The `EMF` of the following cell is found to be `-0.46V:`

If the standard electrode potential of Zince Is `-0.763V`, find the value of `K_(2)` for `H_(2)SO_(4)`, where `K_(2)=([H^(o+)][SO_(4)^(2-)])/([HSO_(4)^(c-)])`

Explanation

At anode `:` `H_(2)hArr2H^(o+)+2e^(-)`
At cathode `:` `Zn^(2+)+2e^(-)hArrZn` `" "E=-0.46` volt
`[` negative value shows cell is not feasible `]`
`overlineunderline("Cell reaction:" H_(2)+Zn^(2+)hArrZn+2H^(o+))`
`E=E^(c-)-(0.059)/(2)log.([H^(o+)]^(2))/([Zn^(2+)])`
`=-0.46=-0.763-(0.059)/(2)log.([H^(o+)]^(2))/(0.3)`
Solve, `[H^(o+)]=4xx10^(-6)`
`HSO_(4)^(c-) rarr H^(o+)+SO^(4)^(2-)`
`K_(2)=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(c-)])=((4xx10^(-6))(6.44xx10^(-3)))/((0.4))`
`=6.44xx10^(-8)`