The zinc/silver oxide cell is used in electric current reaction is as following:
`Zn^(2+) + 2e^(-) to Zn, E^(@) = - 0.760 V`
`Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH, E^(@) = 0.344 V` If F is 96,500 C `mol^(-1)`, `DeltaG^(@)` of the cell will be:

Options

A. `413.21 kJ mol^(-1)`
B. `113.072 kJ mol^(-1)`
C. `213.072 kJ mol^(-1)`
D. `313.082 kJ mol^(-1)`

Answer

C

Explanation

`Zn^(2+) + 2e^(-) to Zn, E^(@) =-.760 V`
`Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH^(-), E^(@) = +0.344 V`
Both are reducing potential,
As, `E_(Ag_(2)Cl//2Ag)^(@) gt E_(Zn^(2+)//Zn)^(@)`
`therefore` Cell reaction will be:
At anode, `Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH^(-), E^(@) = +0.344 V`
At Cathode: `Zn + Ag_(2)O + H_(2)O to Zn^(2+) + 2Ag + 2OH^(-)`
n =2 `E_("cell")^(@) = -nFE_("cell")^(@)`
`DeltaG^(@) =-2 xx 96500 xx 1.10`
`=-213072 J mol^(-1)`
`|DeltaG^(@)| = 213.072 kJ mol^(-1)`