Consider the following cell reaction
`2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq),2H_(2)O(l),E^(@)=1.67V`
At `25^(@)C` `|Fe^(2+)| =10^(-3) M,p(O_(2))=0.1` atm, pH =3, the cell potential at `25^(@)C` is

Options

A. `1.47V`
B. `1.77V`
C. `1.87V`
D. `1.57V`

Answer

D

Explanation

Applying Nernst equation of the given reaction
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)"log"([Fe ^(+2)]^ (2))/(P_(O_(2))xx[H^(+)]^(4))`
For the given reaction, n=4 pH=3 means
`[H^(+)]=10^(-3)M`
`therefore E_("cell")=1.67-(0.059)/(4)"log"((10^(-3))^(2))/(0.1xx(10^(-3))^(4))`
`=1.67-(0.059)/(4)"log"((10^(-1))^(2))/(0.1xx(10^(-3))^(4))`
`=1.69-(0.059)/(4)"log"10^(7)`
`=1.67-0.10=1.57V`