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Predict the products of electrolysis 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) An aqueous solution of H_2SO_4 with platinumelectrodes (ii)An aqueous solution of CuCl_2 with platinum electrodes |
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Answer» Solution :(i) Electrolysis of aqueous solution of `AgNO_3` using SILVER electrodes : `AgNO_3(s)+nH_2O rarr AG^(+)(aq) +NO_3^(-) (aq)` `""H_2O hArr H^(+) +OH^(-)` At 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 rarr Ag^(+) +e^(-)` (II) Electrolysis of aqueous solution of `AgNO_3` using platinum electrodes : (ii) At cathode : same as above At anode : Since silver is not attackable , 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) rarr OH+e^(-)` `4OH^(-) (aq) rarr 2H_2O(l) +O_2(g)` (iii) Electrolysis of `H_2SO_4` with Pt electrodes : `H_2SO_4(aq) rarr 2H^(+) (aq) +SO_4^(2-) (aq)` `H_2OhArr H^(+) +OH^(-)` At cathode : `H^(+) + e^(-) rarr H` `H+H rarr H_2(g)` At anode : `OH^(-) rarr OH+e^(-)` `4OH rarr 2H_2O+O_2(g)` (iv) Electrolysis of aqueous solution of `CuCl_2` with platinum electrodes : `CuCl_2(s) +(aq) rarr Cu^(2+)(aq) +2Cl^(-) (aq)` `H_2O hArr H^(+) +OH^(-)` At cathode : `Cu^(+)` will be reduced in preference to `H^+` is `Cu^(2+)+2e^(-) rarr Cu` At anode : `Cl^(-)` ions will be oxidised in preference to `OH^-` ions `Cl^(-) rarr Cl+e^(-)` `Cl+Cl rarr Cl_2` Thus copper will be deposited on the cathode and `Cl_2` will be liberated at anode . |
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