A lithium battery, which is different from a lithiumion battery, uses lithium metal as one electrode and carbon in contact with \(\mathrm{MnO}_{2}\) in a paste of \(\mathrm{KOH}\) as the other electrode. The electrolyte is lithium perchlorate in a nonaqueous solvent, and the construction is similar to the silver battery. The half-cell reactions involve the oxidation of lithium and the reaction $$\begin{aligned}\mathrm{MnO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}(\mathrm{OH})_{3}(\mathrm{s})+& \\\\\mathrm{OH}^{-}(\mathrm{aq}) &E^{\circ}=-0.20 \mathrm{V}\end{aligned}$$ Draw a cell diagram for the lithium battery, identify the negative and positive electrodes, and estimate the cell potential under standard conditions.

Step by step solution

01

Recognizing the Half-Cell Reactions

First, identify the half-cell reactions. From the problem, the oxidation of lithium is given as \(Li(s) \rightarrow Li^{+}(aq) + e^−\), which happens at the anode (negative electrode) with an E° value of 0. Through the process of oxidation, lithium loses an electron. The reduction of manganese dioxide is given as \(MnO_2(s) + 2H_2O(l) + e^− \rightarrow Mn(OH)_3(s) + OH^{−}(aq)\) which occurs at the cathode (positive electrode) with an E° value of -0.20 V. In this reaction, manganese gains an electron.

02

Drawing the Cell Diagram

Now, using the half-cell reactions, the cell can be diagramed. Anodes are always written on the left and cathodes on the right. The cell diagram is: \(Li(s) | Li^{+}(aq) || OH^{−}(aq), Mn(OH)_3(s) | MnO_2(s)\). The double line represents the salt bridge, and the single lines represent phase boundaries.

03

Identifying the Negative and Positive Electrodes

The negative electrode is the anode, where oxidation occurs. In this case, it's the lithium metal. The positive electrode is the cathode, where reduction happens, which here is the carbon in contact with MnO2.

04

Calculating the Cell Potential

Finally, calculate the cell potential under standard conditions using the Nernst equation. The E° of the cell is the sum of the E° values for the cathode and the anode. Therefore, \(E°_{cell}\ = E°_{cathode} - E°_{anode} = -0.20V - 0V = -0.20V\). The negative sign indicates that the reaction as written is not spontaneous; the reaction would need to proceed in the opposite direction to be spontaneous.

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