1. Tardigrade
  2. Question
  3. Chemistry
  4. During the preparation of H2S2O8 (per disulphuric acid) O2 gas is also released at anode as byproduct. When 9.72 L of H2 releases at cathode and 2.35 L O2 at anode, the weight H2S2O8 produced in gram is

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Q. During the preparation of $H_{2}S_{2}O_{8}$ (per disulphuric acid) $O_{2}$ gas is also released at anode as byproduct. When $9.72\, L$ of $H_{2}$ releases at cathode and $2.35\, L$ $O_{2}$ at anode, the weight $H_{2}S_{2}O_{8}$ produced in gram is

Electrochemistry

Solution:

$2HSO^{-}_{4} \to H_{2}S_{2}O_{8} +2e$
$2O^{2-} \to O_{2} +4e$
$\frac{2.35}{22.4}=0.105 0.42$ mole
$2H^{+} +2e \to H_{2}$
$0.87$ mole $\frac{9.72}{22.4}$ mole
mole of e for $H_{2}S_{2}O_{8} = 0.87 - 0.42 = 0.45$ mole
$\therefore $ mole of $H_{2}S_{2}O_{8} = 0.225$ mole $= 43.65$ g