Question

The electrolysis of $Ba(OH{)}_2$ yields at the anode.

• A. $C{O}_2$
• B. $BaO$
• C. $H_2$
• D. $O_2$

Answer 1

The correct option is D $O_2$

The equation for the electrolysis of $Ba(OH{)}_2$ is:

Anode: $4O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)+4e^{-}$
Cathode: $2H_{2}O\left(l\right)+2e^{-}\to H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

Thus, $O_2$ is liberated at the anode.