1.

Consider the following cell reaction: 2Fc(s) +O_(2)(g) +4H^(+) (aq)to 2Fe^(2+) (aq)+ 2H_(2) O (l)E ^(@)=1.67V At [Fe^(2+)]=10^(-3)M, p(O_(2))=0.1 atm and pH =3, the cell poteintial at 25^(@)C is

Answer»

`1.47 V`
`1.77 V`
`1.87 V`
`1.57 V`

Solution :Here `n=4, and [H^(4)]=10^(-3)(aspH=3)`
Applying Nernst equation
`E=E^(@)-(0.059)/(n)log""([FE^(2+)]^(2))/([H^(+)]^(4) (po_(2)))`
`=1.67 -(0.059)/(4) log ""((10^(-3))^(2))/((10^(-3))^(4) xx0.1)`
`1.67 (0.059)/(4) log 10^(7) =1.67 =0.103=1.567V`


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