Question

Which of the following on electrolysis would does not evolve oxygen at the anode?

• A. Dilute H${}_2$SO${}_4$ with Pt electrodes
• B. Aqueous silver nitrate using Pt electrodes
• C. Aqueous Na${}_2$SO${}_4$ with Pt electrodes
• D. 50% H${}_2$SO${}_4$ with Pt electrodes

Answer 1

The correct option is D 50% H${}_2$SO${}_4$ with Pt electrodes

On electrolysis of 50% $H_{2}S{O}_4$

$H_{2}S{O}_4\to H^{+}HS{O}_4^{-}$

$2H^{+}+2e^{-}\to H_2\uparrow$ (cathode)

$2HS{O}_4^{-}\to H_2{S}_2{O}_8+2e^{-}$ (anode)

During electrolysis of 50% sulphuric acid solution with $Pt$ electrodes, oxygen will not be evolved at anode. Instead $H_2{S}_2{O}_8$ will be formed.
During electrolysis of dilute sulphuric acid with $Pt$ electrodes hydrogen is released at cathode and oxygen is released at anode.
During electrolysis of aqueous silver nitrate using $Pt$ electrodes, oxygen is evolved at anode and $Ag$ is formed at cathode.
During electrolysis of aqueous sodium sulfate with $Pt$ electrodes, During electrolysis of dilute sulphuric acid with $Pt$ electrodes hydrogen is released at cathode and oxygen is released at anode.