Question 18: Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3with platinum electrodes. (iii) A dilute solution of H2SO4with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. Chapter 3: Electrochemistry Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3with platinum electrodes. (iii) A dilute solution of H2SO4with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. is solved by our expert teachers. You can get ncert solutions and notes for class 12 chapter 3 absolutely free. NCERT Solutions for class 12 Chemistry Chapter 3: Electrochemistry is very essencial for getting good marks in CBSE Board examinations
Question 18: Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3with platinum electrodes.
(iii) A dilute solution of H2SO4with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Answer:Remember
All ions are in aqueous state
(i)
Reaction in solution
AgNO3 ↔ Ag+ + NO3–
H2O ↔ H+ + OH–
Reaction at cathode
Ag+ + e-→ Ag
Reaction at anode
Ag(s) + NO3– → AgNO3(aq) + e-
Hence Ag will deposit at cathode and dissolve at anode
(ii)
Reaction in solution
AgNO3 ↔ Ag+ + NO3–
H2O ↔ H+ + OH–
Reaction at cathode
Ag+ + e-→ Ag
Reaction at anode
Due to platinum electrode self of ionization of water will take place
H2O → 2H+ + 1/2O2(g) + 2e-
Hence Ag will deposit at cathode and O2 gas will generate at anode
(iii)
Reaction in solution
H2SO4 ↔ 2H+ + SO42–
H2O ↔ H+ + OH–
Reaction at cathode
H+ + e-→ ½ H2
Reaction at anode
Due to platinum electrode self of ionization of water will take place
H2O → 2H+ + 1/2O2(g) + 2e-
Hence H2 gas will generate at cathode and O2 gas will generate at anode
(iv)
Reaction in solution
CuCl2(s) ↔ Cu2+ + 2Cl–
H2O ↔ H+ + OH–
Reaction at cathode
Cu2+ + 2e-→ Cu(S)
Reaction at anode
2Cl– → Cl2 + 2e-
Hence Cu will deposit at cathode and Cl2 gas will generate at anode
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