1.

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) ("Give`E_("Cell")^@=+0.44V)`

Answer» `"At anode":FetoFe^(2+)+2e^(-)" Atcathode ",2H^(+)+2e^(-)toH_2`
So, total number of electrons (n) transferred = 2
`"Given: " E_("Cell")^@=+0.44" Volt"`
`" Temperature " (T) =298K`
`E_("Cell")=E_("Cell")^@-((2.303RT)/(nF))log.(a_("oxi"))/(a_("red"))`
`=E_("Cell")^@-((0.05916V)/n)log. (a_"oxidation")/a_("reduction")`
`0.44-(0.0591V)/2log. (0.001)/1`
`" Therefore," E_("Cell")=0.44(-0.02955xx(-3))`
`=0.44+0.08865=0.53"Volt"`


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