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)`

Explanation

Parts (i) , (ii) and (iii) , (iv) `Pt (s) |Br_(2) (l) |Br^(-) (0.010M) | H^(+) (0.030M) |H_(2) (g)` (1 bar) |Pt (s)
`2 Br^(-) + 2 H^(+) to Br_(2) + H_(2)`
`E = E^(Theta) - (0.059)/(2) "log" (1)/([Br^(-)]^(2) [H^(+)]^(2))`
`E^(Theta) = 0 -1.08 = -1.08`
`therefore E = - 1.08 - (0.059)/(2) "log" (1)/((0.01)^(2) (0.030)^(2))`
`= -1.08 - (0.059)/(2) "log" (1.11 xx 10^(7))`
`= - 1.08 - (0.059)/(2) xx 7.0457`
`= - 1.08 - 0.208`
`- = 1.288 V`
The given reaction is not feasible . thus , oxidation will occur at the hydrogen electrode and reduction on the `Br_2` electrode and `E_(cell)^(Theta) = 1.288`V.