The following half-reactions occur in the mercury battery used in calculators.If E°_cell = 1.357 V,calculate the equilibrium constant for the cell reaction at 25°C.(Assume the stoichiometric coefficients in the cell reaction are all equal to 1. ) HgO(s)+ H₂O(l)+ 2e^- Hg(l)+ 2OH^-(aq) ZnO(s)+ H₂O(l)+ 2e^- Zn(s)+ 2OH^-(aq)

Question

Quizplus · Accepted Answer

The equilibrium constant can be calculated using the Nernst equation:
K = e^(-deltaG/RT) 
where deltaG is the change in free energy, R is the gas constant, T is the temperature in Kelvin, and e is the base of the natural logarithm.

The half-reactions given are:
HgO(s) + H+ + e- -> Hg(l) + H2O(l); E° = 0.097 V
ZnO(s) + H+ + 2e- -> Zn(s) + H2O(l); E° = 0.763 V

The overall cell reaction is: 
HgO(s) + Zn(s) + 2H+ -> Hg(l) + ZnO(s) + H2O(l); E°cell = 1.357 V

Using the equation E°cell = E°(cathode) - E°(anode), we can find the standard reduction potential for the cathode:
E°(cathode) = E°cell + E°(anode) = 0.097 V + 1.357 V = 1.454 V

Using the Nernst equation: 
E = E° - (RT/nF) ln Q 
where Q is the reaction quotient, n is the number of electrons transferred, and F is the Faraday constant. At equilibrium, E = 0 and Q = K, so we can rearrange to solve for K:
K = e^(nF E°/RT)

For the cathode half-reaction: 
n = 1, F = 96485 C/mol 
Kcathode = e^(96485(1.454)/(8.314*298)) = 7.45 x 10^14

For the anode half-reaction: 
n = 2 
Kanode = e^(-(2*96485*0.763)/(8.314*298)) = 1.44 x 10^-5

The equilibrium constant for the overall cell reaction is the product of the equilibrium constants for the two half-reactions:
K = Kcathode/Kanode = 7.45 x 10^14/1.44 x 10^-5 = 5.17 x 10^19 = 7.5 x 10^10 (rounded to one significant figure)

The Following Half-Reactions Occur in the Mercury Battery Used in Calculators.If