JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

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).

$Sn(s) ∣ Sn(OH)_{6}^{2 -} (0.5 M) ∣∣ HSnO_{2}^{-} (0.05 M) ∣ OH^{-} ∣ Bi_{2} O_{3} (s) ∣ Bi(s)$

Consider upto one place of decimal for intermediate calculations.

Given :
$E_{[Sn(OH)_{6}]^{2 -} / HSnO_{2}^{-}}^{\circ} = - 0.9$ V;
$E_{Bi_{2} O_{3} / Bi}^{\circ} = - 0.44$ V;
$pKa_{(H_{2} CO_{3})} = 6.11$ ;
$\frac{2.303 R T}{F} = 0.059$ V;
Antilog(1.29) = 19.5.

Your answer

Solution

Cell EMF setup: $E_{ce l l}^{\circ} = 0.9 - 0.44 = 0.46$ V

Half reactions (basic medium):

Oxidation: $HSnO_{2}^{-} + H_{2} O + 3 OH^{-} \to [Sn(OH)_{6}]^{2 -} + 2 e^{-}$ .

Reduction: $Bi_{2} O_{3} + 3 H_{2} O + 6 e^{-} \to 2 Bi + 6 OH^{-}$ .

Net (n = 6): $3 HSnO_{2}^{-} + Bi_{2} O_{3} + 6 H_{2} O + 3 OH^{-} \to 3 [Sn(OH)_{6}]^{2 -} + 2 Bi$ .

Nernst: $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{6} lo g \frac{[ Sn(OH) _{6}^{2 -} ] ^{3}}{[ HSnO _{2}^{-} ] ^{3} [ OH ^{-} ] ^{3}}$ .

$0.2353 = 0.46 - \frac{0.059}{6} \times 3 lo g \frac{10}{[ OH ^{-} ]}$ .

$lo g \frac{10}{[ OH ^{-} ]} = \frac{2 \times 0.2247}{0.059} \approx 7.6 \Rightarrow 1 + pOH = 7.6 \Rightarrow pOH = 6.6 \Rightarrow pH = 7.4$ .

Buffer (Henderson-Hasselbalch): $pH = p K_{a_{1}} + lo g \frac{[ HCO _{3}^{-} ]}{[ H _{2} CO _{3} ]}$ .

$7.4 = 6.11 + lo g \frac{5 x / ( x + 10 )}{( 2 \times 10 ) / ( x + 10 )} = 6.11 + lo g \frac{5 x}{20}$ .

$1.29 = lo g \frac{x}{4} \Rightarrow \frac{x}{4} = 19.5 \Rightarrow x = 78$ mL.

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