \rightleftharpoons \text{Ag(s)} + \text{Cl}^- $$
At platinum electrode:
$$ \text{Fe}^{3+} + e^- \rightleftharpoons \text{Fe}^{2+} $$
The Nernst equation for the cell potential can be written as
$$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{1}\log\!\left(\frac{[\text{Fe}^{2+}][\text{Cl}^-]}{[\text{Fe}^{3+}]}\right) $$
From the expression, $E_{\text{cell}}$ increases when the reaction quotient decreases.
Now check each condition:
Increasing Fe$^{2+}$ increases the reaction quotient, hence $E_{\text{cell}}$ decreases (A is false).
Decreasing Fe$^{3+}$ increases the reaction quotient, hence $E_{\text{cell}}$ decreases (B is false).
Decreasing Fe$^{2+}$ decreases the reaction quotient, hence $E_{\text{cell}}$ increases (C is true).
Increasing Fe$^{3+}$ decreases the reaction quotient, hence $E_{\text{cell}}$ increases (D is true).
Increasing Cl$^-$ decreases the electrode potential of Ag/AgCl electrode in such a way that overall cell potential increases (E is true).
Therefore, the correct conditions are
$$ \boxed{\text{C, D and E only}} $$
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.