Q21.88P - Chemistry: Molecular Nature Of Matter And Change

Question

In the electrolysis of molten BaI₂, 

(a) What product forms at the negative electrode?

 (b) What product forms at the positive electrode?

Step-by-Step Solution

Answer

(a) ${\mathrm{I}}_2$ gas

(b) barium metal

Step 1: 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   ${\mathrm{I}}_{\text{2}}\text{(g)}$

Step 2: 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.