Consider the following cell reaction.
`2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(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

Options

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

Answer

D

Explanation

`E=E^(@)-(0.059)/(4)log([Fe^(2+)]^(2))/([H^(+)]^(4)P_(O_(2)))`
`=1.67-(0.06)/(4)log(((10^(-3))^(2))/((10^(-3))^(4)xx0.1))=1.67-(0.03)/(2)log10^(7)`
`=1.67-(0.03)/(2)xx7=1.67-0.105=1.565=1.57V`