IChO 2022 · Theory problem 4

A New Journey for Ancient Sulfur

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Figure 1

Sulfur has been known and used since the ancient times. Nowadays, sulfur is widely used in chemical production and pharmaceutical industry as an inexpensive chemical raw material.

Pyrite (\(\ce{FeS2}\)) is commonly used as a raw material in the industrial production of elemental sulfur. By heating pyrite in the presence of limited air supply, the theoretical yield of elemental sulfur is 100% and a black magnetic oxide (\(\ce{Fe3O4}\)) is the other product.

4.1 Write the balanced reaction equation that describes the above conversion.

In addition to sulfur, a small amount of \(\ce{SO2}\) is generated as a by-product during the actual process. This reaction can be monitored by measuring the amount of \(\ce{SO2}\). The following recipe is used for the monitoring:

Raw ore powder is heated in a temperature-controlled tube furnace. The generated \(\ce{SO2}\) is absorbed by 2 mol L\(^{-1}\) \(\ce{NaOH}\) solution. When the reaction is completed, the solution is transferred to a 500 mL volumetric flask and diluted by distilled water to the mark. 25.00 mL of this diluted solution is then added into an iodine flask containing 50.00 mL 0.05122 mol L\(^{-1}\) \(\ce{I2}\) standard solution and 5 mL 20% \(\ce{H2SO4}\) solution. After maintaining the iodine flask in dark for 5 min, the solution is titrated with 0.1012 mol L\(^{-1}\) \(\ce{Na2S2O3}\) standard solution. When the color of the solution turns light brown, 3 mL 0.5% starch indicator is added. Continue the titration until the blue color disappears.

4.2 Write the balanced reaction equations involving \(\ce{I2}\) in the above measurement.

4.3 In such a test experiment starting from pyrite, 17.6 g of elemental sulfur was collected. The analysis of the by-product gases according to the above procedure resulted in the consumption of 18.47 mL of \(\ce{Na2S2O3}\) standard solution. Assuming that no other sulfur containing species were produced, calculate what percentage of the sulfur in pyrite was lost as the by-product?

The lithium-sulfur battery is a compelling energy storage system because its high theoretical energy density exceeds conventional Li-ion batteries. The net reaction of a lithium-sulfur battery can be simplified as: \(\ce{16Li + S8 -> 8Li2S}\). Sulfur is the cathode and metallic lithium is the anode active material during discharge.

4.4 Write the balanced equations for reactions that take place at cathode (a) and anode (b) during discharge.

4.5 Calculate the mass ratio of the cathode active material to the anode active material, according to the net battery reaction.

A lithium-ion battery (LIB) that possesses an average operating voltage of 3.8 V and a deliverable capacity of 3110 mAh can provide energy for a mobile phone to play videos continuously for 22 hours upon a full charge.

4.6 If the LIB is replaced by an ideal lithium-sulfur battery pack, which has an average operating voltage of 4.2 V and contains 23 g sulfur as the active electrode material that can react stoichiometrically during discharge, calculate how many hours the new battery pack will provide energy for the phone to play videos continuously after a full charge.

Elemental sulfur generally exists as \(\ce{S8}\) molecules. In real lithium-sulfur batteries, \(\ce{S8}\) is not directly reduced to \(\ce{Li2S}\) during discharge, but it undergoes stepwise reactions generating different soluble lithium polysulfides (\(\ce{Li2S_n}\), \(n=3–8\)). These lithium polysulfides could diffuse to the anode and corrode it, which results in the loss of active electrode materials. This phenomenon is called "shuttle effect".

4.7 Write the balanced reaction equation for the corrosion of anode by soluble lithium polysulfides (\(\ce{Li2S_n}\)) that produces \(\ce{Li2S}\).

In order to suppress the "shuttle effect", the forms of polysulfides in an electrolyte have been extensively studied. \(\ce{Li2S6}\) is one of the most representative intermediate products:

Figure 2

A theoretical study shows that two conformers with comparable energies, \(\ce{Li2S6}(I)\) and \(\ce{Li2S6}(II)\) coexist in 1,2-dimethoxyethane (DME), a common electrolyte solvent in lithium-sulfur batteries. The dissociation of \(\ce{Li2S6}\) in DME is shown below:

Figure 3

Chemical equilibria of \(\ce{Li2S6}\), \(\ce{LiS6^-}\), \(\ce{S6^{2-}}\) and \(\ce{LiS3^{\bullet}}\) in DME

Table 4.1 Dissociation Gibbs energies (\(\mathrm{kJ\,mol^{-1}}\)) of different reactions in DME (298.15 K, 1 bar)

\(\Delta G_{\mathrm{d1}}^{\ominus}(I)\) \(\Delta G_{\mathrm{d1}}^{\ominus}(II)\) \(\Delta G_{\mathrm{d2}}^{\ominus}\) \(\Delta G_{\mathrm{dr}}^{\ominus}(I)\) \(\Delta G_{\mathrm{dr}}^{\ominus}(II)\)
20.68 18.92 100.55 45.13 43.37

4.8 Using the data from Table 4.1, calculate the equilibrium concentration ratio of two conformers in DME (298.15 K, 1 bar), \(\dfrac{\ce{Li2S6}(II)}{\ce{Li2S6}(I)}\).

4.9 Using the data from Table 4.1, calculate the apparent dissociation constant of \(\ce{Li2S6 -> Li+ + LiS6^-}\) in DME (298.15 K, 1 bar).

4.10 Sort the following equilibrium concentrations in decreasing order: \([\ce{Li2S6}]\), \([\ce{LiS6^-}]\), \([\ce{S6^{2-}}]\), \([\ce{LiS3^{\bullet}}]\) in DME.

4.11 The standard reduction potential of metallic lithium in water at 298.15 K and 1 bar is:

\[ E^{\ominus}(\ce{Li+}/\ce{Li})=-3.040\ \mathrm{V} \]

The standard solvation Gibbs energies of gaseous \(\ce{Li+(g)}\) to \(\ce{Li+(sol)}\) in different solvents.

\(\ce{Li+}(\ce{H2O})\) \(\ce{Li+}(\mathrm{DME})\)
\(\Delta G^{\ominus}/\mathrm{kJ\,mol^{-1}}\) −116.9 −114.6

Calculate the standard reduction potential of metallic lithium electrode in DME.

Studies have shown that electromotive force of a lithium-sulfur battery would be increased by replacing DME with dimethyl sulfoxide (DMSO). Therefore, the forms of polysulfides in DMSO also attract researchers' attention.

In a test, a certain amount of \(\ce{Li2S}\) and 4.81 mg sulfur powder were added into 10.00 mL DMSO, then heated and stirred until completely dissolved (ignoring the change in volume). Suppose that only the following polysulfides were present in the DMSO: \(\ce{S3^{\bullet-}}\), \(\ce{S4^{2-}}\), \(\ce{S5^{2-}}\), \(\ce{S6^{2-}}\), \(\ce{S7^{2-}}\), \(\ce{S8^{2-}}\). The equilibrium concentration ratio of the sulfur containing species was:

\[ [\ce{S3^{\bullet-}}]:[\ce{S4^{2-}}]:[\ce{S5^{2-}}]: [\ce{S6^{2-}}]:[\ce{S7^{2-}}]:[\ce{S8^{2-}}] =17.50:1.00:4.50:55.00:5.00:0.75 \]

4.12 Calculate the original mass \(m\) (in mg) of \(\ce{Li2S}\) added to DMSO.

Official English text of the problem.

Answers

  • 4.1 \[ \boxed{\ce{3FeS2 + 2O2 -> Fe3O4 + 6S}} \]

  • 4.2 \[ \boxed{\ce{SO3^{2-} + I2 + H2O -> SO4^{2-} + 2I^- + 2H+}} \] \[ \boxed{\ce{2S2O3^{2-} + I2 -> S4O6^{2-} + 2I^-}} \]

  • 4.3 \(\boxed{5.59\%}\) of the original sulfur was lost as \(\ce{SO2}\).

  • 4.4 Cathode (a): \[ \boxed{\ce{S8 + 16Li+ + 16e^- -> 8Li2S}} \] Anode (b): \[ \boxed{\ce{16Li -> 16Li+ + 16e^-}} \]

  • 4.5 \(\boxed{m_{\mathrm{cathode}}:m_{\mathrm{anode}}=2.31:1}\).

  • 4.6 \(\boxed{t\approx3.0\times10^2\ \mathrm{h}}\), with an unrounded result of \(300.7\ \mathrm{h}\).

  • 4.7 \[ \boxed{\ce{Li2S_n + 2(n-1)Li -> nLi2S}\qquad(n=3–8)} \]

  • 4.8 \[ \boxed{\frac{[\ce{Li2S6}(II)]}{[\ce{Li2S6}(I)]}=0.492} \]

  • 4.9 \(\boxed{K_{\mathrm{app}}=1.60\times10^{-4}}\) using dimensionless activities. The corresponding concentration quotient is \(1.60\times10^{-4}\ \mathrm{mol\,L^{-1}}\) for \(c^{\ominus}=1\ \mathrm{mol\,L^{-1}}\).

  • 4.10 The intended ordering at ordinary concentrations is \[ \boxed{[\ce{Li2S6}]>[\ce{LiS6^-}]>[\ce{LiS3^{\bullet}}]>[\ce{S6^{2-}}]}. \] A concentration-independent ordering is not determined by the supplied data; see the qualification below.

  • 4.11 \(\boxed{E_{\mathrm{DME}}^{\ominus}=-3.016\ \mathrm{V}}\), on the same reference scale as the aqueous value.

  • 4.12 \(\boxed{m(\ce{Li2S})=1.38\ \mathrm{mg}}\).

Names and SMILES for the species in these answers are collected at the end of the Solution.

Solution

The numerical calculations use \[ R=8.31446\ \mathrm{J\,mol^{-1}\,K^{-1}},\qquad F=96485.33\ \mathrm{C\,mol^{-1}}, \] \[ M(\ce{S})=32.06\ \mathrm{g\,mol^{-1}},\qquad M(\ce{Li})=6.94\ \mathrm{g\,mol^{-1}}. \] For the equilibrium calculations, activities are approximated by \(a_i=[i]/c^{\ominus}\), where \(c^{\ominus}=1\ \mathrm{mol\,L^{-1}}\).

4.1

One formula unit of magnetite contains three iron atoms and four oxygen atoms. It therefore requires three formula units of pyrite and two oxygen molecules. The six sulfur atoms from the pyrite become elemental sulfur:

\[ \boxed{\ce{3FeS2 + 2O2 -> Fe3O4 + 6S}}. \]

Here \(\ce{S}\) denotes elemental sulfur counted by sulfur atoms. Written using its common molecular form, cyclooctasulfur, the equivalent equation is

\[ \ce{12FeS2 + 8O2 -> 4Fe3O4 + 3S8}. \]

4.2

The alkaline absorber converts sulfur dioxide into sulfite:

\[ \ce{SO2 + 2OH^- -> SO3^{2-} + H2O}. \]

Iodine then oxidizes sulfur from oxidation state \(+4\) in sulfite to \(+6\) in sulfate. Each sulfur atom transfers two electrons, while one iodine molecule accepts two electrons:

\[ \boxed{\ce{SO3^{2-} + I2 + H2O -> SO4^{2-} + 2I^- + 2H+}}. \]

Equivalently, expressing the sulfur(IV) species as sulfur dioxide in the acidic solution,

\[ \ce{SO2 + I2 + 2H2O -> SO4^{2-} + 2I^- + 4H+}. \]

The excess iodine is determined by thiosulfate, which is oxidized to tetrathionate:

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

Thus one mole of sulfur dioxide consumes one mole of iodine, and each mole of residual iodine consumes two moles of thiosulfate.

4.3

The initial iodine amount in the flask is

\[ n_0(\ce{I2}) =(0.05000\ \mathrm{L})(0.05122\ \mathrm{mol\,L^{-1}}) =2.5610\times10^{-3}\ \mathrm{mol}. \]

The thiosulfate amount used is

\[ n(\ce{S2O3^{2-}}) =(0.01847\ \mathrm{L})(0.1012\ \mathrm{mol\,L^{-1}}) =1.869164\times10^{-3}\ \mathrm{mol}. \]

Therefore, the residual iodine amount is

\[ n_{\mathrm{remaining}}(\ce{I2}) =\frac12n(\ce{S2O3^{2-}}) =9.34582\times10^{-4}\ \mathrm{mol}. \]

The iodine consumed by sulfur(IV) in the 25.00 mL aliquot is

\[ n_{\mathrm{aliquot}}(\ce{SO2}) =2.5610\times10^{-3}-9.34582\times10^{-4} =1.626418\times10^{-3}\ \mathrm{mol}. \]

Scaling to the entire 500 mL solution,

\[ n_{\mathrm{total}}(\ce{SO2}) =\frac{500}{25.00}(1.626418\times10^{-3}) =0.03252836\ \mathrm{mol}. \]

The sulfur mass lost in this sulfur dioxide is

\[ m_{\mathrm{lost}}(\ce{S}) =(0.03252836)(32.06) =1.04286\ \mathrm{g}. \]

Because the collected elemental sulfur and sulfur dioxide account for all original sulfur,

\[ \text{percentage lost} =\frac{1.04286}{17.6+1.04286}\times100\% =\boxed{5.59\%}. \]

4.4

During discharge, sulfur is reduced at the cathode. Each sulfur atom gains two electrons, so one \(\ce{S8}\) molecule gains sixteen:

\[ \boxed{\text{Cathode (a): }\ce{S8 + 16Li+ + 16e^- -> 8Li2S}}. \]

Lithium is oxidized at the anode:

\[ \boxed{\text{Anode (b): }\ce{16Li -> 16Li+ + 16e^-}}. \]

Adding these half-reactions cancels the lithium ions and electrons and gives the stated overall reaction.

4.5

The overall reaction consumes one mole of \(\ce{S8}\) per sixteen moles of lithium:

\[ \frac{m_{\mathrm{cathode}}}{m_{\mathrm{anode}}} =\frac{8M(\ce{S})}{16M(\ce{Li})} =\frac{32.06}{2(6.94)} =2.3098. \]

Therefore,

\[ \boxed{m_{\mathrm{cathode}}:m_{\mathrm{anode}}=2.31:1}. \]

4.6

The original battery supplies energy

\[ E_{\mathrm{LIB}} =(3.8\ \mathrm{V})(3.110\ \mathrm{Ah}) =11.818\ \mathrm{Wh}. \]

Hence the phone’s average power consumption is

\[ P=\frac{11.818\ \mathrm{Wh}}{22\ \mathrm{h}} =0.53718\ \mathrm{W}. \]

Each mole of sulfur atoms accepts two moles of electrons. Complete discharge of 23 g sulfur therefore transfers charge

\[ Q=2F\frac{m(\ce{S})}{M(\ce{S})} =2(96485.33)\frac{23}{32.06} =1.38438\times10^5\ \mathrm{C}. \]

Using \(1\ \mathrm{Ah}=3600\ \mathrm{C}\),

\[ Q=38.455\ \mathrm{Ah}. \]

The ideal replacement battery supplies

\[ E_{\mathrm{Li-S}} =(4.2\ \mathrm{V})(38.455\ \mathrm{Ah}) =161.511\ \mathrm{Wh}. \]

At the same phone power consumption,

\[ t=\frac{161.511\ \mathrm{Wh}}{0.53718\ \mathrm{W}} =300.66\ \mathrm{h}. \]

Thus, to the precision of the supplied data,

\[ \boxed{t\approx3.0\times10^2\ \mathrm{h}}. \]

4.7

The \(n\) sulfur atoms in one formula unit of \(\ce{Li2S_n}\) form \(n\) formula units of lithium sulfide. These products contain \(2n\) lithium atoms; two are already present in the polysulfide, so \(2n-2\) must come from the lithium anode:

\[ \boxed{\ce{Li2S_n + 2(n-1)Li -> nLi2S}}. \]

This reaction consumes metallic lithium without delivering its electrons through the external circuit, explaining the loss of active material.

4.8

Both conformers dissociate to the same products, \(\ce{Li+ + LiS6^-}\). Consequently,

\[ G^{\ominus}_{II}-G^{\ominus}_{I} =\Delta G_{\mathrm{d1}}^{\ominus}(I) -\Delta G_{\mathrm{d1}}^{\ominus}(II) =20.68-18.92 =1.76\ \mathrm{kJ\,mol^{-1}}. \]

The radical dissociation energies give the same difference: \(45.13-43.37=1.76\ \mathrm{kJ\,mol^{-1}}\).

For conformer interconversion,

\[ \frac{[\ce{Li2S6}(II)]}{[\ce{Li2S6}(I)]} =\exp\left(-\frac{G^{\ominus}_{II}-G^{\ominus}_{I}}{RT}\right). \]

Therefore,

\[ \frac{[\ce{Li2S6}(II)]}{[\ce{Li2S6}(I)]} =\exp\left[-\frac{1760}{(8.31446)(298.15)}\right] =\boxed{0.492}. \]

Conformer I has the lower standard Gibbs energy and is therefore more abundant.

4.9

The separate dimensionless dissociation constants are

\[ K_I=\exp\left[-\frac{20680}{RT}\right] =2.3824\times10^{-4}, \]

\[ K_{II}=\exp\left[-\frac{18920}{RT}\right] =4.8457\times10^{-4}. \]

Let \(a_I\) and \(a_{II}\) denote the conformer activities, and let \(p=a_{\ce{Li+}}a_{\ce{LiS6^-}}\). Then

\[ K_I=\frac{p}{a_I},\qquad K_{II}=\frac{p}{a_{II}}. \]

The apparent constant uses the total undissociated population:

\[ K_{\mathrm{app}} =\frac{p}{a_I+a_{II}} =\frac{1}{K_I^{-1}+K_{II}^{-1}} =\frac{K_IK_{II}}{K_I+K_{II}}. \]

Numerically,

\[ \boxed{K_{\mathrm{app}}=1.5972\times10^{-4} \approx1.60\times10^{-4}}. \]

In concentration form,

\[ \frac{[\ce{Li+}][\ce{LiS6^-}]} {[\ce{Li2S6}(I)]+[\ce{Li2S6}(II)]} =K_{\mathrm{app}}c^{\ominus} =\boxed{1.60\times10^{-4}\ \mathrm{mol\,L^{-1}}}. \]

4.10

Write

\[ C_0=[\ce{Li2S6}(I)]+[\ce{Li2S6}(II)]. \]

For the second ionic dissociation,

\[ \ce{LiS6^- <=> Li+ + S6^{2-}}, \]

\[ K_{\mathrm{d2}} =\exp\left[-\frac{100550}{RT}\right] =2.42\times10^{-18}. \]

Thus the free hexasulfide dianion is strongly suppressed at ordinary lithium-ion concentrations.

For radical formation,

\[ \ce{Li2S6 <=> 2LiS3^{\bullet}}, \]

the apparent constant for the combined conformer pool is

\[ K_{\mathrm r,app} =\frac{\exp[-45130/(RT)]}{1+0.491656} =8.32\times10^{-9}. \]

If these dissociations are the only source of free lithium ions, electroneutrality gives

\[ [\ce{Li+}]=[\ce{LiS6^-}]+2[\ce{S6^{2-}}] \approx[\ce{LiS6^-}], \]

because the second dissociation is negligible at ordinary concentrations. Consequently,

\[ [\ce{LiS6^-}] \approx\sqrt{K_{\mathrm{app}}C_0c^{\ominus}}, \]

\[ [\ce{LiS3^{\bullet}}] =\sqrt{K_{\mathrm r,app}C_0c^{\ominus}}. \]

Their ratio is approximately

\[ \frac{[\ce{LiS6^-}]}{[\ce{LiS3^{\bullet}}]} \approx \sqrt{\frac{1.5972\times10^{-4}}{8.3157\times10^{-9}}} =139. \]

Thus the intended ordering is

\[ \boxed{[\ce{Li2S6}]>[\ce{LiS6^-}]> [\ce{LiS3^{\bullet}}]>[\ce{S6^{2-}}]}. \]

However, the first inequality requires approximately

\[ C_0>\sqrt{K_{\mathrm{app}}C_0c^{\ominus}} \quad\Longleftrightarrow\quad C_0>K_{\mathrm{app}}c^{\ominus} =1.60\times10^{-4}\ \mathrm{mol\,L^{-1}}. \]

At sufficiently low concentration, ionic dissociation becomes extensive and that inequality reverses. Therefore, the stated ordering is conditional on concentration; it is not a universal consequence of the standard Gibbs energies alone.

4.11

Transfer of a lithium ion from DME to water has standard Gibbs energy

\[ \Delta G_{\mathrm{tr}}^{\ominus} =\Delta G_{\mathrm{solv,H_2O}}^{\ominus} -\Delta G_{\mathrm{solv,DME}}^{\ominus} =-116.9-(-114.6) =-2.3\ \mathrm{kJ\,mol^{-1}}. \]

Reduction in DME can be represented by the following thermodynamic cycle:

  1. Transfer \(\ce{Li+}\) from DME to water.
  2. Reduce the aqueous lithium ion to metallic lithium.

Hence

\[ \Delta G_{\mathrm{red,DME}}^{\ominus} =\Delta G_{\mathrm{tr}}^{\ominus} +\Delta G_{\mathrm{red,H_2O}}^{\ominus}. \]

Using \(\Delta G^{\ominus}=-FE^{\ominus}\) for this one-electron reaction,

\[ -FE_{\mathrm{DME}}^{\ominus} =-2300-FE_{\mathrm{H_2O}}^{\ominus}, \]

so

\[ E_{\mathrm{DME}}^{\ominus} =E_{\mathrm{H_2O}}^{\ominus}+\frac{2300}{F} =-3.040+\frac{2300}{96485.33}. \]

Therefore,

\[ \boxed{E_{\mathrm{DME}}^{\ominus}=-3.016\ \mathrm{V}}. \]

The sign is consistent with the weaker stabilization of the lithium ion in DME: its reduction is slightly more favorable.

4.12

Let the concentrations be the stated numerical ratios multiplied by a common factor \(x\), with units \(\mathrm{mol\,L^{-1}}\).

The relative sulfur-atom concentration is

\[ \begin{aligned} &3(17.50)+4(1.00)+5(4.50)\\ &\qquad+6(55.00)+7(5.00)+8(0.75) =450.00. \end{aligned} \]

Thus the total amount of sulfur atoms is \(450xV\).

The trisulfur radical anion is singly charged; all the other listed anions are doubly charged. Their total negative-charge concentration is therefore

\[ 17.50+2(1.00+4.50+55.00+5.00+0.75) =150.00. \]

Let \(n_0\) be the initial amount of lithium sulfide. Each mole supplies two moles of lithium ions, so charge balance gives

\[ 2n_0=150xV, \qquad n_0=75xV. \]

Sulfur balance gives

\[ n_0+n_{\mathrm{powder}}=450xV. \]

Subtracting the sulfur initially supplied by lithium sulfide,

\[ n_{\mathrm{powder}}=(450-75)xV=375xV. \]

Consequently,

\[ \frac{n_{\mathrm{powder}}}{n_0}=\frac{375}{75}=5. \]

The common volume cancels. Therefore,

\[ n_0 =\frac{1}{5}\frac{4.81\ \mathrm{mg}} {32.06\ \mathrm{mg\,mmol^{-1}}} =0.030006\ \mathrm{mmol}. \]

Since

\[ M(\ce{Li2S})=2(6.94)+32.06 =45.94\ \mathrm{mg\,mmol^{-1}}, \]

\[ m(\ce{Li2S}) =(0.030006)(45.94) =\boxed{1.38\ \mathrm{mg}}. \]

Species names and SMILES

The salt representations below give formal ionic components. They do not encode crystal lattices, solvent coordination, or lithium–sulfur coordination geometry. In particular, the two \(\ce{Li2S6}\) conformers have the same formula-unit SMILES; their different spatial arrangements are shown in the supplied figure. The \(\ce{S8}\) SMILES specifies an eight-membered ring but does not specify its crown conformation.

Species Name SMILES
\(\ce{FeS2}\) Pyrite; iron(II) disulfide, formal ionic representation [Fe+2].[S-][S-]
\(\ce{Fe3O4}\) Magnetite; iron(II,III) oxide, formal ionic representation [Fe+2].[Fe+3].[Fe+3].[O-2].[O-2].[O-2].[O-2]
\(\ce{S8}\) Cyclooctasulfur S1SSSSSSS1
\(\ce{O2}\) Dioxygen O=O
\(\ce{SO2}\) Sulfur dioxide O=S=O
\(\ce{H2O}\) Water O
\(\ce{OH^-}\) Hydroxide ion [OH-]
\(\ce{H+}\) Hydrogen ion; proton [H+]
\(\ce{SO3^{2-}}\) Sulfite ion [O-]S(=O)[O-]
\(\ce{SO4^{2-}}\) Sulfate ion [O-]S(=O)(=O)[O-]
\(\ce{I2}\) Iodine II
\(\ce{I^-}\) Iodide ion [I-]
\(\ce{S2O3^{2-}}\) Thiosulfate ion [O-]S(=O)(=O)[S-]
\(\ce{S4O6^{2-}}\) Tetrathionate ion [O-]S(=O)(=O)SSS(=O)(=O)[O-]
\(\ce{Li}\) Lithium; atomic representation of the metallic element [Li]
\(\ce{Li+}\) Lithium ion [Li+]
\(\ce{S^{2-}}\) Sulfide ion [S-2]
\(\ce{Li2S}\) Lithium sulfide [Li+].[Li+].[S-2]
\(\ce{LiS6^-}\) Lithium hexasulfide monoanion [Li+].[S-]SSSS[S-]
\(\ce{S3^{\bullet-}}\) Trisulfur radical anion [S-]S[S]
\(\ce{LiS3^{\bullet}}\) Lithium trisulfide radical [Li+].[S-]S[S]
DME 1,2-Dimethoxyethane COCCOC
DMSO Dimethyl sulfoxide CS(C)=O

The radical SMILES show one localized charge/radical representation; they do not describe the actual delocalization of spin and charge.

The polysulfides in the general corrosion equation and final equilibrium calculation have the following formal representations:

\(n\) \(\ce{S_n^{2-}}\): name and SMILES \(\ce{Li2S_n}\): name and SMILES
3 Trisulfide ion: [S-]S[S-] Lithium trisulfide: [Li+].[Li+].[S-]S[S-]
4 Tetrasulfide ion: [S-]SS[S-] Lithium tetrasulfide: [Li+].[Li+].[S-]SS[S-]
5 Pentasulfide ion: [S-]SSS[S-] Lithium pentasulfide: [Li+].[Li+].[S-]SSS[S-]
6 Hexasulfide ion: [S-]SSSS[S-] Lithium hexasulfide: [Li+].[Li+].[S-]SSSS[S-]
7 Heptasulfide ion: [S-]SSSSS[S-] Lithium heptasulfide: [Li+].[Li+].[S-]SSSSS[S-]
8 Octasulfide ion: [S-]SSSSSS[S-] Lithium octasulfide: [Li+].[Li+].[S-]SSSSSS[S-]

The solution was checked against the official solution.