Question 20.5: Calculating the Quantity of Work from a Given Amount of Cell...
Calculating the Quantity of Work from a Given Amount of Cell Reactant
The emf of a particular voltaic cell with the cell reaction
\mathrm{Hg}_{2}^{2+}(a q)+\mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Hg}(l)+2 \mathrm{H}^{+}(a q)
is 0.650 \mathrm{~V}. Calculate the maximum electrical work of this cell when 0.500 \mathrm{~g} \mathrm{H}_{2} is consumed.
PROBLEM STRATEGY
The maximum work of a cell equals -n F E_{\text {cell }}. To get n, you split the cell reaction into half-reactions, multiplying each by a factor so that when added together they give the cell reaction. Then n equals the number of moles of electrons in either half-reaction.
The half-reactions are
\begin{gathered} \mathrm{Hg}_{2}{ }^{2+}(a q)+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Hg}(l) \\ \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \end{gathered}
n equals 2 , and the maximum work for the reaction as written is
\begin{aligned} w_{\text {max }}=-n F E_{\text {cell }} & =-2 \mathrm{~mol} \mathrm{e}^{-} \times 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} \times 0.650 \mathrm{~V} \\ & =-1.25 \times 10^{5} \mathrm{~V} \cdot \mathrm{C} \\ & =-1.25 \times 10^{5} \mathrm{~J} \end{aligned}
(Remember that a joule is equal to a volt-coulomb. Also, note that the negative sign means work is done by the cell-that is, energy is lost by the cell.) For 0.500 \mathrm{~g} \mathrm{H}_{2}, the maximum work is
0.500 \cancel{\mathrm{g H}_{2}} \times \frac{1 \cancel{\mathrm{mol H}_{2}}}{2.02 \cancel{\mathrm{g H}_{2}}} \times \frac{-1.25 \times 10^{5} \mathrm{~J}}{1 \cancel{\mathrm{mol H}_{2}}}=\mathbf{- 3 . 0 9} \times \mathbf{1 0}^4 \mathbf{J}
Note that the conversion factor -1.25 \times 10^{5} \mathrm{~J} / 1 \mathrm{~mol} \mathrm{H}_{2} is determined by the chemical equation as written. In the equation, 1 \mathrm{~mol} \mathrm{H}_{2} reacts, and the maximum work produced is -1.25 \times 10^{5} \mathrm{~J}.