Question

In the electrolysis of molten BaI₂,

(a) What product forms at the negative electrode?

(b) What product forms at the positive electrode?

Short answer

(a)${\mathrm{I}}_2$ gas

(b) barium metal

Step-by-step solution

To find the product forms at the negative electrode

${\text{BaI}}_{\text{2}}\to {\text{Ba}}^{\text{2+}}{\text{+I}}^{\text{-}}$

(a)In electrolysis of molten salts, the anion turns into a free element. Anions form free elements in the anode. In molten ${\text{BaI}}_2$, the anion is ${\text{I}}^{\text{-}}$.

$\text{2I}\to {\text{2e}}^{\text{-}}{\text{+I}}_{\text{2(g)}}$

Hence This I turns into role="math" localid="1659623165940" ${\mathrm{I}}_{\text{2}}\text{(g)}$

To find the product forms at the positive electrode

(b) In electrolysis of molten salts, the cation turns into a free element. Cations form free elements in the cathode. In molten ${\text{BaI}}_2,{\text{Ba}}^{2+}$is the cation.

${\text{Ba}}^{\text{2+}}{\text{+2e}}^{\text{-}}\to {\text{Ba}}_{\text{(s)}}$

Hence The ${\text{Ba}}^{\text{2+}}$ions turn into barium metal.