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

Answer

D

Explanation

`pH=3rArr[H^(+)]=10^(-3)M`
`E_("cell")=E_("cell")^(@)-(0.059)/(4)"log"([Fe^(2+)]^(2))/([H^(+)]^(4)p(O_(2)))`
`=1.67-(0.059)/( 4)"log"(10^(-6))/((10^(-3))^(4)xx0.1)`
`=1. 67-(0.059)/(4)"log"(10^(-6+13)`
`=1.67-(0.059xx7)/(4)=1.67-0.1`
`=1.57V`