Question

Complete the following equations which show electrode reactions during certain electrolysis.

$$\begin{gathered} 1.2{\mathrm{O}}^{-2}\to {\mathrm{O}}_2+\mathrm{\_\_\_\_} \\ 2.\mathrm{\_\_\_\_\_\_}+{\mathrm{e}}^{-}\to \mathrm{Ag} \\ 3.\mathrm{\_\_\_\_\_\_}\to {\mathrm{Cl}}_2+2{\mathrm{e}}^{-} \\ 4.2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to \mathrm{\_\_\_\_\_} \end{gathered}$$

Answer 1

1. Here, the two oxide ions give up four electrons to form an oxygen molecule. since the charge of the oxide ion is -2, one oxide ion gives up 2 electrons, so 2 oxide ion gives up 4 electrons. the reaction occurs at the anode

$2{\mathrm{O}}^{-2}\to {\mathrm{O}}_2+\mathbf{}\mathbf{4}{\mathbf{e}}^{\mathbf{-}}$

2. The silver ions are discharged at the cathode by accepting electrons and deposited as a thin film of metallic silver on the cathode

${\mathbf{Ag}}^{\mathbf{+}}+{\mathrm{e}}^{-}\to \mathrm{Ag}$

3. Here, the two chloride ions give up two electrons to form a chlorine molecule. since the charge of chloride ion is -1, one chloride ion gives up 1 electron, so 2 chloride ion gives up 2 electrons

$\mathbf{2}{\mathbf{Cl}}^{\mathbf{-}}\to {\mathrm{Cl}}_2+2{\mathrm{e}}^{-}$

4. Here, the two hydrogen ion accepts 2 electrons to form a hydrogen molecule, since the charge of the hydrogen ion is +1, it needs one electron, since we have two hydrogen ions here, we need 2 electrons to form a hydrogen molecule

$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to \mathbf{}{\mathbf{H}}_{\mathbf{2}}$