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Consider the following cell reation : ...

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

A

1.47 V

B

1.77 V

C

1.87 V

D

1.57 V

Text Solution

Verified by Experts

The correct Answer is:
D
For the given cell reaction, `n=4, E^(@)=1.67 V`
`Q=([Fe^(2+)]^(2))/(p_(O_(2)xx[H^(+)]^(4)))=((10^(-3))^(2))/(0.1xx(10^(-3))^(4))=10^(-6)/10^(-13)=10^(7)`
`E=E^(@)-0.059/n log_(10) Q`
`=+1.67-0.059/4 log10^(7) =+1.67-0.10=1.57 V`
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