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The voltage of the cell Pb(s)|PbSO(4)(...

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`.

Text Solution

Verified by Experts

The correct Answer is:
`1.06 xx 10^(-2)`
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