Question

An electric current is passed through an aqueous solution of lithium bromide.
What are produced at the cathode and anode respectively?

• A. At cathode : $H_2$(g)
  At anode : $O_2$(g)
• B. At cathode : $Li\left(s\right)$
  At anode : $B{r}_2\left(l\right)$
• C. At cathode : $B{r}_2$(l)
  At anode : $Li\left(s\right)$
• D. At cathode : $O_2$(g)
  At anode : $H_2$(g)

Answer 1

The correct option is A At cathode : $H_2$(g)
At anode : $O_2$(g)

Ions present in given soltion are $H^{+}$, $O{H}^{-}$, $L{i}^{+}$and $B{r}^{-}$ and their standard reduction potential are

Ions	Half Reduction reaction	$E^{-}$(ev)
$H^{+}$	$H^{+}(aq.)+e^{-}⟶\frac{1}{2}{H}_2\left(g\right)$	0
$O{H}^{-}$	$2H_{2}O\left(l\right)+O_2\left(g\right)+4e^{-}⟶4O{H}^{-}(aq.)$	0.40
$L{i}^{+}$	$L{i}^{+}(aq.)+e^{-}⟶Li\left(s\right)$	-3.04
$B{r}^{-}$	$B{r}_2+e^{-}⟶\frac{1}{2}B{r}^{-}(aq.)$	1.09

The cation that has a more postive reduction potential ( $E^{-}$) gets reduced.
The anion that has a more negative reduction potential ( $E^{-}$) gets oxidised.
In an aqueous solution, the reaction at anode and cathode can be represented as:
At the anode $H_{2}O⟶2H^{+}+\frac{1}{2}{O}_2\left(g\right)+e^{-}$ oxygen gas is produced

At cathode $H_{2}O+e^{-}⟶\frac{1}{2}{H}_2\left(g\right)+O{H}^{-}$ hydrogen gas is produced