IChO 1976 · Theory problem 7

Lead accumulator

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The density of a sulphuric acid solution in a charged lead accumulator should be equal to \(\rho = 1.28\ \mathrm{g\,cm^{-3}}\) which corresponds to the solution containing 36.87 % of \(\ce{H2SO4}\) by mass. In a discharged lead accumulator it should not decrease under the value of \(\rho = 1.10\ \mathrm{g\,cm^{-3}}\) which corresponds to the 14.35 % solution of sulphuric acid.

(Faraday's constant \(F\) is equal to \(26.8\ \mathrm{Ah\,mol^{-1}}\).)

Problems:

7.1 Write the equation for a total electrochemical reaction that takes place in the lead accumulator when it is charged and discharged.

7.2 Calculate the masses of \(\ce{H2O}\) and \(\ce{H2SO4}\) being consumed or formed according to the equation in No 1.

7.3 Calculate the mass of \(\ce{H2SO4}\) that is required to be added to a led accumulator with a capacity of 120 Ah if the content of \(\ce{H2SO4}\) is to be in the range as given in the task.

7.4 Calculate the difference in volumes of the sulphuric acid solutions in a charged and a discharged lead accumulator with a capacity of 120 Ah.

Official English text of the problem.

Answers

  • 7.1 Discharge: \[ \ce{Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O}. \] Charging proceeds by the reverse reaction.

  • 7.2 Per reaction as written, corresponding to \(53.6\ \mathrm{Ah}\), discharge consumes \(196\ \mathrm{g}\) of \(\ce{H2SO4}\) and forms \(36.0\ \mathrm{g}\) of \(\ce{H2O}\). Equivalently, these amounts are \(3.66\) and \(0.672\ \mathrm{g\,Ah^{-1}}\), respectively. For \(120\ \mathrm{Ah}\), they are \(439\ \mathrm{g}\) and \(80.6\ \mathrm{g}\). Charging reverses consumption and formation.

  • 7.3 The minimum initial quantity of pure \(\ce{H2SO4}\) is \[ \boxed{m(\ce{H2SO4})=634\ \mathrm{g}}, \] contained in approximately \(1.72\ \mathrm{kg}\) of charged electrolyte.

  • 7.4 \[ \boxed{V_{\mathrm{charged}}-V_{\mathrm{discharged}} =1.06\times10^2\ \mathrm{cm^3}=0.106\ \mathrm{L}}. \] The electrolyte volume decreases on discharge.

Solution

7.1

During discharge, metallic lead is oxidized from oxidation state \(0\) to \(+\mathrm{II}\), while lead in lead(IV) oxide is reduced from \(+\mathrm{IV}\) to \(+\mathrm{II}\). Lead(II) sulfate forms at both electrodes.

The discharge half-reactions are:

\[ \begin{aligned} \text{Negative electrode:}\quad& \ce{Pb(s) + SO4^{2-}(aq) -> PbSO4(s) + 2e-},\\ \text{Positive electrode:}\quad& \ce{PbO2(s) + 4H+(aq) + SO4^{2-}(aq) + 2e- -> PbSO4(s) + 2H2O(l)}. \end{aligned} \]

Adding them gives the overall discharge reaction:

\[ \boxed{\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)}}. \]

During charging, an external source drives the reverse reaction:

\[ \boxed{\ce{2PbSO4(s) + 2H2O(l) -> Pb(s) + PbO2(s) + 2H2SO4(aq)}}. \]

The molecular and ionic species used here are sulphuric acid, OS(=O)(=O)O; water, O; sulfate ion, [O-]S(=O)(=O)[O-]; and hydrogen ion, [H+]. The aqueous hydrogen ion denotes a solvated proton. Lead, lead(IV) oxide, and lead(II) sulfate are extended solid phases; molecular SMILES do not describe their crystal lattices.

7.2

Use the molar masses

\[ M(\ce{H2SO4})=98.0\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}. \]

For the overall reaction as written, two moles of electrons pass through the external circuit. The corresponding charge is

\[ Q_0=(2\ \mathrm{mol})(26.8\ \mathrm{Ah\,mol^{-1}}) =53.6\ \mathrm{Ah}. \]

At this reaction extent, discharge consumes

\[ m(\ce{H2SO4})=(2\ \mathrm{mol})(98.0\ \mathrm{g\,mol^{-1}}) =196\ \mathrm{g} \]

and forms

\[ m(\ce{H2O})=(2\ \mathrm{mol})(18.0\ \mathrm{g\,mol^{-1}}) =36.0\ \mathrm{g}. \]

Thus one mole of acid is consumed and one mole of water is formed per mole of electrons transferred. For a charge \(Q\),

\[ m_{\text{acid consumed}}=\frac{Q}{F}M(\ce{H2SO4}), \qquad m_{\text{water formed}}=\frac{Q}{F}M(\ce{H2O}). \]

Per ampere-hour, these amounts are

\[ \frac{98.0}{26.8}=3.66\ \mathrm{g\,Ah^{-1}}, \qquad \frac{18.0}{26.8}=0.672\ \mathrm{g\,Ah^{-1}}. \]

For the \(120\ \mathrm{Ah}\) discharge considered in the following parts,

\[ n(e^-)=\frac{120\ \mathrm{Ah}}{26.8\ \mathrm{Ah\,mol^{-1}}} =4.47761\ \mathrm{mol}. \]

Consequently,

\[ \begin{aligned} a&\equiv m_{\text{acid consumed}} =4.47761(98.0)=438.806\ \mathrm{g},\\ b&\equiv m_{\text{water formed}} =4.47761(18.0)=80.5970\ \mathrm{g}. \end{aligned} \]

Charging consumes the same mass of water and regenerates the same mass of acid.

7.3

Let \(m_c\) be the mass of electrolyte in the charged accumulator. Its initial acid mass is

\[ m_{\mathrm{acid},c}=0.3687m_c. \]

After discharge, the acid mass is \(0.3687m_c-a\). Since acid is consumed and water is formed, the total electrolyte mass becomes

\[ m_d=m_c-a+b=m_c-358.209\ \mathrm{g}. \]

The minimum initial quantity is obtained when the \(120\ \mathrm{Ah}\) discharge reaches exactly the permitted lower acid concentration:

\[ \frac{0.3687m_c-a}{m_c-a+b}=0.1435. \]

Rearranging,

\[ m_c= \frac{a-0.1435(a-b)}{0.3687-0.1435} = \frac{438.806-0.1435(358.209)}{0.2252} =1720.26\ \mathrm{g}. \]

The required initial mass of pure sulphuric acid is therefore

\[ m_{\mathrm{acid},c} =0.3687(1720.26) =634.261\ \mathrm{g}, \]

or

\[ \boxed{m_{\mathrm{acid},c}=634\ \mathrm{g}}. \]

As a check, after discharge the electrolyte contains

\[ 634.261-438.806=195.455\ \mathrm{g} \]

of acid in \(1362.05\ \mathrm{g}\) of solution, giving

\[ \frac{195.455}{1362.05}\times100\%=14.35\%. \]

7.4

Using the given densities and retaining extra digits until the final subtraction,

\[ V_c=\frac{m_c}{\rho_c} =\frac{1720.26\ \mathrm{g}}{1.28\ \mathrm{g\,cm^{-3}}} =1343.95\ \mathrm{cm^3}. \]

The discharged electrolyte mass is

\[ m_d=1720.26-438.806+80.5970 =1362.05\ \mathrm{g}, \]

so its volume is

\[ V_d=\frac{m_d}{\rho_d} =\frac{1362.05\ \mathrm{g}}{1.10\ \mathrm{g\,cm^{-3}}} =1238.23\ \mathrm{cm^3}. \]

Hence

\[ \boxed{ V_c-V_d=105.725\ \mathrm{cm^3} \approx1.06\times10^2\ \mathrm{cm^3} }. \]

The solution was checked against the official solution.