Heart pacemakers are often powered by lithium-silver chromate "button" batteries. The overall cell reaction is $$ 2 \mathrm{Li}(s)+\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \longrightarrow \mathrm{Li}_{2} \mathrm{CrO}_{4}(s)+2 \mathrm{Ag}(s) $$ (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix \(\mathrm{E}\) that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions? (c) The battery generates an emf of \(+3.5 \mathrm{~V}\). How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, \(37^{\circ} \mathrm{C}\). How does this compare to the emf you calculated in part (b)?

In a battery, the anode is where oxidation occurs, and the cathode is where reduction occurs. In this cell reaction, lithium metal is being converted to lithium ions: $$ 2 \mathrm{Li}(s) \longrightarrow 2 \mathrm{Li}^+(aq) + 2e^- $$ Since lithium is losing electrons, oxidation is occurring. Therefore, lithium metal is the reactant at the anode. Answer (a): Lithium metal is the reactant at the anode.

To find the standard emf of the battery, we first need to find the half-reactions involved in the reaction. In the lithium-silver chromate battery, the overall cell reaction is as follows: $$ 2 \mathrm{Li}(s) + \mathrm{Ag}_2 \mathrm{CrO}_4(s) \longrightarrow \mathrm{Li}_2 \mathrm{CrO}_4(s) + 2 \mathrm{Ag}(s) $$ The two half-reactions involved in this battery can be determined from the cell reaction as follows: Oxidation half-reaction (anode): $$ 2 \mathrm{Li}(s) \longrightarrow 2 \mathrm{Li}^+(aq) + 2e^- $$ Reduction half-reaction (cathode): $$ \mathrm{Ag}_2 \mathrm{CrO}_4(s) + 2e^- \longrightarrow 2 \mathrm{Ag}(s) + \mathrm{CrO}_4^{2-}(aq) $$ Now, we will find the standard reduction potentials of both half-reactions in Appendix E or a reduction potential table. The standard reduction potentials for these reactions are: For Lithium: $$ \mathrm{Li}^+(aq) + e^- \longrightarrow \mathrm{Li}(s)\hspace{1cm} E^0_{\mathrm{Li}} = -3.04\,\mathrm{V} $$ For Silver: $$ \mathrm{Ag}^+(aq) + e^- \longrightarrow \mathrm{Ag}(s)\hspace{1cm} E^0_{\mathrm{Ag}}=0.80\,\mathrm{V} $$ Now, we will find the standard cell potential or emf (\(E^0_{cell}\)) for the battery using the Nernst equation: $$ E^0_{\mathrm{cell}}=E^0_{\mathrm{cathode}}-E^0_{\mathrm{anode}} $$ Using the standard reduction potentials: $$ E^0_{\mathrm{cell}}=0.80 - (-3.04) = 3.84\,\mathrm{V} $$ Answer (b): The standard emf of the voltaic cell based on these half-reactions is 3.84 V.

Given that the battery generates an emf of +3.5 V, let's compare the calculated emf from part (b). The difference between the calculated and actual emf values is: $$ \Delta{E} = |E^0_{\mathrm{cell}} - 3.5| = |3.84 - 3.5| = 0.34\,\mathrm{V} $$ Answer (c): The calculated standard emf is 0.34 V higher than the given emf of +3.5 V.

To calculate the emf at body temperature (37°C), we can use the Nernst equation. However, since we don't have the concentrations of the species involved in the cell reaction, we can only compare the calculated emf at 25°C to the emf at 37°C given the same conditions. A more accurate calculation would account for concentration changes, but this information is not provided. Under the assumption that concentration changes are negligible, we can assume the emf at body temperature will be fairly close to the calculated emf: $$ E_{\mathrm{cell}}(37^{\circ}\mathrm{C}) \approx E^0_{\mathrm{cell}} $$ Answer (d): Assuming that concentration changes are negligible, the emf generated at body temperature would be approximately 3.84 V, which is close to the emf calculated in part (b).