Question

One of the methods of preparation of per disulphuric acid, $H_2{S}_2{O}_8$, involve electrolytic oxidation of $H_{2}S{O}_4$ at node $(2H_{2}S{O}_4\to H_{2}S{O}_4+2H^{+}+2e^{-})$ with oxygen and hydrogen as by -products. In such an electrolysis, 9.722 L of $H_2$ and 2.35 L of $O_2$ were generated at STP. What is the weight of $H_2{S}_2{O}_8$ formed in gms? (write the value to the nearest integer).

Answer 1

At anode, $(2H_{2}S{O}_4\to H_{2}S{O}_4+2H^{+}+2e^{-})$
$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$
At cathode, $H_2+2e^{-}\rightleftharpoons 2H^{+}$
The number of moles of hydrogen liberated $=\frac{9.722}{22.4}=0.434$ mol.
The number of moles of electrons required for liberation of hydrogen $=2\times 0.434mol{e}^{-}=0.868mol{e}^{-}$
The number of moles of oxygen liberated $=\frac{2.35}{22.4}=0.1049$ mol.
The number of moles of electrons released during liberation of oxygen $=4\times 0.1049mol{e}^{-}=0.4196mol{e}^{-}$
The number of moles of electrons released during formation of $H_2{S}_2{O}_8=0.868mol{e}^{-}-0.4196mol{e}^{-}=0.448mol{e}^{-}$
The number of moles of $H_2{S}_2{O}_8$ formed $=\frac{0.448}{2}=0.224mol$
Mass of $H_2{S}_2{O}_8$ formed $=0.224mol\times 194.1g/mol=43.45g$.