A volume of x mL of 5 M NaHCO3 — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

We have considered 6 o [Sn(OH) ] /HSnO E 0.9V   Pt|HSnO2 –(aq),[Sn(OH)6] 2–(aq),OH –(aq)|Bi2O3(s)|Bi(s)| 0.5M 0.05M E°cell = + 0.9 – 0.44 = 0.46 V Oxidation Half : HSnO2 – + H2O + 3OH –  [Sn(OH)6] 2– + 2e – Reduction Half : Bi2O3 + 3H2O + 6e –  2Bi + 6OH – _______________________________________ 3HSnO2 –(aq) + Bi2O3(s) + 6H2O + 3OH –(aq) 3[Sn(OH)6] 2–(aq) + 2Bi(s) o cell cell 3 0.059 (0.5) E E log (0.05) [OH ]     0.2353 = 0.46 – 0.059 3log [OH ]         2 0.2247 log OH 0.059         = 7.6 1 + pOH = 7.6 pOH = 6.6 pH = 14 – 6.6 = 7.4 pH = 3 a 3 [HCO ] pK log[H CO ]   7.4 = 6.11 + log 5x 1.29 = x log 4 x 19.5 4  x = 78 Note : In question paper, 2 o HSnO /[Sn(OH) ] E 0.9V   data is given, but NTA has given answer by considering 6 o [Sn(OH) ] /HSnO E 0.9V 