Thallium(III) can oxidise tin(II) to tin(IV) in a two-electron reaction: Tl³⁺ + Sn²⁺ → Tl⁺ + Sn⁴⁺…

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Thallium(III) can oxidise tin(II) to tin(IV) in a two-electron reaction: $\text{Tl}^{3+} + \text{Sn}^{2+} \to \text{Tl}^{+} + \text{Sn}^{4+}$ . The standard potential is $+1.25\ \text{V}$ for Tl $^{3+}$ changing to Tl $^+$ , and $+0.15\ \text{V}$ for Sn $^{4+}$ changing to Sn $^{2+}$ . Find the cell potential, the energy that the reaction releases for each mole in kJ, and $\log K$ at 298 K.

Need a reminder? The key ideaShow

For a redox reaction, $E^\circ_{cell} = E^\circ_{ox} - E^\circ_{red}$ , $\Delta G^\circ = -nFE^\circ_{cell}$ with $F = 96.485\ \text{kJ/(V mol)}$ , and at 298 K, $\log K = \dfrac{nE^\circ_{cell}}{0.05916\ \text{V}}$ . A reaction goes if $E^\circ_{cell}$ is positive. Moles of reactant come from mass divided by molar mass, and the equation gives the ratios.

The worked solution

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