Saved Bookmarks
| 1. |
Predict the products of electrolysis of the following: (i) An aqueous solution of AgNO_(3) with silver electrodes (ii) An aqueous solution of AgNO_(3) with platinum electrodes. (iii) A dilute aqueous solution of H_(2)SO_(4) with platinum electrodes. (iv) An aqueous solution of CuCl_(2) with platinum electrodes. (Given E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Cu^(2+)//Cu)^(@)=+0.34V) |
|
Answer» Solution :(i) Electrolysis of aqueous solution of `AgNO_(3)` with silver electrodes. `AgNO_(3)(s)+aqtoAg^(+)(aq)+NO_(3)^(-)(aq)` `H_(2)OhArrH^(+)+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. Alternatively, we have stanard reduction potential potentials as `Ag^(+)(aq)+e^(-)toAg(s),E^(@)=+0.80V` `H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E^(@)=0.00V` As `Ag^(+)` ions hae HIGHER standard reduction potential than that of `H^(+)` ions, hence `Ag^(+)` ions will be reduced more easily and deposited as Ag. At anode: As Ag anode is attached by `NO_(3)^(-)` ions, Ag of the anode will dissolve to form `Ag^(+)` ions in the solution. `AgtoAg^(+)+e^(-)` Alternatively, out of the three possible oxidation reactions occurring at the anode, i.e., `AgtoAg^(+)+e^(-),2OH^(-)toH_(2)O+(1)/(2)O_(2)+e^(-) and NO_(3)^(-)toNO_(3)+e^(-)`, Ag has highest oxidation potential. hence, Ag of anode is oxidized to `Ag^(+)` ions which pass into the solution. (ii) Electrolysis of aqueous solution of `AgNO_(3)` using platinum electrodes. At cathode: Same as above. ltBrgt At anode: As anode is not attackable, out of `OH^(-) and NO_(3)^(-)` ions, have lower discharge potential. hence, `OH^(-)` ions will be DISCHARGED in preference to `NO_(3)^(-)` ions, which then decompose to give out `O_(2)`. `OH^(-)(aq)toOH+e^(-),4OHto2H_(2)O(L)+O_(2)(g)` (iii) Electrolysis of dilute `H_(2)SO_(4)` with platinum electrodes. `H_(2)SO_(4)(aq)to2H^(+)(aq)+SO_(4)^(2-)(aq)` `H_(2)OhArrH^(+)+OH^(-)` At cathode: `H^(+)+e^(-)toH,H+HtoH_(2)(g)` At anode: `OH^(-)toOH,e,4OHto2H_(2)O+O_(2)(g)` Thus, `H_(2)` is LIBERATED at the cathode and `O_(2)` at the anode. (iv) Electrolysis of aqueous solution of `CuCl_(2)` with platinum electrodes. `CuCl_(2)(s)+aqtoCu^(2+)(aq)+2Cl^(-)(aq)` `H_(2)OhArrH^(+)+OH^(-)` At cathode: `Cu^(2+)` ios will be reduced in preference to `H^(+)` ions `Cu^(2+)+2e^(-)toCu` At anode: `Cl^(-)` ions will be oxidized in preference to `OH^(-)` ions `Cl^(-)toCl+e^(-),Cl+CltoCl_(2)(g)` Thus, Cu will be deposited on the cathode and `Cl_(2)` will be liberated at the anode. |
|