Question

In the electrolysis of molten NaBr,

(a) What product forms at the anode?

(b) What product forms at the cathode?

Short answer

(a) bromine

(b) sodium

Step-by-step solution

To find the product forms at the anode

$\text{NaBr}\to {\text{Na}}^{+}+{\text{Br}}^{-}$

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

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

Hence This${\text{Br}}^{\text{-}}{\text{turnsintoBr}}_{\text{2(g)}}$

To find the product forms at the cathode

(b) In electrolysis of molten salts, the cation turns into a free element. Cations form free elements in the cathode. In molten${\text{NaBr,Na}}^{\text{+}}\text{isthecation.}$

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

Hence The${\text{Na}}^{+}$ ions turn into sodium metal.