JEE Main 2026 — Electrochemistry Question with Solution

The given setup is a concentration cell with identical electrodes (M) and different electrolyte concentrations ($c_1$ and $c_2$).
For a concentration cell, the standard cell potential $E_{cell}^0=0$.
The cell reactions are:
Anode: $M(s)\to M^{+}(c_{anode})+e^{-}$
Cathode: $M^{+}(c_{cathode})+e^{-}\to M(s)$
Net reaction: $M^{+}(c_{cathode})\to M^{+}(c_{anode})$
Using the Nernst equation at $298\text{ K}$:
$E_{cell}=E_{cell}^0-\frac{0.0591}{n}\log \frac{c_{anode}}{c_{cathode}}$
$E_{cell}=0-0.0591\log \frac{c_{anode}}{c_{cathode}}$
For a positive cell potential ($E_{cell}>0$), the term $\log \frac{c_{anode}}{c_{cathode}}$ must be negative.
This requires $\frac{c_{anode}}{c_{cathode}}<1$, which means $c_{cathode}>c_{anode}$.
If $c_1$ is present at the cathode and $c_2$ is at the anode, then for $E_{cell}>0$, we must have $c_1>c_2$.
Checking the options:
(1) If $c_1$ is at cathode, $c_1<c_2$ (Incorrect)
(2) If $c_1$ is at anode, $c_1>c_2$ (Incorrect, would require $c_2>c_1$)
(3) If $c_1$ is at cathode, $c_1>c_2$ (Correct)
(4) If $c_1$ is at anode, $c_1=c_2$ (Incorrect, $E_{cell}$ would be zero)