JEE Main 2026 — 4 April, Evening Shift — Question 64

An electrochemical cell. consist of the following two redox couples, $\mathrm{M}^{\mathrm{x}+}(\mathrm{aq}) / \mathrm{M}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\ominus}=+0.15 \mathrm{~V}\right]$ and $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\ominus}=-0.036 \mathrm{~V}\right]$ The cell EMF ( $\mathrm{E}_{\text {cell }}$ ) is recorded to be 0.2057 V . If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of x is $\_\_\_\_$ (Nearest integer)[0pt] [Given : M is a p-block metal and $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right]$