JEE Main 2026 — Chemistry Physical Chemistry

We have considered

$$E^{\circ}_{[Sn(OH)_6]^{2-}/HSnO_2^-} = -0.9 \text{ V}$$E[Sn(OH)6​]2−/HSnO2−​∘​=−0.9 V

$$Pt | HSnO_2^-(aq), [Sn(OH)_6]^{2-}(aq), OH^-(aq) | Bi_2O_3(s) | Bi(s) |$$Pt∣HSnO2−​(aq),[Sn(OH)6​]2−(aq),OH−(aq)∣Bi2​O3​(s)∣Bi(s)∣

$$0.5M \quad 0.05M$$0.5M0.05M

$$E^{\circ}_{cell} = +0.9 - 0.44 = 0.46 \text{ V}$$Ecell∘​=+0.9−0.44=0.46 V

Oxidation Half :

$$HSnO_2^- + H_2O + 3OH^- \rightarrow [Sn(OH)_6]^{2-} + 2e^-$$HSnO2−​+H2​O+3OH−→[Sn(OH)6​]2−+2e−

Reduction Half :

$$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$$Bi2​O3​+3H2​O+6e−→2Bi+6OH−

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$$3HSnO_2^-(aq) + Bi_2O_3(s) + 6H_2O + 3OH^-(aq) \rightarrow 3[Sn(OH)_6]^{2-}(aq) + 2Bi(s)$$3HSnO2−​(aq)+Bi2​O3​(s)+6H2​O+3OH−(aq)→3[Sn(OH)6​]2−(aq)+2Bi(s)

$$E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{6} \log \dfrac{(0.5)^3}{(0.05)^3 \times [OH^-]^3}$$Ecell​=Ecell∘​−60.059​log(0.05)3×[OH−]3(0.5)3​

$$0.2353 = 0.46 - \dfrac{0.059}{6} \times 3 \log \left[ \dfrac{10}{[OH^-]} \right]$$0.2353=0.46−60.059​×3log[[OH−]10​]

$$\log \left[ \dfrac{10}{OH^-} \right] = \dfrac{2 \times 0.2247}{0.059} = 7.6$$log[OH−10​]=0.0592×0.2247​=7.6

$$1 + pOH = 7.6$$1+pOH=7.6

$$pOH = 6.6$$pOH=6.6

$$pH = 14 - 6.6 = 7.4$$pH=14−6.6=7.4

$$pH = pK_{a_1} + \log \dfrac{[HCO_3^-]}{[H_2CO_3]}$$pH=pKa1​​+log[H2​CO3​][HCO3−​]​

$$7.4 = 6.11 + \log \dfrac{5x}{20}$$7.4=6.11+log205x​

$$1.29 = \log \dfrac{x}{4}$$1.29=log4x​

$$\dfrac{x}{4} = 19.5$$4x​=19.5

$$x = 78$$x=78