IChO 2017 · Theory problem 4

Electrochemistry

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Part A. Galvanic cell

The experiment is performed at 30.00°C. The electrochemical cell is composed of a hydrogen half-cell \([\ce{Pt(s)}\mid\ce{H2(g)}\mid\ce{H+(aq)}]\) containing a metal platinum electrode immersed in a buffer solution under a pressure of hydrogen gas. This hydrogen half-cell is connected to a half-cell of a metal (M) strip dipped in an unknown concentration of \(\ce{M^{2+}(aq)}\) solution. The two half-cells are connected via a salt bridge as shown in Figure 1.

Note: The standard reduction potentials are given in Table 1.

Figure 1

Figure 1 The galvanic cell

Table 1. Standard reduction potential (range 298-308 K)

Half-reaction \(E^\circ\) (V)-
\(\ce{Ba^{2+}(aq) + 2e- -> Ba(s)}\) -2.912
\(\ce{Sr^{2+}(aq) + 2e- -> Sr(s)}\) 2.899
\(\ce{Ca^{2+}(aq) + 2e- -> Ca(s)}\) -2.868
\(\ce{Er^{2+}(aq) + 2e- -> Er(s)}\) -2.000
\(\ce{Ti^{2+}(aq) + 2e- -> Ti(s)}\) -1.630
\(\ce{Mn^{2+}(aq) + 2e- -> Mn(s)}\) -1.185
\(\ce{V^{2+}(aq) + 2e- ->}\ ;\ce{V(s)}\) -1.175
\(\ce{Cr^{2+}(aq) + 2e- -> Cr(s)}\) -0.913
\(\ce{Fe^{2+}(aq) + 2e- -> Fe(s)}\) -0.447
\(\ce{Cd^{2+}(aq) + 2e- -> Cd(s)}\) -0.403
\(\ce{Co^{2+}(aq) + 2e- -> Co(s)}\) -0.280
\(\ce{Ni^{2+}(aq) + 2e- -> Ni(s)}\) -0.257
\(\ce{Sn^{2+}(aq) + 2e- -> Sn(s)}\) -0.138
\(\ce{Pb^{2+}(aq) + 2e- -> Pb(s)}\) -0.126
\(\ce{2H+(aq) + 2e- -> H2(g)}\) 0.000
\(\ce{Sn^{4+}(aq) + 2e- -> Sn^{2+}(aq)}\) +0.151
\(\ce{Cu^{2+}(aq) + e- -> Cu+(aq)}\) +0.153
\(\ce{Ge^{2+}(aq) + 2e- -> Ge(s)}\) +0.240
\(\ce{VO^{2+}(aq) + 2H+(aq) + e- -> V^{3+}(aq) + H2O(l)}\) +0.337
\(\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}\) +0.340
\(\ce{Tc^{2+}(aq) + 2e- -> Tc(s)}\) +0.400
\(\ce{Ru^{2+}(aq) + 2e- -> Ru(s)}\) +0.455
\(\ce{I2(s) + 2e- -> 2I-(aq)}\) +0.535
\(\ce{UO2^{2+}(aq) + 4H+(aq) + 2e- -> U^{4+}(aq) + 2H2O(l)}\) +0.612
\(\ce{PtCl4^{2-}(aq) + 2e- -> Pt(s) + 4Cl-(aq)}\) +0.755
\(\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}\) +0.770
\(\ce{Hg2^{2+}(aq) + 2e- -> 2Hg(l)}\) +0.797
\(\ce{Hg^{2+}(aq) + 2e- -> Hg(l)}\) +0.851
\(\ce{2Hg^{2+}(aq) + 2e- -> Hg2^{2+}(aq)}\) +0.920
\(\ce{Pt^{2+}(aq) + 2e- -> Pt(s)}\) +1.180
\(\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\) +1.224
\(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- -> 2Cr^{3+}(aq) + 7H2O(l)}\) +1.360
\(\ce{Co^{3+}(aq) + e- -> Co^{2+}(aq)}\) +1.920
\(\ce{S2O8^{2-}(aq) + 2e- -> 2SO4^{2-}(aq)}\) +2.010

4.1 If the reaction quotient (\(Q\)) of the whole galvanic cell is equal to \(2.18\times10^{-4}\) at 30.00 °C, the electromotive force is +0.450 V. Calculate the value of standard reduction potential (\(E^\circ\)) and identify the metal “M”.

Note: \(\Delta G=\Delta G^\circ+RT\ln Q\)

4.2 Write the balanced equation of the spontaneous redox reaction of the galvanic cell.

4.3 The unknown concentration of \(\ce{M^{2+}(aq)}\) solution in the cell (Figure 1) can be analyzed by a iodometric titration. A 25.00 \(\mathrm{cm^3}\) aliquot of \(\ce{M^{2+}(aq)}\) solution is added into a conical flask and an excess of \(\ce{KI}\) is added. A volume of 25.05 \(\mathrm{cm^3}\) of a sodium thiosulfate solution with a concentration of 0.800 \(\mathrm{mol\,dm^{-3}}\) is required to reach the equivalent point. Write all the redox reactions associated with this titration and calculate the concentration of \(\ce{M^{2+}(aq)}\) solution.

Note: If a student cannot find the answer, the student can use 0.950 \(\mathrm{mol\,dm^{-3}}\) as the concentration of \(\ce{M^{2+}}\) for further calculations.

4.4 In Figure 1, if the hydrogen half-cell is under 0.360 bar hydrogen gas and the platinum electrode is immersed in a 500 \(\mathrm{cm^3}\) buffer solution containing 0.050 mol lactic acid (\(\ce{HC3H5O3}\)) and 0.025 mol sodium lactate (\(\ce{C3H5O3Na}\)), the electromotive force of the galvanic cell measured is +0.534 V. Calculate the pH of the buffer solution and the dissociation constant (\(K_a\)) of lactic acid at 30.00 °C.

Note: If a student cannot find the answer, the student can use 3.46 as the buffer pH for further calculations.

Official English text of the problem.

Answers

  • 4.1: \(E^\circ=+0.340\ \mathrm V\); the metal M is copper, \(\ce{Cu}\).
  • 4.2: \(\ce{H2(g) + Cu^{2+}(aq) -> 2H+(aq) + Cu(s)}\).
  • 4.3: \(\boxed{[\ce{Cu^{2+}}]=0.802\ \mathrm{mol\,dm^{-3}}}\). The reactions are:

\[ \ce{2Cu^{2+}(aq) + 4I-(aq) -> 2CuI(s) + I2(aq)} \] \[ \ce{I2(aq) + 2S2O3^{2-}(aq) -> 2I-(aq) + S4O6^{2-}(aq)} \]

  • 4.4: \(\boxed{\mathrm{pH}=3.50}\) and \(\boxed{K_a=1.6\times10^{-4}}\).

Solution

Use \(T=303.15\ \mathrm K\), \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), and \(F=96485\ \mathrm{C\,mol^{-1}}\). Thus,

\[ \frac{RT}{2F}=0.0130610\ \mathrm V, \qquad \frac{RT\ln 10}{F}=0.0601483\ \mathrm V. \]

For the concentration calculations, activities are approximated by concentrations divided by \(c^\circ=1\ \mathrm{mol\,dm^{-3}}\); gas pressure is divided by \(p^\circ=1\ \mathrm{bar}\).

4.1

The electron-flow arrows show that the hydrogen electrode is the anode and the metal electrode is the cathode. Reduction of one metal ion requires two electrons:

\[ \ce{M^{2+}(aq) + 2e- -> M(s)}. \]

Therefore \(n=2\). Combining

\[ \Delta G=-nFE,\qquad \Delta G^\circ=-nFE^\circ \]

with the given relation gives the Nernst equation:

\[ E=E^\circ-\frac{RT}{nF}\ln Q. \]

Hence,

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E+\frac{RT}{2F}\ln Q\\ &=0.450\ \mathrm V +(0.0130610\ \mathrm V)\ln(2.18\times10^{-4})\\ &=0.339882\ \mathrm V\\ &\approx\boxed{+0.340\ \mathrm V}. \end{aligned} \]

Since the standard hydrogen reduction potential is zero,

\[ E^\circ_{\mathrm{cell}} =E^\circ_{\mathrm{metal}}-E^\circ_{\mathrm{hydrogen}} =E^\circ_{\mathrm{metal}}. \]

The matching entry is the reduction of copper(II) ions, \(\ce{Cu^{2+}}\), SMILES [Cu+2], to copper:

\[ \ce{Cu^{2+}(aq) + 2e- -> Cu(s)},\qquad E^\circ=+0.340\ \mathrm V. \]

Thus \(\boxed{\mathrm M=\ce{Cu}}\).

4.2

Hydrogen, \(\ce{H2}\), SMILES [H][H], is oxidized to hydrogen ions (protons), \(\ce{H+}\), SMILES [H+].

Anode:

\[ \ce{H2(g) -> 2H+(aq) + 2e-}. \]

Cathode:

\[ \ce{Cu^{2+}(aq) + 2e- -> Cu(s)}. \]

Adding the half-reactions gives

\[ \boxed{\ce{H2(g) + Cu^{2+}(aq) -> 2H+(aq) + Cu(s)}}. \]

Atoms and charge are balanced, and the positive cell potential confirms that this direction is spontaneous under the stated conditions. Platinum conducts electrons and catalyzes the hydrogen electrode reaction.

4.3

Iodide ions, \(\ce{I-}\), SMILES [I-], reduce copper(II) to copper(I), \(\ce{Cu+}\), SMILES [Cu+]. Copper(I) precipitates as sparingly soluble copper(I) iodide, which drives the reaction:

\[ \boxed{\ce{2Cu^{2+}(aq) + 4I-(aq) -> 2CuI(s) + I2(aq)}}. \]

Copper(I) iodide can be represented compositionally as [Cu+].[I-]; this represents an ionic formula unit, not the solid’s crystal lattice.

Each copper(II) ion accepts one electron. Two iodide ions lose two electrons to form one iodine molecule, while two further iodide ions enter the precipitate.

The liberated iodine, \(\ce{I2}\), SMILES II, is titrated with thiosulfate, \(\ce{S2O3^{2-}}\), SMILES [O-]S(=O)(=O)[S-]. Thiosulfate is oxidized to tetrathionate, \(\ce{S4O6^{2-}}\), SMILES [O-]S(=O)(=O)SSS(=O)(=O)[O-]:

\[ \boxed{\ce{I2(aq) + 2S2O3^{2-}(aq) -> 2I-(aq) + S4O6^{2-}(aq)}}. \]

In excess iodide, iodine forms triiodide, \(\ce{I3-}\), SMILES I[I-]I:

\[ \ce{I2(aq) + I-(aq) <=> I3-(aq)}. \]

This is a complex-formation equilibrium, not an additional redox reaction. Triiodide is linear; the SMILES does not specify that geometry. Equivalently, the two redox steps can be written as

\[ \ce{2Cu^{2+}(aq) + 5I-(aq) -> 2CuI(s) + I3-(aq)} \]

and

\[ \ce{I3-(aq) + 2S2O3^{2-}(aq) -> 3I-(aq) + S4O6^{2-}(aq)}. \]

Both representations give the same stoichiometry:

\[ n(\ce{Cu^{2+}}) =2n(\text{liberated iodine}) =n(\ce{S2O3^{2-}}). \]

The amount of thiosulfate used is

\[ n(\ce{S2O3^{2-}}) =(0.800\ \mathrm{mol\,dm^{-3}}) (25.05\times10^{-3}\ \mathrm{dm^3}) =0.02004\ \mathrm{mol}. \]

Therefore,

\[ [\ce{Cu^{2+}}] =\frac{0.02004\ \mathrm{mol}} {25.00\times10^{-3}\ \mathrm{dm^3}} =0.8016\ \mathrm{mol\,dm^{-3}}, \]

or, to three significant figures,

\[ \boxed{[\ce{Cu^{2+}}]=0.802\ \mathrm{mol\,dm^{-3}}}. \]

4.4

For the overall reaction in 4.2, the activity of solid copper is unity, so

\[ Q=\frac{a_{\ce{H+}}^2} {a_{\ce{Cu^{2+}}}(p_{\ce{H2}}/p^\circ)}. \]

Using the tabulated copper potential, \(E^\circ=0.340\ \mathrm V\), and retaining the unrounded concentration from 4.3,

\[ 0.534 =0.340 -0.0130610\ln \left(\frac{a_{\ce{H+}}^2}{0.8016(0.360)}\right). \]

Since \(\ln a_{\ce{H+}}=-(\ln 10)\mathrm{pH}\),

\[ E =E^\circ+\frac{RT}{2F} \ln\!\left[a_{\ce{Cu^{2+}}}(p_{\ce{H2}}/p^\circ)\right] +\frac{RT\ln 10}{F}\mathrm{pH}. \]

Thus,

\[ \begin{aligned} \mathrm{pH} &=\frac{0.534-0.340 -0.0130610\ln(0.8016\times0.360)} {0.0601483}\\ &=3.49523. \end{aligned} \]

Therefore,

\[ \boxed{\mathrm{pH}=3.50}, \qquad [\ce{H+}]\approx3.1972\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

Lactic acid, \(\ce{HC3H5O3}\), is 2-hydroxypropanoic acid, SMILES CC(O)C(=O)O. Its conjugate base is lactate, \(\ce{C3H5O3-}\), SMILES CC(O)C(=O)[O-]. Sodium lactate has the component SMILES CC(O)C(=O)[O-].[Na+]. The problem specifies no stereoisomer, so no stereochemical tags are assigned.

The acid dissociation is

\[ \ce{HC3H5O3(aq) <=> H+(aq) + C3H5O3-(aq)}. \]

The formal concentrations in \(0.500\ \mathrm{dm^3}\) are

\[ C_{\mathrm{acid}} =\frac{0.050}{0.500} =0.100\ \mathrm{mol\,dm^{-3}}, \]

\[ C_{\mathrm{lactate}} =\frac{0.025}{0.500} =0.050\ \mathrm{mol\,dm^{-3}}. \]

Using the usual buffer approximation,

\[ \begin{aligned} K_a &\approx \frac{[\ce{H+}]}{c^\circ} \frac{[\ce{C3H5O3-}]}{[\ce{HC3H5O3}]}\\ &=(3.1972\times10^{-4})\frac{0.050}{0.100}\\ &=1.5986\times10^{-4}. \end{aligned} \]

Retaining the small additional dissociation of lactic acid, with \(h=[\ce{H+}]\) and negligible water autoionization, gives

\[ K_a\approx \frac{h}{c^\circ} \frac{0.050\ \mathrm{mol\,dm^{-3}}+h} {0.100\ \mathrm{mol\,dm^{-3}}-h} =1.6140\times10^{-4}. \]

Both treatments give, at the precision of the buffer amounts,

\[ \boxed{K_a=1.6\times10^{-4}}. \]

Here \(K_a\) is dimensionless, using concentrations normalized by \(c^\circ\).

The solution was checked against the official solution.