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Calculate the emf of the following cell ...

Calculate the emf of the following cell at `25^(@)C`
`Fe|Fe^(@+) ( 0.001 M ) || H^(+) ( 0.01M ) || H_(2) ( g) ( 1 "bar" ) | Pt (s)`
`E_((Fe^(@+) //Fe))^(@) = - 0.44 V , E_((H^(+) //H_(2)))^(@) = 0.00V`

Text Solution

Verified by Experts

For the given cell representation the cell reaction will be
`Fe(s) + 2H^(+) ( aq) rarr Fe^(@+) ( aq) + H_(2)( g)`
The standard emf of the cell will be,
`E_("cell")^(@) = E_(H^(+)//H_(2))^(@) - E_(Fe^(@+) //Fe)^(@)`
`E_("cell")^(@) = 0 - ( - 0.44) = 0.44 \V`
The Nernst equation for the cell reaction at `25^(@)C`
`E_("cell") = E_("cell")^(@) - (0.0591)/( n ) log""([Fe^(@+)])/( [ H^(+)]^(2))`
`= 0.44- ( 0.0591 )/( 2) log""( [ 0.001])/( [ 0.01]^(2))`
`= 0.44-0.02955 ( log 10)`
`= 0.44 - 0.02955 (1)` `[ :. log 10 =1]`
`= 0.41045 V = 0.41 V`
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