Question

Calculate `E_("cell")` for this reaction when [`H^+`] = 4.5 M and [`HSO_4^-`] = 4.5 M, if `E°_("cell")` = 2.04 V?

`Pb"(s)"+PbO_2"(s)"+2H^+"(aq)"+2HSO_4^-"(aq)"\rightleftharpoons2PbSO_4"(s)"+2H_2O"(l)"`

I can't find `n` in the formula.
`E_(cell)=E°_(cell)-0.0592/nV\logQ`

`Q=1/(4.5)^4`

Answer 1

`E_("cell") = 2.12 color(white)(l) "V"`

Explanation:

`n` stands for the number of electrons transferred per mole of atom in the reaction.

Find `n` by writing the half reactions going on in each cell at each electrode:

Anode/Oxidation:

`stackrel(0)("Pb")(s)to stackrel(+2color(white)(D-))("Pb"^(2+))(aq) + color(blue)(2)color(white)(l) "e"^(-)`

Cathode/Reduction:

`stackrel(+4)("Pb")"O"_2(s) +2color(white)(l) "H"^(+) (aq) + color(blue)(2)color(white)(l) "e"^(-) to stackrel(+2color(white)(D-))("Pb"^(2+))(aq) + "H"_2"O"`

Thus each mole of the reaction involves the transfer of two moles of electrons, `n = 2`. The lead (II) ion would then combine with sulfate ions to form the lead (II) sulfate precipitate.

The Nernst equation:

`E_(cell)=E°_(cell)-(0.0592 color(white)(l) "V")/n\logQ`

(`"V"` as in voltage)

`E_("cell") = "2.04 V" - ("0.0592 V"/("2 mol e"^(-)"/mol atoms")) * log(1/((4.5)^2(4.5)^2))`

`= 2.12 color(white)(l) "V"`