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

Applying Nernst equation to 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 and pH = 3 means `[H^(+)] = 10^(-3) M , P_(O_(2)) = 0.1 ` atm , `[Fe^(3+)] = 10^(-3)` ,
`therefore E_(cell) = 1.67 - (0.0591)/(4) "log" ((10^(-3))^(2))/(0.1 xx (10^(-3))^(4))`
`= 1.67 - (0.0591)/(4) log 10^(7) = 1.67 - 0.10 = 1.57 V `