Consider the following cell reation `:`
`2Fe(s)+O_(2)(g)+4H^(o+)(aq) rarr 2Fe^(2+)(aq)+2H_(2)O(l)" "E^(c-)=1.67V`
`At[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) rarr 2Fe^(2+)(aq) +4e^(-)" "[:.n_(cell)=4]`
`E=E^(c-)-(0.059)/(n_(cell))log.([Fe^(2+)]^(2))/((p_(O_(2)))[H^(o+)]^(4))`
`[pH=3implies[H^(o+)]=10^(-3)M]`
`=1.67-(0.059)/(4)log.((10^(-3))^(2))/((0.1atm)(10^(-3))^(4))`
`=1.67-(0.059)/(4)log10^(7)`
`=(1.67-90.059)/(4)xx7`
`=1.67-0.103=1.567~~1.57V`