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For the cell reaction Fe(s)|Fe^(2+)(0.1M...

For the cell reaction `Fe(s)|Fe^(2+)(0.1M)"||"H^(+)(1M)|H_(3)"(1 atm)"`, Pt `E^(@)=0.44V`. The cell e.m.f. is :

A

0.41 V

B

0.47 V

C

1.26 V

D

1.20 V

Text Solution

Verified by Experts

The correct Answer is:
B
`Fe(s)+2H^(+)(aq) rarr Fe^(2+)(aq)+H_(2)`
Nernst equation is :
`E^(@)=E^(@)-(0.059)/(2) log. ([Fe^(2+)])/([H^(+)]^(2))`
`E^(@)=E^(@)(H^(+)"|"H_(2))-E^(@)(Fe^(2+)"|"Fe)`
`=0.0-(-0.44)=0.44V`
`:.E=0.44-(0.059)/(2) log. (0.1)/(1)`
or `E=0.44+0.0295=0.47V`
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