JEE Main 2026 — Electrochemistry Question with Solution

Given standard reduction potentials:
$E_{M^{x+}/M}^{\ominus }=+0.15$ V
$E_{F{e}^{3+}/Fe}^{\ominus }=-0.036$ V

Since $E_{M^{x+}/M}^{\ominus }>E_{F{e}^{3+}/Fe}^{\ominus }$, reduction occurs at the $M^{x+}/M$ electrode (cathode) and oxidation occurs at the $F{e}^{3+}/Fe$ electrode (anode).

Cathode reaction: $(M^{x+}+x{e}^{-}\to M)\times 3$
Anode reaction: $(Fe\to F{e}^{3+}+3e^{-})\times x$

Overall balanced cell reaction:
$3M^{x+}+xFe\to 3M+xF{e}^{3+}$

The number of electrons transferred in the balanced reaction is $n=3x$.

Standard cell potential $E_{cell}^{\ominus }$ is:
$E_{cell}^{\ominus }=E_{cathode}^{\ominus }-E_{anode}^{\ominus }$
$E_{cell}^{\ominus }=0.15-(-0.036)=0.186$ V

Using the Nernst equation:
$E_{cell}=E_{cell}^{\ominus }-\frac{2.303RT}{nF}\log Q$

Substitute the given values ($E_{cell}=0.2057$ V, $Q=10^{-2}$, and $\frac{2.303RT}{F}=0.059$ V):
$0.2057=0.186-\frac{0.059}{3x}\log (10^{-2})$

$0.2057-0.186=-\frac{0.059}{3x}\times (-2)$

$0.0197=\frac{0.118}{3x}$

$3x=\frac{0.118}{0.0197}$

$3x\approx 6$

$x=2$

Answer: $2$