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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`

A

`1.47V`

B

`1.73V`

C

`1.87V`

D

`1.57V`

Text Solution

Verified by Experts

The correct Answer is:
D

`Q_(c)= ([Fe^(+2)]^2)/([H^(+)]^(4) Po_(2))= ([10^(-3)]^2)/(10^(-4)0.1) = (10^(-6))/(10^(-5) = 10^(-1)`
`logQ_(c)= -1`
`E=E_("RXN")^(@)-(0.0591)/(4) (logQ_(c))= 1.67-(0.0591)/(4)=([-1.0])= 1.67+ (0.0591)/(4) = =1.685V`
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