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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

`1.27V`

B

`1.77V`

C

`1.87V`

D

`1.57V`

Text Solution

Verified by Experts

The correct Answer is:
D
`[H^(+)]=10^(-3),Q=([Fe^(2+)])/([H^(+)]^(4)P_(O_(2))),n=4`
`E=1.67-(0.06)/(4)log.((10^(-3))^(2))/((10^(-3))^(4)(10^(-1)))`
`=1.67-(0.06)/(4)xx7`
`=1.57V`
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