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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

1.47V

B

1.77V

C

1.87V

D

1.57V

Text Solution

Verified by Experts

The correct Answer is:
D
`E=E^(o)-(0.059)/(4)"log"([Fe^(2+)]^(2))/([H^(+)]^(4)P_(O_(2)))`
`=1.67-(0.06)/(4)"log"((10^(-3))^(2))/((10^(-3))^(4)xx0.1)=1.67-(0.03)/(2)"log"10^(7)`
`=1.67-(0.03)/(2)xx7=1.67-0.105=1.565=1.57V`
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