Predict the products of electrolysis in eaCHM of the following `:`
`a.` An aqueous solution of `AgNO_(3)` with silver electrodes.
`b.` An aqeous solution of `AgNO_(3)` with platinum electrodes,
`c.` A dilute solution of `H_(2)SSO_(4)` with platinum electrodes.
`d.` An aqueous solution of `CuCl_(2)` with platinum electrodes.

Explanation

(i) `AgNO_(3)` ionizes in aqueous solutions to form `Ag^(+)` and `NO_(3)^(-)` ions .
On On electrolysis, either `Ag_(+)` ions or `H_(2)O` molecules can be reduced at the cathode. But the reduction potential of `Ag_(+)` ions is higher than that of `H_(2)O`.
`Ag_((aq))^(+) + e^(-) to Ag_((s)) , E^(@) = +0.80 V `
`2H_(2)O_((l)) + 2 e^(-) to H_(2 (g)) + 2 OH_((aq))^(-) , E^(@) = - 0.83 V`
Hence `Ag^(+)` ions are reduced at the cathode . Similarly , Ag metal or `H_(2)O` molecules can be oxidized at the anode . But the oxidation potential of Ag is higher than that of `H_(2)O ` molecules .
`Ag_((s)) to Ag_((aq))^(+) + e^(-) , E^(@) = - 0.80 V`
`2H_(2)O_((l)) to O_(2 (g)) + 4 H_((aq))^(+) + 4 e^(-) , E^(@) = -1.23 V`
Therefore , Ag metal gets oxidized at the anode .
(ii) Pt cannot be oxidized easily . Hence , at the anode , oxidation of water occurs to liberate `O_(2)` . At the cathode , `Ag^(+)` ions are reduced and get deposited .
(iii) `H_(2)SO_(4)` ionizes in aqueous solutions to give `H^(+)` and `SO_(4)^(-)` ions .
`H_(2)SO_(4 (aq)) to 2H_((aq))^(+) + SO_(4 (aq))^(2-)`
On electrolysis , either of `H^(+)` ions or `H_(2)O` molecules can get reduced at the cathode .
But the reduction potential of `H^(+)` ions is higher than that of `H_(2)O` molecules .
`2H_((ag))^(+) + 2e^(-) to H_(2 (g)) , E^(@) = 0.0V`
`2H_(2)O_((aq)) + 2e^(-) to H_(2 (g)) + 2OH_((aq))^(-) , E^(@) = -0.83 V`
Hence , at the cathode , `H^(+)` ions are reduced to liberate `H_(2)` gas .
On the other hand , at the anode , either of `SO_(4)^(2-)` ions or `H_(2)O` molecules can get oxidized . But the oxidation of `SO_(4)^(2-)` involves breaking of more bonds than that of `H_(2)O` molecules . Hence , `SO_(4)^(2-)` ions have a lower oxidation potential than `H_(2)O` . Thus `H_(2)O` is oxidized at the anode to liberate `O_(2)` molecules .
(iv) In aqueous solutions , `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as :
`CuCl_(2 (aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)`
On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `H^(+)` ions is higher than that of `H_(2)O` molecules . `2H_((aq))^(+) + 2e^(-) to H_(2(g)) , E^(@) = 0.0V`
`2H_(2)O_((aq))k + 2e^(-) to H_(2 (g)) + 2OH_((aq))^(-) , E^(@) =- 0.83 V` .
Hence , at the cathode , `H^(+)` ions are reduced to liberate `H_(2)` gas .
On other hand , at the anode , either of `SO_(4)^(2-)` ions or `H_(2)O` molecules can get oxidized .
But the oxidation of `SO_(4)^(2-)` involves breaking of more bonds than that of `H_(2)O` molecules .
Hence , `SO_(4)^(2-)` ions have a lower oxidation potential than `H_(2)O` . Thus , `H_(2)O` is oxidized at the anode to liberate `O_(2)` molecules .
(iv ) In aqueous solutions , `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as :
`CuCl_(2(aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)`
On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `Cu^(2+)` is more than that of `H_(2)O` molecules .
`Cu_((aq))^(2+) + 2e^(-) to Cu_((aq)) , E^(@) = - 0.34 V`
`H_(2)O_((l)) + 2e^(-) to H_(2(g)) + 2OH^(-) , E^(@) = - 0.83 V`
Hence `Cu^(2+)` ions are reduced at the cathode and get deposited .
Similarly , at the anode , either of `Cl^(-)` or `H_(2)O` is oxidized . The oxidation potential of `H_(2)O` is higher than that of `Cl^(-)` .
`2Cl_((aq))^(-) to Cl_(2 (aq)) + 2e^(-) , E^(@) = - 1.36V`
`2H_(2)O_((l)) to O_(2 (g)) + 4H_((aq))^(-) + 4e^(-) , E^(@) = -1.23 V `
But the oxidation of `H_(2)O` molecules occurs at a lower electrode potential than that of `Cl^(-)` ions because of over- voltage (extra voltage required to liberate gas ). As a result , `Cl^(-)` ions are oxidized at the anode to liberate `Cl_(2)` gas .