1.

Predict the products of electrolysis in each 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 solution of H_2SO_4 with platinum electrodes. (iv) An aqueous solution of CuCl_(2) , with platinum electrodes.

Answer»

Solution :Electrolysis of AQUEOUS solution of `AgNO_(3)` with silver electrodes:
The ions produced by the dissociation are as under:
`AgNO_(3) (s) + aq to Ag^(+) + NO_(3)^(-3) (aq)`
`H_(2)O At cathode : `Ag^+` ions have lower DISCHARGE potential than H+ ions. Hence, `Ag^+` ions will be deposited in preference to `H^+` ions.
At anode : As Ag anode is attacked by `NO_(3)^(-)`ions, Ag of the anode will dissolve to form `Ag^+` ions in the solution.
`Agto Ag^(+) + E^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_3` using platinum electrodes : At cathode : Same as in (i).
At anode : Out of `OH^(-)`and `NO_3^(-)`ions, `OH^(-)`ions have lower discharge potential. Hence, `OH^(-)` ions will be discharged in preference to `NO_3^(-)` ions. OH ions decompose to give out `O_(2)`as under:
`OH^(-)(aq) to OH + e^(-), 4OH to 2H_(2)O (l) +O_(2)(g)`
(III) Electrolysis of dilute `H_(2)SO_(4)` with platinum electrodes:
`H_(2)SO_(4)` and `H_(2)O` ionise as under:
`H_(2)SO_(4)(aq) to 2H^(+) (aq) + SO_(4)^(2-) (aq)`
`H_(2)O At cathode: `H^(+) + e^(-) to H, H + H to H_(2)(g)`
At cathode: `OH^(-) to OH +e^(-), 4OH to 2H_(2)O + O_(2)(g)` (because of lower discharge potential of `OH^(-)`)
(iv) Electrolysis of aqueous solution of `CuCl_2` with platinum electrodes :
`CuCl_(2)(s) + aq to Cu^(2+) (aq) + 2Cl^(-)(aq)`
`H_(2) O At cathode: `Cu^(2+)` ions will be reduced in preference to `H^(+)` ions
`Cu^(2+) + 2e^(-) to Cu`
At anode: `Cl^(-)` ions will be oxidised in preference to `OH^(-)` ions.
`Cl^(-) to Cl + e^(-), Cl + Cl to Cl_(2)(g)`
THUS, Cu will be deposited on the cathode and `Cl_2` will be liberated at the anode.


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