Calculating the Quantity of Work from a Given Amount of Cell Reactant The emf of a particular voltaic cell with the cell reaction Hg2^2+(aq) + H2(g) ⇌ 2Hg(l) + 2H^+(aq) is 0.650 V. Calculate the maximum electrical work of this cell when 0.500 g H2 is consumed. PROBLEM STRATEGY The maximum work of a

Question

Calculating the Quantity of Work from a Given Amount of Cell Reactant

The emf of a particular voltaic cell with the cell reaction

[latex]\mathrm{Hg}_{2}^{2+}(a q)+\mathrm{H}_{2}(g) ightleftharpoons 2 \mathrm{Hg}(l)+2 \mathrm{H}^{+}(a q)[/latex]

is [latex]0.650 \mathrm{~V}[/latex]. Calculate the maximum electrical work of this cell when [latex]0.500 \mathrm{~g} \mathrm{H}_{2}[/latex] is consumed.

PROBLEM STRATEGY

The maximum work of a cell equals [latex]-n F E_{ ext {cell }}[/latex]. To get [latex]n[/latex], you split the cell reaction into half-reactions, multiplying each by a factor so that when added together they give the cell reaction. Then [latex]n[/latex] equals the number of moles of electrons in either half-reaction.

Accepted Answer

The half-reactions are

[latex] \begin{gathered} \mathrm{Hg}_{2}{ }^{2+}(a q)+2 \mathrm{e}^{-} ightleftharpoons 2 \mathrm{Hg}(l) \ \mathrm{H}_{2}(g) ightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \end{gathered} [/latex]

[latex]n[/latex] equals 2 , and the maximum work for the reaction as written is

[latex] \begin{aligned} w_{ ext {max }}=-n F E_{ ext {cell }} & =-2 \mathrm{~mol} \mathrm{e}^{-} imes 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} imes 0.650 \mathrm{~V} \ & =-1.25 imes 10^{5} \mathrm{~V} \cdot \mathrm{C} \ & =-1.25 imes 10^{5} \mathrm{~J} \end{aligned} [/latex]

(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 [latex]0.500 \mathrm{~g} \mathrm{H}_{2}[/latex], the maximum work is

[latex] 0.500 \cancel{\mathrm{g H}_{2}} imes \frac{1 \cancel{\mathrm{mol H}_{2}}}{2.02 \cancel{\mathrm{g H}_{2}}} imes \frac{-1.25 imes 10^{5} \mathrm{~J}}{1 \cancel{\mathrm{mol H}_{2}}}=\mathbf{- 3 . 0 9} imes \mathbf{1 0}^4 \mathbf{J} [/latex]

Note that the conversion factor [latex]-1.25 imes 10^{5} \mathrm{~J} / 1 \mathrm{~mol} \mathrm{H}_{2}[/latex] is determined by the chemical equation as written. In the equation, [latex]1 \mathrm{~mol} \mathrm{H}_{2}[/latex] reacts, and the maximum work produced is [latex]-1.25 imes 10^{5} \mathrm{~J}[/latex].

Question 20.5: Calculating the Quantity of Work from a Given Amount of Cell...

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