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Consider the following cell reaction: ...

Consider the following cell reaction:
`2Fc(s) +O_(2)(g) +4H^(+) (aq)to 2Fe^(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 poteintial at `25^(@)C` is

A

`1.47 V`

B

`1.77 V`

C

`1.87 V`

D

`1.57 V`

Text Solution

Verified by Experts

The correct Answer is:
D
Here `n=4, and [H^(4)]=10^(-3)(aspH=3)`
Applying Nernst equation
`E=E^(@)-(0.059)/(n)log""([Fe^(2+)]^(2))/([H^(+)]^(4) (po_(2)))`
`=1.67 -(0.059)/(4) log ""((10^(-3))^(2))/((10^(-3))^(4) xx0.1)`
`1.67 (0.059)/(4) log 10^(7) =1.67 =0.103=1.567V`
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