MediumChemistryHigh School (11–12)The d- and f-Block Elements › Lanthanoids and Actinoids (f-Block)Numerical

A 25.00 mL sample of an iron(II) solution is titrated with ceric ammonium sulfate, 0.0500 mol/L…

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A $25.00\ \text{mL}$ sample of an iron(II) solution is titrated with ceric ammonium sulfate, $0.0500\ \text{mol/L}$ . The reaction is $\text{Ce}^{4+} + \text{Fe}^{2+} \rightarrow \text{Ce}^{3+} + \text{Fe}^{3+}$ . The end point needs $21.40\ \text{mL}$ . Find the concentration of iron(II), the mass of iron in $250.0\ \text{mL}$ of the same solution (molar mass $55.85\ \text{g/mol}$ ) and the standard emf of the reaction. Use $E^\circ(\text{Ce}^{4+}/\text{Ce}^{3+}) = 1.74\ \text{V}$ and $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\ \text{V}$ .

Need a reminder? The key ideaShow

The 3+ ion of a lanthanoid is $[\text{Xe}]\,4f^{n}$ with $n = Z - 57$ . Unpaired electrons: $n$ up to 7, then $14 - n$ . Spin-only moment: $\mu = \sqrt{n(n+2)}\ \text{BM}$ . An empty, half-full or full 4f subshell ( $4f^0$ , $4f^7$ , $4f^{14}$ ) makes an ion unusually stable. That explains $\text{Ce}^{4+}$ , $\text{Tb}^{4+}$ , $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$ . For a cell, $\Delta G^\circ = -nFE^\circ$ with $F = 96.5\ \text{kJ/(V mol)}$ .

The worked solution

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