Question

In the electrolytic refining of copper using $CuS{O}_4$ as the electrolyte, the anode reaction is

• A. $4O{H}^{-}\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}$
• B. $C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$
• C. $2S{O}_4^{2-}\left(aq\right)\to S_2{O}_8^{2-}\left(aq\right)+2e^{-}$
• D. $Cu\left(s\right)\to C{u}^{2+}\left(aq\right)+2e^{-}$

Answer 1

The correct option is A $4O{H}^{-}\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}$

Anode is positively charged and hence, anions migrate towards it. We also know that oxidation occurs at anode. Out of $O{H}^{-}$ and $S{O}_4^{2-}$, the former is easier to oxidise.
Hence, the following reaction occurs at anode:
$4O{H}^{-}\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}$