Question

When $CuSO_4$ 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)

$CuSO _4 \rightleftharpoons Cu ^{2+}+ SO _4^{2-}$
$H _2 O \rightleftharpoons H ^{+}+ OH ^{-}$
At cathode : $Cu ^{2+}+2 e ^{-} \rightarrow Cu$
At anode : $4 OH ^{-} \rightarrow 2 H _2 O + O _2+4 e ^{-}$
Sulphate ion requires higher voltage, thus, $OH ^{-}$get oxidized in preference to sulphate ions.