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Consider the following cell reaction. ...

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

A

1.47

B

1.77

C

1.87

D

1.57

Text Solution

Verified by Experts

The correct Answer is:
D
Applying nernst equation to the given reaction,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Fe^(2+)]^(2))/(P_(O_(2))xx[H^(+)]^(4))`
Fot the given reaction, n=4 and pH=3 means `[H^(+)]=10^(-3)M`
`thereforeE_(cell)=1.67-(0.0591)/(2)"log"((10^(-3))^(2))/(0.1xx10^(-3))^(4)`
`=1.67-(0.0591)/(4)log10^(7)=1.67-0.10=1.57V`
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