Q25 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry
A volume of $x\,\mathrm{mL}$ of $5\,\mathrm{M}\,\mathrm{NaHCO_3}$ solution was mixed with $10\,\mathrm{mL}$ of $2\,\mathrm{M}\,\mathrm{H_2CO_3}$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3\,\mathrm{mV}$, then the value of $x = \_\_\_\_$ mL (nearest integer).
$\mathrm{Sn(s)} \; \big| \; \mathrm{Sn(OH)_6^{2-}}(0.5\,\mathrm{M}) \; \big| \; \mathrm{HSnO_2^-}(0.05\,\mathrm{M}) \; \big| \; \mathrm{OH^-} \; \big| \; \mathrm{Bi_2O_3(s)} \; \big| \; \mathrm{Bi(s)}$
Consider up to one place of decimal for intermediate calculations.
$\begin{array}{ll}\text{Given:} & E^\circ_{\mathrm{[Sn(OH)_6]^{2-}/HSnO_2^-}} = -0.90\,\mathrm{V} \\[4pt]& E^\circ_{\mathrm{Bi_2O_3/Bi}} = -0.44\,\mathrm{V} \\[4pt]& \mathrm{p}K_a(\mathrm{H_2CO_3}) = 6.11 \\[4pt]& \dfrac{2.303\,RT}{F} = 0.059\,\mathrm{V} \\[6pt]& \text{Antilog}(1.29) = 19.5\end{array}$
$\mathrm{Sn(s)} \; \big| \; \mathrm{Sn(OH)_6^{2-}}(0.5\,\mathrm{M}) \; \big| \; \mathrm{HSnO_2^-}(0.05\,\mathrm{M}) \; \big| \; \mathrm{OH^-} \; \big| \; \mathrm{Bi_2O_3(s)} \; \big| \; \mathrm{Bi(s)}$
Consider up to one place of decimal for intermediate calculations.
$\begin{array}{ll}\text{Given:} & E^\circ_{\mathrm{[Sn(OH)_6]^{2-}/HSnO_2^-}} = -0.90\,\mathrm{V} \\[4pt]& E^\circ_{\mathrm{Bi_2O_3/Bi}} = -0.44\,\mathrm{V} \\[4pt]& \mathrm{p}K_a(\mathrm{H_2CO_3}) = 6.11 \\[4pt]& \dfrac{2.303\,RT}{F} = 0.059\,\mathrm{V} \\[6pt]& \text{Antilog}(1.29) = 19.5\end{array}$
Correct answer: 78
Solution
We have considered
$E^{\circ}_{[Sn(OH)_6]^{2-}/HSnO_2^-} = -0.9 \text{ V}$
$Pt | HSnO_2^-(aq), [Sn(OH)_6]^{2-}(aq), OH^-(aq) | Bi_2O_3(s) | Bi(s) |$
$0.5M \quad 0.05M$
$E^{\circ}_{cell} = +0.9 - 0.44 = 0.46 \text{ V}$
Oxidation Half :
$HSnO_2^- + H_2O + 3OH^- \rightarrow [Sn(OH)_6]^{2-} + 2e^-$
Reduction Half :
$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$
---------------------------------------
$3HSnO_2^-(aq) + Bi_2O_3(s) + 6H_2O + 3OH^-(aq) \rightarrow 3[Sn(OH)_6]^{2-}(aq) + 2Bi(s)$
$E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{6} \log \dfrac{(0.5)^3}{(0.05)^3 \times [OH^-]^3}$
$0.2353 = 0.46 - \dfrac{0.059}{6} \times 3 \log \left[ \dfrac{10}{[OH^-]} \right]$
$\log \left[ \dfrac{10}{OH^-} \right] = \dfrac{2 \times 0.2247}{0.059} = 7.6$
$1 + pOH = 7.6$
$pOH = 6.6$
$pH = 14 - 6.6 = 7.4$
$pH = pK_{a_1} + \log \dfrac{[HCO_3^-]}{[H_2CO_3]}$
$7.4 = 6.11 + \log \dfrac{5x}{20}$
$1.29 = \log \dfrac{x}{4}$
$\dfrac{x}{4} = 19.5$
$x = 78$
$E^{\circ}_{[Sn(OH)_6]^{2-}/HSnO_2^-} = -0.9 \text{ V}$
$Pt | HSnO_2^-(aq), [Sn(OH)_6]^{2-}(aq), OH^-(aq) | Bi_2O_3(s) | Bi(s) |$
$0.5M \quad 0.05M$
$E^{\circ}_{cell} = +0.9 - 0.44 = 0.46 \text{ V}$
Oxidation Half :
$HSnO_2^- + H_2O + 3OH^- \rightarrow [Sn(OH)_6]^{2-} + 2e^-$
Reduction Half :
$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$
---------------------------------------
$3HSnO_2^-(aq) + Bi_2O_3(s) + 6H_2O + 3OH^-(aq) \rightarrow 3[Sn(OH)_6]^{2-}(aq) + 2Bi(s)$
$E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{6} \log \dfrac{(0.5)^3}{(0.05)^3 \times [OH^-]^3}$
$0.2353 = 0.46 - \dfrac{0.059}{6} \times 3 \log \left[ \dfrac{10}{[OH^-]} \right]$
$\log \left[ \dfrac{10}{OH^-} \right] = \dfrac{2 \times 0.2247}{0.059} = 7.6$
$1 + pOH = 7.6$
$pOH = 6.6$
$pH = 14 - 6.6 = 7.4$
$pH = pK_{a_1} + \log \dfrac{[HCO_3^-]}{[H_2CO_3]}$
$7.4 = 6.11 + \log \dfrac{5x}{20}$
$1.29 = \log \dfrac{x}{4}$
$\dfrac{x}{4} = 19.5$
$x = 78$
Hence, the correct answer is 78.