A volume of x mL of 5 M NaHCO 3 solution was mixed with 10 mL of 2 M H 2 CO 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 mV , then the value of x = \ \ \ \ mL (nearest integer). Sn ( s ) Sn ( OH ) 6 2- (0.5 M ) HSnO 2 - (0.05 M ) mid OH - Bi 2 O 3 ( s ) Bi ( s ) Consider upto one place of decimal for intermediate calculations [begin array ll Given : & E HSnO 2 - mid Sn ( OH ) 6 2- o =-0.9 V \\& E Bi 2 O 3 mid Bi o =-0.44 V \\& pKa ( H 2 CO 3 ) =6.11 \\& frac 2.303 RT F =0.059 V \\& Anti log (1.29)=19.5end array ]

Chemistry · Electrochemistry

JEE Main 2026 — 28 January, Evening Shift — Question 68

A volume of x mL of $5 \mathrm{M} \mathrm{NaHCO}_{3}$ solution was mixed with 10 mL of $2 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}$ 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 mV , then the value of $\mathrm{x}=$ $\_\_\_\_$ mL (nearest integer). $\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}(\mathrm{OH})_{6}^{2-}(0.5 \mathrm{M})\right| \mathrm{HSnO}_{2}^{-}(0.05 \mathrm{M}) \mid \mathrm{OH}^{-} \left|\mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s})\right| \operatorname{Bi}(\mathrm{s})$ Consider upto one place of decimal for intermediate calculations $\left[\begin{array}{ll}\text { Given : } & \mathrm{E}_{\mathrm{HSnO}_{2}^{-} \mid \mathrm{Sn}(\mathrm{OH})_{6}^{2-}}^{\mathrm{o}}=-0.9 \mathrm{~V} \\& \mathrm{E}_{\mathrm{Bi}_{2} \mathrm{O}_{3} \mid \mathrm{Bi}}^{\mathrm{o}}=-0.44 \mathrm{~V} \\& \mathrm{pKa}_{\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)}=6.11 \\& \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V} \\& \text { Anti } \log (1.29)=19.5\end{array}\right]$

Answer: 78

Numerical answer — enter this value.

Step-by-step solution

We have considered $\begin{gathered} \mathrm{E}_{\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-} / \mathrm{HSnO}_{2}^{-}}^{\mathrm{o}}=-0.9 \mathrm{~V} \mathrm{Pt}\left|\mathrm{HSnO}_{2}^{-}(\mathrm{aq}),\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-}(\mathrm{aq}), \mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s})|\mathrm{Bi}(\mathrm{~s})| 0.5 \mathrm{M} 0.05 \mathrm{M} \mathrm{E}_{{ }_{\text {cell }}^{\circ}}^{\circ}=+0.9-0.44=0.46 \mathrm{~V} \end{gathered}$ \section*{Oxidation Half :} $\mathrm{HSnO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O}+3 \mathrm{OH}^{-} \rightarrow\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-}+2 \mathrm{e}^{-}$ \section*{Reduction Half :} $\mathrm{Bi}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Bi}+6 \mathrm{OH}^{-}$ $\begin{aligned} & 3 \mathrm{HSnO}_{2}^{-}(\mathrm{aq})+\mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s})+6 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{OH}^{-}(\mathrm{aq}) \\& \rightarrow 3\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-}(\mathrm{aq})+2 \mathrm{Bi}(\mathrm{~s}) \\& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.059}{6} \log \frac{(0.5)^{3}}{(0.05)^{3} \times\left[\mathrm{OH}^{-}\right]^{3}} \\& 0.2353=0.46-\frac{0.059}{6} \times 3 \log \left[\frac{10}{\left[\mathrm{OH}^{-}\right]}\right] \\& \log \left[\frac{10}{\mathrm{OH}^{-}}\right]=\frac{2 \times 0.2247}{0.059}=7.6 \\& 1+\mathrm{pOH}=7.6 \\& \mathrm{pOH}=6.6 \\& \mathrm{pH}=14-6.6=7.4 \\& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}_{1}}+\log \frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]} \\& 7.4=6.11+\log \frac{5 \mathrm{x}}{20} \\& 1.29=\log \frac{\mathrm{x}}{4} \\& \frac{\mathrm{x}}{4}=19.5 \\& \mathrm{x}=78 \end{aligned}$ Note : In question paper, $\mathrm{E}_{\mathrm{HSnO}_{2}^{-} /\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-}}^{\mathrm{o}}=-0.9 \mathrm{~V}$ data is given, but NTA has given answer by considering $\mathrm{E}_{\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]^{2-} / \mathrm{HSnO}_{2}^{-}}^{\mathrm{o}}=-0.9 \mathrm{~V}$ therefore this question should be BONUS.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Nernst Equation and Electrochemical Series

For strong electrolyte Lambda m increases slowly with dilution and can be represented by the equation Lambda m=Lambda^circ m- Ac^1 / 2…

Answer: 78

Step-by-step solution

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