Consider the following half reactions `:`
`PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)9s)+2H_(2)O" "E^(c-)=+1.70V`
`PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V`
`a.` Calculate the value of `E^(c-)` for the cell.
`b.` Calculate the voltage generated by the cell if `[H^(o+)]=0.10M` and `[SO_(4)^(2-)]=2.0M`
`c.` What voltage is generated by the cell when it is at chemical equilibrium ?

Explanation

`a. E^(c-)._(cell)=E^(c-)._(Cathode)-E^(c-)._(Anode )`
`=1.70-(-0.31)=2.01V`
`PbO_(2)+4H^(o+)+SO_(4)^(2-)+2e^(-)rarr PbSO_(4)+2H_(2)O`
`Pb+SO_(4)^(2-)rarr PbSO_(4)`
`ulbar(PbO_(2)+Pb+4H^(o+)+2SO_(4)^(2-)rarr 2PbSO_(4)+2H_(2)O)`
`b.` `E_(cell)=E^(c-)._(cell)-(0.059)/(2)log.(1)/([H^(o+)][SO_(4)^(2-)]^(2))`
`=2.01-(0.059)/(2)log.(1)/((0.14)^(4)(2.0)^(2))`
`c.` When cell is at equilibrium , `E_(cell)=0`