If the passage of current liberates $H_2$ at cathode and $Cl_2$ at anode then the solution is

  • A
    $CuSO_{4(aq)}$
  • B
    $CdCl_{2(aq)}$
  • C
    $KCl_{(aq)}$
  • D
    $AgCl_{(aq)}$

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