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. 1.77 V
C. 1.87 V
D. 1.57 V

Answer

D

Explanation

(d) The reaction taking place in the cell are :
Anode :`Fe(s) to Fe^(2+)(aq)+2e^(-)]xx2`
Cathode : `O_(2)(g)+4H^(+)(aq)+4e^(-) to 2H_(2)O(aq)`
`2Fe(s)+O_(2)(g)+4H^(+)(aq) to 2Fe^(2+)(aq)+H_(2)O(aq)`
`Q=([Fe^(2+)(aq)]^(2))/([H_((aq))^(+)]^(4)PO_(2(g)))=((10^(-3))^(2))/((10^(-3))^(4)xx0.1)`
`[H^(+)]=-10^(-pH)=10^(-3)]`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)logQ`
From Nernst's equation,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)logQ`
`=1.67-(0.0591)/(4)"log"(10^(-6))/(10^(-13))`
(n=no. of electrons involved`=1.67-(0.0591)/(4)xx7`
`=1.67-0.1034=1.57" V"`