Question

When $CuS{O}_4$ is electrolysed using platinum electrodes,

• A. copper is liberated at cathode, sulphur at anode
• B. copper is liberated at cathode, oxygen at anode
• C. sulphur is liberated at cathode, oxygen at anode
• D. oxygen is liberated at cathode, copper at anode

Answer 1

Answer: (B)

$CuS{O}_4\rightleftharpoons C{u}^{2+}+S{O}_4^{2-}$
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
At cathode : $C{u}^{2+}+2e^{-}\to Cu$
At anode : $4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$
Sulphate ion requires higher voltage, thus, $O{H}^{-}$get oxidized in preference to sulphate ions.