The relationship between ΔG and E_{cell} is:

$\overline{){\mathbf{\u2206}}{\mathbf{G}}{\mathbf{=}}{\mathbf{-}}{\mathbf{nFEcell}}}$

**Calculate E****°**_{cell}.

**Fe ^{3+}_{(aq)} + 3 e^{-} → Al_{(s)} E**

**Sn ^{2+}_{(aq)} + 2 e^{-} → Pb_{(s)} E**

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cathode}}}{\mathbf{-}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{anode}}}}\phantom{\rule{0ex}{0ex}}\mathbf{E}{\mathbf{\xb0}}_{\mathbf{cell}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{}\mathbf{V}\mathbf{}\mathbf{-}\left(\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{}\mathbf{V}\right)\phantom{\rule{0ex}{0ex}}{\mathbf{E}}_{\mathbf{cell}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{\hspace{0.17em}}\mathbf{V}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{}\mathbf{V}$

**E****° _{cell} = 0.1 V**

Calculate ΔG:

Use tabulated half-cell potentials to calculate ΔG°rxn for each of the following reactions at 25 °C.

2Fe^{3+}(aq) + 3Sn(s) → 2Fe(s) + 3Sn^{2+}(aq)

Express the energy change in kilojoules to two significant figures.

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