Question

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)?

Short answer

(a) Lithium metal is the reactant at the anode. (b) The standard emf of the voltaic cell based on these half-reactions is 3.84 V. (c) The calculated standard emf is 0.34 V higher than the given emf of +3.5 V. (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).

Step-by-step solution

Identify the anode and cathode

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.

Determine the half-reactions and standard emf

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.

Compare calculated and given emf values

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.

Calculate the emf at body temperature

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).

Key concepts

lithium-ion batteries

Lithium-ion batteries are popular power sources for various devices, known for their high energy density and the ability to recharge. They contain lithium ions that move between the anode and cathode during discharging and charging. In the context of the heart pacemaker battery, lithium metal acts as the anode, where it undergoes oxidation. The reaction involves lithium releasing electrons and transforming into lithium ions.

• Anode: The negative electrode where oxidation occurs.
• Cathode: The positive electrode where reduction takes place.

Lithium-ion batteries offer several advantages, such as longevity and lightweight portability, making them ideal for small medical devices like pacemakers. They maintain a stable voltage until nearly depletion, ensuring reliable performance as desired for critical applications.
When used in devices like pacemakers, it is essential to understand the electrochemical reactions at play, as these reactions determine the battery's efficiency and lifespan. This understanding helps in designing batteries that meet the specific needs of various devices.

standard reduction potential

The concept of standard reduction potential is crucial in electrochemistry to predict the direction of redox reactions and calculate cell potentials. It is measured in volts and reflects how readily a substance can gain electrons. The higher the reduction potential, the more likely it is for the reaction to occur at the cathode.
In our exercise, lithium and silver are involved, each with distinct standard reduction potentials. Let's break these down:

• Lithium has a low standard reduction potential of -3.04 V, indicating it easily gives up electrons, hence acts as the anode.
• Silver, with a higher reduction potential of 0.80 V, gains electrons more readily, thus serves as the cathode.

The calculation of the cell potential \( E^0_{\mathrm{cell}} \) is derived by subtracting the anode's reduction potential from the cathode's. This method determines the cell's ability to do electrical work, important for applications requiring precise energy delivery, like medical devices.
Understanding these potentials helps in selecting appropriate materials for electrodes, ensuring efficient operation of the cell.

Nernst equation

The Nernst equation is a pivotal tool in electrochemistry for understanding how the cell potential changes with varying conditions. Specifically, it accounts for temperature, concentration, and pressure effects on the cell's emf.
For a simple electrochemical reaction, the Nernst equation is written as:\[E = E^0 - \frac{RT}{nF} \ln{Q}\]Here:

• \(E\) is the cell potential under non-standard conditions.
• \(E^0\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred.
• \(F\) is Faraday's constant.
• \(Q\) is the reaction quotient.

This equation helps us understand how body temperature might affect cell potential. In our exercise, while specific concentrations are unknown, under typical body conditions (37°C), the emf would remain similar to room temperature calculations if concentrations don't shift significantly.
Therefore, the Nernst equation enables predicting cell behavior in real-world conditions, ensuring reliable expectations for devices like pacemakers.