Choose the two half-reactions from a table of standard reduction potentials that most closely approximate the reaction that occur within the battery. What is the standard voltage generated by a cell operating with these half-reactions?
2Li(s) + Ag2CrO4(s) --------> Li2CrO4(s) + 2Ag(s)
Calculate the voltage at body temperature, 37°C
From the limited description you give I would pick
Li ==> Li^+ + e and
Ag^+ + e ==> Ag
You mention "the" battery so the use of the battery and the other conditions not listed may affect the results.
For the second part,
Ecell = Eocell - (RT/nF)ln Q
and you plug in T = 37 C but change that to kelvin. R is 8.314 and F = 96,485. If you want to convert ln to log, that is 2.303*ln. Q is the reaction quotient.
To determine the half-reactions that most closely approximate the reaction within the battery, we need to look up the standard reduction potentials for each half-reaction involved. The half-reactions in this cell are:
1. Li+ (aq) + e- → Li(s) (Reduction potential: -3.05 V)
2. Ag2CrO4(s) + 2e- → 2Ag(s) + CrO4^2- (Reduction potential: +0.15 V)
Now, we need to compare these reduction potentials to determine the most suitable half-reactions. The closest approximation will be a reduction half-reaction with a reduction potential close to -3.05 V and an oxidation half-reaction with a potential close to +0.15 V.
Looking at the available half-reactions, we can see that the closest oxidation half-reaction is the reduction of CrO4^2-. However, we'll need to divide the reaction by 2 to match the stoichiometry of the given chemical equation:
CrO4^2-(aq) + 4H2O(l) → Cr(OH)3(s) + 5O2(g) + 4e- (Reduction potential: +1.33 V)
Now, we have the most appropriate half-reactions:
Li+ (aq) + e- → Li(s) (Reduction potential: -3.05 V)
2Ag2CrO4(s) + 4e- → 2Ag(s) + 2Cr(OH)3(s) + 5O2(g) (Reduction potential: +1.33 V)
To calculate the voltage of the cell at body temperature (37°C), we need to know the temperature dependence of the standard reduction potentials. The Nernst equation relates the standard reduction potential (E°), actual reduction potential (E), temperature (T), and the Faraday constant (F):
E = E° - (RT / nF) * ln(Q)
Where:
E = actual reduction potential
E° = standard reduction potential
R = gas constant (8.314 J/(K*mol))
T = temperature in Kelvin (37°C + 273.15 = 310.15 K)
n = number of moles of electrons transferred
F = Faraday constant (96485 C/mol)
Q = reaction quotient (concentration of products divided by reactants)
Since this calculation involves the concentration of reactants and products, we'll assume that they are at standard concentration (1M) unless otherwise stated.
Now, let's substitute the values into the Nernst equation for both half-reactions and calculate the voltages:
For the reduction of Li+:
E = -3.05 V - ((8.314 J/(K*mol)) * 310.15 K / (1 * 96485 C/mol)) * ln(1)
E = -3.05 V
For the reduction of Ag2CrO4:
E = 1.33 V - ((8.314 J/(K*mol)) * 310.15 K / (4 * 96485 C/mol)) * ln(1)
E = 1.33 V
The overall voltage of the cell is the difference in potential between the reduction and oxidation half-reactions:
Cell voltage = E(red) - E(ox)
= -3.05 V - 1.33 V
= -4.38 V
Therefore, the standard voltage generated by the cell operating with these half-reactions is -4.38 V at body temperature (37°C).