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

## $P b \text{(s)"+PbO_2"(s)"+2H^+"(aq)"+2HSO_4^-"(aq)"\rightleftharpoons2PbSO_4"(s)"+2H_2O"(l)}$ I can't find $n$ in the formula. E_(cell)=E°_(cell)-0.0592/nV\logQ $Q = \frac{1}{4.5} ^ 4$

Jul 28, 2018

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

($\text{V}$ as in voltage)

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

$= 2.12 \textcolor{w h i t e}{l} \text{V}$