Write the Nernst equation and `EMF` of the following cells at `298K :`
`a.` `Mg(s)|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu(s)`
`b.` `Fe(s)|Fe^(2+)(0.001M)|H^(o+)(1M)|H_(2)(g)(1 ba r)|Pt(s)`
`c.` `Sn(s)|Sn^(2+)(0.050M)||H^(o+)(0.020M)|H_(2)(g)(1 ba r)|Pt(s)`
`d.` `Pt(s)|Br_(2)(1)|Br^(c-)(0.010M)||H^(o+)(0.030M)|H_(2)(g)(1 ba r)|Pt(s)`

Answer

`a. 2.68 V`, `b. 0.523V`, ` c. 0.078V` , `d. -1.288V`

Explanation

`a.` Cell reaction `: Mg+Cu^(2+) rarr mg^(2+)=Cu(n=2)`
`E_("cell")=E_("cell")^(c-)-(0.0591)/(2)log``([Mg^(2+)])/([Cu^(2+)])`
`=0.34-(-2.37)-(0.591)/(2)log``(10^(-3))/(10^(-4))`
`=2.71-0.02955=2.68`
`b.` Cell reaction `Fe+2H^(o+) rarr Fe^(2+)+H_(2) (n=2)`
`E_("cell")=E_("cell")^(c-)-(0.0591)/(2)log ([Fe^(2+)])/([H^(o+)])`
`=0-(-0.44)-(0.0591)/(2)log``(10^(-3))/((1)^(2))`
`=0.44+0.0887=0.523V`
`c. Cell reaction : `Sn+2H^(o+) rarr Sn^(2+)+H_(2) (n=2)`
`E_("cell")=E_("cell")^(c-)-(0.0591)/(2)log ([Sn^(2+)])/([H^(o+)])`
`=0-(-0.14)-(0.0591)/(2)log (0.05)/((0.02)^(2))`
`=0.14-(0.0591)/(2)log125`
`=0.14-(0.0591)/(2)(2.0969)=0.078V` ltbr. d. Cell reaction `:2Br^(c-)+2H^(o+) rarr Br_(2)+H_(2)`
`E_("cell")=E_(cell)^(c-)-(0.0591)/(2)log (1)/([Br^(c-)][H^(o+)])`
`=(0-1.08)-(0.591)/(2)log (1)/((0.01)^(2)(0.03)^(2))`
`=-1.08-(0.0591)/(2)log(1.111xx10^(7))`
`=-1.08-(0.0591)/(2)(7.0457)`
`=-1.08-0.208=-1.288V`
Thus, oxidation will occur at the hydrogen electrode and reduction on the `Br_(2)` electrode. `e_("cell")=1.288V`