Question

Question 8.27 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.

Class XI Redox Reactions Page 275

Answer 1

(i) an aqueous solution of AgNO3 with silver electrode.
here two oxidation and two reduction half reactions must be considered .
now, oxidation (at anode )
(a) Ag (s) ---------> Ag+ (aq) + e ; E° = -0.8V
(b) 2H2O(l) ------->O2 ↑ + 4H+(aq) + 4e; E° = -1.23V .

here you can see that oxidation potential of Ag is greater than H2O molecule so, at anode silver get oxidised more readily .

again, reduction (at cathode)
(c) Ag+(aq) + e -----> Ag(s) ; E° = + 0.8V
(d) 2H2O(l) + 2e ------> H2↑ + 2OH- (aq) ; E° = -0.83 V
similarly in above reaction you can see that reduction potential of Ag+ ion is greater than H2O molecule so, at cathode silver ions get reduced more readily.

Therefore , on electrolysis of aqueous solution of AgNO3 with silver electrode , Ag from silver anode dissolves while Ag+ ions deposited at cathode.


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

as you know, Platinum is an inert electrode so, at anode oxidation of water takes place. as a result of this O2 is released at anode according to below equation
2H2O(l) -------> O2 + 4H+ + 4e
at cathode reduction of Ag+ ions takes place

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 .
H2SO4 will break in such a way :
H2SO4 <=> 2H+ + SO4^2-

when current is passed , either H+ ions are reduced at cathode or H2O molecule.
2H+(aq) + 2e -------> H2(g) ;E° = 0.0 V
2H2O(aq) + 2e -------> H2↑ + 2OH-(aq) ; E° = -0.83V

here we see that , reduction potential of H+ ions is greater than H2O molecule so, H+ ions get reduced more readily .

similarly on passing current, either SO4^2- ions are oxidised at anode or H2O molecule.
SO4^2- ions are resistant to oxidation and are not discharge at anode . so, at anode H2O molecule are oxidised to released O2 gas.
Like : 2H2O(l) -----> O2↑ + 4H+(aq) + 4e ; E° = -1.23V
Therefore on electrolysis of an aqueous solution of H2SO4 , only electrolysis of water occurs releasing HE at cathode and O2 at anode.


(iv) An aqueous solution of CuCl2 with platinum electrode .
CuCl2 <=> Cu²+ (aq) + 2Cl-(aq) [ Ionisation of CuCl2 in water ]
two oxidation and two reduction half reactions must be considered.

oxidation (at anode);
2Cl-(aq) -----> Cl2(g) + 2e ; E° = -1.35V
2H2O(l) ------> O2↑ + 4H+(aq) + 4e ;E° = -1.23V

here you can see that oxidation potential of H2O molecule is greater than Cl- ions . oxidation of Cl- ions takes place in preference to H2O molecule . it is because of the over voltage of O2. much lower potential is required for the oxidation of H2O molecules.

reduction (at cathode);
Cu²+(aq) + 2e ------> Cu ; E° = 0.34V
2H2O(l) + 2e ------> H2↑ + 2OH-(aq); E° = -0.83V
you can see that reduction potential of Cu²+ ions are greater than H2O molecule so, Cu²+ ions is reduced at cathode.

Therefore, on electrolysis of an aqueous solution of CuCl2 , Cu metal is released at cathode and Cl2 gas is evolved at anode.