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[" Q.17The voltage of the cell "],[qquad...

[" Q.17The voltage of the cell "],[qquad [quad Pb(s)|PbSO_(4)(s)|NaIISO_(4)(0.600M)|Pb^(2+)(2.50times10^(-5)M)|Pb(s)],[" is "E-+0.061V" .Calculate "K_(2)=[H^(+)][SO_(4)^(2-)]/[HSO_(4)]," the dissociation constant for HSO "_(4)^(-)" ."],[" Given : "Pb(s)+SO_(4)^(2-)(aq)|PbSO_(4)(s)+2e^(-)(E^(0)=0.356)quad E^(0)(Pb^(2+/Pb)=-0.126V)]]

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The voltage of the cell Pb(s)|PbSO_(4)(s)|NaHSO_(4)(0.600M)||Pb^(2+)(2.50 xx 10^(-5)M)|Pb(s) is E = +0.061V . Calculate K_(2) = [H^(+)] [SO_(4)^(2-)]//[HSO_(4)^(-)] the dissociation constant for HSO_(4)^(-) . Given Pb(s) +SO_(4)^(2-) rarr PbSO_(4) +2e^(-) (E^(@) = 0.356V),E^(@) (Pb^(2+)//Pb) =- 0.126V .

The voltage of the cell Pb(s)|PbSO_(4)(s)|NaHSO_(4)(0.600M)||Pb^(2+)(2.50 xx 10^(-5)M)|Pb(s) is E = +0.061V . Calculate K_(2) = [H^(+)] [SO_(4)^(2-)]//[HSO_(4)^(-)] the dissociation constant for HSO_(4)^(-) . Given Pb(s) +SO_(4)^(2-) rarr PbSO_(4) +2e^(-) (E^(@) = 0.356V),E^(@) (Pb^(2+)//Pb) =- 0.126V .

During the working of the following cell Pb-Hg(1M)|Pb^(2+)(aq)(0.1M)| |Pb^(2+)(aq)(0.1M)|Pb-Hg(0.5M)

Consider the following cell Pb_((s))|Pb_((aq))^(2+)(0.5M)||Cu_((aq)^(+2)(0.001M)|Cu E^@Pb= -0.126V, E°Cu = +0.337V: Calculate E_(cell) and E_(cell)^@ at 25^@C .

Pb(NO_(3))_(2)+H_(2)SO_(4) to PbSO_(4)darr+2HNO_(3)