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Consider the following cell reaction 2...

Consider the following cell reaction
`2Fe_((s))+O_(2(g))+4H_((aq.))^(o+)rarr2Fe_((aq.))^(+2)+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 …….
Use `(2.303RT)/(F)=0.059`

Text Solution

Verified by Experts

The correct Answer is:
1.57

`2Fe_((s))+O_(2(g))+4H^(o+)to2Fe_((aq.))^(+2)+2H_(2)O(l)`
`n=4" "impliespH=3implies[H^(+)]=10^(-3)`
`E=E^(@)-(0.059)/(4)xx log""([Fe^(+2)]^(2))/([H^(+)]^(4)(Po_(2)))=1.67-(0.059)/(4)xxlog""((10^(-3))^(2))/((10^(-3))^(4)(0.1))=1.57" V"`
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Knowledge Check

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