Mark the correct Nernst equation for the given cell. `F_((s))|Fe^(2+)(0.001 M) ||H^(+)(1M) | H_(2(g)) ( 1 bar ) | Pt_((s))` is

Explanation

`Fe (s) | Fe^(2+) (0.001M) || H^(+) (1M) |H_(2) (1 atm) | Pt`
The electrode reactions and overall cell reactions are :

Since the reaction involves 2 moles of electrons , therefore , n = 2 and the Nernst equation at 298 K is :
`E_(cell) = E_(cell)^(Theta) - (0.059)/(2) "log" ([Fe^(2+)])/([H^(+)]^(2))`
`E_(cell)^(Theta) = E^(Theta) (H^(+) | H_(2)) - E^(Theta) (Fe^(2+) |Fe)`
`E^(Theta) (H^(+) |H_(2)) =0.0 V , E^(Theta) (Fe^(2+) |Fe) = -0.44V`
`E_(cell)^(Theta) = 0.0 - (-0.44) = 0.44 V`
`E_(cell) = 0.44 - (0.059)/(2) "log" (0.001)/((1)^(2))`
= 0.44 + 0.0885 = 0.5285 V .