Calculate the emf of the following cell at `298K: Fe(s) abs(Fe^(2+)(0.001M))abs(H^+(1M))H_2(g)(1"bar"),Pt(s) ("Given"E_("Cell")^@=+0.44V)`

Explanation

the electrode reactions in this cell are
At anode: `Fe_((s))toFe^(2+)(0.001M)+2e^(-)`
At cathode: `2H^(+)(1M)+2e^(-)toH_(2)(1atm)`
Net reaction `Fe_((s))+2H^(+)(1M)toFe^(2+)(0.001M)+H_(2)(1atm)`
The nearest equation of this cell at `25^(@)C`
`E_(cell)-E_(cell)^(0)-(0.0591)/(2)log((|Fe^(2+)|[H_(2)])/([Fe_((s))][H^(+)]^(2)))` ltbr. The cell emf is then given by
`E_(cell)=0.44-(0.591)/(2)log((0.001xx1)/((1)^(2)))`
`=0.44-0.0296log((1)/(100))`
`=0.44-0.02%log(10^(-3))`
`=0.44+(3xx0.0296)=0.44+0.0888`
Therefore `E_(cell)=+0.53V`.