(i) In aqueous solution, $\mathrm{AgNO_3}$ ionises to give $\mathrm{Ag}^{+}_{(aq)}$ and $\mathrm{NO_3}^{-}_{(aq)}$ ions.$\mathrm{AgNO_3}_{(aq)}$ → $\mathrm{Ag}^{+}_{(aq)} + \mathrm{NO_3}^{-}_{(aq)}$ Thus, when electricity is passed Ag+(aq) ions move towards the cathode while $\mathrm{NO_3}^{-}$ ions move towards the anode. $\mathrm{Ag}^{+}_{(aq)} + e^{-}$ → $\mathrm{Ag}_{(s)}$ ; E° = + 0.80 V ...(i) $2\mathrm{H_2O}_{(l)} + 2e^{-}$ → $\mathrm{H_2}_{(g)} + 2\mathrm{OH}^{-}_{(aq)}$ ; E° = – 0.83 V ...(ii) Since the electrode potential (i.e., reduction potential of $\mathrm{Ag}^{+}_{(aq)}$ ions is higher than that of $\mathrm{H_2O}$ molecules, therefore, at the cathode, $\mathrm{Ag}^{+}_{(aq)}$ ions (rather than $\mathrm{H_2O}$ molecules) are reduced. Similarly, at the anode, either Ag metal of the anode or $\mathrm{H_2O}$ molecules may be oxidised. Their electrode potentials are : $\mathrm{Ag}_{(s)}$ → $\mathrm{Ag}^{+}_{(aq)} + e^{-}$ ; E° = – 0.80 V ...(iii) $2\mathrm{H_2O}_{(l)}$ → $\mathrm{O_2}_{(g)} + 4\mathrm{H}^{+}_{(aq)} + 4e^{-}$ ; E° = – 1.23 V ...(iv) Since the oxidation potential of Ag is much higher than that of $\mathrm{H_2O}$ therefore, at the anode, Ag of the silver anode gets oxidised and not the $\mathrm{H_2O}$ molecules. It may, however, be mentioned here that the oxidation potential of $\mathrm{NO_3}^{-}$ ions is even lower than that of $\mathrm{H_2O}$ since more bonds are to be broken during reduction of $\mathrm{NO_3}^{-}$ ions than those in $\mathrm{H_2O}$. Thus, when an aqueous solution of $\mathrm{AgNO_3}$ is electrolysed, Ag from Ag anode dissolves while $\mathrm{Ag}^{+}_{(aq)}$ ions present in the solution gets reduced and gets deposited on the cathode. (ii) When electrolysis of $\mathrm{AgNO_3}$ solution is carried out using platinum electrodes, instead of silver electrodes, oxidation of water occurs at the anode since Pt being a noble metal does not undergo oxidation easily. As a result, $\mathrm{O_2}$ is liberated at the anode according to equation (iv). (iii) In aqueous solution, $\mathrm{H_2SO_4}$ ionises to give $\mathrm{H}^{+}_{(aq)}$ and $\mathrm{SO_4}^{2-}_{(aq)}$ ions. $\mathrm{H_2SO_4}_{(aq)}$ → $2\mathrm{H}^{+}_{(aq)} + \mathrm{SO_4}^{2-}_{(aq)}$ Thus, when electricity is passed, H+(aq) ions move towards cathode while SO42–(aq) ions move towards anode. $2\mathrm{H}^{+}_{(aq)} + 2e^{-}$ → $\mathrm{H_2}_{(g)}$; E° = 0.0 V $2\mathrm{H_2O}_{(l)} + 2e^{-}$→ $\mathrm{H_2}_{(g)} + 2\mathrm{OH}^{-}_{(aq)}$ ; E° = – 0.83 V Since the electrode potential (i.e. reduction potential) of $\mathrm{H}^{+}_{(aq)}$ ions is higher than that of $\mathrm{H_2O}$, therefore, at the cathode, $\mathrm{H}^{+}_{(aq)}$ ions (rather than $\mathrm{H_2O}$ molecules) are reduced to evolve $\mathrm{H_2}$ gas. Similarly at the anode, either $\mathrm{SO_4}^{2-}_{(aq)}$ ions or $\mathrm{H_2O}$ molecules are oxidised. Since the oxidation potential of $\mathrm{SO_4}^{2-}$ is expected to be much lower (since it involves cleavage of many bonds as compared to those in $\mathrm{H_2O}$) than that of $\mathrm{H_2O}$ molecules, therefore, at the anode, it is $\mathrm{H_2O}$ molecules (rather than $\mathrm{SO_4}^{2-}$ ions) which are oxidised to evolve $\mathrm{O_2}$ gas. From the above discussion, it follows that during electrolysis of an aqueous solution of $\mathrm{H_2SO_4}$ only the electrolysis of $\mathrm{H_2O}$ occurs liberating $\mathrm{H_2}$ at the cathode and $\mathrm{O_2}$ at the anode. (iv) In aqueous solution, $\mathrm{CuCl_2}$ ionises as follows : $\mathrm{CuCl_2}_{(aq)}$ → $\mathrm{Cu}^{2+}_{(aq)} + 2\mathrm{Cl}^{-}_{(aq)}$ On passing electricity, $\mathrm{Cu}^{2+}_{(aq)}$ ions move towards cathode and $\mathrm{Cl}^{-}_{(aq)}$ ions move towards anode. Thus, at cathode, either $\mathrm{Cu}^{2+}_{(aq)}$ or $\mathrm{H_2O}$ molecules are reduced. Their electrode potentials are : $\mathrm{Cu}^{2+}_{(aq)} + 2e^{-}$ → $\mathrm{Cu}_{(s)}$; E° = + 0.34 V $2\mathrm{H_2O}_{(l)} + 2e^{-}$ → $\mathrm{H_2}_{(g)} + 2\mathrm{OH}^{-}_{(aq)}$; E° = – 0.83 V Since the electrode potential of $\mathrm{Cu}^{2+}_{(aq)}$ ions is much higher than that of $\mathrm{H_2O}$, therefore, at the cathode, $\mathrm{Cu}^{2+}_{(aq)}$ ions are reduced and not $\mathrm{H_2O}$ molecules. Similarly, at the anode, either $\mathrm{Cl}^{-}_{(aq)}$ ions or $\mathrm{H_2O}$ molecules are oxidized. Their oxidation potentials are : $2\mathrm{Cl}^{-}_{(aq)}$ → $\mathrm{Cl_2}_{(g)} + 2e^{-}$ ; E° = – 1.36 V $2\mathrm{H_2O}_{(l)}$ → $\mathrm{O_2}_{(g)} + 4\mathrm{H}^{+}_{(aq)} + 4e^{-}$ ; E° = – 1.23 V Although oxidation potential of $\mathrm{H_2O}$ molecules is higher than that of $\mathrm{Cl}^{-}$ ions, nevertheless, oxidation of $\mathrm{Cl}^{-}_{(aq)}$ ions occurs in preference to $\mathrm{H_2O}$ since due to overvoltage of $\mathrm{O_2}$, much more potential than – 1.36 V is required for the oxidation of chloride ions. Thus, when an aqueous solution of $\mathrm{CuCl_2}$ is electrolysed, Cu metal is liberated at the cathode while $\mathrm{Cl_2}$ gas is evolved at the anode.