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.

Explanation

(i) Electrolysis of aqueous solution of `AgNO_(3)` using silver electrodes :
`AgNO_(3)(s)overset("water")toAg^(+)(aq)+NO_(3)^(-)(aq)`
`H_(2)OhArrH^(+)OH^(-)`
cathode. `Ag^(+)` ions have lower discharge potential than `H^(+)` ions. Hence `Ag^(+)` ions will be deposited as silver (in preference to `H^(+)` ions)
At anode. Since silver electrode is attacked by `NO_(3)^(-)` ions, Ag anode will dissolve to form `Ag^(+)` ions in the solution.
`Ag""toAg^(+)+e^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_(3)` using platinum electrodes :
At cathode. Same as above.
At anode. Since silver is not attacked, out of `OH^(-)` and `NO_(3)^(-)` ions, `OH^(-)` ions have lower discharge potential and hence `OH^(-)` ions will be discharged in preference to `NO_(3)^(-)`. The `OH^(-)` will decompose to give `O_(2)`.
`OH^(-)(aq)toOH+e^(-)`
`4OHto2H_(2)O(l)+O_(2)(g)`