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). $\mathrm{Sn}(\mathrm{s})|\mathrm{Sn}(\mathrm{OH})_{6}^{2-}(0.5\,\mathrm{M})|\mathrm{HSnO}_{2}^{-}(0.05\,\mathrm{M})|\mathrm{OH}^{-}|\mathrm{Bi}_{2}\mathrm{O}_{3}(\mathrm{s})|\mathrm{Bi}(\mathrm{s})$ Consider upto one place of decimal for intermediate calculations $\left[\begin{array}{ll}\text{Given: } & \mathrm{E}_{\mathrm{HSnO}_{2}^{-}|\mathrm{Sn}(\mathrm{OH})_{6}^{2-}}^{\mathrm{o}}=-0.9\,\mathrm{V} \\ & \mathrm{E}_{\mathrm{Bi}_{2}\mathrm{O}_{3}|\mathrm{Bi}}^{\mathrm{o}}=-0.44\,\mathrm{V} \\ & \mathrm{pKa}_{(\mathrm{H}_{2}\mathrm{CO}_{3})}=6.11 \\ & \frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\,\mathrm{V} \\ & \text{Anti log}(1.29)=19.5\end{array}\right]$

Redox Reactions and Electrochemistry — JEE Main · 2026 · 28 Jan 2026 · Shift 2 Question with Solution

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_{HSnO_{2}^{-} ∣ Sn (OH)_{6}^{2 -}}^{o} = - 0.9 V E_{Bi_{2} O_{3} ∣ Bi}^{o} = - 0.44 V pKa_{(H_{2} CO_{3})} = 6.11 \frac{2.303 RT}{F} = 0.059 V Anti log (1.29) = 19.5

Answer: 78

Solution

Cathode (Bi):

B i_{2} O_{3} + 3 H_{2} O + 6 e^{-} \to 2Bi + 6O H^{-}

E^{\circ} = - 0.44 V

. Anode (reverse of the

- 0.9 V

couple):

HSn O_{2}^{-} + H_{2} O + 3O H^{-} \to Sn (OH)_{6}^{2 -} + 2 e^{-}

. So

E_{ce l l}^{\circ} = - 0.44 - (- 0.9) = 0.46 V

. Balanced (

n = 6

3HSn O_{2}^{-} + B i_{2} O_{3} + 6 H_{2} O + 3O H^{-} \to 3Sn (OH)_{6}^{2 -} + 2Bi

, giving

Q = \frac{[ Sn ( OH ) _{6}^{2 -} ] ^{3}}{[ HSn O _{2}^{-} ] ^{3} [ O H ^{-} ] ^{3}} = \frac{( 0.5/0.05 ) ^{3}}{[ O H ^{-} ] ^{3}} = \frac{1000}{[ O H ^{-} ] ^{3}}

. Then

E = 0.46 - \frac{0.059}{6} lo g \frac{1000}{[ O H ^{-} ] ^{3}} = 0.46 - 0.0295 (1 + pOH)

. With

E = 0.2353 V

1 + pOH = 7.6 \Rightarrow pOH = 6.6 \Rightarrow pH = 7.4

. For the buffer

pH = p K_{a} + lo g \frac{n _{HC O_{3}^{-}}}{n _{H_{2} C O_{3}}} = 6.11 + lo g (0.25 x)

, so

lo g (0.25 x) = 1.29 \Rightarrow 0.25 x = 19.5 \Rightarrow x = 78 mL

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

Solution

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