Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at `25^(@)C`
`H_(2) (1 atm)|0.5 M HCOOH|| 1M CH_(3)COOH | (1 atm) H_(2)`
`K_(a) " for " HCOOH and CH_(3) COOH` are `1.77 xx 10^(-4) and 1.8 xx 10^(-5)` respectively.

Explanation

`[H^(+)] " in " HCOOH = sqrt(C xx K_(a)) = sqrt(0.5 xx 1.77 xx 10^(-4))`
`[H^(+)]` in `CH_(3)COOH = sqrt(C xx K_(a)) = sqrt(1 xx 1.8 xx 10^(-5))`
`= 4.2426 xx 10^(-3)M`
`E_("cell") = 0.059 log. ([H^(+)]_(RHS))/([H^(+)]_(LHS)) = 0.059 log. (4.2426 xx 10^(-3))/(0.9407 xx 10^(-2))`
`= -0.0204` volt