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#

1 Answer

Answer:

#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"#