(a) Calculate the charge in coulombs required for oxidation of 2 moles of water to oxygen ? [Given: 1F = 96,500 C mol^(-1)] (b) Zinc/silver oxide cell is used in hearing aids and electric watches. The following reactions occur : Zn(s) to Zn^(2+) (aq) + 2e^(-), E_(Zn^(2+)//Zn)^(@) =-0.76 V Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH^(-), E_(Ag^(+)//Ag)^(@) = 0.344 V Calculate (i) standard potential of the cell, (ii) standard Gibbs energy.

(a) Calculate the charge in coulombs required for oxidation of 2 moles

1.	(a) Calculate the charge in coulombs required for oxidation of 2 moles of water to oxygen ? [Given: 1F = 96,500 C mol^(-1)] (b) Zinc/silver oxide cell is used in hearing aids and electric watches. The following reactions occur : Zn(s) to Zn^(2+) (aq) + 2e^(-), E_(Zn^(2+)//Zn)^(@) =-0.76 V Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH^(-), E_(Ag^(+)//Ag)^(@) = 0.344 V Calculate (i) standard potential of the cell, (ii) standard Gibbs energy.
Answer» Solution :`2H_(2)O (l) to O_(2)(g) + 4H^(+) (aq) + 4e^(-)` 2 molecules of water release 4 electrons for OXIDATION to oxygen or 4 Faradays of electricity is required for the oxidation of 2 moles of water to oxygen. Charge in coulombs required for the process = 4 x 96500 coulombs = 386000 coulombs (b)(i) Zn electrode is the anode while silver electrode is the cathode. `E_("CELL")^(@) = E_("cathode") - E_("anode")` `=0.344 V - (-0.76 V) = 1.104 V` (ii) `DeltaG^(@) =-nFE_("cell")^(@)` n=2 (Because 2 electrons each are INVOLVED in the anode and cathode reactions) F = 96500 COLUMBS `E^(@) = 1.104 `V Substituting the values, we have `DeltaG^(@) =-2 xx 96500 xx 1.104 = -213072 J mol^(-1) = -213.072 kJ mol^(-1)`.