Consider the following cell reaction `:`
`2Fe(s)+O_(2)(g)+4H^(o+)(aq) rarr 2Fe^(2+)(aq)+2H_(2)O(l)" "E^(0)=1.67V`
at `25^(@)C, [Fe^(2+)]=10^(-3)M,p(O_(2))=0.1atm` and `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

`2Fe_((s)) = O_(2(g)) +4H_((aq))^(+) rarr 2Fe_((aq))^(2+) +2H_(2)O(l)`
`E = E^(@) -(0.06)/(n)log.([Fe^(+2)])/(P_(O_(2))[H^(+)]^(4))`
`= 1.67 - (0.06)/(4) log.((10^(-3))^(2))/(0.1xx(10^(-3))^(4))`
`E = 1.67 - (0.06)/(4) log 10^(7) = 1.67 - (0.06)/(4) xx 7`
`= 1.67 - 0.105 = 1.565 V = 1.57V`