During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed at the anode in equimolar amounts. The moles of $H_2$ that will form simultaneously at the other electrode will be (Given: $2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-$)

  • A
    thrice of $O_2$
  • B
    twice of $O_2$
  • C
    equal to $O_2$
  • D
    half of $O_2$

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