Calculate `e.m.f.` of half cells given below :
(a) `{:(Pt_(2)),(4atm):}|{:(H_(2)SO_(4)),(0.02 M):} E_(OP)^(@) = 0 V`
(b) `{:Fe:}|{:(FeSO_(4)),(0.2 M):} E_(OP)^(@) = 0.44 V`
(c) `{:(Pt_(Cl_(2))),(10atm):}|{:(HCl),(0.02 M):} E_(OP)^(@) = -1.36 V`

Explanation

`H_(2) rarr 2H^(+)+2e`
`:. E_(OP) = E_(OP)^(@) - (0.059)/(2)"log"([H^(+)]^(2))/(P_(H_(2)))`
`[H^(+) = 0.02 xx 2M]`
`= 0 - (0.059)/(2)"log"((0.02 xx 2)^(2))/(4)`
`E_(OP_(H_(2//H^(+)))) = +0.100V`
(b) `Fe rarr Fe^(2+)+2e`
`:. E_(OP) = E_(OP)^(@) - (0.059)/(2)"log"[Fe^(2+)]`
`= 0.44 - (0.059)/(2)"log"[0.2]`
`E_(OP_(Fe//Fe^(2+))) = +0.4606 "volt"`
(c) `2Cl^(-) rarr Cl_(2)+2e`
`E_(OP)=E_(OP)^(@)-(0.059)/(2)"log"(P_(Cl_(2)))/([Cl^(-)]^(2)) = -1.36`
`- (0.059)/(2)"log"(10)/(0.02)^(2)`
`= E_(OP_(Cl^(-)//Cl_2)) = -1.49 "volt"`