1.

A fuel cell involves combustion of butane at 1 atm and 298 K C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(l),DeltaG^(@)=-2746" kJ "mol^(-1) The E_(cell)^(@) will be

Answer»

0.545 V
1.09 V
0.922 V
0.755 V

Solution :`overset(-10)(C_(4))overset(+10)(H_(10))(g)+(13)/(2)O_(2)(g)to4overset(+4)(C)overset(-4)(O_(2))+5H_(2)O(L)`
Change in oxidation number of carbon
`=4(+4)-(-10)=26`
or `(13)/(2)O_(2)^(@)+26e^(-)to13O^(2-)`
`(8O^(2-)` in `4CO_(2)` and `5O^(2-)` in `5H_(2)O)`
Thus, CELL reaction involves 26 ELECTRONS, i.e.,
n=26
`E_(cell)^(@)=(DeltaG^(@))/(nF)=(-(-2746)xx1000)/(26xx96500)=+1.09V`


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