Question

Construct a coulometric titration curve of 100.0 mL of a 1 M H₂SO₄ solution containing Fe(II) titrated with Ce(IV) generated from 0.075 M Ce(III). The titration is monitored by potentiometry. The initial amount of Fe(II) present is 0.05182 mmol. A constant current of 20.0 mA is used. Find the time corresponding to the equivalence point. Then, for about ten values of time before the equivalence point, use the stoichiometry of the reaction to calculate the amount of Fe³⁺ produced and the amount of Fe²⁺remaining. Use the Nernst equation to find the system potential. Find the equivalence point potential in the usual manner for a redox titration. For about ten times after the equivalence point, calculate the amount of Ce⁴⁺ produced from the electrolysis and the amount of Ce³⁺ remaining. Plot the curve of system potential versus electrolysis time.

Short answer

A coulometric titration curve needs to be constructed containing 100.0 mL of a 1 M H₂SO₄ solution in which Fe(II) is titrated against Ce(IV) which is generated from 0.075 M Ce(III).

Step-by-step solution

Given Information

A coulometric titration curve needs to be constructed containing 100.0 mL of a 1 M H₂SO₄ solution in which Fe(II) is titrated against Ce(IV) which is generated from 0.075 M Ce(III).

Explanation

The electrode potential and reactions are represented as follows:

$$\begin{gathered} {\mathrm{Ce}}^{4+}+\mathrm{e}\to {\mathrm{Ce}}^{3+}{\mathrm{E}}^0=1.44\mathrm{V} \\ {\mathrm{Fe}}^{3+}+\mathrm{e}\to {\mathrm{Fe}}^{2+}{\mathrm{E}}^0=0.68\mathrm{V} \end{gathered}$$

Overall: ${\mathrm{Ce}}^{4+}+{\mathrm{Fe}}^{2+}\to {\mathrm{Ce}}^{3+}+{\mathrm{Fe}}^{3+}$

Equivalence point volume = $0.05182\times {10}^{-3}\mathrm{mol}\times \frac{1\mathrm{mol}}{1\mathrm{mol}}\times \frac{1000\mathrm{mL}}{0.075\mathrm{mol}}=0.691\mathrm{mL}$

E_system before the equivalence point can be calculated by applying Nernst equation to Fe²⁺/Fe³⁺ half-cell reaction

$$\begin{gathered} {\mathrm{E}}_{\mathrm{system}}={\mathrm{E}}^0-\frac{0.0592}{\mathrm{n}}\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]} \\ \left[{\mathrm{Fe}}^{2+}\right]=\frac{\mathrm{initial}\mathrm{moles}\mathrm{of}{\mathrm{Fe}}^{2+}-\mathrm{moles}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}}{100.0\mathrm{mL}+\mathrm{Volume}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}} \\ \left[{\mathrm{Fe}}^{3+}\right]=\frac{\mathrm{moles}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}}{100.0\mathrm{mL}+\mathrm{Volume}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}} \end{gathered}$$

At the equivalence point,$\left[{\mathrm{Fe}}^{2+}\right]=\left[{\mathrm{Ce}}^{4+}\right]\mathrm{and}\left[{\mathrm{Fe}}^{3+}\right]=\left[{\mathrm{Ce}}^{3+}\right]$.

${\mathrm{E}}_{\mathrm{system}}=\frac{{{\mathrm{E}}^0}_{\mathrm{Fe}}+{{\mathrm{E}}^0}_{\mathrm{Cr}}}{2}$

After the equivalence point Esystem can be determined by applying Nernst equation to Cerium half-cell reaction.

${\mathrm{E}}_{\mathrm{system}}={\mathrm{E}}^0-\frac{0.0592}{\mathrm{n}}\log \frac{\left[{\mathrm{Ce}}^{2+}\right]}{\left[{\mathrm{Ce}}^{4+}\right]}$

$$\begin{gathered} \left[{\mathrm{Ce}}^{4+}\right]=\frac{\mathrm{initial}\mathrm{moles}\mathrm{of}{\mathrm{Ce}}^{4+}-\mathrm{moles}\mathrm{of}{\mathrm{Fe}}^{2+}\mathrm{added}}{100.0\mathrm{mL}+\mathrm{Volume}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}} \\ \left[{\mathrm{Ce}}^{3+}\right]=\frac{\mathrm{initial}\mathrm{moles}\mathrm{of}{\mathrm{Fe}}^{2+}\mathrm{added}}{100.0\mathrm{mL}+\mathrm{Volume}\mathrm{of}{\mathrm{Ce}}^{4+}\mathrm{added}} \end{gathered}$$

Electrolysis time at each point can be determined by

$t=\frac{n_{C{e}^{4+}}nF}{I}$

nCe⁴⁺ - moles of Ce⁴⁺

n – number of moles of electrons

F – faraday constant

I – current

I – current

Time corresponding to the equivalence point

$$\begin{gathered} \mathrm{t}=\frac{{\displaystyle \frac{0.075\mathrm{mol}}{1000\mathrm{mol}}}\times 0.691\mathrm{mL}\times 1\times 96485\mathrm{C}/\mathrm{mol}}{0.02\mathrm{A}} \\ \mathrm{t}=250.02\mathrm{s} \end{gathered}$$

t, s	It	[Fe3+]	[Fe2+]	[Ce3+]	[Ce4+]	Esystem
100	2	0.020729	0.031093			0.669575
150	3	0.031093	0.020729			0.690424
200	4	0.041457	0.010365			0.715641
220	4.4	0.045603	0.006219			0.731224
230	4.6	0.047676	0.004146			0.74279
240	4.8	0.049749	0.002073			0.761703
249	4.98	0.051614	0.000208			0.821796
249.5	4.99	0.051718	0.000104			0.83961
249.6	4.992	0.051739	8.34E-05			0.845329
249.7	4.994	0.051759	6.27E-05			0.852687
249.8	4.996	0.05178	4.19E-05			0.863025
249.9	4.998	0.051801	2.12E-05			0.880567
250	5					1.06
250.1	5.002			0.023158	0.051842	1.460719
250.2	5.004			0.023137	0.051863	1.460753
250.3	5.006			0.023116	0.051884	1.460786
250.4	5.008			0.023096	0.051904	1.460819
250.5	5.01			0.023075	0.051925	1.460853
260	5.2			0.021106	0.053894	1.464103
270	5.4			0.019033	0.055967	1.467731
280	5.6			0.01696	0.05804	1.471631
290	5.8			0.014887	0.060113	1.475885
300	6			0.012814	0.062186	1.480611