Answer to Question #42832 in Physical Chemistry for anuragasaksena

Question

The cathodic reaction of a dry cell is represented by 2MnO2(s) + Zn2+ + 2e- → ZnMn2O4(s) If, there are 8 g of MnO2 in the cathodic compartment then the time for which the dry cell will continue to give a current of 2 milliampere is:

Answer 1

According to the Faraday's law:
m = (It/F)*M/z
where m is the mass of substance, I is the current strength, t is the time of experiment, F is the Faraday's constant, M is the molar mass of substance, z is the factor of equivalence.
The time of the electrolysis can be expressed as follows:
t = mzF/IM
The factor of equivalence can be found dividing the number of electrons by the number of reacted particles. In case of MnO₂ the factor is:
z = 2/2 = 1
The molar mass of MnO₂ is:
M(MnO₂) = 55 + 16*2 = 87 g/mol
Now we have enough data to calculate the time:
t = 8*1*96485/(0.002*87) = 4.436*10⁶ s = 1232 hours = 51.3 days
These calculations were made in assumption that the current strength will be constant until the full amount of MnO₂ is consumed. However in real life the current strength will gradually drop with the consumption of reactants, so the real time of use will be lower.