(i) Electrolysis of aqueous solution of $\mathrm{AgNO}_3$ with silverelectrodes, $\mathrm{AgNO}_{3(s)} + \mathrm{aq} \rightarrow \mathrm{Ag}^{+}_{(aq)} + \mathrm{NO}^{\,-}_{3(aq)}$ $\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{\,-}$ At cathode: Ag$^{+}$ions have lower discharge potential than H$^{+}$ions. Hence,Ag$^{+}$ions will be deposited as Ag in preference to H$^{+}$ions. At anode: As Ag anode is attacked by NO$^{\,-}$$_{3}$ ions, Ag of the anode willdissolve to form Ag$^{+}$ions in the solution. $\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^{\,-}$ (ii) Electrolysis of aqueous solution of $\mathrm{AgNO}_3$using platinum electrodes, At cathode: Ag will be deposited. At anode: As anode is not attacked, out of OH$^{\,-}$and NO$^{\,-}$$_{3}$ ions, OH$^{\,-}$ionshave lower discharge potential. Hence, OH$^{\,-}$ions will be discharged inpreference to NO$^{\,-}$$_{3}$ ions, and OH$^{\,-}$will then decompose to give out O$_{2}$. $\mathrm{OH}^{\,-}_{(aq)} \rightarrow \mathrm{OH} + e^{\,-}$, $4\mathrm{OH} \rightarrow 2\mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)}$ (iii) Electrolysis of dilute $\mathrm{H}_2\mathrm{SO}_4$ with platinum electrodes, $\mathrm{H}_2\mathrm{SO}_{4(aq)} 2\mathrm{H}^{+}_{(aq)} + \mathrm{SO}^{2-}_{4(aq)}$ $\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{\,-}$ At cathode:$\mathrm{H}^{+} + e^{\,-} \rightarrow \mathrm{H},$ $\mathrm{H} + \mathrm{H} \rightarrow \mathrm{H}_{2(g)}$ At anode: $\mathrm{OH}^{\,-} \rightarrow \mathrm{OH} + e^{\,-},$$4\mathrm{OH} \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_{2(g)}$Thus, $\mathrm{H}_2$ is liberated at the cathode and $\mathrm{O}_2$ at the anode.(iv) Electrolysis of aqueous solution of $\mathrm{CuCl}_2$ with platinum electrodes,$\mathrm{CuCl}_{2(s)} + \mathrm{aq} \rightarrow \mathrm{Cu}^{2+}_{(aq)} + 2\mathrm{Cl}^{\,-}_{(aq)}$$\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{\,-}$At cathode: $\mathrm{Cu}^{2+}$ ions will be reduced in preference to $\mathrm{H}^{+}$ions.$\mathrm{Cu}^{2+} + 2e^{\,-} \rightarrow \mathrm{Cu}$At anode: Cl$^{\,-}$ions will be oxidized in preference to OH$^{\,-}$ions.$\mathrm{Cl}^{\,-} \rightarrow \mathrm{Cl} + e^{\,-},$$\mathrm{Cl} + \mathrm{Cl} \rightarrow \mathrm{Cl}_{2(g)}$Thus, Cu will be deposited on the cathode and $\mathrm{Cl}_2$ will be liberated at theanode.