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Predict the product of electolysis in each of the following (i) An aquesous solution of AgNO_(3) with silver electrodes (ii) An Aqueous solution fo silver nitrate with platinum elctrodes (iii) A dilute soution of H_(2)SO_(4) with platinum electrodes (iv) An aqueoius solution of CuCI_(2) with platinum electordes |
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Answer» Solution :(i) In aqueous solution `AgNO_(3)` ionises ot give `Ag^(+)(Aq)and NO_(3)^(-)(aq)`ions `AgNO_(3)(aq)rarrAg^(+)(aq)+NO_(3_^(-) (aq)` thus when electricity is passed `Ag^(+) (aq)+e^(-)rarrAg(s), E^(@)=+_0.80 v` `2H_(2)O(l)+2e^(-)rarrH_(2)(g)+2O^(-)(aq),e^(@)=-0.83 v` since the elctrode potential i.e reductin potential of `Af^(+)` (aq) ions is higher than that of `H_(2)O` molecules therefore at the CATHODE it is the `Ag^(+)` (aq) ions (RATHER than `H_(2)O` molecules ) which are reduced similary at the naode either ag metal of the anode ofr `H_(2)O` molecules may be oxidised their electrode thus when an queous soluton of `AgNO_(3)` is electrolysed As from As anode dissolved while `Ag^(+)` (aq) INS present in the solution get reduced and get depostied on the cathode (iii) if howver elctrolysis of `AgNO_(3)`is electrolysed Ag from Ag anode dissolves wile `Ag^(+)` ions present in the solutoin get reduced and get depostited on the cathode since the elctron potential of `H^(+)` (aq) ions is higher than that of `H_(2)O` therefore at the cathode it is `H^(+)` (aq) ioons which are reduced to evolve `H_(2)O` gasof `SO_(4)^(2-)` is expected to be much lower than that of `H_(2)O` molecuels therefore at the anode it is `H_(2)O` mleculeswhich are oxidised to evolve `O_(2)` gas from the abvoe discussion it followthat during electrolysis of an aqueous solution of `H_(2)O` only the electrolysis of `H_(2)O` occurs liberating `H_(2)` at the cathode and `O_(2)` at the anode thus as cathode either `Cu^(2+)`(aq) or `H_(2)O` molecuelsare reduced their electrode potential are `Cu^(2+)(aq)+2e^(-)rarrCu(s) , =+ 0.34 V` `H_(2)O(l)+2e^(-)rarrH_(2)(g)+2OH^(-), E^(@)=-0.83V` since the electrode potential of `Cu^(2+)`(aq) ionsis much higher than that of `H_(2)O` therefore at the cathode it is `Cu^(2+)` (aq) ions whihc are reduced and not 1 `H_(2)O` molecules similary at the anode either `CI^(-)` (aQ) ions or `H_(32)O` molecules are oxidised their oxidation potentialsare `2CI^(-)(aq)rarrCI_(2)(g)+2e^(-), E^(@)=-1.36 v` `2H_(2)O(l)rarrO_(2)+4H^(+)(aq)+4e^(-),E^(@)=-1.23 v` Although oxidation potential of `H_(2)O` molecules is higher than of `CI^(-)` ions neverthelss oxidation of `CI^(-)(aq)` ions occurs is prefrence to `H_(2)O` since due to `"overvoltage"^(*)` of `O_(2)` much lower potential than than -1.36 v is NEEDED for the oxidation of `H_(2)O` molecules thus when aqueous solution of `CuCI_(2)` is electrolysed Cu metal is liberated at the cathode while `CI_(2)` gas is evolved at the anode |
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