Question

Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes (ii) An aqueous solution AgNO3 with platinum electrodes (iii) A dilute solution of H2SO4 with platinum electrodes (iv) An aqueous solution of CuCl2 with platinum electrodes.

Answer 1

Answer:

Therefore , on electrolysis of aqueous solution of AgNL3 solution with platinum electrode , O2 is released at anode and Ag+ ions deposited at cathode. (iii)A dilute solution of H2SO4 with platinum electrode . when current is passed , either H+ ions are reduced at cathode or H2O molecule.

Answer 2

$\huge{\fcolorbox{white}{pink}{QUESTION:}}$

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes.

(ii) An aqueous solution of AgNO3 with platinum electrodes.

(iii) A dilute solution of H2SO4 with platinum electrodes.

(iv) An aqueous solution of CuCl2  with platinum electrodes

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(i) An aqueous solution of AgNO3 with silver electrodes.

• At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

• At anode: Silver anode will dissolve to form silver ions in the solution.

Ag→Ag++e−

(ii) An aqueous solution of AgNO3 with platinum electrodes.

• At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

• At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.

4OH−(aq)→2H2O(l)+O2(g)+4e−

(iii) A dilute solution of H2SO4 with platinum electrodes.

• At cathode: 2H++2e−→H2(g)

• At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.

4OH−(aq)→2H2O(l)+O

(iv) An aqueous solution of CuCl2  with platinum electrodes

• At cathode: Cupric ions will be reduced in preference to protons

Cu2++2e−→Cu

• At anode: Chloride ions will be oxidized in preference to hydroxide ions

2Cl−→Cl2+2e−