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:

Explanation

(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`
`=-213072 J mol^(-1) = -213.072 kJ mol^(-1)`