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#
I can't find
1 Answer
Explanation:
Find
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,
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"#