Question

Heart pacemakers are often powered by lithium–silver chromate “button” batteries.

The overall cell reaction is

2 Li(s) + Ag₂CrO₄(s)→ Li₂CrO₄(s) + 2Ag(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 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 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 °C. How does this compare to the emf you calculated in part (b)?

One Class Solution:-a) Li(s) is oxidized at the anode.

Ag₂CrO₄(s)+ 2e-→ 2Ag(s) + CrO₄^2-(aq), E= 0.446V

Li(s) → Li+(aq)+e^- E=–3.05V

2 Li(s) + Ag₂CrO₄(s)→ Li₂CrO₄(s) + 2Ag(s)

b) E⁰=0.446V-(-3.05V) =3.496=3.50V

c) The emf of the battery, 3.5 V, is exactly the standard cell potential calculated in part (b).

d) For this battery at ambient conditions, E⁰= E⁰, so logQ=O. This makes sense because all reactants and products in the battery are solids and thus present in their standard states. Assuming that E⁰ is relatively constant with temperature, the value of the second termin the Nernst equation is= 0 at 37°C, and E=3.5V.