1.

Consider the following cell reaction: 2Fe(2)+O_(2)+4H^(+)(aq)to2Fe^(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 potential at 25^(@)C is,

Answer»

1.47
1.77
1.87
1.57

Solution :Applying nernst equation to the given reaction,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([FE^(2+)]^(2))/(P_(O_(2))XX[H^(+)]^(4))`
Fot the given reaction, n=4 and pH=3 MEANS `[H^(+)]=10^(-3)M`
`thereforeE_(cell)=1.67-(0.0591)/(2)"log"((10^(-3))^(2))/(0.1xx10^(-3))^(4)`
`=1.67-(0.0591)/(4)log10^(7)=1.67-0.10=1.57V`


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