Marshalls acid is prepared by the electrolytic oxidation of H2SO4 as 2H2SO4→H2S2O8+2H⊕+2e−
Oxygen and hydrogen are byproducts. In such Electrolysis 2.24 L of H2 amd 0.56 L O2 were produced at STP. The weight of H2S2O8 formed is:
Marshalls acid is prepared by the electrolytic oxidation of H2SO4 as 2H2SO4→H2S2O8+2H⊕+2e−
Oxygen and hydrogen are byproducts. In such Electrolysis 2.24 L of H2 amd 0.56 L O2 were produced at STP. The weight of H2S2O8 formed is:
The correct option is A 9.7g
H2SO4→H⊕+HSO⊖4
These two reactions are competing at anode:
2HSO⊖4→2S2O8+2e−
Anode reaction : (Oxidation of H2O)
⎡⎢
⎢
⎢
⎢
⎢⎣2⊖OH→H2O+12O2+2e−orH2O→2H⊕+12O2+2e−⎤⎥
⎥
⎥
⎥
⎥⎦
Cathode reaction : 2H⊕+2e−(fromH2SO4)⇌H2(g)
Since one reaction at cathode and two reactions ar anode are taking place, therefore, the equivalent of H2(g) Produced at cathode should be equal to the equivalent of O2 produced and the equivalent of H2S2O8 formed.
∴ Equivalent of H2=Equivalent of O2+EquivalentofH2S2O82.24×222.4=0.56×422.4+W194/2
(MwofH2S2O8=194,Ew1942)
After solving, we get W=9.7g
H2SO4→H⊕+HSO⊖4
These two reactions are competing at anode:
2HSO⊖4→2S2O8+2e−
Anode reaction : (Oxidation of H2O)
⎡⎢ ⎢ ⎢ ⎢ ⎢⎣2⊖OH→H2O+12O2+2e−orH2O→2H⊕+12O2+2e−⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
Cathode reaction : 2H⊕+2e−(fromH2SO4)⇌H2(g)
Since one reaction at cathode and two reactions ar anode are taking place, therefore, the equivalent of H2(g) Produced at cathode should be equal to the equivalent of O2 produced and the equivalent of H2S2O8 formed.
∴ Equivalent of H2=Equivalent of O2+EquivalentofH2S2O82.24×222.4=0.56×422.4+W194/2
(MwofH2S2O8=194,Ew1942)
After solving, we get W=9.7g
