ChemistryJEE Main 28 Jan 2026, Shift 2·Q24·Numerical·Tough

Electrochemistry · Nernst equation with buffer pH

A volume of x mL of 5 M $\ce{NaHCO3}$ solution was mixed with 10 mL of 2 M $\ce{H2CO3}$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x $=$ mL (nearest integer).

$\ce{Sn(s) | Sn(OH)6^2-}$ (0.5 M) $|$ $\ce{HSnO2-}$ (0.05 M) $|$ $\ce{OH-}$ $|$ $\ce{Bi2O3(s) | Bi(s)}$

Consider upto one place of decimal for intermediate calculations $\left[\begin{array}{ll} \text{Given :}\qquad & E^\circ_{\ce{Sn(OH)6^2-}|\ce{HSnO2-}} = -0.9\ \mathrm{V} \\[2pt] & E^\circ_{\ce{Bi2O3}|\ce{Bi}} = -0.44\ \mathrm{V} \\[2pt] & \mathrm{p}K_{\mathrm{a}(\ce{H2CO3})} = 6.11 \\[2pt] & \dfrac{2.303RT}{F} = 0.059\ \mathrm{V} \\[6pt] & \text{Antilog}\,(1.29) = 19.5 \end{array}\right]$

We corrected a printing error in the original paper

Source printed the first datum as “ $E^\circ$ with subscript HSnO $_2^-$ $|$ Sn(OH) $_6^{2-}$ $= -0.9$ V” (couple written reduced form first, readable as an oxidation potential); corrected to the reduction couple Sn(OH) $_6^{2-}$ $|$ HSnO $_2^-$ , the value the key uses.

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

See the full 28 Jan 2026, Shift 2 paper