a) Calculate the charge in coulombs required for the oxidation of 2 moles of water to oxygen ? [Given 1 F = 96500 C mol^(-1) ] (b) Zinc/Silver oxide cell is used in hearing aids and 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 + OH^(-), E_(Ag^(+)//Ag)^(@) = 0.344 V Calculate (i) Standard potential of the cell, (ii) Gibb's free energy:

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

1.	a) Calculate the charge in coulombs required for the oxidation of 2 moles of water to oxygen ? [Given 1 F = 96500 C mol^(-1) ] (b) Zinc/Silver oxide cell is used in hearing aids and 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 + OH^(-), E_(Ag^(+)//Ag)^(@) = 0.344 V Calculate (i) Standard potential of the cell, (ii) Gibb's free energy:
Answer» Solution :(a) `2H_(2)O (l) to O_(2)(g) + 4H^(+) (aq) + 4e^(-)` Two molecules of water release four electrons for oxidation to oxygen. Thus four Faradays of electricity is required for the oxidation of two moles of water. Charge in coulombs = 4 x 96500 coulombs = 386000 coulombs (b)(i) Zinc electrode is the anode while silver electrode is the cathode: `E_("cell")^(@) = 0.344 V - (-0.76 V) = 1.104 V` ii) Gibbs free energy can be CALCULATED as under : `DeltaG^(@) = -NFE^(@)` Substituting the values, we have `DeltaG^(@) =-2 XX 9650 C MOL^(-1) xx 1.104 V` `=-213072J mol^(-1) = -213.072 kJ mol^(-1)`